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diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch1.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch1.ipynb new file mode 100644 index 00000000..7a41be48 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch1.ipynb @@ -0,0 +1,860 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1 : Introduction : Economics of Power Generation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.1 Page23" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)Load factor of the plant is 0.65\n", + "(b)Load factor of a standby equipment of 30 capacity if it takes up all the loads above 70 MW is 0.75\n", + "(c)Use factor is 0.75\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f5c10894090>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "from __future__ import division\n", + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, title, xlabel, ylabel, show\n", + "#Input data\n", + "C=30#Capacity in MW\n", + "M=70#Loads are taken above 70 MW\n", + "t1=[0,6]#Time range in hours\n", + "t2=[6,10]#Time range in hours\n", + "t3=[10,12]#Time range in hours\n", + "t4=[12,16]#Time range in hours\n", + "t5=[16,20]#Time range in hours\n", + "t6=[20,22]#Time range in hours\n", + "t7=[22,24]#Time range in hours\n", + "L=[30,70,90,60,100,80,60]#Load in MW\n", + "\n", + "#Calculations\n", + "E=((L[0]*(t1[1]-t1[0]))+(L[1]*(t2[1]-t2[0]))+(L[2]*(t3[1]-t3[0]))+(L[3]*(t4[1]-t4[0]))+(L[4]*(t5[1]-t5[0]))+(L[5]*(t6[1]-t6[0]))+(L[6]*(t7[1]-t7[0])))#Energy generated in MWh\n", + "AL=(E/24)#Average load in MW\n", + "PL=max(L[0],L[1],L[2],L[3],L[4],L[5],L[6])#Peak load in MW\n", + "LF=(AL/PL)#Load factor of the plant\n", + "E1=((L[2]-M)*(t3[1]-t3[0]))+((L[4]-M)*(t5[1]-t5[0]))+((L[5]-M)*(t6[1]-t6[0]))#Energy generated if the load above 70 MW is supplied by a standby unit of 30 MW capacity in MWh\n", + "T=(t3[1]-t3[0])+(t5[1]-t5[0])+(t6[1]-t6[0])#Time during which the standby unit remains in operation in h\n", + "AL1=(E1/T)#Average load in MW\n", + "LF1=(AL1/C)#Load factor \n", + "U=(E1/(C*T))#Use factor\n", + "\n", + "#Output\n", + "t=[0,0,6,6,10,10,12,12,16,16,20,20,22,22,24,24]#Time for plotting load curve in hours\n", + "l=[0,30,30,70,70,90,90,60,60,100,100,80,80,60,60,0]#Load for plotting load curve in MW\n", + "plot(t,l)#Load curve taking Time in hours on X-axis and Load in MW on Y- axis\n", + "title('Load Curve')\n", + "xlabel('Time hours')\n", + "ylabel('Load MW')\n", + "print \"(a)Load factor of the plant is %3.2f\\n(b)Load factor of a standby equipment of %3.0f capacity if it takes up all the loads above %3.0f MW is %3.2f\\n(c)Use factor is %3.2f\"%(LF,C,M,LF1,U)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.2 Page25" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The average load on the power plant is 30 MW \n", + "(b) The energy supplied per year is 262.8 *10**6 kWh \n", + "(c) Demand factor is 0.811 \n", + "(d) Diversity factor is 1.233\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "P=60#Peak load on power plant in MW\n", + "L=[30,20,10,14]#Loads having maximum demands in MW\n", + "C=80#Capacity of the power plant in MW\n", + "A=0.5#Annual load factor\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "AL=(P*A)#Average load in MW\n", + "E=(AL*1000*Y)/10**6#Energy supplied per year in kWh*10**6\n", + "DF=(P/(L[0]+L[1]+L[2]+L[3]))#Demand factor \n", + "DIF=((L[0]+L[1]+L[2]+L[3])/P)#Diversity factor\n", + "\n", + "#Output\n", + "print \"(a) The average load on the power plant is %3.0f MW \\n(b) The energy supplied per year is %3.1f *10**6 kWh \\n(c) Demand factor is %3.3f \\n(d) Diversity factor is %3.3f\"%(AL,E,DF,DIF)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.3 Page25" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The annual revenue earned by the power plant is Rs 46.25 crore \n", + "(b) Capacity factor is 0.457\n" + ] + } + ], + "source": [ + "#Input data\n", + "C=210#Capacity of thermal power plant in MW\n", + "P=160#Maximum load in MW\n", + "L=0.6#Annual load factor \n", + "m=1#Coal consumption per kWh of energy generated\n", + "Rs=450#Cost of coal in Rs per tonne\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "AL=(L*P)#Average load in MW\n", + "E=(AL*Y)#Energy generated per year in MWh\n", + "CL=(E*1000)#Coal required per year in kg\n", + "CY=(E*Rs)#Cost of coal per year\n", + "CE=CL#Cost of energy sold in Rs\n", + "RY=(CE-CY)/10**7#Revenue earned by the power plant per year in Rs crore\n", + "CF=(AL/C)#Capacity factor\n", + "\n", + "#Output\n", + "print \"(a) The annual revenue earned by the power plant is Rs %3.2f crore \\n(b) Capacity factor is %3.3f\"%(RY,CF)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.4 Page 26" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Annual energy production is 394.2 * 10**6 kWh \n", + "(b) Reserve capacity over and above the peak load is 15 MW \n", + "(c) The hours during which the plant is not in service per year is 674 hrs\n" + ] + } + ], + "source": [ + "#Input data\n", + "L=0.75#Load factor\n", + "C=0.60#Capacity factor\n", + "U=0.65#Use factor\n", + "M=60#Maximum power demand in MW\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "A=(L*M)#Average load in MW\n", + "P=((A*1000)*Y)/10**6#Annual energy production in kWh *10**6\n", + "PC=(A/C)#Plant capacity in MW\n", + "R=(PC-M)#Reserve capacity in MW\n", + "HIO=(P*1000/(U*PC))#Hours in operation in hrs\n", + "NH=(Y-HIO)#Hours not in service in a year in hrs\n", + "\n", + "#Output\n", + "print \"(a) Annual energy production is %3.1f * 10**6 kWh \\n(b) Reserve capacity over and above the peak load is %3.0f MW \\n(c) The hours during which the plant is not in service per year is %3.0f hrs\"%(P,R,NH)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.5 Page26" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Overall cost per kWh in Steam power plant is 91 paise \n", + "(ii)Overall cost per kWh in Hydroelectric power plant is 67 paise \n", + "(iii)Overall cost per kWh in Nuclear power plant is 99 paise\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "Dd=500#Maximum demand in MW\n", + "L=0.7#Load factor \n", + "#1)Steam power plant 2)Hydroelectric power plant 3)Nuclear power plant\n", + "CC=[0,3,4,5]#Capital cost per MW installed in Rs. crore\n", + "I=[0,6,5,5]#Interest in percent\n", + "D=[0,6,4,5]#Depreciation in percent\n", + "OP=[0,30,5,15]#Operating cost (including fuel) per kWh\n", + "TD=[0,2,3,2]#Transmission and distribution cost per kWh\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "#1)Steam power plant\n", + "CCX=(CC[(1)]*Dd*10**7)#Capital cost in Rs\n", + "IX=((I[(1)]/100)*CCX)#Interest in Rs\n", + "DX=((D[(1)]/100)*CCX)#Depreciation in Rs\n", + "AFCX=IX+DX#Annual fixed cost in Rs\n", + "EX=(L*Dd*1000*Y)#Energy generated per year in kWh\n", + "RX=(OP[(1)]+TD[(1)])#Running cost/kWh in paise\n", + "OX=((AFCX/EX)+(RX/100))*100#Overall cost/kWh in paise\n", + "\n", + "#2)Hydroelectric Power plant\n", + "CCY=(CC[(2)]*Dd*10**7)#Capital cost in Rs\n", + "IY=((I[(2)]/100)*CCY)#Interest in Rs\n", + "DY=((D[(2)]/100)*CCY)#Depreciation in Rs\n", + "AFCY=IY+DY#Annual fixed cost in Rs\n", + "EY=(L*Dd*1000*Y)#Energy generated per year in kWh\n", + "RY=(OP[(2)]+TD[(2)])#Running cost/kWh in paise\n", + "OY=((AFCY/EY)+(RY/100))*100#Overall cost/kWh in paise\n", + "\n", + "#3)Nuclear power plant\n", + "CCZ=(CC[(3)]*Dd*10**7)#Capital cost in Rs\n", + "IZ=((I[(3)]/100)*CCZ)#Interest in Rs\n", + "DZ=((D[(3)]/100)*CCZ)#Depreciation in Rs\n", + "AFCZ=IZ+DZ#Annual fixed cost in Rs\n", + "EZ=(L*Dd*1000*Y)#Energy generated per year in kWh\n", + "RZ=(OP[(3)]+TD[(3)])#Running cost/kWh in paise\n", + "OZ=((AFCZ/EZ)+(RZ/100))*100#Overall cost/kWh in paise\n", + "\n", + "#Output\n", + "print \"(i)Overall cost per kWh in Steam power plant is %3.0f paise \\n(ii)Overall cost per kWh in Hydroelectric power plant is %3.0f paise \\n(iii)Overall cost per kWh in Nuclear power plant is %3.0f paise\"%(OX,OY,OZ)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.6 Page28" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The cost of power generation per kWh is 70 paise \n", + "(b) The reserve capacity is 21 MW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "C=210#Capacity in MW\n", + "ID=12#Interest and depreciation in percent\n", + "CC=18000#Capital cost/kW installed in Rs\n", + "L=0.6#Annual load factor\n", + "AC=0.54#Annual capacity factor\n", + "RC=(200*10**6)#Annual running charges in Rs\n", + "E=6#Energy consumed by power plant auxiliaries in percent\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "MD=(C/L)*AC#Maximum demand in MW\n", + "RSC=(C-MD)#Reserve Capacity in MW\n", + "AL=(L*MD)#Average load in MW\n", + "EP=(AL*1000*Y)#Energy produced per year in kWh\n", + "NE=((100-E)/100)*EP#Net energy delivered in kWh\n", + "AID=((ID/100)*CC*C*1000)#Annual interest and depreciation in Rs\n", + "T=(AID+RC)#Total annual cost in Rs\n", + "CP=(T/NE)*100#Cost of power generation in paise\n", + "\n", + "#Output\n", + "print \"(a) The cost of power generation per kWh is %3.0f paise \\n(b) The reserve capacity is %3.0f MW\"%(CP,RSC)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.7 Page28" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The economic loading of two units when the total load supplied by the power plants is 200 MW are 75.86 MW and 124.14 MW\n", + "(b)The loss in fuel cost per hour if the load is equally shared by both units is Rs.42.24 per hour\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#Input data\n", + "L=200#The total load supplied by the plants in MW\n", + "#The incremental fuel costs for generating units a and b of power plant are given by\n", + "#dFa/dPa=0.065Pa+25\n", + "#dFb/dPb=0.08Pa+20\n", + "\n", + "#Calculations\n", + "#Solving two equations\n", + "#Pa+Pb=200\n", + "#0.065Pa+25=0.08Pb+20\n", + "A=mat([[1, 1],[0.065, -0.08]])#Coefficient matrix\n", + "B=mat([[L],[(20-25)]])#Constant matrix\n", + "X=(A**-1)*B#Variable matrix\n", + "P=100#If load is shared equally then Pa=Pb=100MW\n", + "a=(((0.065*P**2)/2)+(25*P))-(((0.065*X[0]**2)/2)+(25*X[0]))#increase in fuel cost for unit a in Rs. per hour\n", + "b=(((0.08*P**2)/2)+(20*P))-(((0.08*X[1]**2)/2)+(20*X[1]))#increase in fuel cost for unit a in Rs. per hour\n", + "x=a+b#Net increase in fuel cost due to departure from economic distribution of load in Rs. per hour\n", + "\n", + "#Output\n", + "print \"(a)The economic loading of two units when the total load supplied by the power plants is 200 MW are %3.2f MW and %3.2f MW\\n(b)The loss in fuel cost per hour if the load is equally shared by both units is Rs.%3.2f per hour\"%(X[0],X[1],x)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.8 page29" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Cost of generation per kWh is 61 paise \n", + " Saving in cost per kWh if the annual load factor is raised to 60 percent is 11 paise\n" + ] + } + ], + "source": [ + "from math import ceil\n", + "#Input data\n", + "C=200#Installed capacity of the plant in MW\n", + "CC=400#Capital cost in Rs crores\n", + "ID=12#Rate of interest and depreciation in percent\n", + "AC=5#Annual cost of fuel, salaries and taxation in Rs. crores\n", + "L=0.5#Load factor\n", + "AL2=0.6#Raised Annual load\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "AvL=(C*L)#Average Load in MW\n", + "E=(AvL*1000*Y)#Energy generated per year in kWh\n", + "IDC=((ID/100)*CC*10**7)#Interest and depreciation (fixed cost) in Rs\n", + "T=(IDC+(AC*10**7))#Total annual cost in Rs\n", + "CP1=(T/E)*100#Cost per kWh in paise\n", + "AvL2=(C*AL2)#Average Load in MW\n", + "E2=(AvL2*1000*Y)#Energy generated per year in kWh\n", + "CP2=(T/E2)*100#Cost per kWh in paise\n", + "S=((CP1)-(CP2))#Saving in cost per kWh in paise\n", + "S1=ceil(S)#Rounding off to next higher integer\n", + "\n", + "#Output\n", + "print \" Cost of generation per kWh is %3.0f paise \\n Saving in cost per kWh if the annual load factor is raised to 60 percent is %3.0f paise\"%(CP1,S1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.9 Page30" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Load factor is 0.875 \n", + "(b) Capacity factor is 0.70\n" + ] + } + ], + "source": [ + "#Input data\n", + "C=300#Capacity of power plant in MW\n", + "MXD=240#Maximum demand in MW in a year\n", + "MND=180#Minimum demand in MW in a year\n", + "#Assuming the load duration curve shown in Figure E1.9 on page no 30 to be straight line\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "E=((MND*Y)+0.5*(MXD-MND)*Y)*1000#Energy supplied per year in kWh\n", + "AL=(E/Y)#Average load in kW\n", + "L=((AL/1000)/MXD)#Load factor\n", + "CF=((AL/1000)*Y)/(C*Y)#Capacity factor\n", + "\n", + "#Output\n", + "print \"(a) Load factor is %3.3f \\n(b) Capacity factor is %3.2f\"%(L,CF)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.10 Page 31" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Revenue earned by the power plant = 3.679e+08 Rs./year\n" + ] + } + ], + "source": [ + "#Input data\n", + "C=60#Capacity of power plant in MW\n", + "MXD=50#Maximum demand in MW in a year\n", + "L=60/100#Load factor\n", + "cc = 1 # kg/unit (Coal consumption)\n", + "c_cost = 600 # Rs/Tonne\n", + "e_cost = 2 # Rs/kWh\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "AL=(MXD*L)#Average load in MW\n", + "E=AL*10**3*Y #Energy generated per year in kWh\n", + "Coal = E*cc/10**3 # Tonnes (Coal required per year)\n", + "CC = Coal*c_cost # rupees (Coal cost / year)\n", + "CE = E*e_cost # Rs (Cost of energy sold)\n", + "Rev = CE-CC\n", + "#Output\n", + "print \"Revenue earned by the power plant = %0.3e Rs./year\"%(Rev)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.11 Page31" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f5bf6f06d50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(c)Suitable generating units to supply the load are\n", + "i)One unit of 30 MW will run for 24 hours\n", + "ii)One unit of 30 MW will run for 18 hours\n", + "iii)One unit of 30 MW will run for 10 hours\n", + "iv)One unit of 10 MW will run for 4 hours\n", + "\n", + "(d)Load factor is 0.64\n", + "\n", + "(e)Capacity of the plant is 130 MW and Capacity factor is 0.494\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot,subplot,title,xlabel,ylabel,show\n", + "#Input data\n", + "t1x=[0,6]#Time range in hours\n", + "t2x=[6,12]#Time range in hours\n", + "t3=[12,14]#Time range in hours\n", + "t4=[14,18]#Time range in hours\n", + "t5=[18,24]#Time range in hours\n", + "L=[30,90,60,100,50]#Load in MW\n", + "\n", + "#Calculations\n", + "t1=[0,6,6,12,12,14,14,18,18,24,24]#Time in hours for Load curve\n", + "L1=[30,30,90,90,60,60,100,100,50,50,0]#Load in MW for Load curve\n", + "t2=[0,4,4,10,10,12,12,18,18,24,24]#Time in hours for Load duration curve\n", + "L2=[100,100,90,90,60,60,50,50,30,30,24]#Load in MW for Load duration curve\n", + "E=((L[0]*(t1x[1]-t1x[0]))+(L[1]*(t2x[1]-t2x[0]))+(L[2]*(t3[1]-t3[0]))+(L[3]*(t4[1]-t4[0]))+(L[4]*(t5[1]-t5[0])))#Energy generated in MWh\n", + "AL=E/24#Average load in MW\n", + "MD=max(L[0],L[1],L[2],L[3],L[4])#Maximum demand in MW\n", + "LF=(AL/MD)#Load factor\n", + "Lx=[30,10]#Loads for selecting suitable generating units in MW\n", + "tx=[24,18,10,4]#Time for selecting suitable generating units in hrs\n", + "PC=(Lx[0]*tx[3]+Lx[1]*1)#Plant capacity in MW\n", + "CF=(E/(PC*24))#Capacity factor \n", + "\n", + "#Output\n", + "plot(t1,L1)#Load curve taking Time in hrs on X- axis and Load in MW on Y- axis\n", + "title('Load curve')\n", + "xlabel('Time hrs')\n", + "ylabel('Load MW')\n", + "show()\n", + "plot(t2,L2)#Load duration curve taking Time in hrs on X- axis and Load in MW on Y- axis\n", + "title('Load duration curve')\n", + "xlabel('Time hrs')\n", + "ylabel('Load MW')\n", + "show()\n", + "print \"(c)Suitable generating units to supply the load are\\ni)One unit of %3.0f MW will run for %3.0f hours\\nii)One unit of %3.0f MW will run for %3.0f hours\\niii)One unit of %3.0f MW will run for %3.0f hours\\niv)One unit of %3.0f MW will run for %3.0f hours\\n\\n(d)Load factor is %3.2f\\n\\n(e)Capacity of the plant is %3.0f MW and Capacity factor is %3.3f\"%(Lx[0],tx[0],Lx[0],tx[1],Lx[0],tx[2],Lx[1],tx[3],LF,PC,CF)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.12 Page32" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Overall cost of energy per kWh for:\n", + "(a)Domestic consumers is 70 paise\n", + "(b)Industrial consumers is 36 paise\n", + "(c)Street-lighting load is 51 paise\n" + ] + } + ], + "source": [ + "#Input data\n", + "C=10#Capacity of generating unit in MW\n", + "MD=[6,3.6,0.4]#Maximum demand for domestic consumers, industrial consumers and street-lighting load respectively in MW\n", + "L=[0.2,0.5,0.3]#Load factor for domestic consumers, industrial consumers and street-lighting load respectively\n", + "CC=10000#Capital cost of the plant per kW in Rs\n", + "RC=3600000#Total rumming cost per year in Rs\n", + "AID=10#Annual interest and depreciation on capital cost in percent\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "E=((MD[0]*L[0])+(MD[1]*L[1])+(MD[2]*L[2]))*Y*1000#Energy supplied per year to all three consumers in kWh\n", + "OC=(RC/E)#Operating charges per kWh in Rs\n", + "CCP=(C*1000*CC)#capital cost of the plant in Rs\n", + "FCY=((AID/100)*CCP)#Fixed charges per year in Rs\n", + "FCkW=(FCY/CC)#Fixed charges per kW in Rs\n", + "#a) For domestic consumers\n", + "TC1=((FCkW*MD[0]*1000)+(OC*MD[0]*L[0]*Y*1000))#Total chrges in Rs\n", + "OC1=(TC1/(MD[0]*L[0]*Y*1000))*100#Overall cost per kWh in paise\n", + "#b)For industrial consumers\n", + "TC2=((FCkW*MD[1]*1000)+(OC*MD[1])*L[1]*Y*1000)#Total chrges in Rs\n", + "OC2=(TC2/(MD[1]*L[1]*Y*1000))*100#Overall cost per kWh in paise\n", + "#c) For street-lighting load\n", + "TC3=((FCkW*MD[2]*1000)+(OC*MD[2])*L[2]*Y*1000)#Total chrges in Rs\n", + "OC3=(TC3/(MD[2]*L[2]*Y*1000))*100#Overall cost per kWh in paise\n", + "\n", + "#Output\n", + "print \"Overall cost of energy per kWh for:\\n(a)Domestic consumers is %3.0f paise\\n(b)Industrial consumers is %3.0f paise\\n(c)Street-lighting load is %3.0f paise\"%(OC1,OC2,OC3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.13 Page32" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of money to be saved annually is Rs.961317/-\n" + ] + } + ], + "source": [ + "#Input data\n", + "CC=(80*10**6)#Capital cost in Rs\n", + "L=30#Useful life in years\n", + "S=5#Salvage value of the capital cost in percent\n", + "i=0.06#Yearly rate of compound interest\n", + "\n", + "#Calculations\n", + "A=((100-S)/100)*CC#Difference of capital cost and salvage value in Rs\n", + "P=((A*i)/((1+i)**L-1))#The amount of money to be saved annually in Rs\n", + "\n", + "#Output\n", + "print \"The amount of money to be saved annually is Rs.%3.0f/-\"%(P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.14 Page34" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Present worth of the payments at the time of commissioning is Rs.5994.39 crores\n" + ] + } + ], + "source": [ + "#Input data\n", + "i=4000#Initial investment in Rs crore\n", + "Y=4#Period in years\n", + "A=1200#Amount added in Rs crore\n", + "B=400#Amount paid from 5th year onwards to the 12th year in Rs crore\n", + "a=5#5th year\n", + "b=12#12th year\n", + "y=30#Period in years\n", + "C=600#Salvage value in Rs crore\n", + "I=0.1#Interest rate \n", + "\n", + "#Calculations\n", + "X=(1/(1+I))#X value for calculations\n", + "PW=(i+(A*X**Y)+((B/I)*X**b*((I+1)**b-1))-((B/I)*X**a*((I+1)**a-1))-(C*X**y))#Present worth of the payments at the time of commissioning in Rs. crores\n", + "\n", + "#Output\n", + "print \"Present worth of the payments at the time of commissioning is Rs.%3.2f crores\"%(PW)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.15 Page 35" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Incremental heat transfer rate at which the combined output of the two units is 1000 MW is IR = (IR)P = (IR)Q = 9293 kJ/kWh\n" + ] + } + ], + "source": [ + "#Input data\n", + "O=1000#Combined output of two units in MW\n", + "#Two coal generating units P and Q have the incremental heat rate defined by\n", + "#(IR)P=0.4818*10**-7.LP**4 - 0.9089*10**-4.LP**3 + 0.6842*10**-1.LP**2 - 0.2106*10.LP + 9860\n", + "#(IR)R=0.9592*10**-7.LQ**4 - 0.7811*10**-4.LQ**3 + 0.2625*10**-1.LQ**2 - 0.2189*10.LQ + 9003\n", + "\n", + "#Calculations\n", + "#LP+LQ=1000\n", + "#By making (IR)P=(IR)Q and solving the above three equations by a numerical methos such as Newton-Raphson algorithm, we get \n", + "LP=732.5#Heat rate in MW\n", + "LQ=(O-LP)#Heat rate in MW\n", + "IR=0.4818*10**-7*LP**4 - 0.9089*10**-4*LP**3 + 0.6842*10**-1*LP**2 - 0.2106*100*LP + 9860\n", + "IR1=0.9592*10**-7*LQ**4 - 0.7811*10**-4*LQ**3 + 0.2625*10**-1*LQ**2 - 0.2189*10*LQ + 9003\n", + "\n", + "#Output\n", + "print \"Incremental heat transfer rate at which the combined output of the two units is %3.0f MW is IR = (IR)P = (IR)Q = %d kJ/kWh\"%(O,IR)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex1.16 Page 36" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "==============================================================================\n", + "Load Energy produced Fixed cost Fuel and Total cost Cost per\n", + "factor in 1hr with per hr operating cost per hr kWh\n", + "(percent) 1kW plant(kWh) (paise) (paise) (paise) (paise)\n", + "==============================================================================\n", + "100 1 31 40 71 71\n", + " 75 0.75 31 30 61 81\n", + " 50 0.50 31 20 51 102\n", + " 25 0.25 31 10 41 163\n", + "==============================================================================\n" + ] + } + ], + "source": [ + "#Input data\n", + "F=2700#Fixed cost of the thermal station per kW of installed capacity per year in Rs,\n", + "FO=40#Fuel and operating costs per kWh generated in paise\n", + "L=[100,75,50,25]#Load factors\n", + "Y=8760#Number of hours in a year of 365 days\n", + "\n", + "#Calculations\n", + "FC=(F/Y)*100#Fixed costs per kW per hour in paise\n", + "E1=(L[0]/100)#Energy produced in 1 hr with 1 kW plant in kWh\n", + "FOC1=(E1*FO)#Fuel and operating cost in paise\n", + "TC1=(FC+FOC1)#Total cost per hr in paise\n", + "C1=(TC1/E1)#Cost per kWh in paise\n", + "E2=(L[1]/100)#Energy produced in 1 hr with 1 kW plant in kWh\n", + "FOC2=(E2*FO)#Fuel and operating cost in paise\n", + "TC2=(FC+FOC2)#Total cost per hr in paise\n", + "C2=(TC2/E2)#Cost per kWh in paise\n", + "E3=(L[2]/100)#Energy produced in 1 hr with 1 kW plant in kWh\n", + "FOC3=(E3*FO)#Fuel and operating cost in paise\n", + "TC3=(FC+FOC3)#Total cost per hr in paise\n", + "C3=(TC3/E3)#Cost per kWh in paise\n", + "E4=(L[3]/100)#Energy produced in 1 hr with 1 kW plant in kWh\n", + "FOC4=(E4*FO)#Fuel and operating cost in paise\n", + "TC4=(FC+FOC4)#Total cost per hr in paise\n", + "C4=(TC4/E4)#Cost per kWh in paise\n", + "\n", + "#Output\n", + "print \"==============================================================================\\nLoad Energy produced Fixed cost Fuel and Total cost Cost per\\nfactor in 1hr with per hr operating cost per hr kWh\\n(percent) 1kW plant(kWh) (paise) (paise) (paise) (paise)\\n==============================================================================\\n%3.0f %3.0f %3.0f %3.0f %3.0f %3.0f\\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\\n%3.0f %3.2f %3.0f %3.0f %3.0f %3.0f\\n==============================================================================\"%(L[0],E1,FC,FOC1,TC1,C1,L[1],E2,FC,FOC2,TC2,C2,L[2],E3,FC,FOC3,TC3,C3,L[3],E4,FC,FOC4,TC4,C4)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch10.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch10.ipynb new file mode 100644 index 00000000..9acec3f7 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch10.ipynb @@ -0,0 +1,957 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 10 : Hydroelecric power plant" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.1 Pg: 709" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the efficiency of the runner is 93.17 percent \n", + " (b) the diameter of each jet is 0.1614 m\n" + ] + } + ], + "source": [ + "from math import sqrt,cos,pi\n", + "from __future__ import division\n", + "#Input data\n", + "P=4000#Power in kW\n", + "N=400#Speed in r.p.m\n", + "h=200#Head in m\n", + "e=90#Efficiency in percent\n", + "d=1.5#Diameter in m\n", + "vd=10#Percentage decrease in velocity\n", + "a=165#Angle with which jet is deflected in degrees\n", + "\n", + "#Calculations\n", + "V1=sqrt(2*9.81*h*(e/100))#Velocity in m/s\n", + "Vb=(3.14*d*N)/60#Velocity in m/s\n", + "nn=((2*(1-((e/100)*cos(pi/180*a)))*(V1-Vb)*Vb)/V1**2)*100#Efficiency in percent\n", + "p=(P/(nn/100))#Power developed in kW\n", + "pj=(p/2)#Power developed per jet in kW\n", + "dx=sqrt((pj*8)/(3.14*V1**3))#Diameter of each jet in m\n", + "\n", + "#Output\n", + "print \" (a) the efficiency of the runner is %3.2f percent \\n (b) the diameter of each jet is %3.4f m\"%(nn,dx)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.2 Pg: 710" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the number of jets are 3 \n", + " (b) diameter of each jet is 0.123 m \n", + " (c) diameter of the wheel is 1.23 m \n", + " (d) the quantity of water required is 2.4 m**3/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "P=6000#Power in kW\n", + "h=300#Net head availabe in m\n", + "N=550#Speed in r.p.m\n", + "rd=(1/10)#Ratio of jet diameter to wheel diameter\n", + "nh=0.85#Hydraulic efficiency \n", + "Cv=0.98#Coefficient of velocity\n", + "f=0.46#Speed ratio\n", + "d=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "V1=Cv*sqrt(2*9.81*h)#Velocity in m/s\n", + "Vb=f*sqrt(2*9.81*h)#Velocity in m/s\n", + "Q=((P*10**3)/(nh*d*9.81*h))#Discharge in m**3/s\n", + "D=((Vb*60)/(3.14*N))#Diameter in m\n", + "d=(D/10)#Diameter of jet in m\n", + "n=(Q/((V1*(3.14/4)*d**2)))#Number of jets\n", + "\n", + "#Output\n", + "print \" (a) the number of jets are%3.0f \\n (b) diameter of each jet is %3.3f m \\n (c) diameter of the wheel is %3.2f m \\n (d) the quantity of water required is %3.1f m**3/s\"%(n,d,D,Q)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.3 Pg: 711" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Diameter of jet is 0.183 m \n", + " Diameter of bucket wheel is 3.61 m\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "from __future__ import division\n", + "#Input data\n", + "P=10#Capacity in MW\n", + "h=500#Head in m\n", + "Ns=10#Specific speed of the turbine\n", + "on=80#Overall efficiency in percent\n", + "Cv=0.98#Coefficient of velocity\n", + "x=0.46#Speed of the bucket wheel to the velocity of jet\n", + "da=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "N=(Ns*h**(5/4))/sqrt(P*10**3)#Speed in r.p.m\n", + "V=(Cv*sqrt(2*9.81*h))#Velocity in m/s\n", + "Vb=(x*V)#Speed of bucket wheel in m/s\n", + "D=((60*Vb)/(3.14*N))#Diameter in m\n", + "d=sqrt((P*10**6)/((on/100)*(3.14/4)*da*V*9.81*h))#Diameter in m\n", + "\n", + "#Output\n", + "print \" Diameter of jet is %3.3f m \\n Diameter of bucket wheel is %3.2f m\"%(d,D)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.4 Pg: 711" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The specific speed of a single jet Pelton wheel is about 202 (d/D) where d and D represent the jet and bucket wheel diameters respectively\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "Cv=0.97#Coefficient of velocity\n", + "f=0.45#Friction coefficient\n", + "h=0.85#Head in m\n", + "d=1000#Density in kg/m**3\n", + "n=1#For a single jet turbine\n", + "\n", + "#Calculations\n", + "Ns=((60/3.14)*(f*sqrt(2*9.8))*sqrt(n*(3.14/4)*Cv*sqrt(2*9.8)*9.8*h))#Specific speed in terms of d/D \n", + "\n", + "#Output\n", + "print \"The specific speed of a single jet Pelton wheel is about %3.0f (d/D) where d and D represent the jet and bucket wheel diameters respectively\"%(Ns)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.5 Pg: 712" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Velocity of the jet for maximum efficiency is 90 m/s \n", + " Power developed is 4050 kW \n", + " Hydraulic efficiency is 98.3 percent\n" + ] + } + ], + "source": [ + "from math import cos,pi\n", + "#Input data\n", + "n=4#Number of jets\n", + "d=60#Diameter of each jet in mm\n", + "a=165#Angle in degrees\n", + "v=45#Speed of the bucket wheel in m/s\n", + "de=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "v1=(2*v)#Jet velocity in m/s\n", + "Q=(3.14/4)*(d/1000)**2*v1#Discharge in m**3/s\n", + "P=(1-cos(pi/180*a))*(v1**2/4)*Q*de*10**-3#Power developed in kW\n", + "P4=(P*4)#For four jets in kW\n", + "nd=((1-cos(pi/180*a))/2)*100#Maximum efficiency in percent\n", + "\n", + "#Output\n", + "print \" Velocity of the jet for maximum efficiency is %3.0f m/s \\n Power developed is %d kW \\n Hydraulic efficiency is %3.1f percent\"%(v1,P4,nd)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.6 Pg: 713" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Head on the wheel is 43.3 m \n", + " The power generated by the turbine is 238 kW \n", + " Eit angle of guide vanes is 173.29 degrees and Inlet blade angle is 33.7 degrees\n" + ] + } + ], + "source": [ + "from math import atan, degrees\n", + "#Input data\n", + "v=20#Peripheral velocity in m/s\n", + "vw=17#Velocity of whirl in m/s\n", + "vr=2#Radial velocity in m/s\n", + "Q=0.7#Flow in m**3/s\n", + "hn=80#Hydraulic efficiency in percent\n", + "d=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "H=((vw*v)/(9.81*(hn/100)))#Head on the wheel in m\n", + "P=(d*Q*9.81*H*(hn/100)*10**-3)#Power generated in kW\n", + "al=180-degrees(atan(vr/vw))#Angle of guide vanes in degrees\n", + "bl=degrees(atan(vr/(v-vw)))#Inlet blade angle in degrees\n", + "\n", + "#Output\n", + "print \" Head on the wheel is %3.1f m \\n The power generated by the turbine is %3.0f kW \\n Eit angle of guide vanes is %3.2f degrees and Inlet blade angle is %3.1f degrees\"%(H,P,al,bl)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.7 Pg: 714" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Inlet blade angle is 35 degrees 38 minutes \n", + " Outlet blade angle is 17 degrees 16 minute\n" + ] + } + ], + "source": [ + "from math import atan,tan,pi,sqrt,degrees\n", + "#Input data\n", + "od=1.5#Outer diameter in m\n", + "id=0.75#Inner diameter in m\n", + "h=150#Head in m\n", + "P=14000#Power in kW\n", + "Ns=120#Specific speed \n", + "vw2=0#Velocity in m/s\n", + "a=(11+(20/60))#Angle in degrees\n", + "hn=92#Hydraulic efficiency in percent\n", + "\n", + "#Calculations\n", + "N=(Ns*h**(5/4))/sqrt(P)#Speed in rpm\n", + "vb1=(3.14*od*N)/60#velocity in m/s\n", + "vw1=(((hn/100)*9.81*h)/vb1)#velocity in m/s\n", + "vf1=(tan(pi/180*a)*vw1)#Velocity in m/s\n", + "vf2=vf1#Velocity in m/s\n", + "b1=degrees(atan(vf1/(vb1-vw1)))#Angle in degrees\n", + "b1x=(b1-int(b1))*60#For output\n", + "vb2=(vb1/2)#Velocity in m/s\n", + "b2=degrees(atan(vf1/(vb2-vw2)))#Angle in degrees\n", + "b2x=(b2-int(b2))*60#For output\n", + "\n", + "#Output\n", + "print \" Inlet blade angle is %3.0f degrees %3.0f minutes \\n Outlet blade angle is %3.0f degrees %3.0f minute\"%(b1,b1x,b2,b2x)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.8 Pg: 715" + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)the guide vane angle is 13.3 degrees \n", + " (b)the runner vane angle at inlet is 26.6 degrees and outlet is 41.72 degrees \n", + " (c)the diametres of the runner at inlet is 0.5 metre and outlet is 0.25 metre\n", + " (d)the width of the wheel at inlet is 0.05 metre\n" + ] + } + ], + "source": [ + "from math import sqrt,pi,atan,degrees\n", + "#Input data\n", + "h=70#net head in m\n", + "N=700#speed in rpm\n", + "o=85#over all efficiency in %\n", + "P=350#shaft power in kW\n", + "he=92#hydraulic efficiency in %\n", + "fr=.22#flow ratio\n", + "b=.1#breadth ratio\n", + "s=2#outer diameter in terms of inner diametre\n", + "#Calculations\n", + "vf1=fr*sqrt(2*9.81*h)#velocity in m/s\n", + "q=(P/(9.81*h*(o/100)))#discharge in m**3/s\n", + "d1=sqrt(q/(.94*b*vf1*3.14))#diameter in metre\n", + "b1=d1*b#breadth in metre\n", + "d2=d1/2#diametre in metre\n", + "vb1=(3.14*d1*N)/60#velocity in m/s\n", + "vw1=((he/100)*9.81*h)/vb1#velcity in m/s\n", + "a=degrees(atan(vf1/vw1))#angle in degrees\n", + "bet=degrees(atan(vf1/(vw1-vb1)))#angle in degrees\n", + "vb2=(d2/d1)*vb1#velocity in m/s\n", + "bet2=degrees(atan(vf1/vb2))#angle in degrees\n", + "\n", + "#Output\n", + "print \" (a)the guide vane angle is %3.1f degrees \\n (b)the runner vane angle at inlet is %3.1f degrees and outlet is %3.2f degrees \\n (c)the diametres of the runner at inlet is %3.1f metre and outlet is %3.2f metre\\n (d)the width of the wheel at inlet is %3.2f metre\"%(a,bet,bet2,d1,d2,b1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.9 Pg: 717" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the discharge of the turbine is 15.34 m**3/s \n", + " (b) the jet diameter is 0.227 m \n", + " (c) the nozzle tip diameter is 0.284 m \n", + " (d) the pitch circle diameter of the wheel is 2.89 m \n", + " (e) the specific speed is 69.04 \n", + " (f) the number of buckets on the wheel are 22 \n", + " (g) the workdone per kg of water on the wheel is 443.02 kg.m/kg \n", + " (h) the hydraulic efficiency is 93 percent\n" + ] + } + ], + "source": [ + "from math import ceil,sqrt,cos,pi\n", + "#Input data\n", + "n=4#Number of units\n", + "P=70000#Power in kVA\n", + "f=50#Frequency in Hz\n", + "p=10#No.of pair of poles\n", + "h=505#Gross head in m\n", + "tn=94#Transmission efficiency in percent\n", + "po=260#Power in MW\n", + "e=91#Efficiency in percent\n", + "nn=0.98#Nozzle efficiency\n", + "Cv=0.98#Coefficient of velocity\n", + "x=0.48#Vb=0.48 V\n", + "dd=25#Nozzle diameter is 25% bigger than jet diameter\n", + "a=165#Angle of buckets in degrees\n", + "de=99.75#Discharge efficiency in percent\n", + "\n", + "#Calculations\n", + "N=(120*f)/(p*2)#Synchronous speed in r.p.m\n", + "nh=((tn/100)*h)#Net head in m\n", + "pt=(po*10**3)/n#Power developed per turbine in kW\n", + "ip=(pt/(e/100))#Input water power in kW\n", + "Q=(ip/(9.81*nh))#Discharge in m**3/s\n", + "Qj=(Q/n)#Discharge per jet in m**3/s\n", + "V1=Cv*sqrt(2*9.81*nh)#Velocity in m/s\n", + "d=sqrt((4/3.14)*(Qj/V1))#Diameter of jet in m\n", + "nd=(1+(dd/100))*d#Nozzle tip diameter in m\n", + "Vb=(x*V1)#Velocity in m/s\n", + "D=((Vb*60)/(3.14*N))#Pitch circle diameter of the wheel in m\n", + "Ns=((N*sqrt(po*10**3))/nh**(5/4))#Specific speed\n", + "jr=(D/d)#Jet ratio\n", + "nob=(jr/2)+15#Number of buckets\n", + "nobb=ceil(nob)#Rounding off to next integer\n", + "W=((V1-Vb)*(1-(nn*cos(pi/180*a)))*Vb)/9.81#Workdone per kg in kg.m/kg\n", + "nth=((W/nh)*de)#Hydraulic efficiency in percent\n", + "\n", + "#Output\n", + "print \" (a) the discharge of the turbine is %3.2f m**3/s \\n (b) the jet diameter is %3.3f m \\n (c) the nozzle tip diameter is %3.3f m \\n (d) the pitch circle diameter of the wheel is %3.2f m \\n (e) the specific speed is %3.2f \\n (f) the number of buckets on the wheel are %3.0f \\n (g) the workdone per kg of water on the wheel is %3.2f kg.m/kg \\n (h) the hydraulic efficiency is %3.0f percent\"%(Q,d,nd,D,Ns,nobb,W,nth)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.10 Pg: 718" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Blade angle at inlet is 62 degrees \n", + " Hydraulic efficiency is 81 percent\n" + ] + } + ], + "source": [ + "from math import sin,cos,degrees,atan,sqrt,pi\n", + "#Input data\n", + "gh=35#Gross head in m\n", + "md=2#Mean diameter in m\n", + "N=145#Speed in rpm\n", + "a=30#Angle in degrees\n", + "oa=28#Outlet angle in degrees\n", + "x=7#Percentage of gross head lost\n", + "y=8#Reduction in relative velocity in percent\n", + "\n", + "#Calculations\n", + "H=((100-x)/100)*gh#Net haed in m\n", + "V1=sqrt(2*9.81*H)#Velocity in m/s\n", + "Vb=(3.14*md*N)/60#Velocity in m/s\n", + "b1=degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Angle in degrees\n", + "Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s\n", + "Vr2=((100-y)/100)*Vr1#Velocity in m/s\n", + "Vw1=(V1*cos(pi/180*a))#Velocity in m/s\n", + "Vw2=(Vb-(Vr2*cos(pi/180*oa)))#Velocity in m/s\n", + "E=((Vb*(Vw1-Vw2))/9.81)#Workdone in kg.m/kg\n", + "nb=(E/gh)*100#Hydraulic efficiency in percent\n", + "\n", + "#Output\n", + "print \" Blade angle at inlet is %3.0f degrees \\n Hydraulic efficiency is %3.0f percent\"%(b1,nb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.11 Pg: 719" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the speed is 166.7 rpm \n", + " (b) the diameter of the runner is 3.56 m \n", + " (c) the specific speed is 746\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "P=10000#Power in kW\n", + "h=12#Head in m\n", + "Nr=2#Speed ratio\n", + "Fr=0.65#Flow ratio\n", + "x=0.3#Diameter of hub is 0.3 times the eternal diameter of the vane \n", + "on=94#Overall efficiency in percent\n", + "\n", + "#Calculations\n", + "Q=(P/(9.81*h*(on/100)))#Discharge in m**3/s\n", + "Vr1=(Fr*sqrt(2*9.81*h))#Velocity in m/s\n", + "Ab=(Q/Vr1)#Area of flow in m**2\n", + "D=sqrt(((Ab*4)/3.14)/(1-x**2))#Diameter of runner in m\n", + "Vb=(Nr*sqrt(2*9.81*h))#Velocity in m/s\n", + "N=((Vb*60)/(3.14*D))#Speed in rpm\n", + "f=50#Taking frequency as 50 Hz\n", + "p=(120*50)/N#Number of poles\n", + "N1=(120*f)/int(p)#Speed in rpm\n", + "Ns=(N1*sqrt(P))/h**(5/4)#Specific speed\n", + "\n", + "#Output\n", + "print \" (a) the speed is %3.1f rpm \\n (b) the diameter of the runner is %3.2f m \\n (c) the specific speed is %3.0f\"%(N1,D,Ns)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.12 Pg: 721" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Specific speed is 241.5 \n", + " Normal speed is 120.7 rpm \n", + " Output under a head of 20 m is 7155 kW\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "P=10000#Power in kW\n", + "h=25#Head in m. In textbook it is given wrong as 2 m\n", + "N=135#Speed in rpm\n", + "h1=20#Head in m\n", + "\n", + "#Calculations\n", + "Ns=((N*sqrt(P))/h**(5/4))#Specific speed\n", + "N1=sqrt(h1/h)*N#Speed in rpm\n", + "P2=P/(h/h1)**(3/2)#Power in kW\n", + "\n", + "#Output\n", + "print \" Specific speed is %3.1f \\n Normal speed is %3.1f rpm \\n Output under a head of %d m is %3.0f kW\"%(Ns,N1,h1,P2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.13 Pg: 721" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The number of turbines in \n", + " (a) Francis turbine are 2 \n", + " (b) Kaplan turbine are 4\n" + ] + } + ], + "source": [ + "from math import ceil\n", + "#Input data\n", + "Q=175#Discharge in m**3/s\n", + "h=18#Head in meter\n", + "N=150#Speed in rpm\n", + "oe=82#Overall efficiency in percent\n", + "Ns1=460#Maximum specific speed\n", + "Ns2=350#Maximum specific speed\n", + "d=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "P=(d*Q*9.81*h*(oe/100)*10**-3)#power in kW\n", + "P1=((Ns1*h**(5/4))/N)**2#Power in kW\n", + "n1=P/P1#No.of turbains\n", + "P2=((Ns2*h**(5/4))/N)**2#Power in kW\n", + "n2=ceil(P/P2)#No.of turbains\n", + "\n", + "#Output\n", + "print \"The number of turbines in \\n (a) Francis turbine are%3.0f \\n (b) Kaplan turbine are %d\"%(n1,n2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.14 Pg: 722" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Speed is 285 rpm \n", + " Power is 23.309 kW \n", + " Scale ratio is 5.518 \n", + " Flow through the turbine is 63.6 m**3/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "Ns=210#Specific speed \n", + "P=30#Power in MW\n", + "N=180#Speed in rpm\n", + "Q=0.6#Discharge in m**3/s\n", + "h=4.5#Head in m\n", + "e=88#Efficiency in percent\n", + "d=1000#Density in kg/m**3\n", + "\n", + "#Calculations\n", + "Pm=(d*Q*9.81*h*(e/100)*10**-3)#Power in kW\n", + "Nm=(Ns*h**(5/4))/sqrt(Pm)#Speed in rpm\n", + "Hp=((N*sqrt(P*1000))/Ns)**(4/5)#Head in m\n", + "Dpm=(Nm/N)*sqrt(Hp/h)#Scale ratio\n", + "Qp=(P*10**6)/(d*9.81*Hp*(e/100))#Discharge in m**3/s\n", + "\n", + "#Output\n", + "print \" Speed is %3.0f rpm \\n Power is %3.3f kW \\n Scale ratio is %3.3f \\n Flow through the turbine is %3.1f m**3/s\"%(Nm,Pm,Dpm,Qp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.15 Pg: 723" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Speed is 402 rpm \n", + " Power is 11180 kW\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "x=1/5#Scale model\n", + "h=1.5#Head in m\n", + "P=5#Power in kW\n", + "N=450#Speed in rpm\n", + "h1=30#Head in m\n", + "\n", + "#Calculations\n", + "N1=(x*N)/sqrt(h/h1)#Speed in rpm\n", + "Ns=(N*sqrt(P))/h**(5/4)#Specific speed\n", + "P1=((Ns*h1**(5/4))/N1)**2#Power in kW\n", + "\n", + "#Output\n", + "print \" Speed is %3.0f rpm \\n Power is %3.0f kW\"%(N1,P1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.16 Pg: 723" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Speed of the turbine is 307.8 rpm \n", + " Maximum flow rate is 1.5 m**3/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "h=19#Head in m\n", + "Q=3#Flow rate in m**3/s\n", + "N=600#Speed in rpm\n", + "h1=5#Head in m\n", + "\n", + "#Calculations\n", + "N1=N/sqrt(h/h1)#Speed in rpm\n", + "Q1=Q/sqrt(h/h1)#Discharge in m**3/s\n", + "\n", + "#Output\n", + "print \" Speed of the turbine is %3.1f rpm \\n Maximum flow rate is %3.1f m**3/s\"%(N1,Q1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.17 Pg: 724" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Least number of machines required if using \n", + " (a) Francis turbines are 13 \n", + " (b) Kaplan turbines are 2\n" + ] + } + ], + "source": [ + "from math import ceil\n", + "#Input data\n", + "Q=350#Discharge in m**3/s\n", + "h=30#Head in m\n", + "e=87#Turbine efficiency in percent\n", + "f=50#Frequency in Hz\n", + "p=24#Number of poles\n", + "Ns1=300#Specific speed\n", + "Ns2=820#Specific speed\n", + "d=1000#Dnsity of water in kg/m**3\n", + "\n", + "#Calculations\n", + "N=(120*f)/p#Speed in rpm\n", + "P=d*Q*9.81*h*(e/100)*10**-3#Power in kW\n", + "P1=((Ns1*h**(5/4))/N)**2#Power in kW\n", + "n1=P/P1#No.of turbines\n", + "P2=((Ns2*h**(5/4))/N)**2#Power in kW\n", + "n2=ceil(P/P2)#No.of turbines\n", + "\n", + "#Output\n", + "print \"Least number of machines required if using \\n (a) Francis turbines are %3.0f \\n (b) Kaplan turbines are %3.0f\"%(n1,n2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.18 Pg: 725" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Power developed is 2302 kW \n", + " (b) As the available head is low, Kaplan turbines are suggested.\n", + " Two turbines each of 3000kW capacity may be installed.\n" + ] + } + ], + "source": [ + "#Input data\n", + "h=27#Head in m\n", + "A=430#Area in sq.km\n", + "R=150#Rainfall in cm/year\n", + "pr=65#Percentage of rainfall utilised\n", + "pe=95#Penstock efficiency in percent\n", + "te=80#Turbine efficiency in percent\n", + "ge=86#Generator efficiency in percent\n", + "lf=0.45#Load factor\n", + "d=1000#Density of water in kg/m**3\n", + "\n", + "#Calculations\n", + "Q=A*10**6*(R/100)*(pr/100)#Discharge in m**3 per year\n", + "Qs=(Q/(365*24*3600))#Quantity of water per second in m**3\n", + "P=(pe/100)*(te/100)*(ge/100)*d*Qs*9.81*h*10**-3#Power in kW\n", + "plc=(P/lf)#Peak load capacity in kW\n", + "C=(plc/(2*(ge/100)))#Capacity of each unit in kW\n", + "\n", + "#Output\n", + "print \" (a) Power developed is %3.0f kW \\n (b) As the available head is low, Kaplan turbines are suggested.\\n Two turbines each of 3000kW capacity may be installed.\"%(P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 10.19 Pg: 725" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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K+P73U9L46U9dXZjV4sJFTwLvVdqAWS1JsN9+6XoXU6emQQYnT653VGa9WzkV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\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f04cd2ebed0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed is 13.79 MW\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, subplot, title,show,xlabel,ylabel\n", + "#Input data\n", + "q=[0,30,25,20,0,10,50,80,100,110,65,45,30]#Mean discharge in millions of cu.m per month respectively\n", + "h=90#Head in m\n", + "n=86#Overall efficiency in percent\n", + "\n", + "#Calculations\n", + "Qm=(q[(1)]+q[(2)]+q[(3)]+q[(4)]+q[(5)]+q[(6)]+q[(7)]+q[(8)]+q[(9)]+q[(10)]+q[(11)]+q[(12)])/12#Mean discharge in millions m**3/s\n", + "Q=[0,30,30,25,25,20,20,0,0,10,10,50,50,80,80,100,100,110,110,65,65,45,45,30,30,0]#Discharge(million m**3/month) on y-axis\n", + "y=[0,0,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10,11,11,12,12]#Months on x-ais\n", + "D=[0,110,100,90,80,70,60,50,40,30,25,20,10,0]#Discharge per month in million m**3\n", + "pt=[0,8.3,16.7,25,25,25,33.3,41.7,50,66.7,75,83.3,91.7,100]#Percentage time \n", + "Po=((Qm*10**6*9.81*h*(n/100))/(30*24*3600*1000))#Power developed in MW\n", + "\n", + "#Output\n", + "#subplot(131)\n", + "plot(y[1:],Q[1:])#Graph Discharge(million m**3/month) vs Month\n", + "title(\"Discharge(million m**3/month) vs Month\")\n", + "xlabel(\"Months\")\n", + "ylabel(\"Discharge(million m**3/month)\")\n", + "show()\n", + "#subplot(133)\n", + "plot(pt[1:],D[1:])#Graph percentage time vs Discharge(million m**3/month)\n", + "title(\"percentage time vs Discharge(million m**3/month)\")\n", + "xlabel(\"percentage time\")\n", + "ylabel(\"Discharge(million m**3/month)\")\n", + "show()\n", + "print \"Power developed is %3.2f MW\"%(Po)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch11.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch11.ipynb new file mode 100644 index 00000000..84b289a0 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch11.ipynb @@ -0,0 +1,529 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 11 : Diesel engine and Gas Turbine Power Plants" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.1 Pg: 766" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Net increase in brake power is 28.67 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "C=3.5#Capacity in litres\n", + "P=13.1#Indicated power in kW/m**3\n", + "N=3600#Speed in rpm\n", + "ve=82#Volumetric efficiency in percent\n", + "p1=1.013#Pressure in bar\n", + "T1=25+273#Temperature in K\n", + "rp=1.75#Pressure ratio\n", + "ie=70#Isentropic efficiency in percent\n", + "me=80#Mechanical efficiency in percent\n", + "g=1.4#Ratio of specific heats\n", + "R=0.287#Gas constant in kJ/kg.K\n", + "Cp=1.005#Specific heat in kJ/kg.K\n", + "\n", + "#Calculations\n", + "EC=(C/1000)#Engine capacity in m**3\n", + "Vs=(N/2)*EC#Swept volume in m**3\n", + "Vui=(ve/100)*Vs#Unsupercharged induced volume in m**3/min\n", + "dp=(rp*p1)#Blower delivery pressure in bar\n", + "T2sT1=(rp)**((g-1)/g)#Ratio of temperatures\n", + "T2s=(T2sT1*T1)#Temperature in K\n", + "dT21=(T2s-T1)/(ie/100)#Difference in temperature in K\n", + "T2=dT21+T1#Temperature in K\n", + "EV=(Vs*dp*T1)/(p1*T2)#Equivalent volume in m**3/min\n", + "iiv=EV-Vui#Increase in induced volume in m**3/min\n", + "iip=(P*iiv)#Increase in indicated power in kW\n", + "iipi=((dp-p1)*100*Vs)/60#Increase in induced power due to increase in induction pressure in kW\n", + "tiip=iip+iipi#Total increase in indicated power in kW\n", + "tibp=tiip*(me/100)#Total increase in brake power in kW\n", + "ma=(dp*100*Vs)/(60*R*T2)#Mass of air in kg/s\n", + "WI=(ma*Cp*(T2-T1))#Work input to heater in kW\n", + "Pb=(WI/(me/100))#Power required in kW\n", + "NI=tibp-Pb#Net increase in brake power in kW\n", + "\n", + "#Output\n", + "print \"Net increase in brake power is %3.2f kW\"%(NI)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.2 Pg: 768" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the temperature of air leaving the compressor is 129.76 degree C \n", + " (b) the temperature of gases leaving the turbine is 482.26 degree C \n", + " (c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is 15.08 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "p1=0.97#Pressure in bar\n", + "t1=30+273#Temperature in K\n", + "p2=2.1#Pressure in bar\n", + "af=18#Air fuel ratio\n", + "t3=580+273#Temperature in K\n", + "p3=1.9#Pressure in bar\n", + "p4=1.06#Pressure in bar\n", + "iec=0.75#Isentropic efficiency of compressor\n", + "iet=0.85#Isentropic efficiency of turbine\n", + "cpa=1.01#Specific heat for air in kJ/kg.K\n", + "ga=1.4#Ratio of specific heats\n", + "cpex=1.15#Specific heat in kJ/kg.K\n", + "gex=1.33#Ratio of specific heats\n", + "\n", + "#Calculations\n", + "t2s=t1*(p2/p1)**((ga-1)/ga)#Tempeature in K\n", + "t21=(t2s-t1)/iec#Temperature in K\n", + "t2=t21+t1#Temperature in K\n", + "T2=t2-273#Temperature in degree C\n", + "t3t4s=(p3/p4)**((gex-1)/gex)#Ratio of temperatures\n", + "t4s=(t3/t3t4s)#Temperature in K\n", + "t4=t3-((t3-t4s)*iet)#Temperature in K\n", + "T4=t4-273#Temperature in degree C\n", + "mp=(((cpex*(1+(1/af))*(t3-t4))-(cpa*(t2-t1)))/(cpex*(1+(1/af))*(t3-t4)))*100#Percentage of mechanical power loss\n", + "\n", + "#Output\n", + "print \" (a) the temperature of air leaving the compressor is %3.2f degree C \\n (b) the temperature of gases leaving the turbine is %3.2f degree C \\n (c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is %3.2f percent\"%(T2,T4,mp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.3 Pg: 770" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " ENERGY BALANCE SHEET \n", + " (in kW) (in percent)\n", + " 1. Brake power 53.12 37.69 \n", + " 2. Heat carried away by exhaust gases 29.09 20.64 \n", + " 3. Heat lost to jacket cooling water 45.10 32.00 \n", + " 4. Heat loss unaccounted 13.64 9.67 \n", + " Total 140.94 100.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "a=215#Current in A\n", + "v=210#Voltage in V\n", + "e=85#Efficiency in percent\n", + "q=11.8#Quantity of fuel supplied in kg/h\n", + "cv=43#Calorific value in MJ/kg\n", + "af=18#Air fuel ratio\n", + "w=560#Water in litres/h\n", + "tw=38#Temeparature in degree C\n", + "te=97#Temeparature in degree C\n", + "cp=1.04#Specific heat in kJ/kg.K\n", + "ta=30#Temeparature in degree C\n", + "l=32#Percentage lost \n", + "sw=4.187#Specific heat in kJ/kg.K\n", + "\n", + "#Calculations\n", + "P=(a*v)/1000#Power in kW\n", + "BP=(P/(e/100))#Brake power in kW\n", + "E=(q/3600)*cv*1000#Energy supplied in kW\n", + "mg=(q/3600)*(1+af)#Rate of gases in kg/s\n", + "he=(mg*cp*(te-ta))+((w/3600)*sw*tw)#Heat carried away by exhaust gases in kW\n", + "hj=(l/100)*E#Heat lost to jacket cooling water in kW\n", + "pBP=(BP/E)*100#Percentage\n", + "pE=(E/E)*100#Percentage\n", + "phe=(he/E)*100#Percenatge\n", + "phj=(hj/E)*100#Percenatge \n", + "\n", + "#Output\n", + "print \" ENERGY BALANCE SHEET \\n (in kW) (in percent)\\n 1. Brake power %3.2f %3.2f \\n 2. Heat carried away by exhaust gases %3.2f %3.2f \\n 3. Heat lost to jacket cooling water %3.2f %3.2f \\n 4. Heat loss unaccounted %3.2f %3.2f \\n Total %3.2f %3.2f\"%(BP,pBP,he,phe,hj,phj,(E-(BP+he+hj)),(((E-(BP+he+hj))/E)*100),E,(pBP+phe+phj+(((E-(BP+he+hj))/E)*100)))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.4 Pg: 771" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Indicated power is 18.07 kW \n", + " Brake power is 12.811 kW \n", + "\n", + " ENERGY BALANCE SHEET \n", + " (in kJ) (in percent)\n", + " 1. Energy equivalent in ip 21685 31.59 \n", + " 2. Energy carried away by cooling water 16748 24.40 \n", + " 3. Energy carried away by exhaust gases 18887 27.52 \n", + " 4. Unaccounted for energy loss 11320 16.49 \n", + " Total 68640 100.00\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "t=20#Trial time in minutes\n", + "NL=680#Net brake load in N\n", + "mep=3#Mean effective pressure in bar\n", + "N=360#Speed in rpm\n", + "Fc=1.56#Fuel consumption in kg\n", + "cw=160#Cooling water in kg\n", + "Tw=32#Temperature of water at inlet in degree C\n", + "Wo=57#Water outlet temperature in degree C\n", + "a=30#Air in kg\n", + "Ta=27#Room temperature in degree C\n", + "Te=310#Exhaust gas temperature in degree C\n", + "d=210#Bore in mm\n", + "l=290#Stroke in mm\n", + "bd=1#Brake diameter in m\n", + "cv=44#Calorific value in MJ/kg\n", + "st=1.3#Steam formed in kg per kg fuel in the exhaust\n", + "cp=2.093#Specific heat of steam in exhaust in kJ/kg.K\n", + "cpx=1.01#Specific heat of dry exhaust gases in kJ/kg.K\n", + "cpw=4.187#Specific heat of water in kJ/kg.K\n", + "\n", + "#Calculations\n", + "ip=(mep*100*(l/1000)*(3.14/4)*(d/1000)**2*N)/60#Indicated Power in kW\n", + "bp=((2*3.14*N*(NL*(1/2)))/60)/1000#Brake power in kW\n", + "nm=(bp/ip)*100#Mechanical efficiency in percent\n", + "qs=(Fc*cv*10**3)#Heat supplied in kJ\n", + "qip=(ip*t*60)#Heat equivalent of ip in kJ\n", + "qcw=(cw*cpw*(Wo-Tw))#Heat carried away by cooling water in kJ\n", + "tm=(Fc*a)#Toatl mass of exhaust gas in kg\n", + "ms=(st*Fc)#Mass of steam formed in kg\n", + "mde=(tm-ms)#Mass of dry exhaust gas in kg\n", + "Ed=(mde*cpx*(Te-Ta))#Energy carried away by dry exhaust gases in kJ\n", + "Es=(ms*((cpw*(100-Ta))+2257.9+(cp*(Te-100))))#Energy carried away by steam in kJ\n", + "TE=(Ed+Es)#Total energy carried away by exhaust gases in kJ\n", + "ue=(qs-(qip+qcw+TE))#Unaccounted energy in kJ\n", + "pqip=(qip/qs)*100#Percentage\n", + "pqcw=(qcw/qs)*100#Percentage\n", + "pTE=(TE/qs)*100#Percentage\n", + "pue=(ue/qs)*100#Percentage\n", + "\n", + "#Output\n", + "print \" Indicated power is %3.2f kW \\n Brake power is %3.3f kW \\n\\n ENERGY BALANCE SHEET \\n (in kJ) (in percent)\\n 1. Energy equivalent in ip %3.0f %3.2f \\n 2. Energy carried away by cooling water %3.0f %3.2f \\n 3. Energy carried away by exhaust gases %3.0f %3.2f \\n 4. Unaccounted for energy loss %3.0f %3.2f \\n Total %3.0f %3.2f\"%(ip,bp,qip,pqip,qcw,pqcw,TE,pTE,ue,pue,qs,(pqip+pqcw+pTE+pue))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.5 Pg: 796" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The flow velocity is 393 m/s \n", + " (b) The blade angles at the tip are : \n", + " Fixed blades (root) are 57.28 degrees and 23.88 degrees \n", + " Moving blades (root) are 38.78 degrees and 50.10 degrees \n", + " Fixed blades (tip) are 47.38 degrees and 17.17 degrees \n", + " Moving blades (tip) are 0.45 degrees and 54.23 degrees \n", + " (c) The degree of reaction at : \n", + " the tip is 64 percent \n", + " the root is 26 percent\n" + ] + } + ], + "source": [ + "from math import tan,sin,atan,degrees,pi,cos,sqrt\n", + "#Input data\n", + "Vbm=360#Blade velocity in m/s\n", + "b1=20#Blade angle at inlet in degrees\n", + "a2=b1#Angle in degrees\n", + "b2=52#Blade angle at exit in degrees\n", + "a1=b2#Angle in degrees\n", + "R=50#Degree of reaction in percent\n", + "dm=0.45#Mean diameter of the blade in m\n", + "bh=0.08#Mean blade height in m\n", + "\n", + "#Calculations\n", + "Vf=(Vbm/(tan(pi/180*b2)-tan(pi/180*b1)))#Velocity in m/s\n", + "rt=(dm/2)+(bh/2)#Mean radius in m\n", + "Vbt=(Vbm*(rt/(dm/2)))#Velocity in m/s\n", + "Vw1m=Vf*tan(pi/180*a1)#Velocity in m/s\n", + "Vw1t=(Vw1m*((dm/2)/rt))#Velocity in m/s\n", + "dVw1=(Vf*(tan(pi/180*b1)+tan(pi/180*b2))*Vbm)/Vbt#Velocity in m/s\n", + "rr=(dm/2)-(bh/2)#Radius in m\n", + "Vbr=(Vbm*(rr/(dm/2)))#Velocity in m/s\n", + "Vw1r=(Vw1m*((dm/2)/rr))#Velocity in m/s\n", + "Vr2=Vf/cos(pi/180*b2)#Velocity in m/s\n", + "dVwr=((Vw1m+((Vr2*sin(pi/180*b2))-Vbm))*Vbm)/Vbr#Velocity in m/s\n", + "a1r=degrees(atan(Vw1r/Vf))#Angle in degrees\n", + "a2r=degrees(atan((dVwr-Vw1r)/Vf))#Angle in degrees\n", + "b1r=degrees(atan((Vw1r-Vbr)/Vf))#Angle in degrees\n", + "b2r=degrees(atan((Vbr+(Vf*tan(pi/180*a2r)))/Vf))#Angle in degrees\n", + "a1t=degrees(atan(Vw1t/Vf))#Angle in degrees\n", + "a2t=degrees(atan((dVw1-Vw1t)/Vf))#Angle in degrees\n", + "b1t=degrees(atan((Vw1t-Vbt)/Vf))#Angle in degrees\n", + "b2t=degrees(atan((Vbt+(Vf*tan(pi/180*a2t)))/Vf))#Angle in degrees\n", + "Rt=((Vf*(tan(pi/180*b2t)-tan(pi/180*b1t)))/(2*Vbt))*100#Degree of reaction at the tip in percent\n", + "Rr=((Vf*(tan(pi/180*b2r)-tan(pi/180*b1r)))/(2*Vbr))*100#Degree of reaction at the root in percent\n", + "\n", + "#Output\n", + "print \" (a)The flow velocity is %3.0f m/s \\n (b) The blade angles at the tip are : \\n Fixed blades (root) are %3.2f degrees and %3.2f degrees \\n Moving blades (root) are %3.2f degrees and %3.2f degrees \\n Fixed blades (tip) are %3.2f degrees and %3.2f degrees \\n Moving blades (tip) are %3.2f degrees and %3.2f degrees \\n (c) The degree of reaction at : \\n the tip is %3.0f percent \\n the root is %3.0f percent\"%(Vf,a1r,a2r,b1r,b2r,a1t,a2t,b1t,b2t,Rt,Rr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.6 Pg: 799" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Impeller tip diameter is 549 mm\n" + ] + } + ], + "source": [ + "from math import tan,sin,atan,degrees,pi,cos,sqrt\n", + "#Input data\n", + "N=16000#Speed in rpm\n", + "T1=17+273#Temperature in K\n", + "rp=4#Pressure ratio\n", + "In=82#Isentropic efficiency in percent\n", + "s=0.85#Slip factor\n", + "a=20#Angle in degrees\n", + "d=200#Diameter in mm\n", + "V=120#Velocity in m/s\n", + "cp=1.005#Specific heat in kJ/kg.K\n", + "g=1.4#Ratio of specific heats\n", + "\n", + "#Calculations\n", + "T2sT1=(rp)**((g-1)/g)#Temperature ratio\n", + "T2s=T1*T2sT1#Temeprature in K\n", + "dTs=(T2s-T1)#Temperature difference in K\n", + "dT=dTs/(In/100)#Temperature difference in K\n", + "Wc=(cp*dT)#Power input in kJ/kg\n", + "Vb1=(3.14*(d/1000)*N)/60#Velocity in m/s\n", + "Vw1=(V*sin(pi/180*a))#Pre-whirl velocity in m/s\n", + "Vb2=sqrt(((Wc*1000)+(Vb1*Vw1))/s)#Velocity in m/s\n", + "d2=((Vb2*60)/(3.14*N))*1000#Tip diameter in mm\n", + "\n", + "#Output\n", + "print \"Impeller tip diameter is %3.0f mm\"%(d2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.7 Pg: 801" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Workdone factor of the compressor is 0.86\n" + ] + } + ], + "source": [ + "from math import tan,sin,atan,degrees,pi,cos,sqrt\n", + "#Input data\n", + "T1=25+273#Temperature in K\n", + "rp=6#Pressure ratio\n", + "Vb=220#Mean velocity in m/s\n", + "b1=45#Angle in degrees\n", + "a2=b1#Angle in degrees\n", + "b2=15#Angle in degrees\n", + "a1=b2#Angle in degrees\n", + "R=50#Degree of reaction in percent\n", + "n=10#Number of stages\n", + "In=83#Isentropic efficiency in percent\n", + "cp=1.005#Specific heat in kJ/kg.K\n", + "g=1.4#Ratio of specific heats\n", + "\n", + "#Calculations\n", + "V1=(Vb/(sin(pi/180*b2)+(cos(pi/180*a1)*tan(pi/180*a2))))#Velocity in m/s\n", + "V2=(V1*cos(pi/180*b2))/cos(pi/180*b1)#Velocity in m/s\n", + "dVw=(V2*sin(pi/180*a2))-(V1*sin(pi/180*a1))#Velocity in m/s.Textbook answer is wrong. Correct answer is 127 m/s\n", + "T2sT1=rp**((g-1)/g)#Temperature ratio\n", + "T2s=(T2sT1*T1)#Temperature in K\n", + "dTs=(T2s-T1)#Temperature difference in K\n", + "dT=(dTs/(In/100))#Temperature difference in K\n", + "W=(cp*dT)#Workdone in kJ/kg\n", + "w=(W*10**3)/(Vb*dVw*n)#Work done factor\n", + "\n", + "#Output\n", + "print \"Workdone factor of the compressor is %3.2f\"%(w)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 11.8 Pg: 802" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the air fuel ratio is 75.55 \n", + " (b) the cycle efficiency is 41.8 percent \n", + " (c) the power supplied by the plant is 41 MW \n", + " (d) the specific fuel consumption of the plant is 0.244 kg/kW.h and the fuel consumption per hour is 10007.26 kg\n" + ] + } + ], + "source": [ + "from math import sqrt,pi\n", + "#Input data\n", + "p1=1#Pressure in bar\n", + "T1=20+273#Temperature in K\n", + "Tm=900+273#Maximum temperature in K\n", + "rp=6#Pressure ratio\n", + "e=0.7#Effectiveness of regenerator\n", + "ma=210#Rate of air flow in kg/s\n", + "CV=40800#Calorific value in kJ/kg\n", + "ic=0.82#Isentropic efficiencies of both the compressors\n", + "it=0.92#Isentropic efficiencies of both the turbine\n", + "cn=0.95#Combustion efficiency \n", + "mn=0.96#Mechanical efficiency\n", + "gn=0.95#Generator efficiency\n", + "cp=1.005#Specific heat of air in kJ/kg.K\n", + "cpg=1.08#Specific heat of gas in kJ/kg.K\n", + "g1=1.4#Ratio of specific heats for air\n", + "g=1.33#Ratio of specific heats for gas\n", + "\n", + "#Calculations\n", + "pi=sqrt(p1*rp)#Intermediate pressure in bar\n", + "T2sT1=(pi/p1)**((g1-1)/g1)#Temperature ratio\n", + "T2s=(T2sT1*T1)#temperature in K\n", + "T2=((T2s-T1)/ic)+T1#Temperature in K\n", + "T4s=(T1*(rp/pi)**((g1-1)/g1))#Temperature in K\n", + "T4=((T4s-T1)/ic)+T1#Temperature in K\n", + "T7s=(Tm/(rp/p1)**((g-1)/g))#Temperature in K\n", + "T7=Tm-(it*(Tm-T7s))#Temperature in K\n", + "T5=(e*(T7-T4))+T4#Temperature in K\n", + "mf=1/((cp*(Tm-T5))/((CV*cn)-(cp*(Tm-T5))))#Air fuel ratio\n", + "Wgt=((1+(1/mf))*cpg*(Tm-T7))#Workdone by turbine in kJ/kg of air\n", + "Wc=(cp*((T2-T1)+(T4-T1)))#Workdone by compressor in kJ/kg of air\n", + "Wnet=(Wgt-Wc)#Net workdone in kJ/kg of air\n", + "Q=(CV*cn)/mf#Heat supplied in kJ/kg of air\n", + "ncy=(Wnet/Q)*100#Cycle efficiency in percent\n", + "PO=(Wnet*ma*mn*gn)/10**3#Power output in MW\n", + "Fc=(ma*3600*(1/mf))#Fuel consumption per hour in kg\n", + "SFC=(Fc/(PO*10**3))#Specific fuel consumption in kg/kW.h\n", + "\n", + "#Output\n", + "print \" (a) the air fuel ratio is %3.2f \\n (b) the cycle efficiency is %3.1f percent \\n (c) the power supplied by the plant is %3.0f MW \\n (d) the specific fuel consumption of the plant is %3.3f kg/kW.h and the fuel consumption per hour is %3.2f kg\"%(mf,ncy,PO,SFC,Fc)\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch12.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch12.ipynb new file mode 100644 index 00000000..cb66198b --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch12.ipynb @@ -0,0 +1,145 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 : Flywheel Energy storage" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 12.1 Pg: 849" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the compressed sir temperature is 1162 degree C \n", + " (b) the storage time is 7.71 hour \n", + " (c) the total energy storage is 2426 MWh \n", + " (d) the total energy delivered by the peaking turbine is 1455 MWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "V=64000#Volume in m**3\n", + "Q=8300#Discharge in m**3/hr\n", + "p1=1#Pressure in bar\n", + "T1=20+273#Temperature in K\n", + "p2=100#Pressure in bar\n", + "pn=70#Polytropic efficiency in percent\n", + "pt=60#Peaking turbine efficiency in percent\n", + "g=1.4#Ratio of specific heats\n", + "cp=1.005#Specific heat in kJ/kg.K\n", + "R=0.287#Gas constant in kJ/kg.K\n", + "\n", + "#Calculations\n", + "T2sT1=(p2/p1)**((g-1)/g)#Temperature ratio \n", + "T2s=(T1*T2sT1)#Temperature in K\n", + "T21=(T2s-T1)/(pn/100)#Difference in Temperatures in K\n", + "T2=(T21+T1)-273#Temperature in degree C\n", + "v=(R*T1)/(p2*100)#Specific volume in m**3/kg\n", + "mf=(Q/(v*3600))#Mass flow rate in kg/s\n", + "E=(mf*cp*T21)/1000#Rate of energy storage in MW\n", + "t=(V/Q)#Storage time in hour\n", + "tE=(E*t)#Total energy storage in MWh\n", + "Ed=(tE*(pt/100))#Total energy delivered by the peaking turbine in MWh\n", + "\n", + "#Output\n", + "print \" (a) the compressed sir temperature is %3.0f degree C \\n (b) the storage time is %3.2f hour \\n (c) the total energy storage is %3.0f MWh \\n (d) the total energy delivered by the peaking turbine is %3.0f MWh\"%(T2,t,tE,Ed)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 12.2 Pg: 850" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the storage time is 13.969 h \n", + " (b) the total energy stored in the accumulator is 16678.8 MWh \n", + " (c) the total energy that can be delivered by the peaking turbine is 4002.92 MWh\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "#Input data\n", + "V=175000#Volume in m**3\n", + "d=4#Diameter in m\n", + "U=1.5#Overall heat transfer coefficient in W/m**2.K\n", + "p2=2#Pressure in bar\n", + "p1=20#Pressure in bar\n", + "Ta=20#Ambient temperature in degree C\n", + "cp=4.35#Specific heta of water in kJ/kg.K\n", + "e=96#Efficiency in percent\n", + "ppe=25#Peaking plant efficiency in percent\n", + "\n", + "#Calculations\n", + "#At 20 bar\n", + "T1=212.37#Saturation temperature in degree C\n", + "hf1=908.5#Enthalpy in kJ/kg\n", + "vf1=0.0011766#Specific volume in m**3/kg\n", + "#At 2 bar\n", + "T2=120.23#Saturation temperature in degree C\n", + "hf2=504.8#Enthalpy in kJ/kg\n", + "vf2=0.0010605#Specific volume in m**3/kg\n", + "ad=(1/2)*((1/vf1)+(1/vf2))#Average density of water in kg/m**3\n", + "tc=(d*ad*cp*1000)/(4*U*3600)#Time constant in h\n", + "ts=(log(1/(1-((1-((e/100)))/((T1-Ta)/(T1-T2))))))*tc#Storage time in h\n", + "m=(V/vf1)#Mass of water needed in kg\n", + "E=(m*(hf1-hf2))/(3600*10**3)#Total energy stored in MWh\n", + "Ed=(E*(e/100)*(ppe/100))#Energy delivered in MWh\n", + "\n", + "#Output\n", + "print \" (a) the storage time is %3.3f h \\n (b) the total energy stored in the accumulator is %3.1f MWh \\n (c) the total energy that can be delivered by the peaking turbine is %3.2f MWh\"%(ts,E,Ed)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch2.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch2.ipynb new file mode 100644 index 00000000..1f75887b --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch2.ipynb @@ -0,0 +1,600 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 2 : Analysis of Steam Cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.1 Pg: 82" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)Power output of the cycle is 1746.3 kW \n", + " Efficiency of the cycle is 35.1 percent \n", + "\n", + " (b)Without geothermal heat supply \n", + " Power output of the cycle is 955.17 kW \n", + " Efficiency of the cycle is 19.22 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "p1=40#Initial pressure of steam in bar\n", + "T1=500#Initial temperature of steam in degree C\n", + "m1=5500#Rate of steam in kg/h\n", + "p2=2#Pressure of steam after expansion in bar\n", + "n1=0.83#Isentropic efficiency \n", + "q=0.87#Quality\n", + "m2=2700#Mass flow rate in kg/h\n", + "p3=0.1#Pressure of steam after expansion in l.p turbine in bar\n", + "n2=0.78#Isentropic efficiency\n", + "\n", + "#Calculations\n", + "h1=3445.3#Enthalpy in kJ/kg\n", + "s1=7.0901#Entropy in kJ/kg.K which is 1.5301+x2s*5.5970\n", + "x2s=(5.5600/5.5970)#dryness fraction\n", + "h2s=(504.7+(x2s*2201.9))#Enthalpy in kJ/kg\n", + "h2=h1-(n1*(h1-h2s))#Enthalpy in kJ/kg\n", + "h3=(504.7+(q*2201.9))#Enthalpy in kJ/kg\n", + "h4=((m2*h3+m1*h2)/(m1+m2))#Enthalpy in kJ/kg\n", + "x4=(2183.78/2201.9)#dryness fraction\n", + "s4=(1.5301+x4*5.5970)#Entropy in kJ/kg.K\n", + "x5s=0.8574#dryness fraction\n", + "h5s=(191.84+x5s*2392.5)#Enthalpy in kJ/kg\n", + "dh4h5=(n2*(h4-h5s))#Difference in enthalpy (h4-h5) in kJ/kg\n", + "h6=191.83#Enthalpy in kJ/kg\n", + "W1=((m1*(h1-h2))+((m1+m2)*dh4h5))/3600#Power output of the plant in kW\n", + "Q1=(m1*(h1-h6))/3600#Heat input in kW\n", + "n1=(W1/Q1)*100#Efficiency in percent\n", + "WT=(m1*(h1-h2))/3600#Power output without the geothermal heat supply in kW\n", + "Q2=(m1*(h1-h6))/3600#Heat input without the geothermal heat supply in kW\n", + "n2=(WT/Q2)*100#Efficiency of the cycle without the geothermal heat supply in percent\n", + "\n", + "#Output\n", + "print \"(a)Power output of the cycle is %3.1f kW \\n Efficiency of the cycle is %3.1f percent \\n\\n (b)Without geothermal heat supply \\n Power output of the cycle is %3.2f kW \\n Efficiency of the cycle is %3.2f percent\"%(W1,n1,WT,n2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.2 Pg: 83" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)Mass flow rate of steam at turbine inlet is 452 kg/s \n", + " (b)The cycle efficiency is 38.82 percent \n", + " (c)Work ratio is 0.999\n" + ] + } + ], + "source": [ + "#Input data\n", + "p1=90#Initial pressure of steam in bar\n", + "T1=500#Initial temperature of steam in degree C\n", + "O=(500*1000)#Output in kW\n", + "T2=40#Condensation temperature in degree C\n", + "nhp=0.92#Efficiency of h.p turbine\n", + "nlp=0.9#Efficiency of l.p turbine\n", + "np=0.75#Isentropic efficiency of the pump \n", + "TTD=-1.6#Temperature in degree C\n", + "\n", + "#Calculations\n", + "p2=(0.2*p1)#Optimum reheat pressure in bar\n", + "h1=3386.1#Enthalpy in kJ/kg\n", + "s1=6.6576#Entropy in kJ/kg.K\n", + "s2s=s1#Entropy in kJ/kg.K\n", + "h2s=2915#Enthalpy in kJ/kg\n", + "h3=3469.8#Enthalpy in kJ/kg\n", + "s3=7.4825#Entropy in kJ/kg.K\n", + "x4s=(s3-0.5725)/7.6845#Dryness fraction\n", + "h4s=(167.57+x4s*2406.7)#Enthalpy in kJ/kg\n", + "h5=167.57#Enthalpy in kJ/kg\n", + "h7=883.42#Enthalpy in kJ/kg\n", + "Wps=(0.001008*p1*10)#Workdone by the pump in kJ/kg\n", + "h6s=176.64#Enthalpy in kJ/kg\n", + "dh1h2=(nhp*(h1-h2s))#Difference in enthalpy (h1-h2) in kJ/kg\n", + "h2=h1-dh1h2#Enthalpy in kJ/kg\n", + "dh3h4=(nlp*(h3-h4s))#Difference in enthalpy (h3-h4) in kJ/kg\n", + "h4=h3-dh3h4#Enthalpy in kJ/kg\n", + "Wp=(Wps/np)#Workdone by the pump in kJ/kg\n", + "h6=(Wp+h5)#Enthalpy in kJ/kg\n", + "tsat=207.15#Saturation temperature at 18 bar in degree C\n", + "t9=(tsat-TTD)#Temperature in degree C\n", + "h9=875#Enthalpy in kJ/kg\n", + "m=((h9-h6)/(h2-h7))#Mass of steam in kg\n", + "WT=(dh1h2+(1-m)*dh3h4)#Workdone by the turbine in kJ/kg\n", + "Wnet=(WT-Wp)#Net workdone in kJ/kg\n", + "ws=(O/Wnet)#Mass flow rate of steam at turbine inlet in kg/s\n", + "Q1=((h1-h9)+(1-m)*(h3-h2))#Heat input in kJ/kg\n", + "n=(Wnet/Q1)*100#Efficiency of the cycle in percent\n", + "Wr=(Wnet/WT)#Work ratio\n", + "\n", + "#Output\n", + "print \"(a)Mass flow rate of steam at turbine inlet is %3.0f kg/s \\n (b)The cycle efficiency is %3.2f percent \\n (c)Work ratio is %3.3f\"%(ws,n,Wr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.3 Pg: 86" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The optimum pressure and temperature at different heaters are: \n", + " Heater 1: t1 = 255.2 degree C and p1 = 4.33 MPa\n", + " Heater 2: t2 = 224.5 degree C and p2 = 2.5318 MPa\n", + " Heater 3: t3 = 193.8 degree C and p3 = 1.367 MPa\n", + " Heater 4: t4 = 163.1 degree C and p4 = 0.6714 MPa\n", + " Heater 5: t5 = 132.4 degree C and p5 = 0.2906 MPa\n", + " Heater 6: t6 = 101.7 degree C and p6 = 0.108 MPa\n", + " Heater 7: t7 = 71.0 degree C and p7 = 32.65 kPa\n" + ] + } + ], + "source": [ + "#Input data\n", + "p1=70#Pressure at which an ideal seam power plant operates in bar\n", + "T1=550#Temperature at which an ideal seam power plant operates in degrees C\n", + "p2=0.075#Pressure at which an ideal seam power plant operates in bar\n", + "\n", + "#Calculations\n", + "TB=285.9#Saturation temperature at 70 bar in degree C\n", + "TC=40.3#Saturation temperature at 0.075 bar in degree C\n", + "Tr=(TB-TC)/(7+1)#Temperature rise per heater for maximum cycle efficiency in degree C\n", + "t1=(TB-Tr)#Temperature at heater 1 in degree C\n", + "P1=4.33#Pressure at heater 1 in MPa\n", + "t2=(t1-Tr)#Temperature at heater 2 in degree C\n", + "P2=2.5318#Pressure at heater 2 in MPa\n", + "t3=(t2-Tr)#Temperature at heater 3 in degree C\n", + "P3=1.367#Pressure at heater 3 in MPa\n", + "t4=(t3-Tr)#Temperature at heater 4 in degree C\n", + "P4=0.6714#Pressure at heater 4 in MPa\n", + "t5=(t4-Tr)#Temperature at heater 5 in degree C\n", + "P5=0.2906#Pressure at heater 5 in MPa\n", + "t6=(t5-Tr)#Temperature at heater 6 in degree C\n", + "P6=0.108#Pressure at heater 6 in MPa\n", + "t7=(t6-Tr)#Temperature at heater 7 in degree C\n", + "P7=32.65#Pressure at heater 7 in kPa\n", + "\n", + "#Output\n", + "print \"The optimum pressure and temperature at different heaters are: \\n Heater 1: t1 = %3.1f degree C and p1 = %3.2f MPa\\n Heater 2: t2 = %3.1f degree C and p2 = %3.4f MPa\\n Heater 3: t3 = %3.1f degree C and p3 = %3.3f MPa\\n Heater 4: t4 = %3.1f degree C and p4 = %3.4f MPa\\n Heater 5: t5 = %3.1f degree C and p5 = %3.4f MPa\\n Heater 6: t6 = %3.1f degree C and p6 = %3.3f MPa\\n Heater 7: t7 = %3.1f degree C and p7 = %3.2f kPa\"%(t1,P1,t2,P2,t3,P3,t4,P4,t5,P5,t6,P6,t7,P7)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.4 Pg: 87" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage of total electricity generated which is consumed in running the auxiliaries is 7.32 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "ng=0.97#Efficiency of electric generator \n", + "nt=0.95#Efficiency of turbine\n", + "nb=0.92#Efficiency of boiler\n", + "nc=0.42#Efficiency of cycle\n", + "no=0.33#Efficiency of overall plant\n", + "\n", + "#Calculations\n", + "na=(no/(ng*nt*nb*nc))#Efficiency of auxiliaries\n", + "n=(1-na)*100#Percentage of total electricity generated which is consumed in running the auxiliaries\n", + "\n", + "#Output\n", + "print \"Percentage of total electricity generated which is consumed in running the auxiliaries is %3.2f percent\"%(n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.5 Pg: 87" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Efficiency of steam generator is 91.54 percent \n", + "\n", + " Heat transfer per kg fuel in \n", + " (i)economiser is 5.3155 MJ/kg \n", + " (ii)boiler is 13.086 MJ/kg \n", + " (iii)superheater is 4.665 MJ/kg \n", + " (iv)air pre-heater is 3.392 MJ/kg \n", + "\n", + " Percentage of total heat absorption taking place in \n", + " (i)economiser is 23.04 percent \n", + " (ii)boiler is 56.73 percent \n", + " (iii)superheater is 20.23 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "T1=140#Temperature with which feed water enters into economiser in degree C\n", + "T2=[25,250]#Temperature from air is preheated to in degree C\n", + "P1=60#Pressure with which steam leaves the drum in bar\n", + "x1=0.98#Dryness fraction\n", + "T3=450#Temperature with which steam leaves the superheater in degree C\n", + "cc=25.2#Calorific value of coal in MJ/kg\n", + "r=8.5#Rate of evaporation of steam per kg coal \n", + "wf=1#Mass of coal in kg\n", + "R=15#Air fuel ratio by mass\n", + "Cpa=1.005#Specific heat of air at constant pressure in kJ/kg.K\n", + "Cpw=4.2#Specific heat of water at constant pressure in kJ/kg.K\n", + "\n", + "#Calculations\n", + "h1=(T1*Cpw)#Enthalpy in kJ/kg\n", + "hf=1213.35#Enthalpy in kJ/kg\n", + "h2=hf#Enthalpy in kJ/kg\n", + "hfg=1571#Enthalpy in kJ/kg\n", + "h4=3301.8#Enthalpy in kJ/kg\n", + "h3=(hf+x1*hfg)#Enthalpy in kJ/kg\n", + "n=((r*(h4-h1))/(wf*cc*1000))*100#Efficiency\n", + "he=(r*(h2-h1))/wf*10**-3#Heat transfer in the economiser in MJ/kg\n", + "hb=(r*(h3-h2))/wf*10**-3#Heat transfer in the boiler in MJ/kg\n", + "hs=(r*(h4-h3))/wf*10**-3#Heat transfer in the superheater in MJ/kg\n", + "ha=(R*Cpa*(T2[1]-T2[0]))/wf*10**-3#Heat transfer in the air preheater in MJ/kg\n", + "pe=((h2-h1)/(h4-h1))*100#Percentage of total heat absorbed in the economiser in percent\n", + "pb=((h3-h2)/(h4-h1))*100#Percentage of total heat absorbed in the boiler in percent\n", + "ps=((h4-h3)/(h4-h1))*100#Percentage of total heat absorbed in the superheater in percent\n", + "\n", + "#Output\n", + "print \"Efficiency of steam generator is %3.2f percent \\n\\n Heat transfer per kg fuel in \\n (i)economiser is %3.4f MJ/kg \\n (ii)boiler is %3.3f MJ/kg \\n (iii)superheater is %3.3f MJ/kg \\n (iv)air pre-heater is %3.3f MJ/kg \\n\\n Percentage of total heat absorption taking place in \\n (i)economiser is %3.2f percent \\n (ii)boiler is %3.2f percent \\n (iii)superheater is %3.2f percent\"%(n,he,hb,hs,ha,pe,pb,ps)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.6 Pg: 88" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The cycle efficiency is 48.58 percent \n", + " (b) The feedwater temperature is 264 degree C \n", + " (c) The steam rate is 2.69 kJ/kWh \n", + " (d) The heat rate is 7411 kJ/kWh \n", + " (e) The quality of steam at turbine exhaust is 0.8932 \n", + " (f) The power output is 111.58 MW\n" + ] + } + ], + "source": [ + "#Input data\n", + "p1=150#Pressure of inlet steam in bar\n", + "T1=550#Temperature of steam in degree C\n", + "p2=20#Pressure after expansion in bar\n", + "T2=500#Reheat temperature in degree C\n", + "pc=0.075#Condenser pressure in bar\n", + "php=50#Pressure of steam in h.p turbine in bar\n", + "pip=[10,5,3]#Pressure of steam in i.p turbines in bar\n", + "plp=1.5#Pressure of steam in l.p turbine in bar\n", + "m=300*1000#Steam flow rate in kg/h\n", + "\n", + "#Calculations\n", + "h1=3448.6#Enthalpy in kJ/kg\n", + "h4=3467.6#Enthalpy in kJ/kg\n", + "s1=6.5119#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "s3=s1#Entropy in kJ/kg.K\n", + "s4=7.4317#Entropy in kJ/kg.K\n", + "s5=s4#Entropy in kJ/kg.K\n", + "s6=s5#Entropy in kJ/kg.K\n", + "s7=s6#Entropy in kJ/kg.K\n", + "s8=s7#Entropy in kJ/kg.K\n", + "s9=s8#Entropy in kJ/kg.K\n", + "t2=370#Temperature in degree C\n", + "t3=245#Temperature in degree C\n", + "t5=400#Temperature in degree C\n", + "t6=300#Temperature in degree C\n", + "t7=225#Temperature in degree C\n", + "t8=160#Temperature in degree C\n", + "h2=3112#Enthalpy in kJ/kg\n", + "h3=2890#Enthalpy in kJ/kg\n", + "h5=3250#Enthalpy in kJ/kg\n", + "h6=3050#Enthalpy in kJ/kg\n", + "h7=2930#Enthalpy in kJ/kg\n", + "h8=2790#Enthalpy in kJ/kg\n", + "x9=(s9-0.5764)/7.6751#Dryness fraction\n", + "h9=168.79+x9*2406##Enthalpy in kJ/kg\n", + "h10=168.79#Enthalpy in kJ/kg\n", + "h11=h10+0.001*pip[1]*100#Enthalpy in kJ/kg\n", + "h12=467.11#Enthalpy in kJ/kg\n", + "t14=111.37#Temperature in degree C\n", + "h14=467#Enthalpy in kJ/kg\n", + "h13=h12#Enthalpy in kJ/kg\n", + "h14=h13#Enthalpy in kJ/kg\n", + "h15=h14#Enthalpy in kJ/kg\n", + "h16=561.47#Enthalpy in kJ/kg\n", + "h17=h16#Enthalpy in kJ/kg\n", + "h18=640.23#Enthalpy in kJ/kg\n", + "h19=h18+0.001*(p1-pip[1])*100#Enthalpy in kJ/kg\n", + "h20=762.8#Enthalpy in kJ/kg\n", + "h21=h20#Enthalpy in kJ/kg\n", + "h22=1154.23#Enthalpy in kJ/kg\n", + "h23=h22#Enthalpy in kJ/kg\n", + "m1=((h23-h21)/(h2-h22))#Mass in kg\n", + "m2=((h21-h19)-(m1*(h22-h20)))/(h5-h20)#Mass in kg\n", + "m3=(((1-m1-m2)*(h18-h17))-((m1+m2)*(h20-h18)))/(h6-h18+h18-h17)#Mass in kg\n", + "m4=((1-m1-m2-m3)*(h17-h15))/(h7-h16)#Mass in kg\n", + "m5=(((1-m1-m2-m3-m4)*(h14-h11))-(m4*(h16-h12)))/(h8-h12+h14-h11)#Mass in kg\n", + "WT=(h1-h2)+(1-m1)*(h2-h3)+(1-m1)*(h4-h5)+(1-m1-m2)*(h5-h6)+(1-m1-m2-m3)*(h6-h7)+(1-m1-m2-m3-m4)*(h7-h8)+(1-m1-m2-m3-m4-m5)*(h8-h9)#Workdone by turbine in kJ/kg\n", + "Wp=(0.5+14.5+0.15)#Workdone in kJ/kg\n", + "Wnet=(WT-Wp)#Net workdone in kJ/kg\n", + "Q1=(h1-h23)+(1-m1)*(h4-h3)#Heat supplied in kJ/kg\n", + "ncy=(Wnet/Q1)*100#Cycle efficiency in percent\n", + "t23=264#Temperature in degree C\n", + "sr=(3600/Wnet)#Steam rate in kJ/kWh\n", + "hr=((Q1/Wnet)*3600)#Heat rate in kJ/kWh\n", + "P=((Wnet*m)/3600)/10**3#Power output in MW\n", + "\n", + "#Output\n", + "print \"(a) The cycle efficiency is %3.2f percent \\n (b) The feedwater temperature is %d degree C \\n (c) The steam rate is %3.2f kJ/kWh \\n (d) The heat rate is %3.0f kJ/kWh \\n (e) The quality of steam at turbine exhaust is %3.4f \\n (f) The power output is %3.2f MW\"%(ncy,t23,sr,hr,x9,P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.7 Pg: 92" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The steam condition required at inlet of the turbine: \n", + " Enthalpy is 3085.3 kJ/kg \n", + " Entropy is 6.6192 kJ/kg.K \n", + " Pressure is 37.3 bar \n", + " Temperature is 344 degree C\n" + ] + } + ], + "source": [ + "#Input data\n", + "m=10000#Mass flow rate of steam in kg/h\n", + "p=3#Pressure of steam in bar\n", + "P=1000#Power in kW\n", + "n=0.7#Internal efficiency of turbine\n", + "\n", + "#Calculations\n", + "dh=(P*3600)/m#Change in enthalpy in kJ/kg\n", + "h2=2725.3#Enthalpy in kJ/kg from Fig. E2.7 \n", + "h1=dh+h2#Enthalpy in kJ/kg \n", + "dh1h2s=dh/n#Change in enthalpy in kJ/kg\n", + "h2s=h1-dh1h2s#Enthalpy in kJ/kg\n", + "x2s=(h2s-561.47)/2163.8#Dryness fraction\n", + "s2s=1.6718+x2s*(6.999-1.6718)#Entropy in kJ/kg.K\n", + "s1=s2s#Entropy in kJ/kg.K\n", + "p1=37.3#Pressure in bar from Mollier diagram\n", + "t1=344#Temperature in degree C\n", + "\n", + "#Output\n", + "print \"The steam condition required at inlet of the turbine: \\n Enthalpy is %3.1f kJ/kg \\n Entropy is %3.4f kJ/kg.K \\n Pressure is %3.1f bar \\n Temperature is %d degree C\"%(h1,s1,p1,t1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.8 Pg: 93" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) the steam generation capacity of the bolier is 16.74 t/h \n", + " (b) the heat input to the boiler is 15113.7 kW \n", + " (c) the fuel burning rate of the bolier is 2.473 t/h \n", + " (d) the heat rejected to the condensor is 8369 kW \n", + " (e) the rate of flow of cooling water in the condensor is 0.333 m**3/s\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "Pl=5.6#Power load in MW\n", + "Hl=1.163#Heat load in MW\n", + "p1=40#Pressure in bar\n", + "T1=500+273#Temperature in K\n", + "p2=0.06#Pressure in bar\n", + "p3=2#Pressure in bar\n", + "CV=25#Calorific value in MJ/kg\n", + "n=88#Boiler efficiency in percent\n", + "T=6#Temperature rise in degree C\n", + "\n", + "#Calculations\n", + "h1=3445.3#Enthalpy in kJ/kg\n", + "s1=7.0901#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "s3=s1#Entropy in kJ/kg.K\n", + "x2=(s2-1.5301)/5.5970#Dryness fraction\n", + "h2=2706.7#Enthalpy in kJ/kg\n", + "h26=2201.9#Difference in enthalpy in kJ/kg\n", + "w=(Hl*10**3)/h26#Rate of steam extraction in kg/h\n", + "x3=(s1-0.52)/7.815#Dryness fraction\n", + "h3=(149.79+x3*2416)#Enthalpy in kJ/kg\n", + "h4=149.79#Enthalpy in kJ/kg\n", + "ws=((Pl*10**3+(w*(h2-h3)))/((h1-h2)+(h2-h3)))#Steam generation capacity in kg/s\n", + "ws1=(ws*3600)/1000#Steam generation capacity in t/h\n", + "h7=(504.7+(1.061*10**-3*(p1-p3)*100))#Enthalpy in kJ/kg\n", + "h5=(149.79+(1.006*100*p1*10**-3))#Enthalpy in kJ/kg\n", + "Q1=(((ws-w)*(h1-h5))+(w*(h1-h7)))#Heat input in kW\n", + "wf=((Q1/1000)/((n/100)*CV))*(3600/1000)#Fuel burning rate in t/h\n", + "Q2=((ws-w)*(h3-h4))#Heat rejected to the condensor in kW\n", + "wc=(Q2/(4.187*T))/1000#Rate of flow of cooling water in m**3/s\n", + "\n", + "#Output\n", + "print \"(a) the steam generation capacity of the bolier is %3.2f t/h \\n (b) the heat input to the boiler is %3.1f kW \\n (c) the fuel burning rate of the bolier is %3.3f t/h \\n (d) the heat rejected to the condensor is %3.0f kW \\n (e) the rate of flow of cooling water in the condensor is %3.3f m**3/s\"%(ws1,Q1,wf,Q2,wc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 2.9 Pg: 94" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) the steam quality at the exhaust of the h.p turbine is 0.956 \n", + " (b) the power developed by the h.p turbine is 1154.62 kW \n", + " (c) the isentropic efficiency of the h.p turbine is 76.61 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "m=21000#Steam rate in kg/h\n", + "p1=17#Pressure in bar\n", + "T1=230+273#Temperature in K\n", + "P=132.56#Power in kW\n", + "x2=0.957#Dryness fraction\n", + "p2=3.5#Pressure in bar\n", + "Pl=1337.5#Power in l.p turbine in kW\n", + "p3=0.3#Pressure in bar\n", + "x3=0.912#Dryness fraction\n", + "\n", + "#Calculations\n", + "h1=2869.7#Enthalpy in kJ/kg\n", + "s1=6.5408#Entropy in kJ/kg.K\n", + "h2=(870.44+x2*1924.7)#Enthalpy in kJ/kg\n", + "h3=h2#Enthalpy in kJ/kg\n", + "h56=(Pl*3600)/m#Difference in Enthalpy in kJ/kg\n", + "h6=(289.23+x3*2336.1)#Enthalpy in kJ/kg\n", + "h5=2649.04#Enthalpy in kJ/kg\n", + "s4s=s1#Entropy in kJ/kg.K\n", + "x4s=(s4s-1.7275)/5.2130#Dryness fraction\n", + "h4s=584.33+x4s*2148.1#Enthalpy in kJ/kg\n", + "w=(P/(h1-h2))#Flow rate in kg/s\n", + "ws=(m/3600)#Steam flow rate in kg/s\n", + "h4=((ws*h5)-(w*h3))/(ws-w)#Enthalpy in kJ/kg\n", + "x4=(h4-584.33)/2148.1#Dryness fraction\n", + "W=(ws-w)*(h1-h4)#Power developed by h.p turbine in kW\n", + "n=((h1-h4)/(h1-h4s))*100#Isentropic efficiency in percent\n", + "\n", + "#Output\n", + "print \"(a) the steam quality at the exhaust of the h.p turbine is %3.3f \\n (b) the power developed by the h.p turbine is %3.2f kW \\n (c) the isentropic efficiency of the h.p turbine is %3.2f percent\"%(x4,W,n)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch3.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch3.ipynb new file mode 100644 index 00000000..72e7fae0 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch3.ipynb @@ -0,0 +1,529 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 3 : Combined Cycle Power Generation" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.1 Pg: 143" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The amount of mercury circulated per kg of water is 7.4151 kg \n", + " (b) The efficiency of the combined cycle is 48.1 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "p=40#Pressure in bar\n", + "T1=400+273#Temperature in K\n", + "T2=40+273#Temperature in K\n", + "x=[0,10,515.5,72.23,363.0,0.1478,0.5167,80.9*10**-6,0.0333]#Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m**3/kg)\n", + "y=[0,0.2,277.3,38.35,336.55,0.0967,0.6385,77.4*10**-6,1.163]#Property values from table p(bar),t(degree C), hf,hg(kJ/kg),sf,sg(kJ/kg.K),vf,vg(m**3/kg)\n", + "\n", + "#Calculations\n", + "h1=3216#Enthalpy in kJ/kg\n", + "s1=6.7690#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "x2=(s2-0.5725)/(8.2570-0.5725)#Dryness fraction\n", + "h2=167.57+x2*2406.7#Enthalpy in kJ/kg\n", + "h3=167.57#Enthalpy in kJ/kg\n", + "h4=(167.57+p*100*1.008*10**-3)#Enthalpy in kJ/kg\n", + "h5=1087.31#Enthalpy in kJ/kg\n", + "h6=2801.4#Enthalpy in kJ/kg\n", + "ha=x[(4)]#Enthalpy in kJ/kg\n", + "sa=x[(6)]#Entropy in kJ/kg.K\n", + "sb=sa#Entropy in kJ/kg.K\n", + "xb=(sb-y[(5)])/(y[(6)]-y[(5)])#Dryness fraction\n", + "hb=(y[(3)]+xb*(y[(4)]-y[(3)]))#Enthalpy in kJ/kg\n", + "hc=y[(3)]#Enthalpy in kJ/kg\n", + "hd=hc#Enthalpy in kJ/kg\n", + "m=(h6-h5)/(hb-hc)#Mass of mercury circulated per kg of steam\n", + "Q1=m*(ha-hd)+(h1-h6)+(h5-h4)#Heat supplied in kJ/kg\n", + "Q2=(h2-h3)#Heat rejected in kJ/kg\n", + "nc=(1-(Q2/Q1))*100#Efficiency in percent\n", + "\n", + "#Output\n", + "print \" (a) The amount of mercury circulated per kg of water is %3.4f kg \\n (b) The efficiency of the combined cycle is %3.1f percent\"%(m,nc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.2 Pg: 145" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Rate of heat transfer in the steam generator is 14.844 kW \n", + " (b) The net power output of the binary cycle is 4030 kW \n", + " (c) The rate of heat transfer to the industrial process is 5799 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "m=5#Mass flow rate in kg/s\n", + "p1=40#Pressure in bar\n", + "T1=440+273#Temperature in K\n", + "p2=1.5#Pressure in bar\n", + "p3=1#Pressure in bar\n", + "T3=60+273#Temperature in K\n", + "p4=16#Pressure in bar\n", + "T4=100+273#Temperature in K\n", + "p5=9#Pressure in bar\n", + "\n", + "#Calculations\n", + "h1=3307.1#Enthalpy in kJ/kg\n", + "s1=6.9041#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "h2=2570.8#Enthalpy in kJ/kg\n", + "h3=417.46#Enthalpy in kJ/kg\n", + "h6=(251.13+(1.0172*10**-3)*(p3-0.1994)*100)#Enthalpy in kJ/kg\n", + "m3=(m/2)#Mass flow rate in kg/s\n", + "m6=m3#Mass flow rate in kg/s\n", + "h4=(m3*h3+m6*h6)/m#Enthalpy in kJ/kg\n", + "h5=(h4+(1.0291*10**-3)*(p1-p3)*100)#Enthalpy in kJ/kg\n", + "ha=241.58#Enthalpy in kJ/kg\n", + "sa=0.7656#Entropy in kJ/kg.K\n", + "sb=sa#Entropy in kJ/kg.K\n", + "hb=229.43#Enthalpy in kJ/kg\n", + "hc=71.93#Enthalpy in kJ/kg\n", + "hd=hc+(0.7914*10**-3*(p4-p5)*100)#Enthalpy in kJ/kg\n", + "Q1=(m*(h1-h5))/1000#Heat supplied in kW\n", + "Wnets=(m*((h1-h2)-(h5-h4)))#Net workdone by steam in kW\n", + "mR12=(m3*(h2-h3))/(ha-hd)#Mass of R12 in kg/s\n", + "WnetR=(mR12*((ha-hb)-(hd-hc)))#Net workdone by R12 in kW\n", + "T=Wnets+WnetR#Total output in kW\n", + "Qh=(m6*(h2-h6))#Heat rejected in kW\n", + "\n", + "#Output\n", + "print \"(a) Rate of heat transfer in the steam generator is %3.3f kW \\n (b) The net power output of the binary cycle is %d kW \\n (c) The rate of heat transfer to the industrial process is %3.0f kW\"%(Q1,T,Qh)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.3 Pg: 146" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The flow rate of air is 396.33 kg/s and steam is 82.22 kg/s \n", + " (b) The power outputs of the gas turbine is 87.77 MW and steam turbine is 112.23 MW \n", + " (c) The thermal efficiency of the combined plant is 50 percent \n", + " (d) The air fuel ratio is 42.7\n" + ] + } + ], + "source": [ + "from numpy import mat\n", + "#Input data\n", + "rp=7.5#Pressure ratio \n", + "T1=15+273#Inlet air temperature in K\n", + "T3=750+273#Maximum temperature in K\n", + "T6=100+273#Temperature in K\n", + "p1=50#Pressure in bar\n", + "T7=600+273#Temperature in K\n", + "p2=0.1#Pressure in bar\n", + "P=200#Total power in MW\n", + "CV=43.3#calorific value in MJ/kg\n", + "cpg=1.11#Specific heat for gas in kJ/kg.K\n", + "g=1.33#Ratio of specific heats for gas\n", + "cpa=1.005#Specific heat for air in kJ/kg.K\n", + "g1=1.4#Ratio of specific heats for air\n", + "\n", + "#Calculations\n", + "T2=(T1*rp**((g1-1)/g1))#Temperature in K\n", + "T4=(T3/rp**((g-1)/g))#Temperature in K\n", + "ha=3670#Enthalpy in kJ/kg\n", + "hb=2305#Enthalpy in kJ/kg\n", + "hc=192#Enthalpy in kJ/kg\n", + "hd=hc#Enthalpy in kJ/kg\n", + "#ma*cpg*(T3-T6)=ms*(ha-hd)\n", + "#ma*cpg*(T3-T4)-ma*cpa*(T2-T1)+ms*(ha-hb)=P*1000\n", + "#Solving these two equations\n", + "A=mat([[cpg*(T3-T6), (hd-ha)],[cpg*(T3-T4)-cpa*(T2-T1), (ha-hb)]])#Coefficient matrix\n", + "B=mat([[0],[(P*10**3)]])#Constant matrix\n", + "X=(A**-1)*B#Variable matrix\n", + "\n", + "Wgt=(cpg*(T3-T4)-cpa*(T2-T1))*X[0]*10**-3#Net workdone by Gas turbine in MW\n", + "Wst=(P-Wgt)#Net workdone by steam turbine in MW\n", + "Q1=(X[0]*cpg*(T3-T2+T3-T4))#Heat supplied in MW\n", + "nth=(P/(Q1*10**-3))*100#Thermal efficiency in percent\n", + "af=(CV*10**3)/(cpg*(T3-T2+T3-T4))#Air fuel ratio\n", + "\n", + "#Output\n", + "print \" (a) The flow rate of air is %3.2f kg/s and steam is %3.2f kg/s \\n (b) The power outputs of the gas turbine is %3.2f MW and steam turbine is %3.2f MW \\n (c) The thermal efficiency of the combined plant is %3.0f percent \\n (d) The air fuel ratio is %3.1f\"%(X[0],X[1],Wgt,Wst,nth,af)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.4 Pg: 148" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Total power output is 82422.08 kW and overall efficiency is 41.54 percent lost heat coefficient is 0.351\n", + " Exergy efficiency is 40 percent \n", + "\n", + " Input is 212810 kW \n", + " Total Output is 82422 kW \n", + " Total losses is 123309 kW \n", + " Exergy outut + exergy destruction = 205731 kW which is 1.3 percent gretter than the exergy input\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import log\n", + "#Input data\n", + "p1=1#Pressure in bar\n", + "T1=25+273#Temperature in K\n", + "rp=8#Pressure ratio of compressor\n", + "Tm=900+273#Maximum temperature in K\n", + "pd=3#pressure drop in combustion chamber in percent\n", + "nc=0.88#Efficiency of compressor\n", + "nt=0.88#Efficiency of turbine\n", + "CV=44.43#Calorific value of fuel in MJ/kg\n", + "cpa=1.006#Specific heat of air in kJ/kg.K\n", + "cpg=1.148#Specific heat of gas in kJ/kg.K\n", + "g1=1.333#Specific heat ratio of gas\n", + "g=1.4#Specific heat ratio of air\n", + "T3=425+273#Temperature in K\n", + "p2=40#Pressure in bar\n", + "p3=0.04#Condensor pressure in bar\n", + "Th=170.4+273#Temperature of feed water to the HRSG in K\n", + "nst=0.82#Efficiency of steam turbine\n", + "pdh=5#Pressure drop in HRSG in kPa\n", + "m=29.235#Steam flow rate in kg/s\n", + "A=1.0401#si=1.0401+0.1728*(h/c)\n", + "B=0.1728#si=1.0401+0.1728*(h/c)\n", + "\n", + "#Calculations\n", + "#Gas turbine plant\n", + "T2=(rp**((g-1)/(g*nt)))*T1#Temperature in K\n", + "#Combustor\n", + "pc=((pd/100)*rp)#Pressure loss in bar\n", + "pcx=(rp-pc)#Pressure in bar\n", + "f=((cpg*(Tm-T1))-(cpa*(T2-T1)))/((CV*10**3)-(cpa*(T2-T1)))#Fuel flow rate in kg/s\n", + "af=(1-f)/f#Air fuel ratio\n", + "#C8H18+12.5O2->8CO2+9H2O\n", + "afc=(12.5*32)/(0.232*114)#Air fuel ratio for stoichiometric combustion\n", + "ea=((af-afc)/afc)*100#Excess air in percent\n", + "#Gas turbine\n", + "p4=p1+0.05#Pressure in bar\n", + "T4=(Tm/(pcx/p4)**(((g1-1)*nt)/g1))#Temperature in K\n", + "#HRSG\n", + "T5=250+30#Temeprature in K\n", + "ha=3272#Enthalpy in kJ/kg\n", + "hf=1087.31#Enthalpy in kJ/kg\n", + "ws=(cpg*((T4-273)-T5))/(ha-hf)#Flow rate in kg/s\n", + "he=721.1#Enthalpy in kJ/kg\n", + "T6=(T4-273)-((ws*(ha-he))/cpg)#Temperature in degree C\n", + "#Power output\n", + "sa=6.853#Entropy in kJ/kg.K\n", + "sbs=sa#Entropy in kJ/kg.K\n", + "xbs=(sbs-0.4266)/8.052#Dryness fraction\n", + "hbs=(121.46+xbs*2432.9)#ENthalpy in kJ/kg\n", + "Wst=(m*(ha-hbs)*nst)#Workdone in kW\n", + "wg=(m/ws)#gas flow rate in kg/s\n", + "wa=(1-f)*wg#Air flow rate entering the compressor in kg/s\n", + "Wgt=(wg*cpg*(Tm-T4))-(wa*cpa*(T2-T1))#Power output of gas turbine in kW\n", + "TO=Wst+Wgt#Total power output in kW\n", + "wf1=(f*wa)#Fuel mass flow rate in kg/s\n", + "wf=4.466#Rounding off of wf1 for exact answers\n", + "no=(TO/(wf*(CV*10**3)))*100#Overall efficiency of the combined plant in percent\n", + "ns=((ha-hbs)/(ha-he))*nst#Efficiency of steam plant\n", + "ngtp=(Wgt/(wf*(CV*10**3)))#Efficiency of the GT plant\n", + "xL=((wg*cpg*(T6-(T1-273)))/(wf*(CV*10**3)))#Lost heat coefficient\n", + "nov=(ns+ngtp-ns*ngtp-ngtp*xL)#The overall efficiency\n", + "#Energy fluxes and irreversibilities\n", + "si=(A+B*((18*1)/(8*12)))#si for octane C8H18\n", + "dHo=(wf*CV*10**3)#Power in kW\n", + "dGo=(si*dHo)#Power in kW\n", + "TS=(dGo-dHo)#Power in kW\n", + "#Compressor\n", + "dS=(cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))#change in entropy in kJ/kg.K\n", + "Ic=(wa*T1*dS)#power in kW\n", + "Icx=((wg*T1*((cpg*log(Tm/T1))-(((cpg*(g1-1))/g1)*log(pcx))))-(wa*T1*((cpa*log(T2/T1))-(((cpa*(g-1))/g)*log(rp))))+TS)#Compressor in kW\n", + "Icg=(-cpg*log(Tm/T4))-(((cpg*(g1-1))/g1)*log(p4/pcx))#Difference in entropy in kJ/kg.K\n", + "IGT=(Icg*T1*wg)#Gas turbine in kW\n", + "se=2.046#Enntropy in kJ/kg.K\n", + "sae=(sa-se)#Difference in entropy in kJ/kg.K\n", + "s64=(cpg*log((T6+273)/T4))-(((cpg*(g1-1))/g1)*log(p4/p1))#Difference in entropy in kJ/kg.K\n", + "Ih=(T1*m*sae)+(wg*T1*s64)#For HRSG in kW\n", + "hb=(ha-(nst*(ha-hbs)))#Enthalpy in kJ/kg\n", + "xb=(hb-121.46)/2432.9#Dryness Fraction\n", + "sb=(0.4226+xb*8.052)#Entropy in kJ/kg.K\n", + "Ist=(m*(sb-sa)*T1)#For steam turbine in kW\n", + "Iexh=(wg*cpg*((T6-(T1-273))-(T1*log((T6+273)/T1))))#For exhaust in kW\n", + "Tl=Icx+Icg+IGT+Ih+Ist+Iexh#Exergy losses in kW\n", + "T=Tl+Wgt+Wst#Total exergy output and exergy destruction in kW\n", + "ee=((Wst+Wgt)/T)*100#Exergy efficiency in percent\n", + "\n", + "#Output\n", + "print \"(a) Total power output is %3.2f kW and overall efficiency is %3.2f percent lost heat coefficient is %3.3f\\n Exergy efficiency is %3.0f percent \\n\\n Input is %3.0f kW \\n Total Output is %3.0f kW \\n Total losses is %3.0f kW \\n Exergy outut + exergy destruction = %3.0f kW which is 1.3 percent gretter than the exergy input\"%(TO,no,xL,ee,dGo,(Wgt+Wst),Tl,T)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.5 Pg: 154" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The overall efficiency of the combined cycle is 77.5 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "n1=0.5#Efficiency of mercury\n", + "n2=0.4#Efficiency of steam\n", + "n3=0.25#Efficiency of composite cycle\n", + "\n", + "#Calculations\n", + "n=(1-(1-n1)*(1-n2)*(1-n3))*100#Overall efficiency of the combined cycle in percent\n", + "\n", + "#Output\n", + "print \"The overall efficiency of the combined cycle is %3.1f percent\"%(n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.6 Pg: 156" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The overall efficiency of the combined plant is 58 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "z=30.0#Percentage of total energy of fuel\n", + "n=40.0#Cycle efficiency in percent\n", + "\n", + "#Calculations\n", + "on=((z/100)+(1-(z/100))*(n/100))*100#Overall efficiency in percent\n", + "\n", + "#Output\n", + "print \"The overall efficiency of the combined plant is %3.0f percent\"%(on)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.7 Pg: 158" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The output voltage is 1.1629 V \n", + " (b) The current density in the cathode is 4.239 A/cm**2 and anode is 1.092 A/cm**2 \n", + " (c) Power output per unit area is 3.66 W/cm**2 \n", + " (d) Thermal efficiency is 44.3 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import exp\n", + "#Input data\n", + "Tc=1250+273#Cathode temperature in K\n", + "Ta=500+273#Anode temperature in K\n", + "e=1.602*10**-19#Charge in coloumb\n", + "K=1.38*10**-23#Boltzmann constant in J/molecule.K\n", + "b=18#Constant\n", + "\n", + "#Calculations\n", + "Va=((b*K*Ta)/e)#Voltage of anode in V\n", + "Vc=((b*K*Tc)/e)#Voltage of cathode in V\n", + "Vo=Vc-Va#Output voltage in V\n", + "Ja=(120*Ta**2*exp(-b))#Current density in Cathode in A/cm**2\n", + "Jc=(120*Tc**2*exp(-b))#Current density in Anode in A/cm**2\n", + "P=Vo*(Jc-Ja)#Power output per unit area in /cm**2\n", + "nth=(((Tc-Ta)/Tc)*(b/(b+2)))*100#Thermal efficiency in percent\n", + "\n", + "#Output\n", + "print \"(a) The output voltage is %3.4f V \\n (b) The current density in the cathode is %3.3f A/cm**2 and anode is %3.3f A/cm**2 \\n (c) Power output per unit area is %3.2f W/cm**2 \\n (d) Thermal efficiency is %3.1f percent\"%(Vo,Jc,Ja,P,nth)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 3.8 Pg: 159" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) The thermal efficiency of thermocouple generator is 9.1 percent \n", + " (b) The number of thermo couples in series is 309 \n", + " (c) The lenght of the thermal elements is 0.519 cm and area is 43.48 cm**2 \n", + " (d) The output open-circuit voltage is 0.6 V \n", + " (e) At full load: \n", + " The heat input is 3.558 kW \n", + " The heat rejected is 3.236 kW \n", + " At no load: \n", + " The heat input is 2.093 kW \n", + " The heat rejected is 2.093 kW \n", + " (f) The overall efficiency of the combined thermo-electric steam power plant is 34.18 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt\n", + "#Input data\n", + "P=100#Power in kW\n", + "V=115#Voltage in V\n", + "To=1500#Outer temperature in K\n", + "Te=1000#Exit temperature in K\n", + "Ta=350#Ambient temperature in K\n", + "nth=30#Thermal efficiency in percent\n", + "nge=92#Generator efficiency in percent\n", + "#Properties of thermoelectrons \n", + "a=0.0012#At 1250K in V/K\n", + "kp=0.02#In W/cm.K\n", + "kn=0.03#In W/cm.K\n", + "dp=0.01#In ohm.cm\n", + "dn=0.012#In ohm.cm\n", + "J=20#Current density in A/cm**2\n", + "\n", + "#Calculations\n", + "zmax=(a**2/(sqrt(dp*kp)+sqrt(dn*kn))**2)#Maximum value of figure of merit in K**-1\n", + "mo=sqrt(1+(zmax*((To+Te)/2)))#Optimum value of the resistance ratio\n", + "nmax=(((To-Te)/To)*((mo-1)/(mo+(Te/To))))*100#Maximum thermal efficiency in percent\n", + "Vl=(a*(To-Te)*(mo/(mo+1)))#Voltage per couple in V\n", + "nc=(V/Vl)#Number of couples in series\n", + "L=((a*(To-Te))/((1+mo)*(dp+dn)))/J#Length in cm\n", + "A=((P*Te)/V)/J#Area in cm**2\n", + "I=(J*A)#Current in A\n", + "Vo=(a*(To-Te))#Voltage in V\n", + "Q1=((a*I*To)-((1/2)*(L/A)*I**2*(dp+dn))+((A/L)*(kp+kn)*(To-Te)))/1000#Heat input to the thermoelectric generator in kW\n", + "Q2=((a*I*Te)+((A/L)*(kp+kn)*(To-Te))+P)/1000#Heat rejected at full load in kW\n", + "Q1n=(((A/L)*(kp+kn)*(To-Te)))/1000#At no load heat input in kW\n", + "Q2n=Q1n#At no load heat rejected in kW\n", + "no=((nmax/100)+(1-(nmax/100))*(nth/100)*(nge/100))*100#Overall efficiency in percent\n", + "\n", + "#Output\n", + "print \"(a) The thermal efficiency of thermocouple generator is %3.1f percent \\n (b) The number of thermo couples in series is %d \\n (c) The lenght of the thermal elements is %3.3f cm and area is %3.2f cm**2 \\n (d) The output open-circuit voltage is %3.1f V \\n (e) At full load: \\n The heat input is %3.3f kW \\n The heat rejected is %3.3f kW \\n At no load: \\n The heat input is %3.3f kW \\n The heat rejected is %3.3f kW \\n (f) The overall efficiency of the combined thermo-electric steam power plant is %3.2f percent\"%(nmax,nc,L,A,Vo,Q1,Q2,Q1n,Q2n,no)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch4.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch4.ipynb new file mode 100644 index 00000000..0658d992 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch4.ipynb @@ -0,0 +1,757 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4 : Fuels and combustion" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.1 Pg: 211" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Mass of air supplied WA = 13.80 kg \n", + " (b)Percentage excess air supplied = 4.68 percentage \n", + " (c)mass of dry flue gas formed = 13.82 kg \n", + " (d) Mass of water vapour formed = 0.90 kg\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "C=84##The mass of carbon present in the fuel in %\n", + "H=10##The mass of hydrogen present in the fuel in %\n", + "S=3.2##The mass of sulphur present in the fuel in %\n", + "O=1.6##The mass of oxygen present in the fuel in %\n", + "I=1.2##The mass of incombustible in the fuel in %\n", + "X=15.72##The flue gas of combined CO2 and SO2 by volume in %\n", + "Og=1##The flue gas of O2 by volume in %\n", + "Y=100##Let us consider the fuel oil in kg\n", + "C1=12##Molecular weight of Carbon \n", + "H1=2##Molecular weight of hydrogen\n", + "S1=32##Molecular weight of sulphur\n", + "O1=32##Molecular weight of oxygen\n", + "Co2=44##Molecular weight of carbondioxide\n", + "So2=64##Molecular weight of sulphurdioxide\n", + "N1=28##Molecular weight of nitrogen \n", + "H2O=18##Molecular weight of water\n", + "\n", + "#Calculations\n", + "b=C/C1##Equating coefficients of the carbon from equation\n", + "g=H/H1##Equating coefficients of the hydrogen from equation\n", + "d=S/S1##Equating coefficients of the sulphur from the equation\n", + "e=(b+d)/(X/Og)##By volumetric analysis\n", + "x=b+d+e+(g/2)-(O/O1)##Moles of oxygen are supplied for combustion\n", + "f=3.76*x##Equating coefficients of the nitrogen from equation\n", + "Mo=x*O1##Mass of oxygen supplied in kg\n", + "Ma=Mo/0.232##Mass of air supplied for 100 kg of fuel in kg\n", + "Wa=Ma/100##Mass of air supplied for 1 kg fuel in kg\n", + "Wrh=((11.5*C)+(34.5*((H)-(O/8))+(4.3*S)))/100##Theoretical air required per kg of fuel in kg\n", + "E=((Wa-Wrh)/Wrh)*100##Percentage of excess air in %\n", + "D=(b*Co2)+(d*So2)+(e*O1)+(f*N1)##Mass of dry flue gas formed for 100 kg fuel in kg\n", + "dfg=D/100##Mass of dry flue gas formed per kg of fuel in kg\n", + "Mw=(g*H2O)/100##Mass of water vapour formed per kg of fuel\n", + "\n", + "#Output\n", + "print \"(a) Mass of air supplied WA = %3.2f kg \\n (b)Percentage excess air supplied = %3.2f percentage \\n (c)mass of dry flue gas formed = %3.2f kg \\n (d) Mass of water vapour formed = %3.2f kg\"%(Wa,E,dfg,Mw)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.2 Pg 212" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage excess air used is = 11.3 percentage\n" + ] + } + ], + "source": [ + "#Input data\n", + "CO2=11.5##Percentage of carbondioxide present in combustion in %\n", + "O2=2.7##Percentage of oxygen present in the combustion in %\n", + "CO=0.7##Percentage of carbonmonoxide present in the combuston in %\n", + "\n", + "#Calculations \n", + "a=85.1/3.76##Equating moles for nitrogen from the equation \n", + "x=(CO2+CO)/3##Equating moles for carbon from the equation \n", + "b=(a-CO2-(CO/2)-O2)*2##Equating moles for oxygen from the equation \n", + "y=a/x##Moles of oxygen supplied for one mole of propane gas\n", + "z=5##Theoretically 5 moles of oxygen are required for reacting\n", + "E=((y-z)/z)*100##The excess of air supplied in %\n", + "\n", + "#Output\n", + "print \"The percentage excess air used is = %3.1f percentage\"%(E)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.3 Pg: 213" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The air fuel ratio during the test = 17.05 \n", + " (b) The excess or deficiency of air used = 18 Percentage of excess air used\n" + ] + } + ], + "source": [ + "#Input data\n", + "CO2=12.1##The amount of carbondioxide released from the combustion in %\n", + "O2=3.8##The amount of oxygen released from the combustion in %\n", + "CO=0.9##The amount of carbonmonoxide released from the combustion in %\n", + "MO=32##Molecular weight of Oxygen\n", + "\n", + "#Calculations\n", + "a=83.2/3.76##Equating moles for nitrogen from the equation \n", + "b=(2*a)-(2*CO2)-(2*O2)-CO##Equating moles for oxygen from the equation\n", + "x=CO2+CO##Equating moles for carbon from the equation\n", + "y=2*b##Equating moles for hydrogen from the equation\n", + "z=18.75##Moles of Oxygen from the stoichiometric equation \n", + "z1=a##Moles of Oxygen from the combustion equation \n", + "E=((z1-z)/z)*100##Percentage of excess air in%\n", + "A=(a*MO)/0.232##Actual air supplied per mole of C13H23\n", + "Mc=179##Molecular weight of C13H23\n", + "Af=A/Mc##Air fuel ratio during the test\n", + "\n", + "#Output\n", + "print \" (a) The air fuel ratio during the test = %3.2f \\n (b) The excess or deficiency of air used = %3.0f Percentage of excess air used\"%(Af,E)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.4 Pg: 214" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The amount of dry flue gas produced per kg fuel = 11.73 kg \n", + " (b)The dry exhaust loss = 1723.7 kJ/kg fuel and incomplete combustion loss per kg fuel = 1648.15 kJ/kg fuel \n", + " (c)The boiler efficiency = 83.12 percentage \n", + " (d) THe fuel burning rate = 5.566 kg/s \n", + " (e)The percentage of excess air used = 31.12 percentage \n", + " (f) The percentage of energy absorbed in the superheater = 24.05 percentage\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "C=61##The mass of carbon present in the coal according to coal analysis on mass basis in %\n", + "H=4##The mass of hydrogen present in the coal according to coal analysis on mass basis in %\n", + "O=3##The mass of oxygen present in the coal according to coal analysis on mass basis in %\n", + "N=2##The mass of nitrogen present in the coal according to coal analysis on mass basis in %\n", + "S=1##The mass of sulphur present in the coal according to coal analysis on mass basis in %\n", + "M=4##The mass of moisture present in the coal according to coal analysis on mass basis in %\n", + "A=25##The mass of ash present in the coal according to coal analysis on mass basis in %\n", + "HHV=24.3##The high heating value of the coal i.e energy released by complete combustion of 1 kg fuel in MJ/kg\n", + "CO2=12##The amount of carbondioxide by volume according to dry flue gas analysis in %\n", + "CO=1.5##The amount of carbonmonoxide by volume according to dry flue gas analysis in %\n", + "O2=7##The amount of oxygen by volume according to dry flue gas analysis in %\n", + "N2=79.5##The amount of nitrogen by volume according to dry flue gas analysis in %\n", + "Te=170##Exhaust gas temperature in degree centigrade\n", + "L=0.03##Energy loss other than dry exhaust loss and incomplete combustion is 3% of HHV\n", + "R=150##Steam generation rate in t/h\n", + "Po=100##Steam condition at boiler outlet in bar\n", + "To=500##Steam condition at boiler outlet in degree centigrade\n", + "Ti=160##Feed water inlet temperature in degree centigrade\n", + "HCO2=33083##Heat of reaction in kJ/kg carbon\n", + "HCO=9500##Heat of reaction in kJ/kg carbon\n", + "cp=1.05##Heat capacity of dry flue gas (dfg) in kJ/kgK\n", + "Ta=30##The ambient temperature of air in degree centigrade\n", + "Mc=44##Molecular weight of Carbondioxide\n", + "Mco=28##Molecular weight of carbonmonoxide\n", + "Mo=32##Molecular weight of oxygen\n", + "Mn=28##Molecular weight of nitrogen\n", + "Mx=12##Molecular weight of carbon\n", + "h1=3373.7##Enthalpy at 100 bar and 500 degree centigrade in kJ/kg\n", + "hf=675.55##Enthalpy at 160 degree centigrade in kJ/kg\n", + "hg=2724.7##Enthalpy at 100 bar in kJ/kg\n", + "\n", + "#Calculations\n", + "Mdfg=(((C/100)*((Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(N2/100))))/(Mx*((CO2/100)+(CO/100))))##Mass of dry flue gas produced per kg of fuel in kg\n", + "Ed=Mdfg*cp*(Te-Ta)##Energy loss due to dry exhaust gas in kJ/kg fuel\n", + "#Since Mdfg is 11.73kg through sciab calculation, there is a variation in Ed value and Ei value\n", + "Ei=((Mdfg)*(HCO2-HCO)*(Mx/Mco))*((Mco*(CO/100))/((Mc*(CO2/100))+(Mco*(CO/100))+(Mo*(O2/100))+(Mco*(N2/100))))##Energy loss due to incomplete combustion in kJ/kg fuel\n", + "El=L*HHV##Energy loss other than dry exhaust loss and incomplete combustion loss in MJ/kg fuel\n", + "TEl=(Ed/1000)+(Ei/1000)+El##Total energy loss in MJ/kg fuel\n", + "Be=((HHV-TEl)/(HHV))*100##Boiler efficiency in %\n", + "Wf=(((R*1000)*(h1-hf))/((Be/100)*HHV*1000))/3600##The fuel burning rate in kg/s\n", + "Wth=(11.5*(C/100))+(34.5*((H/100)-(O/800)))+(4.3*(S/100))##Thearetical air required per kg of fuel in kg\n", + "WA=(((3.04*(N2/100)*(C/100)))/((CO2/100)+(CO/100)))-((N/100)*(1/0.768))##Actual air supplied per kg of fuel in kg\n", + "per=((WA-Wth)/Wth)*100##Percentage excess air used in %\n", + "pea=((h1-hg)/(h1-hf))*100##Percentage of energy absorbed in the superheater\n", + "\n", + "#Output\n", + "print \" (a)The amount of dry flue gas produced per kg fuel = %3.2f kg \\n (b)The dry exhaust loss = %3.1f kJ/kg fuel and incomplete combustion loss per kg fuel = %3.2f kJ/kg fuel \\n (c)The boiler efficiency = %3.2f percentage \\n (d) THe fuel burning rate = %3.3f kg/s \\n (e)The percentage of excess air used = %3.2f percentage \\n (f) The percentage of energy absorbed in the superheater = %3.2f percentage\"%(Mdfg,Ed,Ei,Be,Wf,per,pea)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.5 Pg: 216" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The total volume of combustion products at 200 degee centigrade and 1.013 bar = 26.26 m**3 \n", + " (b)The dry flue gas analysis based on carbondioxide,oxygen and nitrogen is \n", + " Carbondioxide = 11.64 percent \n", + " Oxygen = 5.07 percent \n", + " Nitrogen = 83.29 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "C=83.7##The amount of carbon present in the fuel oil according to ultimate analysis of a fuel oil in %\n", + "H=12.7##The amount of hydrogen present in the fuel oil according to ultimate analysis of a fuel oil in %\n", + "O=1.2##The amount of oxygen present in the fuel oil according to ultimate analysis of a fuel oil in %\n", + "N=1.7##The amount of nitrogen present in the fuel oil according to ultimate analysis of a fuel oil in %\n", + "S=0.7##The amount of sulphur present in the fuel oil according to ultimate analysis of a fuel oil in %\n", + "td=27##The dry bulb temperature of combustion air in degree centigrade\n", + "tw=21##The wet bulb temperature of combustion air in degree centigrade\n", + "E=0.3##Excess air and assuming complete combustion in %\n", + "t=200##Temperature to find total volume of combustion products in degree centigrade\n", + "p=1.013##Pressure to find total volume of combustion procucts in bar\n", + "\n", + "#Calculations\n", + "Wth=(11.5*(C/100))+(34.5*((H/100)-(O/100)*(1/8)))+(4.3*(S/100))##Theoretical air required per kg of fuel in kg\n", + "WA=(1+E)*Wth##Actual air required per kg of fuel in kg/kg fuel\n", + "sh=0.0132##Specific humidity at DBT and WBT in kg moisture/kg dry air\n", + "W=WA*sh##Water vapour entering with air per kg fuel in kg vap/kg fuel\n", + "Tw=(9*(H/100))+WA##Total water vapour formed per kg fuel in kg\n", + "CO2=(44/12)*(C/100)##mass of carbondioxide gas per kg of fuel\n", + "O2=0.232*E*Wth##Mass of oxygen gas per kg of fuel\n", + "N2=0.768*(1+E)*Wth+(N/100)##Mass of nitrogen gas per kg of fuel\n", + "SO2=(64/32)*(S/100)##Mass of nitrogen gas per kg of fuel\n", + "H2O=1.383##Mass of water per kg of fuel\n", + "M=(CO2/44)+(O2/32)+(N2/28)+(SO2/64)+(H2O/18)##Moles of combustion gases formed per kg fuel\n", + "VG=M*22.4*((273+t)/273)*(1.013/1.013)##Volume of flue gases at 200 degree centigrade and 1.013 bar per kg fuel\n", + "CO21=((CO2/44)/((CO2/44)+(O2/32)+(N2/28)))*100##Composition of dry flue gas CO2 by volume\n", + "O21=((O2/32)/((CO2/44)+(O2/32)+(N2/28)))*100##Composition of dry flue gas O2 by volume\n", + "N21=((N2/28)/((CO2/44)+(O2/32)+(N2/28)))*100##Composition of dry flue gas N2 by volume\n", + "\n", + "#Output\n", + "print \" (a)The total volume of combustion products at 200 degee centigrade and 1.013 bar = %3.2f m**3 \\n (b)The dry flue gas analysis based on carbondioxide,oxygen and nitrogen is \\n Carbondioxide = %3.2f percent \\n Oxygen = %3.2f percent \\n Nitrogen = %3.2f percent\"%(VG,CO21,O21,N21)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.6 Pg: 217" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The molecular weight of the combustion products M = 28.36 kg/kg mol \n", + " (b) The total gas volume for complete combustion at 260 degree centigrade and 1.013 bar is 31.49 m**3/kg fuel \n", + " (c)The dry flue gas analysis on \n", + " carbondioxide = 9.7 percent \n", + " oxygen = 4.6 percent \n", + " nitrogen = 85.7 percent \n" + ] + } + ], + "source": [ + "#Input data\n", + "C2H6=22.6##The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %\n", + "CH4=73.6##The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %\n", + "CO2=2.4##The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %\n", + "N2=1.4##The amount of gas present in the fuel gas according to volumetric analysis of fuel gas by volume in %\n", + "E=0.25##Assuming combustion air to be dry and in excess\n", + "t=260##The temperature for the total gas volume for complete combustion in degree centigrade\n", + "p=1.013##The pressure for the total gas volume for complete combustion in bar\n", + "Mch=30##Molecular weight of C2H6\n", + "Mc=16##Molecular weight of CH4\n", + "Mco=44##Molecular weight of CO2\n", + "Mn=28##Molecular weight of N2\n", + "Mo=32##Molecular weight of O2\n", + "Mh=18##Molecular weight of H2O\n", + "\n", + "#Calculations\n", + "x=100##Assuming 100 moles of fuel gas \n", + "Mf=((C2H6/100)*Mch)+((CH4/100)*Mc)+((N2/100)*Mn)+((CO2/100)*Mco)##Molecular weight of fuel gas\n", + "Ma=((226.3*(Mo+(3.76*Mn))*(1+E)))/28.96##Moles of air supplied\n", + "Mc=1440##Moles of combustion gas from the equation \n", + "Mr=x+Ma+Mc##Total reaction molecules\n", + "Mwc=((121.2*Mco)+(215*Mh)+(56.6*Mo)+(1065.4*Mn))/Mc##Molecular weight of combustion gas in kg/kgmol\n", + "Mt=Mc/(x*20)##Total number of moles of combustion gas per kg fuel gas\n", + "VG=Mt*22.4*((273+t)/273)##Volume of combustion products per kg fuel gas\n", + "CO21=(121.2/(121.2+56.6+1065.4))*100##Gas analysis of CO2 by volume\n", + "O21=(56.6/1243.2)*100##Gas analysis of O2 by volume\n", + "N2=(1065.4/1243.2)*100##Gas analysis of N2 by volume\n", + "\n", + "#Output\n", + "print \" (a)The molecular weight of the combustion products M = %3.2f kg/kg mol \\n (b) The total gas volume for complete combustion at 260 degree centigrade and 1.013 bar is %3.2f m**3/kg fuel \\n (c)The dry flue gas analysis on \\n carbondioxide = %3.1f percent \\n oxygen = %3.1f percent \\n nitrogen = %3.1f percent \"%(Mwc,VG,CO21,O21,N2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.7 Pg: 218" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The air leakage into the air preheater per kg of coal fired is 1 kg air/kg fuel\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "CO21=9.7##Carbondioxide gas analysis before the air preheater \n", + "CO22=9.2##Carbondioxide gas analysis after the air preheater\n", + "O21=4.0##Oxygen gas analysis before the air preheater\n", + "O22=4.9##Oxygen gas analysis after the air preheater\n", + "N21=86.3##Nitrogen gas analysis before the air preheater\n", + "N22=85.9##Nitrogen gas analysis after the air preheater\n", + "C=72.0##The coal used shows the carbon percentage by mass in %\n", + "\n", + "#Calculations\n", + "W1=((3.04)*(N21/100)*(C/100))/((CO21/100))##Before air preheater in kg air/kg fuel\n", + "W2=((3.04)*(N22/100)*(C/100))/((CO22/100))##After air preheater in kg air/kg fuel\n", + "A=W2-W1##Air leakage in kg air/kg fuel\n", + "\n", + "#Output\n", + "print \"The air leakage into the air preheater per kg of coal fired is %3.0f kg air/kg fuel\"%(A)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.8 Pg: 218" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The height of the stack H = 36.44 m \n", + " (b)The diameter at its base D = 1.06 m \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "n=6##Total lancashire boilers in a textile factory\n", + "Ws=6##Each boiler supplying steam in t/h\n", + "p=16##Pressure at which steam is supplied in bar\n", + "t=250##Temperature at which steam is supplied in degree centigrade\n", + "CV=43960##Calorific value of the fuel oil in kJ/kg\n", + "no=75##Overall efficiency of the boiler in %\n", + "a=16##The amount of air required for efficient burning of the fuel inkg\n", + "H=20##Drought of water gauge required at the base of chimney in mm\n", + "tf=320##The flue gases leave the boiler in degree centigrade\n", + "ts=300##The average temperature of the gases in the stack in degree centigrade\n", + "ta=30##The atmospheric temperature in degree centigrade\n", + "R=0.287##Real Gas constant in kJ/kgK\n", + "h1=2919.2##enthalpy at the entrance of the boiler in kJ/kgK\n", + "hf=125.8##Enthalpy at the feed in kJ/kgK\n", + "pi=3.1412##Mathematical constant\n", + "g=9.81##gravitational fore constant in m/s**2\n", + "P=1.013##Atmospheric pressure in bar\n", + "\n", + "#Calculations\n", + "H1=((H*R*(273+ta)*(273+ts)))/(P*100*((273+ts)-(273+ta)))##The draught produced in m\n", + "Wf=(((Ws*1000)*6*(h1-hf))/((no/100)*CV))/3600##Air fuel ratio in kg/s\n", + "Wa=a*Wf##Actual air fuel ratio in kg/s\n", + "Wfg=17*Wf##Air fuel ratio in kg/s\n", + "D=(((Wfg*R*(273+ts)*(4/pi)))/((101.3)*(2*g*H1)**(1/2)))**(1/2)##Diameter at its base in m\n", + "\n", + "#Output\n", + "print \"(a)The height of the stack H = %3.2f m \\n (b)The diameter at its base D = %3.2f m \"%(H1,D)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.9 Pg: 219" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required motor capacity needed for the FD fan is 90.49 kW \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "Wf=10##Coal rate in t/h\n", + "C=78##The mass of carbon present in the coal according to coal analysis on mass basis in %\n", + "H=3##The mass of hydrogen present in the coal according to coal analysis on mass basis in %\n", + "O=3##The mass of oxygen present in the coal according to coal analysis on mass basis in %\n", + "S=1##The mass of sulphur present in the coal according to coal analysis on mass basis in %\n", + "M=7##The mass of moisture present in the coal according to coal analysis on mass basis in %\n", + "A=8##The mass of ash present in the coal according to coal analysis on mass basis in %\n", + "E=0.3##Excess air in percentage\n", + "p=180##Plenum chamber pressure in mm water gauge\n", + "nm=0.6##Mechanical efficiency of the fan\n", + "ta=30##Room temperature in degree centigrade\n", + "R=0.287##Real gas constant\n", + "P=101.325##Atmospheric pressure in kPa\n", + "g=9.812##gravitational force constant m/s**2\n", + "\n", + "#Calculations\n", + "Wth=(11.5*(C/100))+(34.5*((H/100)-(O/(8*100))))+(4.3*(S/100))##Theoretical air required per kg fuel in kg air/kg fuel\n", + "WA=Wth*(1+0.3)##Actual air required per kg fuel in kg air/kg fuel\n", + "Va=(R*(273+ta))/P##Volume flow rate of air in m**3/kg\n", + "FD=((WA*Wf*1000*Va*p*g)/(3600*nm))/1000##FD fan motor capacity in kW\n", + "\n", + "#Output\n", + "print \"The required motor capacity needed for the FD fan is %3.2f kW \"%(FD)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.10 Pg: 220" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The motor capacity of the ID fan is 233.64 kW \n" + ] + } + ], + "source": [ + "#Input data\n", + "tg=180##The gas temperature in degree centigrade\n", + "p=250##The draught produced by the ID fan in mm\n", + "nf=0.52##The efficiency of the fan\n", + "Va=0.858##Volume flow rate of air in m**3/kg\n", + "g=9.812##gravitational force constant in m/s**2\n", + "Wf=10##Coal rate in t/h\n", + "Wa=12.9##Actual air required per kg fuel in kg air/kg fuel\n", + "ta=30##Room temperature in degree centigrade\n", + "\n", + "#Calculations\n", + "Wfg=((Wf+(Wa*10))*1000)/3600##Fuel gas required in kg/s\n", + "Vfg=(Va*(tg+273))/(ta+273)##Volume flow rate of fuel gas in m**3/kg\n", + "ID=((Wfg*Vfg*p*g)/(nf))/1000##ID fan motor capacity in kW\n", + "\n", + "#Output\n", + "print \"The motor capacity of the ID fan is %3.2f kW \"%(ID)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.11 Pg: 220" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The complete volumetric composition of the dry flue gas is \n", + " Carbondioxide by volume = 13.20 percentage \n", + " Oxygen by volume = 3.20 percentage \n", + " Carbonmonoxide by volume = 4.21 percentage \n", + " Nitrogen by volume = 79.39 percentage \n", + " (b) The actual amount of air supplied per kg coal = 12.27 kg \n", + " (c) Mass of water vapour formed per kg coal = 0.40 kg \n", + " (d) The dew point temperature of the flue gas = 32.90 degree centigrade \n" + ] + } + ], + "source": [ + "#Input data\n", + "CO2=13.2##The volume of carbondioxide present in the partial analysis of dry flue gas in %\n", + "O2=3.2##The volume of oxygen present in the partial analysis of dry flue gas in %\n", + "C=88##The mass of carbon present in the coal according to coal analysis on mass basis in %\n", + "H=4.4##The mass of hydrogen present in the coal according to coal analysis on mass basis in %\n", + "A=7.6##The mass of ash present in the coal according to coal analysis on mass basis in %\n", + "M=0##Moisture present in the fuel was nil\n", + "Mc=12##Molecular weight of the carbon \n", + "Mh=2##Molecular weight of the hydrogen\n", + "Mo=32##Molecular weight of the oxygen\n", + "Mho=18##Molecular weight of water\n", + "p=101.325##Atmospheric pressure in kPa\n", + "\n", + "#Calculations\n", + "c=C/Mc##Equating coefficients of the carbon from the equation \n", + "g=H/Mh##Equating coefficients of the hydrogen from the equation \n", + "x=(CO2/100)/(O2/100)##From dry fuel gas analysis (dfg) \n", + "d=(((CO2/100)*(47.5))-7.333)/(((CO2/100)*(3.032))-1)##Coefficient of the carbonmonoxide in the equations product side\n", + "b=c-d##Coefficient of the carbondioxide in the equation product side\n", + "a=10.21-(0.742*d)##Coefficient of the oxygen in the reactant side of the equation \n", + "e=b/x##Coefficient of the oxygen in the product side of the equation \n", + "f=3.76*a##Equating coefficients of the nitrogen from the equation \n", + "ma=(a*Mo)/0.232##Mass of air supplied for 100 kg coal in kg\n", + "ma1=ma/100##Mass of air supplied per kg coal in kg\n", + "T=b+d+e+f##Total number of moles of dry flue gas (dfg)\n", + "CO21=(b/T)*100##Carbondioxide by volume in percentage\n", + "O21=(e/T)*100##Oxygen by volume in percentage\n", + "CO1=(d/T)*100##Carbonmonoxide by volume in percentage\n", + "N21=(f/T)*100##Nitrogen by volume in percentage\n", + "Mwv=(g*Mho)/100##Mass of watervapour formed per kg coal in kg\n", + "Mf=(g)/(b+d+e+f+g)##Mole fraction of water vapour in flue gas\n", + "P=Mf*p##Partial pressure of water vapour in kPa\n", + "D=32.9##Dew point temperature from steam tables in degree centigrade\n", + "\n", + "#Output\n", + "print \"(a)The complete volumetric composition of the dry flue gas is \\n Carbondioxide by volume = %3.2f percentage \\n Oxygen by volume = %3.2f percentage \\n Carbonmonoxide by volume = %3.2f percentage \\n Nitrogen by volume = %3.2f percentage \\n (b) The actual amount of air supplied per kg coal = %3.2f kg \\n (c) Mass of water vapour formed per kg coal = %3.2f kg \\n (d) The dew point temperature of the flue gas = %3.2f degree centigrade \"%(CO21,O21,CO1,N21,ma1,Mwv,D)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.12 Pg: 222" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The height of the gas plume is H = 1842.5 m \n" + ] + } + ], + "source": [ + "#Input data\n", + "H=200##Height of the stack in m\n", + "D=4##Diameter of the stack in m\n", + "m=1000##Mass flow rate of gas in kg/s\n", + "Ts=100##Stack exit gas temperature in degree centigrade\n", + "Ta=5##Ambient air temperature in degree centigrade\n", + "Vw=50##Wind velocity in Km/h\n", + "Cp=1.005##Specific heat of the gas in kJ/kgK\n", + "pi=3.142##Mathematical constant the value of pi\n", + "\n", + "#Calculations\n", + "Vw1=(50*1000)/(60*60)##Wind velocity in m/s\n", + "Qe=m*Cp*(Ts-Ta)##Heat emission from plume in kW\n", + "Qe1=Qe/1000##Heat emission from the plume in MW\n", + "p=(101.325)/(0.287*373)##Density of the gas in kg/m**3\n", + "A=(pi*D**2)/4##Area of the stack in m**2\n", + "Vs=m/(p*A)##Stack gas exict velocity in m/s\n", + "H1=((2.62*(Qe1**(1/2))*1000)/Vw1)-((0.029*Vs*D)/Vw1)##The height of the gas plume in m\n", + "\n", + "#Output\n", + "print \"The height of the gas plume is H = %3.1f m \"%(H1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 4.13 Pg: 222" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The maximum temperature of the cycle T3 = 1168 K \n", + " (b)Thermal efficiency of the plant = 0.492 or 49.23 percentage\n", + " (c) Specific fuel consumption = 0.641 kg/kWh \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "CV=20##Calorific value of the fuel in MJ/kg\n", + "C=65##The amount of carbon present in the fuel according to gravimetric analysis in %\n", + "H=25##The amount of hydrogen present in the fuel according to gravimetric analysis in %\n", + "O=10##The amount of oxygen present in the fuel according to gravimetric analysis in %\n", + "p1=1##Pressure at the inlet of the compressor in bar\n", + "t1=27##Temperature at the inlet of the compressor in degree centigrade\n", + "p2=4##The pressure which compressor compresses it isentropically in bar\n", + "Re=78##The regenerator effectiveness in %\n", + "CO2=6##The amount of carbondioxide according to the analysis of dry exhaust gas in %\n", + "CO=1.5##The amount of carbonmonoxide according to the analysis of dry exhaust gas in %\n", + "Cp=1.005##Specific heat capacity of the air in kJ/kgK\n", + "i=1.44##Isentropic index for the air\n", + "Cp1=1.15##Specific heat capacity of the air in kJ/kgK\n", + "i1=1.33##Isentropic index for the combustion products\n", + "Mc=12##Molecular weight of the carbon \n", + "Mh=2##Molecular weight of the hydrogen\n", + "Mo=32##Molecular weight of the oxygen\n", + "Mho=18##Molecular weight of water\n", + "T0=288##Datum temperature in K (Assumed)\n", + "\n", + "#Calculations\n", + "h=(C/100)/(Mc)##Equating coefficients of the carbon from the equation \n", + "e=(H/100)/Mh##Equating coefficients of the hydrogen from the equation \n", + "y=(CO/100)/(CO2/100)##From dry exhaust gas analysis for solving\n", + "a=h/(1+y)##The coefficient of the carbondioxide in the product side of the equation \n", + "b=h-a##The coefficient of the carbonmonoxide in the product side of the equation \n", + "z=b/(CO/100)##The sum of coefficients of the product side of the equation \n", + "x=z-(b/2)+(e/2)##Mol of air supplied in kmol\n", + "wa=x*28.96##Air supplied in kg/kg fuel\n", + "wf=1##Assuming 1 kg of fuel supplied\n", + "T2=(t1+273)*(p2/p1)**((i-1)/i)##Temperature at the outlet of the compressor in K\n", + "T3=(((wa*Cp*(T2-T0))+(wf*CV*1000))/((wa+wf)*(Cp1)))+T0##Maximum temperature of the cycle in K\n", + "T4=T3/((4)**((i1-1)/i1))##Temperature at point of the cycle in K\n", + "T5=((Re/100)*(T4-T2))+T2##Temperature at point of the cycle in K\n", + "Wc=wa*Cp*(T2-(t1+273))##Work done by the compressor in kW\n", + "Wt=23.54*Cp1*(T3-T4)##Work done by the turbine in kW\n", + "Q1=23.54*Cp1*(T3-T5)##Total work done by the system in kW\n", + "nc=(Wt-Wc)/Q1##Efficiency of the cycle \n", + "nc1=nc*100##Efficiency of the cycle in %\n", + "spc=3600/(Wt-Wc)##Specific fuel consumption in kg/kWh\n", + "\n", + "#Output \n", + "print \" (a) The maximum temperature of the cycle T3 = %3.0f K \\n (b)Thermal efficiency of the plant = %3.3f or %3.2f percentage\\n (c) Specific fuel consumption = %3.3f kg/kWh \"%(T3,nc,nc1,spc)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch5.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch5.ipynb new file mode 100644 index 00000000..9eb71916 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch5.ipynb @@ -0,0 +1,290 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5 : Combustion Mechanism, Combustion Equipment And Firing Methods" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.1 Page 308" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The total surface area of the particles in the bed As = 8423 m**2 \n" + ] + } + ], + "source": [ + "#Input data\n", + "Vs=2500##The mass of a bed of solid particles in kg\n", + "p=2650##The density of the solid in kg/m**3\n", + "d=800*10**-6##The mean particle size in m\n", + "s=0.84##The sphericity of the particle\n", + "\n", + "#Calculations\n", + "As=(6*Vs)/(p*d*s)##The total surface area of the particles in the bed\n", + "\n", + "#Output\n", + "print \" The total surface area of the particles in the bed As = %3.0f m**2 \"%(As)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.2 Page 309" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The voidage of the bed = 0.417 \n", + " (b) The minimum fluidization velocity Umf = 0.187 m/s \n" + ] + } + ], + "source": [ + "#Input data\n", + "d=427*10**-6##The mean particle size in m\n", + "pg=1.21##The density of air in kg/m**3\n", + "v=1.82*10**-5##The viscosity of air in kg/ms\n", + "pl=1620##The density of the loosely packed bed in kg/m**3\n", + "ps=2780##The density of the solids in kg/m**3\n", + "c1=27.2##(Grace,1982)constant value.\n", + "c2=0.0408##(Grace,1982)constant value\n", + "g=9.812##Gravitational forc constant in m/s**2\n", + "\n", + "#Calculations\n", + "E=1-(pl/ps)##The voidage of the bed\n", + "Ar=((pg)*(ps-pg)*g*(d**3))/v**2##Archimedes number\n", + "Re=(c1**2+(c2*Ar))**(0.5)-c1##Reynolds number\n", + "Umf=Re*v/(pg*d)##Minimum superficial velocity in m/s\n", + "\n", + "#Output\n", + "print \" (a) The voidage of the bed = %3.3f \\n (b) The minimum fluidization velocity Umf = %3.3f m/s \"%(E,Umf)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.3 Page 309" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The sphericity of particles is = 0.811 \n" + ] + } + ], + "source": [ + "from scipy.optimize import fsolve\n", + "#Input data\n", + "d=427*10**-6##The mean particle size in m\n", + "pg=1.21##The density of air in kg/m**3\n", + "v=1.82*10**-5##The viscosity of air in kg/ms\n", + "Umf=0.14##Minimum superficial velocity in m/s\n", + "Ar=7753##Archimedes number from previous example problem\n", + "\n", + "#Calculations\n", + "\n", + "Re=(Umf*pg*d)/v##Reynolds number\n", + "def F(x):##function definition\n", + " f = 7753*x**2- 381.1*x -4793#\n", + " return f\n", + "x = 100##Initial guss\n", + "y = fsolve(F,x)#\n", + "\n", + "#Output\n", + "print \"The sphericity of particles is = %3.3f \"%(y)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.4 Page 310" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required flow rate of limestone is 2405.3 kg/h \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "O=35##The output of the fluidized bed combustion system in MW\n", + "n=0.80##Efficiency of the fluidized bed combustion system \n", + "H=26##The heating value of coal in MJ/kg\n", + "S=3.6##Sulphur content in the coal in %\n", + "C=3##The calcium sulphur ratio \n", + "Ca=85##The amount of calcium carbonate in the limestone in %\n", + "CaCO3=100##The molecular weight of CaCO3\n", + "\n", + "#Calculations\n", + "Cb=O/(n*H)##Coal burning rate in kg/s\n", + "Cb1=Cb*3600##Coal burning rate in kg/h\n", + "Sf=(Cb1*(S/100))/32##Flow rate of sulphur in Kmol/h\n", + "Cf=Sf*C##The flow rate of calcium in Kmol/h\n", + "Caf=Cf*CaCO3##Mass flow rate of CaCO3 in kg/h\n", + "L=Caf/(Ca/100)##Mass flow rate of limestone in kg/h\n", + "\n", + "#Output\n", + "print \"The required flow rate of limestone is %3.1f kg/h \"%(L)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.5 Page 310" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The rate of heat removal from the bed = 6405 kW \n", + " (b) The rate of heat removal from the above bed zone = 16333 kW \n" + ] + } + ], + "source": [ + "#Input data\n", + "CV=24##The calorific value of the fuel in MJ/kg\n", + "C=0.65##The amount of calorific value released in the bed in %\n", + "to=850##Temperature at which products leave in degree centigrade\n", + "ti=30##The inlet temperature in degree centigrade\n", + "tb=850##The bed temperature in degree centigrade\n", + "A=14.5##The air fuel ratio by mass\n", + "Cp=1.035##The specific heat of the products leaving the bed surface in kJ/kgK\n", + "B=7000##The burning rate of coal in kg/h\n", + "\n", + "#Calculations\n", + "H=(C*CV*1000)-(A*Cp*(to-ti))##Heat removal from the bed per kg fuel in kJ/kg fuel\n", + "Hr=(H*B)/3600##Rate of heat removal from the bed in kW\n", + "Hb=(B/3600)*(1-C)*CV*1000##The rate of heat removal from the above bed zone in kW\n", + "\n", + "#Output\n", + "print \" (a) The rate of heat removal from the bed = %3.0f kW \\n (b) The rate of heat removal from the above bed zone = %3.0f kW \"%(Hr,Hb)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex 5.6 Page 311" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The planform area = 2.4 m**2 \n", + " (b) Fuel burning rate = 0.192 kg/s \n", + " Air flow rate = 2.1888 kg/s \n", + " Planform area = 2.58 m**2 \n" + ] + } + ], + "source": [ + "#Input data\n", + "tb=850##The bed temperature in degree centigrade\n", + "CV=25##The calorific value of the fuel in MJ/kg\n", + "A=9.5##The stoichiometric air fuel ratio by mass\n", + "E=20##The amount of excess air used in %\n", + "F=4.8##The total fueling rate in MW\n", + "p=0.3145##The density of air at bed temperature in kg/m**3\n", + "f=2##The firing rate in MW/m**2\n", + "v=2.7##The fluidizing velocity in m/s\n", + "\n", + "#Calculations\n", + "P=F/f##Planform area in m**2\n", + "m=(F*1000)/(CV*1000)##Fuel burning rate in kg/s\n", + "ma=A*(1+(E/100))*m##Mass flow rate of air in kg/s\n", + "Pa=ma/(p*v)##Planform area in m**2\n", + "\n", + "#Output\n", + "print \" (a) The planform area = %3.1f m**2 \\n (b) Fuel burning rate = %3.3f kg/s \\n Air flow rate = %3.4f kg/s \\n Planform area = %3.2f m**2 \"%(P,m,ma,Pa)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch6.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch6.ipynb new file mode 100644 index 00000000..499e2ad1 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch6.ipynb @@ -0,0 +1,474 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6 : Steam generators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.1 Pg: 411" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The pressure head developed = 35.8 kPa \n", + " (b)The void fraction at riser exit = 0.5519 \n", + " (c) The heat transfer rate per unit projected area = 293.1 kW/m**2 \n" + ] + } + ], + "source": [ + "#Input data\n", + "H=18##The length of furnace wall riser in m\n", + "O=76.2##The outer diameter of the furnace wall riser in mm\n", + "T=6.1##The thickness of the furnace wall riser in mm\n", + "P=80##Pressure at which saturated water is recieved in bar\n", + "V=1.5##The velocity of the saturated water in m/s\n", + "CR=12.5##Assuming circulation ratio\n", + "S=1.2##Assuming slip ratio\n", + "g=9.81##Gravitational force constant in m/s**2\n", + "pi=3.142##Mathematical constant\n", + "\n", + "#Calculations\n", + "xt=1/CR##The quality of steam at the top of the riser\n", + "vf=0.001384##Specific volume of saturated liquid at 80 bar in m**3/kg\n", + "vfg=0.02214##Specific volume of Evaporation gas at 80 bar in m**3/kg\n", + "vg=0.02352##Specific volume of saturated gas at 80 bar in m**3/kg\n", + "pf=1/vf##Density of the saturated liquid at 80 bar in kg/m**3\n", + "vt=vf+(xt*vfg)##Specific volume of the steam at the top of the riser in m**3/kg\n", + "pt=1/vt##Density of steam at the top of the riser in kg/m**3\n", + "pm=(pt+pf)/2##Mean density in kg/m**3\n", + "Ph=(H*g*(pf-pm))/1000##The pressure head developed in kPa\n", + "C=(vf/vg)*S##The part of calculation for the void fraction\n", + "VF=1/(1+((1-xt)*C)/xt)##The void fraction at riser exit\n", + "hfg=1441.3##Enthalpy of the evaporation in kJ/kg\n", + "di=O-12.2##Inner diameter of the furnace wall riser in mm\n", + "A=(pi*di**2)/4##Inner area in m**2\n", + "w=pf*A*V##Mass flow rate of saturated water entering the riser in kg/s\n", + "ws=xt*w##The rate of steam formation in the riser tube in kg/s\n", + "h=((ws*hfg)/(O*H))/1000##Heat transfer rate per unit projected area in kW/m**2\n", + "\n", + "#output\n", + "print \" (a) The pressure head developed = %3.1f kPa \\n (b)The void fraction at riser exit = %3.4f \\n (c) The heat transfer rate per unit projected area = %3.1f kW/m**2 \"%(Ph,VF,h)\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.2 Pg: 413" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The amount of water that must be sprayed is 7.737 t/h or 2.149 kg/s \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "t=60##The temperature of water while supplying it to desuperheater in degree centigrade\n", + "ws=200##The amount of steam carrying in a steam line in t/h\n", + "p=35##The pressure of steam in bar\n", + "ts=400##The temperature to be maintained by the steam in degree centigrade\n", + "to=450##The outlet temperature of the steam from boiler in degree centigrade\n", + "h1=3337.2##The enthalpy of steam at 450 degree C in kJ/kg\n", + "h2=252##The enthalpy of water at 60 degree C in kJ/kg\n", + "h3=3222.3##The enthalpy of steam at 400 degree C in kJ/kg\n", + "\n", + "#Calculations\n", + "w=(ws*(h1-h3))/(h3-h2)##Mass flow rate of water in t/h\n", + "w1=w*(1000/3600)##Mass flow rate of water in kg/s\n", + "\n", + "#Output\n", + "print \"The amount of water that must be sprayed is %3.3f t/h or %3.3f kg/s \"%(w,w1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.3 Pg: 413" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The pressure head developed due to natural circulation is 38163 N/m**2 or 38.163 kPa\n" + ] + } + ], + "source": [ + "from math import log\n", + "#Input data\n", + "\n", + "H=15##The high of downcomer riser circuit in m\n", + "P=160##The pressure at which downcomer riser circuit operates in bar\n", + "xe=0.5##The exit quality of the steam \n", + "S=1.2##Slip factor\n", + "vf=0.001711##Specific volume of saturated liquid in m**3/kg\n", + "vg=0.009306##Specific volume of saturated gas in m**3/kg\n", + "g=9.806##Gravitational force constant in m/s**2\n", + "\n", + "#Calculations\n", + "C=S*(vf/vg)##The part of calculation for the void fraction \n", + "VF=1/(1+((1-xe)*C)/xe)##The void fraction at riser exit\n", + "pf=1/vf##Density of the saturated liquid in kg/m**3\n", + "pg=1/vg##Density of the saturated gas in kg/m**3\n", + "pm=pf-(((pf-pg)/(1-C))*(1-((1/((VF)*(1-C)))-1)*log(1/(1-(VF*(1-C))))))##The average mixture density in the riser in kg/m**3\n", + "P1=g*(pf-pm)*H##Pressure head developed due to natural circulation in N/m**2\n", + "P2=P1/1000##ressure head developed due to natural circulation in kPa\n", + "\n", + "#Output \n", + "print \"The pressure head developed due to natural circulation is %3.0f N/m**2 or %3.3f kPa\"%(P1,P2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.4 Pg: 414" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The steam generation rate = 93.383 kg/s \n", + " (b) The fuel burning rate = 10.898 kg/s \n", + " (c) The evaporation factor = 8.57 \n", + " (d) The pressure head available for natural circulation = 0.6448 bar \n", + " (e) The circulation ratio = 12.5 \n", + " (f)The number of risers required = 713 \n", + " (g) The heat absorbtion rate per unit projected area of the riser = 72.51 kW/m**2\n" + ] + } + ], + "source": [ + "from math import ceil\n", + "#Input data\n", + "\n", + "W=120##The amount of electricity produced in the power plant in MW\n", + "po=100##The pressure of the steam at the outlet of boiler in bar\n", + "to=500##The temperature of steam at the outlet of boiler in degree centigrade\n", + "p=0.1##The condenser pressure in bar\n", + "nb=0.9##The efficiency of the boiler\n", + "CV=25.7##The calorific value of the coal in MJ/kg\n", + "ti=160##The feed water temperature at boiler inlet in degree centigrade\n", + "H=40##The high of the risers in the furnace wall in m\n", + "xt=0.08##The quality of the steam at the top of the riser\n", + "v=2##The exit velocity of the riser and entering the drum in m/s\n", + "Do=60##The outer diameter of the risers in mm\n", + "T=3##The thickness of the wall in mm\n", + "pi=3.142##Mathematical constant\n", + "g=9.806##Gravitational force constant in m/s**2\n", + "\n", + "#Calculations\n", + "h1=3374.8##Enthalpy at point 1 in kJ/kg\n", + "s1=6.6011##Entropy at point 1 in kJ/kgK\n", + "sf=0.6479##Entropy of the saturated liquid at point 1 in kJ/kgK\n", + "sg=7.5055##Entropy of the Saturated vapour at point 1 in kJ/kgK\n", + "x2=(s1-sf)/sg##The quality of the steam\n", + "h2=191.46+(x2*2393.29)##Enthalpy at point 2 in kJ/kg\n", + "h3=191.46##Enthalpy at point 3 in kJ/kg\n", + "h5=675.5##Enthalpy at point 5 in kJ/kg\n", + "ws=(W*1000)/(h1-h2)##Mass flow rate of steam in kg/s\n", + "wf=(ws*(h1-h5))/(nb*CV*1000)##Mass flow rate of fuel in kg/s\n", + "E=ws/wf##Evaporation factor \n", + "vf=0.0014523##The specific volume of saturated liquid in m**3/kg\n", + "vg=0.0165884##The specific volume of saturated vapour in m**3/kg\n", + "vt=vf+(xt*vg)##Specific volume at the top in m**3/kg\n", + "pt=1/vt##Density of the steam at the top in kg/m**3\n", + "pf=1/vf##The density of the steam in kg/m**3\n", + "pm=(pf+pt)/2##The average mixture density in kg/m**3\n", + "H1=(g*H*(pf-pm))/10**5##Pressure head available for natural circulation in bar\n", + "CR=1/xt##Circulation ratio\n", + "di=(Do-(2*T))/1000##The inner diameter of the riser in m\n", + "A=(pi*di**2)/4##Area for the inner diameter in m**2\n", + "w=(A*pt*v*xt)##The rate of steam formation in the riser in kg/s\n", + "Nr=ceil(ws)/w##The number of risers\n", + "hfg=1319.8##Enthalpy of the evaporation in kJ/kg\n", + "Ha=(w*hfg)/((Do/1000)*H)##Heat absorption rate per unit projected area of the riser in kW/m**2\n", + "\n", + "#Output \n", + "print \" (a)The steam generation rate = %3.3f kg/s \\n (b) The fuel burning rate = %3.3f kg/s \\n (c) The evaporation factor = %3.2f \\n (d) The pressure head available for natural circulation = %3.4f bar \\n (e) The circulation ratio = %3.1f \\n (f)The number of risers required = %3.0f \\n (g) The heat absorbtion rate per unit projected area of the riser = %3.2f kW/m**2\"%(ws,wf,E,H1,CR,Nr,Ha)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.5 Pg: 415" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The blowdown required = 0.2796 kg/s \n", + " (b) Heat loss in blowdown as a percentage of total heat released in the furnace = 0.19 percentage \n", + " (c) The deposition of scale in superheater tube = 0.553 kg/day \n" + ] + } + ], + "source": [ + "#Input data\n", + "ws=64##The steam flow rate in kg/s\n", + "p=60##The pressure at which steam leaves the boiler in bar\n", + "m=0.02##Moisture contant in the steam \n", + "wf=62##The feedwater flow rate in kg/s\n", + "Pf=3##concentration of feedwater in ppm\n", + "wm=2##The flow rate of makeup water \n", + "Pm=50##concentration of makeup water in ppm\n", + "Ps=5##Leaving the drum water in ppm\n", + "Pw=1000##The concentration in the drum water in ppm\n", + "mf=7##The fuel burning rate in kg/m\n", + "CV=23##The heating value in MJ/kg\n", + "ta=30##The room temperature in degree centigrade\n", + "hf=1213.35##Enthalpy of saturated liquid at 60 bar in kJ/kg\n", + "ha=125.79##Enthalpy at ambient temperature in kJ/kg\n", + "\n", + "#Calculations\n", + "BD=((wf*Pf)+(wm*Pm)-(m*ws*Ps))/1000##The rate of blowdown in kg/s\n", + "E=((BD*(hf-ha))/(mf*CV*1000))*100##The energy loss in blowdown in percentage\n", + "S=m*ws*Ps*10**-6*3600*24##Scale deposition in superheater tubes\n", + "\n", + "#Output\n", + "print \" (a)The blowdown required = %3.4f kg/s \\n (b) Heat loss in blowdown as a percentage of total heat released in the furnace = %3.2f percentage \\n (c) The deposition of scale in superheater tube = %3.3f kg/day \"%(BD,E,S)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.6 Pg: 416" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The number of coils needed in the economiser = 285 \n", + " (b) The length of one coil = 314.7 m \n", + " (c) The verticle height of the duct occupied by the economiser coils = 5.26 m \n" + ] + } + ], + "source": [ + "from math import log\n", + "#Input data\n", + "ws=600##Mass flow rate of feedwater in kg/s\n", + "p=140##The inlet pressure of the feedwater in bar\n", + "t=170##The inlet temperature of the feedwater in degree centigrade\n", + "wg=1250##The mass flow rate of flue gases in kg/s\n", + "tg2=450##The temperature at which flue gases leave the economisers coils in degree centigrade\n", + "Vf=12##The velocity of the flue gas in m/s\n", + "Vw=1.2##The velocity of the water leaving the coil in m/s\n", + "Do=0.07##The outer diameter of the tube in m\n", + "Di=0.06##The inner diameter of the tube in m\n", + "U=70##The overall heat transfer coefficient in W/m**2K\n", + "Cp=1.12##The specific heat capacity of the flue gases in kJ/kgK\n", + "V=0.08##The vertical pitch of the coil in m\n", + "B=4.8##The width of the duct in m\n", + "C=0.005##The clearence on the both sides of the duct in m\n", + "pi=3.142##Mathematical constant\n", + "\n", + "#Calculations\n", + "hf=1571.1##The enthalpy of the saturated liquid at 140 bar in kJ/Kg\n", + "ts=336.75##The saturated temperature at 140 bar in degree centigrade\n", + "vf=0.001611##The specific volume of the saturated liquid at 140 bar in m**3/kg\n", + "hf1=719.21##The enthalpy of the saturated liquid at 170 degree C in kJ/kg\n", + "vf1=0.001114##The specific volume of the saturated liquid at 170 degree C in m**3/kg\n", + "tg1=((ws*(hf-hf1))/(wg*Cp))+tg2##The temperature at which flue gases enters the economisers coils in degree centigrade\n", + "t1m=(478.25-280)/(log(478.25/280))##The mean temperature for inlet and exit temperature in degree centigrade \n", + "Q=ws*(hf-hf1)##The rate of heat transfer in the economiser in kW\n", + "Ao=(Q/(U*t1m))*10**3##The outer area in m**2\n", + "n=((ws*(vf/Vw)*(4/pi)*(1/Di**2)))##The number of coils needed in the economiser\n", + "l=Ao/(n*pi*Do)##The length of one coil in m\n", + "nt=l/(B-(2*C))##The number of turns in on ecoil \n", + "VH=nt*V##The vertical height of the duct occupied by the economiser coils\n", + "\n", + "#Output\n", + "print \" (a) The number of coils needed in the economiser = %3.0f \\n (b) The length of one coil = %3.1f m \\n (c) The verticle height of the duct occupied by the economiser coils = %3.2f m \"%(n,l,VH)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.7 Pg: 418" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)The length of the tubes = 9.55 m\n", + " (b) The number of tubes = 69647 \n" + ] + } + ], + "source": [ + "from math import log\n", + "#Input data\n", + "\n", + "tg2=160##The temperature to which the flue gases are cooled in degree centigrade\n", + "ta1=35##The ambient temperature of the air in degree centigrade\n", + "wa=1167##The mass flow rate of air in kg/s\n", + "Vg=13##The inlet velocity of the flue gases in m/s\n", + "U=30##The overall heat transfer coefficient in W/m**2K\n", + "Cpg=1.10##The specific heat of the flue gas in kJ/kgK\n", + "Cpa=1.005##The specific heat of the air in kJ/kgK\n", + "R=0.287##Real gas constant in kJ/kgK\n", + "wg=1250##The mass flow rate of gas in kg/s\n", + "tg1=450##The temperature at the inlet of flue gas in degree centigrade\n", + "P=101.325##Atmospheric temperature in kPa\n", + "pi=3.1414##Mathematical constant\n", + "Di=0.06##The inner diameter of the tube in m\n", + "Do=0.065##The outer diameter of the tube in m\n", + "\n", + "#Calculations\n", + "vg1=(R*(273+tg1))/P##Specific volume of the gas in m**3/kg\n", + "ta2=((wg*Cpg*(tg1-tg2))/(wa*Cpa))+ta1##The temperature of the heated air in degree centigrade\n", + "t1m=(75-125)/log(75/125)##The mean temperature of the inlet and exit temperature in degree centigrade\n", + "Q=wg*Cpg*(tg1-tg2)##The rate of heat transfer in the economiser in kW\n", + "Ao=(Q/(U*t1m))*10**3##The outer area in m**2\n", + "n=((wg*(vg1/Vg)*(4/pi)*(1/Di**2)))##The number of coils needed in the economiser\n", + "l=Ao/(n*pi*Do)##The length of one coil in m\n", + "\n", + "#Output\n", + "print \"(a)The length of the tubes = %3.2f m\\n (b) The number of tubes = %3.0f \"%(l,n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 6.8 Pg: 419" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a)The length of the one coil = 8.38 m\n", + " (b) The number of coils = 231 \n" + ] + } + ], + "source": [ + "#Input data\n", + "di=0.05##The inner diameter of the superheater coil in m\n", + "T=0.005##The thickness of the coil in m\n", + "p=60##The pressure of the steam at the exit in bar\n", + "t=500##The temperature of the steam at the exit in degree centigrade\n", + "V2=10##The velocity of the steam at the exit in m/s\n", + "ws=80##The mass flow rate of steam in kg/s\n", + "H=140##The heat flux in the super heated coils in kW/m**2\n", + "pi=3.142##Mathematical constant\n", + "Do=0.06##The outer diameter of the tube in m\n", + "\n", + "#Calculations\n", + "h1=2784.3##The enthalpy of the saturated gas at 60 bar in kJ/kg\n", + "h2=3422.2##The enthalpy of the saturated gas at 500 degreeC in kJ/kg\n", + "v2=0.05665##The specific volume of gas at 500 degreeC in m**3/kg\n", + "Q=ws*(h2-h1)##Heat absorption rate in superheater coil in kW\n", + "Ao=Q/H##Surface area required in m**2\n", + "n=((ws*(v2/V2)*(4/pi)*(1/di**2)))##The number of coils needed in the economiser\n", + "l=Ao/(n*pi*Do)##The length of one coil in m\n", + "\n", + "#Output\n", + "print \" (a)The length of the one coil = %3.2f m\\n (b) The number of coils = %3.0f \"%(l,n)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch7.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch7.ipynb new file mode 100644 index 00000000..5a7f9b93 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch7.ipynb @@ -0,0 +1,1604 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7 : Steam generators" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.1 Pg: 510" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum area of the nozzles is 7.626 sq.cm\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=10#Initial pressure in bar\n", + "T1=300+273#Initial temperature in K\n", + "p2=2#Final pressure in bar\n", + "m=1#Mass flow rate of steam in kg/s\n", + "\n", + "#Calculations\n", + "px=(0.546*p1)#Critical pressure in bar\n", + "ho=3052.2#Enthalpy in kJ/kg\n", + "so=7.1229#Entropy in kJ/kg.K\n", + "sx=so#Entropy in kJ/kg.K\n", + "hx=2905.9#Enthalpy in kJ/kg\n", + "vx=0.4125#Specific volume in m**3/kg\n", + "Vx=(44.72*sqrt(ho-hx))#Critical velocity in m/s\n", + "Ax=(vx/Vx)*10**4#Minimum area of the nozzle in sq.cm\n", + "\n", + "#Output\n", + "print \"Minimum area of the nozzles is %3.3f sq.cm\"%(Ax)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.2 Pg: 511" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Minimum area of the nozzles is 7.98 cm**2 \n", + " the number of nozzles are 4\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=10#Initial pressure in bar\n", + "T1=300+273#Initial temperature in K\n", + "p2=1#Final pressure in bar\n", + "x=0.15#Friction loss of the isentropic enthalpy drop\n", + "ms=1#Steam flow rate in kg/s\n", + "d=25#Exit diameter of the nozzles in mm\n", + "\n", + "#Calculations\n", + "px=(0.546*p1)#Critical pressure in bar\n", + "h1=3052.2#Enthalpy in kJ/kg\n", + "s1=7.1276#Entropy in kJ/kg\n", + "s2s=s1#Entropy in kJ/kg\n", + "h2s=2916.2#Enthalpy in kJ/kg\n", + "Vx=(44.72*sqrt(h1-h2s))#Critical velocity in m/s\n", + "h3s=2605#Enthalpy in kJ/kg\n", + "V1=(44.72*sqrt((h1-h2s)+(0.85*(h2s-h3s))))#Velocity in m/s\n", + "s3s=s1#Entropy in kJ/kg\n", + "x3s=(s3s-1.3025)/6.0579#Dryness fraction\n", + "h3s=(417.46+(x3s*2258.01))#Enthalpy in kJ/kg\n", + "h2s3=((1-x)*(h2s-h3s))#Enthalpy in kJ/kg\n", + "h3=h2s-h2s3#Enthalpy in kJ/kg\n", + "x3=(h3-417.46)/2258.01#Dryness fraction\n", + "v3=(0.001043+(x3*1.694))#Specific volume in m**3/kg\n", + "v2s=0.416#Specific volume in m**3/kg\n", + "vx=v2s#Specific volume in m**3/kg\n", + "Ax=(ms/Vx)*vx*10**4#Minimum area in cm**2\n", + "A1=(ms*v3)/V1*10**4#Area in cm**2\n", + "n=(A1*4)/(3.14*(d/10)**2)#Number of nozzles\n", + "\n", + "#Output\n", + "print \" Minimum area of the nozzles is %3.2f cm**2 \\n the number of nozzles are %3.0f\"%(Ax,n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.3 Pg: 512" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Area of throat is 2430 mm**2 \n", + " Exit area is 4029 mm**2\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=7.8#Pressure in bar\n", + "t1=180+273#Temperature in K\n", + "p2=1.03#pressure in bar\n", + "m=3.6#flow rate of air in kg/s\n", + "g=1.4#Ratio of specific heats\n", + "R=287#Characteristic gas constant in J/kg.K\n", + "cp=1.005#Specific heat in kJ/kg.K\n", + "\n", + "#Calculations\n", + "pxpo=(2/(g+1))**(g/(g-1))#Ratio of pressure\n", + "px=pxpo*p1#Critical pressure in bar\n", + "txto=(2/(g+1))#Ratio of temperatures\n", + "tx=t1*txto#Critical temperature in K\n", + "vx=(R*tx)/(px*10**5)#Critical specific volume in m**3/kg\n", + "Vx=sqrt(g*R*tx)#Critical velocity in m/s\n", + "Ax=((m*vx)/Vx)*10**6#Critical area in mm**2\n", + "tot1=(p1/p2)**((g-1)/g)#Ratio of temperatures\n", + "t1i=t1/tot1#Temperature in K\n", + "v1=(R*t1i)/(p2*10**5)#Specific volume in m**3/kg\n", + "V1=44.72*sqrt(cp*(t1-t1i))#Velocity in m/s\n", + "A1=((m*v1)/V1)*10**6#Area in mm**2\n", + "\n", + "#Output\n", + "print \" Area of throat is %3.0f mm**2 \\n Exit area is %d mm**2\"%(Ax,A1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.4 Pg: 514" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Throat raea is 0.0286 m**2 \n", + " Exit arae is 0.0370 m**2\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=3.8#pressure in bar\n", + "T1=450+273#Tempereture in K\n", + "p2=1#pressure in bar\n", + "m=16#Flow rate in kg/s\n", + "Cd=0.98#coefficient of discharge\n", + "nv=0.93#nozzile effeciency\n", + "cp=1.11#Specific heat in kJ/kg.K\n", + "g=1.333#Ratio of specific heats\n", + "\n", + "#Calculations\n", + "pxpo=(2/(g+1))**(g/(g-1))#Pressure ratio\n", + "px=pxpo*p1#Critical pressure in bar\n", + "TxTo=2/(g+1)#Temperature ratio\n", + "Tx=T1*TxTo#Critical temperature in K\n", + "Vx=44.72*sqrt(cp*(T1-Tx))#critical velocity in m/s\n", + "R=(cp*(g-1)*1000)/g#Characteristic gas constant in J/kg.K\n", + "vx=(R*Tx)/(px*10**5)#Critical specific volume in m**3/kg\n", + "ws=(m/Cd)#Mass flow rate in kg/s\n", + "Ax=(ws*vx)/Vx#Critical area in m**2\n", + "T1sTo=(p2/p1)**((g-1)/g)#Temperature ratio\n", + "T1s=T1*T1sTo#Temperature in K\n", + "T1i=(T1-(nv*(T1-T1s)))#Temperature in K\n", + "v1=(R*T1i)/(p2*10**5)#Specific volume in m**3/kg\n", + "V1=44.72*sqrt(cp*(T1-T1i))#Velocity in m/s\n", + "A1=(ws*v1)/V1#Area in m**2\n", + "\n", + "#Output\n", + "print \" Throat raea is %3.4f m**2 \\n Exit arae is %3.4f m**2\"%(Ax,A1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.5 Pg: 515" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Area of throat is 112.6 mm**2 \n", + " Exit area is 184.5 mm**2 \n", + "\n", + " (b) Area of throat is 143.6 mm**2 \n", + " Exit area is 219.0 mm**2\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=20#pressure in bar\n", + "T1=300+273#Tempereture in K\n", + "p2=3#pressure in bar\n", + "m=0.3#Flow rate in kg/s\n", + "n=1.3#Adiabatic constant\n", + "Cd=0.98#Coefficient of discharge\n", + "Cv=0.92#Coefficient of velocity\n", + "\n", + "#Calculations\n", + "vo=0.1255#Specific volume in m**3/kg\n", + "px=(0.546*p1)#Critical pressure in bar\n", + "vx=(p1/px)**(1/n)*vo#Critical specific volume in m**3/kg\n", + "Vx=sqrt(n*px*10**5*vx)#Critical velocity in m/s\n", + "Ax=((m*vx)/Vx)*10**6#Critical area in m**2\n", + "v1vo=(p1/p2)**(1/n)#Ratio of specific volumes\n", + "v1=(vo*v1vo)#Specific volume in m**3/kg\n", + "V1=sqrt(2*((n/(n-1))*10**5*((p1*vo)-(p2*v1))))#Velocity in m/s\n", + "A1=((m*v1)/V1)*10**6#Area in mm**2\n", + "ho=3050#Enthalpy in kJ/kg\n", + "hx=2920#Enthalpy in kJ/kg\n", + "h1s=2650#Enthalpy in kJ/kg\n", + "ws=(m/Cd)#Flow rate in kg/s\n", + "Vsx=44.72*sqrt(ho-hx)#Velocity in m/s\n", + "V1s=44.72*sqrt(ho-h1s)#Velocity in m/s\n", + "Vo1=(V1s*Cv)#Velocity in m/s\n", + "hoh1=(V1/44.72)**2#Change in enthalpy in kJ/kg\n", + "h1=ho-hoh1#Enthalpy in kJ/kg\n", + "x1=(h1-561.47)/2163.8#Dryness fraction\n", + "vo1=(0.001073+(x1*0.6047))#Specific volume in m**3/kg\n", + "Ao1=((ws*vo1)/Vo1)*10**6#Exit nozzle area in mm**2\n", + "Vox=(Vsx*Cv)#Velocity in m/s\n", + "hohx=(Vox/44.72)**2#Change in enthalpy in kJ/kg\n", + "hox=(ho-hohx)#Enthalpy in kJ/kg\n", + "vox=0.22#Specific volume in m**3/kg\n", + "Aox=((ws*vox)/Vox)*10**6#Critical area in m**2\n", + "\n", + "#Output\n", + "print \"(a) Area of throat is %3.1f mm**2 \\n Exit area is %3.1f mm**2 \\n\\n (b) Area of throat is %3.1f mm**2 \\n Exit area is %3.1f mm**2\"%(Ax,A1,Aox,Ao1)\n", + "#In textbook, Ao1 is given wrong." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.6 Pg: 517" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Mass flow rate is 0.966 kg/s \n", + " Nozzle efficiency is 0.971\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=5#Pressure of steam in bar\n", + "V=100#Velocity in m/s\n", + "p2=1.5#Exit pressure in bar\n", + "At=1280#Throat area in mm**2\n", + "Ae=1600#Exit area in mm**2\n", + "rp=0.58#Critical pressure ratio\n", + "\n", + "#Calculations\n", + "ho=2749#Enthalpy in kJ/kg\n", + "so=6.822#Entropy in kJ/kg.K\n", + "px=(rp*p1)#Critical pressure in bar\n", + "sx=so#Entropy in kJ/kg.K\n", + "xx=(sx-1.660)/5.344#Dryness fraction\n", + "hx=(556+(xx*2168))#Enthalpy in kJ/kg\n", + "Vx=sqrt(((ho+((V**2*10**-3))/2)-hx)*(2/10**-3))#Velocity in m/s\n", + "vx=(xx*0.6253)#Specific volume in m**3/kg\n", + "w=(At*10**-6*Vx)/vx#Mass flow rate in kg/s\n", + "s1s=sx#Entropy in kJ/kg.K\n", + "x1s=(so-1.434)/5.789#Dryness fraction\n", + "h1s=(467+x1s*2226)#ENthalpy in kJ/kg\n", + "z=((Vx**2*10**-3)/2)-hx#z value\n", + "#By iteratio scheme\n", + "x1=0.932#Dryness fraction\n", + "v1=1.080#Specific volume in m**3/kg\n", + "h1=2542#Enthalpy in kJ/kg\n", + "V1=652.2#Velocity in m/s\n", + "nn=((hx-h1)/(hx-h1s))#Nozzle efficiency\n", + "\n", + "#Output\n", + "print \" Mass flow rate is %3.3f kg/s \\n Nozzle efficiency is %3.3f\"%(w,nn)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.7 Pg: 819" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Exit area when the flow is in equilibrium throughout is 398 mm**2 \n", + " (b) Exit area when the flow is supersaturated is 435.4 mm**2 \n", + " (i) The degree of supercooling is 10.38 degree C \n", + " (ii) The degree of supersaturation is 1.396\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=5#Pressure in bar\n", + "T1=200+273#Temperature in K\n", + "p2=2#Pressure in bar\n", + "m=0.3#Mass flow rate in kg/s\n", + "n=1.3#Adiabatic index\n", + "\n", + "#Calculations\n", + "vo=0.4249#Specific volume in m**3/kg\n", + "ho=2855.4#Enthalpy in kJ/kg\n", + "so=7.0592#Entropy in kJ/kg.K\n", + "x1=0.972#Dryness fraction \n", + "h1=(504.7+x1*2201.9)#Enthalpy in kJ/kg\n", + "v1=x1*0.8857#Specific volume in m**3/kg\n", + "V1=44.72*sqrt(ho-h1)#Velocity in m/s\n", + "A1=((m*v1)/V1)*10**6#Area in mm**2\n", + "rp=(p1/p2)**(1/n)#Specific volume ratio\n", + "vR=(vo*rp)#Specific volume in m**3/kg\n", + "VR=sqrt(2*((n/(n-1))*(p1*vo-p2*vR)*10**5))#Velocity in m/s\n", + "AR=((m*vR)/VR)*10**6#Area in mm**2\n", + "TR=T1/(p1/p2)**((n-1)/n)#Temperature in K\n", + "tR=(TR-273)#Temperature in degree C\n", + "ts=120.23#Saturation temperature at pressure p1 in degree C\n", + "ds=ts-tR#Degree of subcooling in degree C\n", + "ps=1.4327#Saturation pressure at tR in bar\n", + "dsu=(p2/ps)#Degree of supersaturation\n", + "\n", + "#Output\n", + "print \" (a) Exit area when the flow is in equilibrium throughout is %3.0f mm**2 \\n (b) Exit area when the flow is supersaturated is %3.1f mm**2 \\n (i) The degree of supercooling is %3.2f degree C \\n (ii) The degree of supersaturation is %3.3f\"%(A1,AR,ds,dsu)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.8 Pg: 520" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Exit area is 403.2 mm**2 which upon checking is 400 mm**2\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "#Input data\n", + "p1=5#Pressure in bar\n", + "T1=200#Temperature in degree C\n", + "p2=2#Pressure in bar\n", + "m=0.3#Mass flow rate in kg/s\n", + "n=1.3#Adiabatic index\n", + "nn=0.92#Nozzle efficiency\n", + "cp=1.925#mean specific heat in kJ/kg.K\n", + "x=[0,2.308,1943]#pv*10**3 = 2.308(h-1943)\n", + "\n", + "#Calculations\n", + "vo=0.4249#Specific volume in m**3/kg\n", + "ho=2855.4#Enthalpy in kJ/kg\n", + "so=7.0592#Entropy in kJ/kg.K\n", + "x1=0.972#Dryness fraction \n", + "h1=(504.7+x1*2201.9)#Enthalpy in kJ/kg\n", + "v1=x1*0.8857#Specific volume in m**3/kg\n", + "V1=44.72*sqrt(ho-h1)#Velocity in m/s\n", + "h=ho-h1#Change in enthalpy in kJ/kg\n", + "hoq=nn*h#Change in enthalpy in kJ/kg\n", + "VQ=44.72*sqrt(hoq)#Velocity in m/s\n", + "toq=(hoq/cp)#Temperature difference in degree C\n", + "tQ=(T1-toq)#Temperature in degree C\n", + "TQ=tQ+273#Temperature in K\n", + "vQ=((p1*100*vo)/(T1+273))*(TQ/T1)#Specific volume in m**3/kg\n", + "A1=((m*vQ)/VQ)*10**6#Area in mm**2\n", + "vQ=(x[(1)]*(ho-hoq-x[(2)]))/(10**3*p2)#Specific volume in m**3/kg\n", + "A11=((m*vQ)/VQ)*10**6#Area in mm**2\n", + "\n", + "#Output\n", + "print \"Exit area is %3.1f mm**2 which upon checking is %3.0f mm**2\"%(A1,A11)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.9 Pg: 422" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Blade Angle is 32.36 degrees \n", + "\n", + " Neglecting the friction effects \n", + " Tangential force is 809.49 N \n", + " Axial thrust is 0 N \n", + " Diagram efficiency is 86.3 percent \n", + "\n", + " Considering the friction effects \n", + " Axial thrust is 51.3 N \n", + " Diagram Power is 291.42 kW \n", + " Diagram efficiency is 77.71 percent\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "V1=1000#Speed in m/s\n", + "Vb=400#Peripheral velocity in m/s\n", + "a=20#Nozzle angle in degree\n", + "m=0.75#Mass flow in kg/s\n", + "f=80#Percentage reduction of relative velocity\n", + "\n", + "#Calculations\n", + "b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Blade angle in degree\n", + "V=342#Velocity from E7.9 in m/s\n", + "Vr1=V/sin(b1*pi/180)#Velocity in m/s\n", + "dVw=(2*Vr1*cos(b1*pi/180))#Velocity in m/s\n", + "Pt=(m*dVw)#Tangential thrust in N\n", + "WD=(Pt*Vb)/1000#Diagram power in kW\n", + "nD=(WD/(0.5*m*V1**2*10**-3))*100#Diagram efficiency in percent\n", + "Pa=0#Axial thrust in N\n", + "Vr2=(f/100)*Vr1#Velocity in m/s\n", + "Pa2=m*sin(b1*pi/180)*(Vr1-Vr2)#Axial thrust in N\n", + "WD2=(m*(Vr1+Vr2)*cos(b1*pi/180)*Vb)/1000#Diagram power in kW\n", + "nD2=(WD2/(0.5*m*V1**2*10**-3))*100#Diagram efficiency in percent\n", + "\n", + "#Output\n", + "print \" Blade Angle is %3.2f degrees \\n\\n Neglecting the friction effects \\n Tangential force is %3.2f N \\n Axial thrust is %d N \\n Diagram efficiency is %3.1f percent \\n\\n Considering the friction effects \\n Axial thrust is %3.1f N \\n Diagram Power is %3.2f kW \\n Diagram efficiency is %3.2f percent\"%(b1,Pt,Pa,nD,Pa2,WD2,nD2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:7.10 Pg: 523" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Work done in the stage per kg of steam is 42.56 kJ/kg \n", + " Stage efficiency is 66.4 percent \n", + "\n", + " (b) Number of stages are 15\n" + ] + } + ], + "source": [ + "from math import atan, sin, cos, degrees, ceil,pi\n", + "#Input data\n", + "a=20#Nozzle angle in degrees\n", + "b2=30#Blade exit angle in degrees\n", + "Vb=130#Mean blade speed in m/s\n", + "V1=330#Velocity of steam in m/s\n", + "f=0.8#Friction factor\n", + "nn=0.85#Nozzle efficiency\n", + "p1=20#Pressure in bar\n", + "T1=250+273#Temperature in K\n", + "p2=0.07#Pressure in bar\n", + "rf=1.06#Reheat factor\n", + "\n", + "#Calculations\n", + "b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Blade angle in degrees\n", + "Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s\n", + "Vr2=(f*Vr1)#Velocity in m/s\n", + "dVw=(Vr1*cos(b1*pi/180))+(Vr2*cos(b2*pi/180))#Vecoity in m/s\n", + "WD=(dVw*Vb)/1000#Workdone in kJ/kg\n", + "nb1=((2*dVw*Vb)/V1**2)*100#Efficiency in percent\n", + "nst=(nn*nb1)#Efficiency in percent\n", + "nin=(nst*rf)*100#Efficiency in percent\n", + "h1=2902.3#Enthalpy in kJ/kg\n", + "s1=6.5466#Entropy in kJ/kg.K\n", + "x2s=(s1-0.5582)/7.7198#Dryness fraction\n", + "h2s=(163.16+x2s*2409.54)#Enthalpy in kJ/kg\n", + "h12=(0.7041*(h1-h2s))#Change in enthalpy in kJ/kg\n", + "n=ceil(h12/WD)#Number of stages\n", + "\n", + "#Output\n", + "print \" (a) Work done in the stage per kg of steam is %3.2f kJ/kg \\n Stage efficiency is %3.1f percent \\n\\n (b) Number of stages are %3.0f\"%(WD,nst,n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.10 Pg: 525" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power deveoped in the blading is 355.89 kW\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import atan, sin, cos, degrees, ceil,pi\n", + "#Input data\n", + "d=800#Diameter in mm\n", + "N=3000#Speed in rpm\n", + "V1=300#Velocity in m/s\n", + "a=20#Nozzle angle in degrees\n", + "f=0.86#Frictional factor\n", + "T=140#Axial thrust in N\n", + "\n", + "#Calculations\n", + "Vb=((3.14*(d/1000)*N)/60)#Velocity in m/s\n", + "b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Blade angle in degrees\n", + "b2=b1#Blade angle in degrees\n", + "Vr1=(V1*sin(a*pi/180))/sin(b1*pi/180)#Velocity in m/s\n", + "Vr2=f*Vr1#Velocity in m/s\n", + "w=(T/((Vr1*sin(b1*pi/180))-(Vr2*sin(b2*pi/180))))#Mass flow rate in kg/s\n", + "dVw=(Vr2*cos(b2*pi/180))+(Vr1*cos(b1*pi/180))#Velocity in m/s\n", + "P=(w*dVw*Vb*10**-3)#Power developed in kW\n", + "\n", + "#Output\n", + "print \"Power deveoped in the blading is %3.2f kW\"%(P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.12 Pg: 525" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Axial thrust is 11.56 N \n", + " Tangential thrust is 146.85 N \n", + "\n", + " (b) Diagram Power is 29.437 kW \n", + "\n", + " (c) Diagram Efficiency is 86.3 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, cos, pi, atan,sin, degrees\n", + "#Input data\n", + "p1=15#Pressure in bar\n", + "T1=300+273#Temperature in K\n", + "p2=10#Pressure in bar\n", + "nn=95#Nozzle efficiency in percent\n", + "a=20#Nozzle angle in degrees\n", + "x=5#The blade exit angle is 5 degrees less than the inlet angle\n", + "f=0.9#Friction factor\n", + "m=1350#Steam flow rate in kg/h\n", + "\n", + "#Calculations\n", + "h1=3038.9#Enthalpy in kJ/kg\n", + "s1=6.9224#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "t2s=250#Temperature in degree C\n", + "h2s=2943.1#Enthalpy in kJ/kg\n", + "V1=44.72*sqrt((nn/100)*(h1-h2s))#Velocity in m/s\n", + "Vb=V1*(cos(a*pi/180)/2)#Velocity in m/s\n", + "b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Blade angle in degrees\n", + "b2=b1-x#Blade angle in degrees\n", + "Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s\n", + "Vr2=(f*Vr1)#Velocity in m/s\n", + "dVw=(Vr1*cos(b1*pi/180))+(Vr2*cos(b2*pi/180))#Velocity in m/s\n", + "dVa=(Vr1*sin(b1*pi/180))-(Vr2*sin(b2*pi/180))#Velocity in m/s\n", + "Pa=(m/3600)*dVa#Axial thrust in N\n", + "Pt=(m/3600)*dVw#Tangential thrust in N\n", + "WD=(Pt*Vb*10**-3)#Diagram Power in kW\n", + "dn=((WD*1000)/((1/2)*(m/3600)*V1**2))*100#Diagram efficiency in percent\n", + "\n", + "#Output\n", + "print \" (a) Axial thrust is %3.2f N \\n Tangential thrust is %3.2f N \\n\\n (b) Diagram Power is %3.3f kW \\n\\n (c) Diagram Efficiency is %3.1f percent\"%(Pa,Pt,WD,dn)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.13 Pg: 527" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the tangential thrust is 5.355 kW \n", + " (b) Axial thrust is 0.29 kN \n", + " (c) Power developed is 642.54 kW \n", + " (d) Diagram efficiency is 71.39 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, cos, pi, atan,sin, degrees\n", + "\n", + "#Input data\n", + "V1=600#Velocity in m/s\n", + "a=16#Nozzle angle in degrees\n", + "Vb=120#Mean blade angle in degrees\n", + "b2=18#Exit angle in degrees\n", + "aa1=22#Exit angle in degrees\n", + "b4=36#Exit angle in degrees\n", + "m=5#Steam flow rate in kg/s\n", + "f=0.85#Friction coefficient\n", + "\n", + "#Calculations\n", + "b1=degrees(atan((V1*sin(a*pi/180))/((V1*cos(a*pi/180))-Vb)))#Exit angle in degrees\n", + "Vr1=((V1*sin(a*pi/180))/sin(b1*pi/180))#Velocity in m/s\n", + "Vr2=(f*Vr1)#Velocity in m/s\n", + "a1=degrees(atan((Vr2*sin(pi/180*b2))/((Vr2*cos(pi/180*b2))-Vb)))#Angle in degrees\n", + "V2=((Vr2*sin(pi/180*b2))/sin(pi/180*a1))#Velocity in m/s\n", + "V3=(f*V2)#Velocity in m/s\n", + "dVw1=(Vr1*cos(pi/180*b1))+(Vr2*cos(pi/180*b2))#Velocity in m/s\n", + "dVa1=(V1*sin(pi/180*a))-(V2*sin(pi/180*a1))#Velocity in m/s\n", + "b3=degrees(atan((V3*sin(pi/180*aa1))/((V3*cos(pi/180*aa1))-Vb)))#Angle in degrees\n", + "Vr3=((V3*sin(pi/180*aa1))/sin(pi/180*b3))#Velocity in m/s\n", + "Vr4=(f*Vr3)#velocity in m/s\n", + "dVw2=(Vr3*cos(pi/180*b3))+(Vr4*cos(pi/180*b4))#Velocity in m/s\n", + "dVa2=(V3*sin(pi/180*aa1))-(Vr4*sin(pi/180*b4))#Velocity in m/s\n", + "udVw=(dVw1+dVw2)#Total velocity in m/s\n", + "udVa=(dVa1+dVa2)#Total velocity in m/s\n", + "Pt=(m*udVw*10**-3)#tangential thrust in kN\n", + "Pa=(m*udVa*10**-3)#Axial thrust in kN\n", + "WD=(Pt*Vb)#Power developed in kW\n", + "nd=((2*udVw*Vb)/V1**2)*100#Diagram efficiency in percent\n", + "\n", + "#Output\n", + "print \" (a) the tangential thrust is %3.3f kW \\n (b) Axial thrust is %3.2f kN \\n (c) Power developed is %3.2f kW \\n (d) Diagram efficiency is %3.2f percent\"%(Pt,Pa,WD,nd)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.14 Pg: 529" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Absolute velocity of steam leaving the stage is 75 m/s \n", + " (b) the diagram work is 130 kJ/kg \n", + " (c) the diagram efficiency is 78.64 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "a=17#Nozzle angle in degrees\n", + "Vb=125#Blade velocity in m/s\n", + "b2=22#Blade angle n degrees\n", + "a1=26#Blade angle n degrees\n", + "b4=30#Blade angle n degrees\n", + "f=0.9#Friction factor\n", + "a2=90#Axial angle in degrees\n", + "\n", + "#Calculations\n", + "dVw=1040#Velocity in m/s from Velocity triangles Fig. E.7.14\n", + "V1=575#Velocity in m/s from Velocity triangles Fig. E.7.14\n", + "V4=75#Velocity of steam exiting stage in m/s from Velocity triangles Fig. E.7.14\n", + "WD=(dVw*Vb)/1000#Diagram work in kJ/kg\n", + "nd=((WD*1000)/((1/2)*V1**2))*100#Diagram efficiency in percent\n", + "\n", + "#Output\n", + "print \" (a) Absolute velocity of steam leaving the stage is %3.0f m/s \\n (b) the diagram work is %3.0f kJ/kg \\n (c) the diagram efficiency is %3.2f percent\"%(V4,WD,nd)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.15 Pg: 530" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Efficiency of first stage is 66.67 percent \n", + " Percentage of the total power developed by the turbine is 22.22 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "p1=35#Pressure in bar\n", + "T1=350+273#Temperature in K\n", + "p2=0.07#Pressure in bar\n", + "x=1/4#Fraction of drop in isentropic enthalpy\n", + "a=20#Nozzle angle in degrees\n", + "nn=88#Nozzle efficiency in percent\n", + "y=0.2#Velocity ratio\n", + "b2=30#Exit blade angle in degrees\n", + "b4=30#Exit blade angle in degrees\n", + "f=0.9#Friction coefficienct\n", + "In=75#Internal efficiency of the turbine in percent\n", + "\n", + "#Calculations\n", + "h1=3106.4#Enthalpy in kJ/kg\n", + "s1=6.6643#Entropy in kJ/kg.K\n", + "x2s=(s1-0.5582)/7.7198#dryness fraction\n", + "h2s=(163.16+x2s*2409.54)#Enthalpy in kJ/kg\n", + "dh=(h1-h2s)#Change in enthalpy in kJ/kg\n", + "h13s=x*dh#Change in enthalpy in kJ/kg\n", + "h13=(nn/100)*h13s#Change in enthalpy in kJ/kg\n", + "V1=(44.72*sqrt(h13))#Velocity in m/s\n", + "Vb=(y*V1)#Velocity in m/s\n", + "b1=degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Angle in degrees\n", + "Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s\n", + "Vr2=(f*Vr1)#Velocity in m/s\n", + "dVw1=(Vr1*cos(pi/180*b1))+(Vr2*cos(pi/180*b2))#Velocity in m/s\n", + "V2=sqrt((Vr2*sin(pi/180*b2))**2+((Vr2*cos(pi/180*b2))-Vb)**2)#Velocity in m/s\n", + "V3=f*V2#Velocity in m/s\n", + "b3=degrees(atan((V3*sin(pi/180*b2))/((V3*cos(pi/180*b2))-Vb)))#Angle in degrees\n", + "Vr3=((V3*sin(pi/180*b2))/sin(pi/180*b3))#Velocity in m/s\n", + "Vr4=f*Vr3#Velocity in m/s\n", + "dVw2=(Vr3*cos(pi/180*b3))+(Vr4*cos(pi/180*b4))#Velocity in m/s\n", + "dVw=(dVw1+dVw2)#Velocity in m/s\n", + "nb1=((2*dVw*Vb)/V1**2)*100#Efficiency in percent\n", + "ns=(nn*nb1)/100#Efficiency in percent\n", + "ht=(In/100)*dh#Total change in enthalpy in kJ/kg\n", + "hc=(ns/100)*h13s#Total change in enthalpy in kJ/kg\n", + "pp=(hc/ht)*100#Percentage of enthalpy\n", + "\n", + "#Output\n", + "print \" Efficiency of first stage is %3.2f percent \\n Percentage of the total power developed by the turbine is %3.2f percent\"%(ns,pp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.16 Pg: 532 " + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the required height of blading is 127.2 mm \n", + " (b) the power developed by the ring is 2.368 kW\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "R=50#Percentage of reaction\n", + "b1=35#Angle in degrees\n", + "q=b1#Angle in degrees\n", + "b2=20#Angle in degrees\n", + "a=b2#Angle in degrees\n", + "N=1500#Speed in rpm\n", + "d=0.67#Mean diameter in m\n", + "p=1.5#Pressure in bar\n", + "x=0.96#Dryness fraction\n", + "w=3.6#Flow rate in kg/s\n", + "\n", + "\n", + "#Calculations\n", + "Vb=(3.14*d*N)/60#Velocity in m/s\n", + "V1=(Vb*(sin(pi/180*180-b1)/sin(pi/180*b1-b2)))#Veocity in m/s\n", + "Vr1=(Vb*(sin(pi/180*b2)/sin(pi/180*b1-b2)))#Velocity in m/s\n", + "dVw=(V1*cos(pi/180*a))+(Vr1*cos(pi/180*q))#Velocity in m/s\n", + "v1=(0.001052+x*1.15937)#Specific volume in m**3/kg\n", + "hb=((w*v1)/(3.14*d*V1*sin(pi/180*a)))*1000#Required height in mm\n", + "P=(w*dVw*Vb)/1000#Power developed in kW\n", + "\n", + "#Output\n", + "print \" (a) the required height of blading is %3.1f mm \\n (b) the power developed by the ring is %3.3f kW\"%(hb,P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.17 Pg: 533" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) Blade height is 138 mm \n", + " (b) the diagram power is 15.42 kW\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "N=400#Speed in rpm\n", + "P=5#Power in MW\n", + "m=6#Flow rate in kg/kWh\n", + "b2=20#Blade angle in degrees\n", + "a=b2#Angle in degrees\n", + "x=1.35#Velocity ratio\n", + "p=1.2#Pressure in bar\n", + "x1=0.95#Steam quality \n", + "Dh=12#Ratio of Dm and hb\n", + "\n", + "#Calculations\n", + "w=(m*P*1000)/3600#Mass flow rate in kg/s\n", + "v1=(0.0010468+x1*1.454)#Specific volume in m**3/kg\n", + "hb=((w*v1)/(Dh*3.14*x*((Dh*N)/60)*3.14*sin(pi/180*a)))**(1/3)*1000#Blade height in mm\n", + "Vb=((3.14*Dh*(hb/1000)*N)/60)#velocity in m/s\n", + "V1=(x*Vb)#Velocity in m/s\n", + "dVw=((2*V1*cos(pi/180*a))-Vb)#velocity in m/s\n", + "WD=(w*dVw*Vb*10**-3)#Diagram power in kW\n", + "\n", + "#Output\n", + "print \" (a) Blade height is %3.0f mm \\n (b) the diagram power is %3.2f kW\"%(hb,WD)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.18 Pg: 534" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Mass flow rate of steam is 4.698 kg/s \n", + " Useful enthalpy drop is 117.80 kJ/kg \n", + " The diagram power is 553.4 kW\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "N=3000#Speed in rpm\n", + "Vb=100#Mean blade speed in m/s\n", + "x=0.56#Velocity ratio\n", + "a=20#Blade angle in degrees\n", + "b2=a#Blade angle in degrees\n", + "v=0.65#Specific volume in m**3/kg\n", + "h=25#Mean height in mm\n", + "n=5#Number of pairs of blades\n", + "\n", + "#Calculations\n", + "V1=(Vb/x)#Velocity in m/s\n", + "Vr2=V1#Velocity in m/s\n", + "Dm=(Vb*60)/(3.14*N)#Diameter in m\n", + "w=((3.14*Dm*h*V1*sin(pi/180*a))/v)/1000#Mass flow rate in kg/s\n", + "ws=(w*3600)#Mass flow rate in kg/hr\n", + "b1=degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Blade angle in degrees\n", + "Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s\n", + "dhmb=(1/2)*(Vr2**2-Vr1**2)/1000#Change in enthalpy in kJ/kg\n", + "dsta=(2*dhmb)#Change in enthalpy of stage in kJ/kg\n", + "dsta5=(n*dsta)#Total Change in enthalpy of stage in kJ/kg\n", + "Dp=(w*dsta5)#Diagram power in kW\n", + "\n", + "#Output\n", + "print \" Mass flow rate of steam is %3.3f kg/s \\n Useful enthalpy drop is %3.2f kJ/kg \\n The diagram power is %3.1f kW\"%(w,dsta5,Dp)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.19 Pg: 535" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the number of impulse stages are 6 \n", + " (b) the nozzle height is 2.4 mm\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "P=8#Power in MW\n", + "N=5000#Speed in rpm\n", + "p=40#pressure in bar\n", + "T=500#Temperature in degree C\n", + "p2=0.1#Pressure in bar\n", + "In=0.85#Internal efficiency of turbine\n", + "nm=0.96#Mechanical efficiency\n", + "nn=0.92#Nozzle efficiency\n", + "a=15#Nozzle angle in degrees\n", + "Vb=300#Blade velocity in m/s\n", + "\n", + "#Calculations\n", + "V1=(2*Vb)/cos(pi/180*a)#Velocity in m/s\n", + "dh=((V1/44.72)**2/nn)#Change in enthalpy in kJ/kg\n", + "h1=3445.3#Enthalpy in kJ/kg\n", + "s1=7.0901#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "x2=(s2-0.6493)/7.5009#Dryness fraction\n", + "h2s=(191.83+x2*2392.8)#Enthalpy in kJ/kg\n", + "h12s=(h1-h2s)#Change in enthalpy in kJ/kg\n", + "n=(h12s/dh)#Number of stages\n", + "w=((P*1000)/(In*nm))/h12s#Mass flow rate in kg/s\n", + "h13=(nn*dh)#Change in enthalpy in kJ/kg\n", + "h3=h1-h13#Enthalpy in kJ/kg\n", + "v3=0.17#Specific volume in m**3/kg\n", + "A=(w*v3)/V1#Area in m**2\n", + "hm=(A/(((Vb*60)/N)*sin(pi/180*a)))*1000#Height in mm\n", + "\n", + "#Output\n", + "print \" (a) the number of impulse stages are%3.0f \\n (b) the nozzle height is %3.1f mm\"%(n,hm)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.20 Pg: 537" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Height of the moving blades at exit is 58.3 mm \n", + " Power developed in the blade row is 166.70 kW\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "from __future__ import division\n", + "#Input data\n", + "p=1.5#Pressure in bar\n", + "x1=0.9#Dryness fraction\n", + "m=7#Steam flow rate in kg/s\n", + "N=3000#Turbine speed in rpm\n", + "x=0.7#Velocity ratio\n", + "y=0.75#Velocity ratio\n", + "a=20#Exit angle in degrees\n", + "b2=a#Angle in degrees\n", + "hx=1/10#Fraction of height\n", + "\n", + "#Calculations\n", + "v=0.001052+x1*1.15937#Specific volume in m**3/kg\n", + "Dm=((m*v*60)/(3.14*hx*y*3.14*N))**(1/3)#Diameter in m\n", + "hb=Dm*1000*hx#Height in mm\n", + "Vb=(3.14*Dm*N)/60#Velocity in m/s\n", + "dVw=((2*x*Vb*cos(pi/180*a)/sin(pi/180*a))-Vb)#Velocity in m/s\n", + "P=(m*dVw*Vb)/1000#Power developed in kW\n", + "\n", + "#Output\n", + "print \" Height of the moving blades at exit is %3.1f mm \\n Power developed in the blade row is %3.2f kW\"%(hb,P)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.21 Pg: 538" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The mean diameter of the wheel if the turbine were of \n", + " (a) single impulse stage is 4.74 m \n", + " (b) single 50 percent reaction stage is 6.7 m \n", + " (c) four pressure (or Rateau) stages is 2.37 m \n", + " (d) one two-row Curtis stage is 2.370 m \n", + " (e) four stage 50 percent reaction stages is 3.35 m\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "p=40#Pressure in bar\n", + "T=500#Temperature in degree C\n", + "p1=0.1#Pressure in bar\n", + "a=16#Nozzle angle in degrees\n", + "N=3000#Speed in rpm\n", + "\n", + "#Calculations\n", + "h1=3445.3#Enthalpy in kJ/kg\n", + "s1=7.0901#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "x2s=(s2-0.6493)/7.5009#Dryness fraction\n", + "h2s=(191.83+x2s*2392.8)#Enthalpy in kJ/kg\n", + "V1=44.72*sqrt(h1-h2s)#Velocity in m/s\n", + "Vb=V1*(cos(pi/180*a)/2)#Velocity in m/s\n", + "Dm=(Vb*60)/(3.14*N)#Diameter in m\n", + "V2=44.72*sqrt((h1-h2s)/2)#Velocity in m/s\n", + "Vb2=V2*cos(pi/180*a)#Velocity in m/s\n", + "Dm2=(Vb2*60)/(3.14*N)#Diameter in m\n", + "V3=44.72*sqrt((h1-h2s)/4)#Velocity in m/s\n", + "Vb3=V3*(cos(pi/180*a)/2)#Velocity in m/s\n", + "Dm3=(Vb3*60)/(3.14*N)#Diameter in m\n", + "V4=44.72*sqrt(h1-h2s)#Velocity in m/s\n", + "Vb4=V4*(cos(pi/180*a)/4)#Velocity in m/s\n", + "Dm4=(Vb4*60)/(3.14*N)#Diameter in m\n", + "V5=44.72*sqrt((h1-h2s)/(2*4))#Velocity in m/s\n", + "Vb5=V5*cos(pi/180*a)#Velocity in m/s\n", + "Dm5=(Vb5*60)/(3.14*N)#Diameter in m\n", + "\n", + "#Output\n", + "print \"The mean diameter of the wheel if the turbine were of \\n (a) single impulse stage is %3.2f m \\n (b) single 50 percent reaction stage is %3.1f m \\n (c) four pressure (or Rateau) stages is %3.2f m \\n (d) one two-row Curtis stage is %3.3f m \\n (e) four stage 50 percent reaction stages is %3.2f m\"%(Dm,Dm2,Dm3,Dm4,Dm5)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.22 Pg: 539" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of stages required in \n", + " (a) all simple impulse stages are 8 \n", + " (b) all 50 percent reaction stages are 13 \n", + " (c) a 2-row Cutris stage follwed by simple impulse stages are 3 \n", + " (d) a 2-row Cutris stage followed by 50 percent reaction stages are 6\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees,ceil\n", + "from __future__ import division\n", + "#Input data\n", + "p=150#Pressure in bar\n", + "T=600#Temperature in degree C\n", + "Vb=300#Velocity in m/s\n", + "nn=95#Nozzle efficiency in percent\n", + "a=15#Nozzle angle in degrees\n", + "a1=25#Angle in degrees\n", + "\n", + "#Calculations\n", + "h1=3582.3#Enthalpy in kJ/Kg\n", + "s1=6.6776#Entropy in kJ/kg.K\n", + "s2=s1#Entropy in kJ/kg.K\n", + "x2s=(s2-0.6493)/7.5009#Dryness fraction\n", + "h2s=(191.83+x2s*2392.8)#Enthalpy in kJ/kg\n", + "h12s=(h1-h2s)#Difference in enthalpy in kJ/kg\n", + "V1=(Vb*2)/cos(pi/180*a)#Velocity in m/s\n", + "dhs=(V1/44.72)**2/(nn/100)#Change in enthalpy in kJ/kg\n", + "n1=ceil(h12s/dhs)#Number of stages\n", + "V2=(Vb/cos(pi/180*a1))#Velocity in m/s\n", + "dhs2=(V2/44.72)**2/(nn/(2*100))#Change in enthalpy in kJ/kg\n", + "n2=h12s/dhs2#Number of stages\n", + "V3=(Vb*4)/cos(pi/180*a)#Velocity in m/s\n", + "dhs3=(V3/44.72)**2/(nn/100)#Change in enthalpy in kJ/kg\n", + "dhhs3=(h12s-dhs3)#Change in enthalpy in kJ/kg\n", + "n3=dhhs3/dhs#Number of stages\n", + "n4=dhhs3/dhs2#Number of stages\n", + "\n", + "#Output\n", + "print \"Number of stages required in \\n (a) all simple impulse stages are %3.0f \\n (b) all 50 percent reaction stages are %3.0f \\n (c) a 2-row Cutris stage follwed by simple impulse stages are %3.0f \\n (d) a 2-row Cutris stage followed by 50 percent reaction stages are %3.0f\"%(n1,n2,n3,n4)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.23 Pg: 541" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The interstage pressures are 8 bar, 2.6 bar, 0.6 bar \n", + " The reheat factor is 1.071 \n", + " The turbine internal efficiency is 80.3 percent\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "p1=20#Pressure in bar\n", + "T=400#Temperature in degree C\n", + "p2=0.1#Pressure in bar\n", + "n=4#Number of stages\n", + "ns=75#Stage efficiency in percent\n", + "\n", + "#Calculations\n", + "h16s=(3250-2282)#Change in enthalpy in kJ/kg\n", + "h12s=(h16s/n)#Change in enthalpy in kJ/kg\n", + "p=[0,8,2.6,0.6]#pressures in bar from Mollier chart\n", + "h12=(ns/100)*h12s#Change in enthalpy in kJ/kg\n", + "h23s=(3060-2800)#Change in enthalpy in kJ/kg\n", + "h23=(ns/100)*h23s#Change in enthalpy in kJ/kg\n", + "h34s=(2870-2605)#Change in enthalpy in kJ/kg\n", + "h34=(ns/100)*h34s#Change in enthalpy in kJ/kg\n", + "h45s=(2680-2410)#Change in enthalpy in kJ/kg\n", + "h45=(ns/100)*h45s#Change in enthalpy in kJ/kg\n", + "h5=2470#Enthalpy in kJ/kg\n", + "rf=(h12s+h23s+h34s+h45s)/h16s#Reheat factor\n", + "nth=((h12+h23+h34+h45)/h16s)*100#Internal efficiency in percent\n", + "nin=(ns*rf)#Internal efficiency in percent\n", + "\n", + "#Output\n", + "print \" The interstage pressures are %d bar, %3.1f bar, %3.1f bar \\n The reheat factor is %3.3f \\n The turbine internal efficiency is %3.1f percent\"%(p[(1)],p[(2)],p[(3)],rf,nin)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.25 Pg: 544" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) the flow rate of steam required is 17.99 kg/s \n", + " (b) the mean blade diameter is 1.302 m \n", + " (c) the speed of the rotor is 2065 rpm\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "n=20#Number of stages\n", + "P=12#Power in MW\n", + "p=15#Pressure in bar\n", + "T=350#Temperature in degree C\n", + "p1=0.14#Pressure in bar\n", + "ns=75#Stage efficiency in percent\n", + "rf=1.04#Reheat factor\n", + "p2=1#Pressure in bar\n", + "a=20#Angle in degrees\n", + "v=0.7#Velocity ratio\n", + "h=1/12#Blade height in terms of mean blade diameter\n", + "\n", + "#Calculations\n", + "nint=(ns/100)*rf#Internal efficiency \n", + "dhs=855#Enthalpy in kJ/kg\n", + "dha=ceil(nint*dhs)#Actual enthalpy change in kJ/kg\n", + "w=(P*1000)/dha#Mass flow rate in kg/s\n", + "Vb=(sqrt((dha/n)/((((2/v)*cos(pi/180*a))-1)*10**-3)))#Velocity in m/s#\n", + "vg=1.694#Specific volume in m**3/kg\n", + "Dm=sqrt((w*vg)/(3.14*h*(Vb/v)*sin(pi/180*a)))#Diameter in m\n", + "N=((Vb*60)/(3.14*Dm))#Speed in rpm\n", + "\n", + "#Output\n", + "print \" (a) the flow rate of steam required is %3.2f kg/s \\n (b) the mean blade diameter is %3.3f m \\n (c) the speed of the rotor is %3.0f rpm\"%(w,Dm,N)\n", + "#In textbook, Vb is given wrong as 141.4 m/s instead of 140.6 m/s. Hence the answers are different." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.26 Pg: 546" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " Blade heights at the exit of each row: \n", + " First row of moving blades is 29.4 mm \n", + " Fixed row of guide blades is 37.8 mm \n", + " Second row of moving blades is 40.6 mm\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import sqrt, atan,sin,cos,pi,degrees\n", + "#Input data\n", + "V1=600#Velocity in m/s\n", + "Vb=120#Mean blade velocity in m/s\n", + "a=16#Nozzle angle in degrees\n", + "b=[0,18,21,35]#Exit angles in degrees\n", + "m=5#Steam flow rate in kg/s\n", + "h=25#Nozzle height in mm\n", + "v=0.375#Specific volume in m**3/kg\n", + "p=25#Pitch in mm\n", + "t=0.5#Thickness in mm\n", + "kb=0.9#Constant\n", + "\n", + "#Calculations\n", + "l=((m*v)/(sin(pi/180*a)*V1*(h/1000)*kb))#Length of the nozzle arc in m #Length of the nozzle arc is calculated wrong as 0.454m instead of 0.5 m\n", + "b1= degrees(atan((V1*sin(pi/180*a))/((V1*cos(pi/180*a))-Vb)))#Angle in degrees\n", + "Vr1=((V1*sin(pi/180*a))/sin(pi/180*b1))#Velocity in m/s\n", + "Vr2=kb*Vr1#Velocity in m/s\n", + "V2=sqrt(Vr2**2+Vb**2-2*Vr2*Vb*cos(pi/180*b[(1)]))#Velocity in m/s\n", + "V3=291#Velocity in m/s\n", + "b3=degrees(atan((V3*sin(pi/180*b[(2)]))/((V3*cos(pi/180*b[(2)]))-Vb)))#Angle in degrees\n", + "Vr3=((V3*sin(pi/180*b[(2)]))/sin(pi/180*b3))#Velocity in m/s\n", + "Vr4=(Vr3*kb)#Velocity in m/s\n", + "hb1=((m*v*(h/1000))/(l*((p/1000)*sin(pi/180*b[(1)])-(t/1000))*Vr2))*1000#Height in mm\n", + "hn=((m*v*(h/1000))/(l*((p/1000)*sin(pi/180*b[(2)])-(t/1000))*V3))*1000#Height in mm\n", + "hb2=((m*v*(h/1000))/(l*((h/1000)*sin(pi/180*b[(3)])-(t/1000))*Vr4))*1000#Height in mm\n", + "\n", + "#Output\n", + "print \" Blade heights at the exit of each row: \\n First row of moving blades is %3.1f mm \\n Fixed row of guide blades is %3.1f mm \\n Second row of moving blades is %3.1f mm\"%(hb1,hn,hb2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.27 Pg: 548" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Number of parallel exhausts in : \n", + " (a)condition a are 4 \n", + " (b)condition b are 8\n" + ] + } + ], + "source": [ + "from math import sqrt, atan,sin,cos,pi,degrees, ceil\n", + "#Input data\n", + "P=200#Power in MW\n", + "p=180#Pressure in bar\n", + "T=550#Temperature in degree C\n", + "P1=600#Power in MW\n", + "p1=300#Pressure in bar\n", + "T1=580#Temperature in degree C\n", + "nt=90#Turbine efficiency in percent\n", + "\n", + "#Calculations\n", + "h1=3430#Enthalpy in kJ/kg\n", + "h2s=3040#Enthalpy in kJ/kg\n", + "h12s=(h1-h2s)#Enthalpy in kJ/kg\n", + "h12=(nt/100)*h12s#Enthalpy in kJ/kg\n", + "h2=3070#Enthalpy in kJ/kg\n", + "v2=0.06#Specific volume in m**3/kg\n", + "h4=3560#Enthalpy in kJ/kg\n", + "h3s=2000#Enthalpy in kJ/kg\n", + "h13s=(h1-h3s)#Enthalpy in kJ/kg\n", + "h13=(nt/100)*h13s#Enthalpy in kJ/kg\n", + "w=(P*10**3)/h13#Mass flow rate in kg/s\n", + "Vbm=350#Velocity in m/s\n", + "N=3000#Speed in rpm\n", + "a=25#Angle in degrees\n", + "Dm=(Vbm*60)/(3.14*N)#Diameter in m\n", + "hD=0.3#Assuming (hb/Dm)max \n", + "hb=(hD*Dm)#Height in m\n", + "Ab=(3.14*Dm*hb*0.9*sin(pi/180*a))#Flow area in m**2\n", + "V1=(Vbm/cos(pi/180*a))#Velocity in m/s\n", + "Vo=(Ab*V1)#Volume flow rate in m**3/s\n", + "v=(Vo/w)#Specific volume in m**3/kg\n", + "h5s=2456#Enthalpy in kJ/kg\n", + "p5=0.36#Pressure in bar\n", + "h45s=(h4-h5s)#Enthalpy in kJ/kg\n", + "h45=(nt/100)*h45s#Enthalpy in kJ/kg\n", + "h5=h4-h45#Enthalpy in kJ/kg\n", + "x5=0.952#Dryness fraction\n", + "h56s=(h5-2340)#Enthalpy in kJ/kg\n", + "h56=(nt/100)*h56s#Enthalpy in kJ/kg\n", + "h6=h5-h56#Enthalpy in kJ/kg\n", + "v6=18#Specific volume in m**3/kg\n", + "mm=(Vo/v6)#Maximum mass flow that one stage can accommodate in kg/s\n", + "np=(w/mm)#Number of parallel exhausts\n", + "rp=(p1/4)#Reheat pressure in bar\n", + "xh1=3410#Enthalpy in kJ/kg\n", + "xh2s=3015#Enthalpy in kJ/kg\n", + "xh12s=xh1-xh2s#Enthalpy in kJ/kg\n", + "xh12=(nt/100)*xh12s#Enthalpy in kJ/kg\n", + "xv2=0.035#Specific volume in m**3/kg\n", + "xh4=3060#Enthalpy in kJ/kg\n", + "xh3s=1960#Enthalpy in kJ/kg\n", + "xh13s=xh1-xh3s#Enthalpy in kJ/kg\n", + "xh3=(xh1-xh13s)#Enthalpy in kJ/kg\n", + "xw=(P1*10**3)/xh13s#Mass flow rate in kg/s\n", + "xvm=(Vo/xw)#Maximum specific volume in m**3/kg\n", + "Vf=(xw*xv2)#Volume flow rate in m**3/s\n", + "xh5s=2300#Enthalpy in kJ/kg\n", + "xh45s=xh4-xh5s#Enthalpy in kJ/kg\n", + "xh45=(nt/100)*xh45s#Enthalpy in kJ/kg\n", + "xh5=xh4-xh45s#Enthalpy in kJ/kg\n", + "xv5=1.25#Specific volume in m**3/kg\n", + "xx5=0.86#Dryness fraction\n", + "xh6s=2050#Enthalpy in kJ/kg\n", + "xh56s=xh5-xh6s#Enthalpy in kJ/kg\n", + "xh56=(nt/100)*xh56s#Enthalpy in kJ/kg\n", + "xh6=(xh5-xh56)#Enthalpy in kJ/kg\n", + "xv6=12#Specific volume in m**3/kg\n", + "xx6=0.792#Dryness fraction\n", + "xmm=(Vo/xv6)#Maximum mass flow in kJ/kg\n", + "xnp=ceil(xw/xmm)#Number of parallel exhausts\n", + "\n", + "#Output\n", + "print \"Number of parallel exhausts in : \\n (a)condition a are %d \\n (b)condition b are %d\"%(np,xnp) " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 7.28 Pg: 552" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a) Steam rate at: \n", + " One-fourth load is 5.75 kg/kWh \n", + " Half load is 5.25 kg/kWh \n", + " Three-fourth load is 5.08 kg/kWh \n", + " Full load is 5.0 kg/kWh \n", + "\n", + " (b) Rankine cycle efficiency is 40.7 percent \n", + " (c) Actual efficiency at full load is 21.7 percent \n", + " (d) The turbogenerator efficiency at full load is 53.2 percent\n" + ] + } + ], + "source": [ + "#Input data\n", + "P=100#Power in MW\n", + "T=550#temperature in degree C\n", + "p=0.1#Pressure in bar\n", + "m=500000#Mass flow rate in kg/h at rated load\n", + "mo=25000#Mass flow rate in kg/h at zero load\n", + "x=[0,1/4,1/2,3/4,1]#Fraction of load\n", + "\n", + "#Calculations\n", + "b=(m-mo)/(P*10**3)#Steam rate in kg/kWh\n", + "y1=(x[(1)]*(P*10**3))#For one-fourth load\n", + "s1=(mo/y1)+b#Steam rate in kg/kWh\n", + "y2=(x[(2)]*(P*10**3))#For one-fourth load\n", + "s2=(mo/y2)+b#Steam rate in kg/kWh\n", + "y3=(x[(3)]*(P*10**3))#For one-fourth load\n", + "s3=(mo/y3)+b#Steam rate in kg/kWh\n", + "y4=(x[(4)]*(P*10**3))#For one-fourth load\n", + "s4=(mo/y4)+b#Steam rate in kg/kWh\n", + "h1=3511#Enthalpy in kJ/kg\n", + "xs1=6.8142#Entropy in kJ/kg.K\n", + "xs2=xs1#Entropy in kJ/kg.K\n", + "x2s=(xs2-0.6493)/7.5009#Dryness fraction\n", + "h2s=191.83+x2s*2392.8#Enthalpy in kJ/kg\n", + "nR=((h1-h2s)/(h1-191.83))*100#Rankine efficiency in percent\n", + "nac=((P*10**3*3600)/(m*(h1-191.83)))*100#Actual efficiency in percent\n", + "nTG=((P*10**3*3600)/(m*(h1-h2s)))*100#Turbogenerator efficiency in percent\n", + "\n", + "#Output\n", + "print \"(a) Steam rate at: \\n One-fourth load is %3.2f kg/kWh \\n Half load is %3.2f kg/kWh \\n Three-fourth load is %3.2f kg/kWh \\n Full load is %3.1f kg/kWh \\n\\n (b) Rankine cycle efficiency is %3.1f percent \\n (c) Actual efficiency at full load is %3.1f percent \\n (d) The turbogenerator efficiency at full load is %3.1f percent\"%(s1,s2,s3,s4,nR,nac,nTG)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch8.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch8.ipynb new file mode 100644 index 00000000..37aa3970 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch8.ipynb @@ -0,0 +1,282 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 : Condenser Feedwater and Circulating water systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 8.1 Pg: 591 " + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The rate of flow of cooling water = 5817.7 kg/s \n", + " (b) The rate of air leakage into the condenser shell = 5.615 kg/s \n", + " (c) The length of tubes = 20.74 m \n", + " (d) The number of tubes = 7849 \n" + ] + } + ], + "source": [ + "from math import pi,log\n", + "#Input data\n", + "ws=250##The amount of steam received by the surface condenser in t/h\n", + "tsat=40##The saturated temperature in degree centigrade\n", + "m=12##The amount of moisture present in the steam in percentage\n", + "tc1=32##The inlet temperature of cooling water in degree centigrade\n", + "tc2=38##The outlet temperature of cooling water in degree centigrade\n", + "p=0.078##The pressure inside the condenser in bar\n", + "V=1.8##velocity of circulating water in m/s\n", + "do=0.0254##The outer diameter of the condenser tubes in m\n", + "T=0.00125##The thickness of the condenser tubes in m\n", + "pi=3.141##Mathematical constant of pi\n", + "U=2.6##The overall heat transfer coefficient in kW/m**2K\n", + "Cpc=4.187##The specific heat of water in kJ/kgK\n", + "R=0.287##Real gas constant in kJ/kgK\n", + "P=1000##The density of water in kg/m**3\n", + "\n", + "#Calculations\n", + "x2=0.88##The quality of the steam\n", + "hfg=2407##The enthalpy of evaporation at 40 degreeC in kJ/kg\n", + "h=x2*hfg##The change in enthalpy in kJ/kg\n", + "di=do-(2*T)##The inner diameter of the condenser tubes in m\n", + "wc=(((ws*1000)/3600)*h)/(Cpc*(tc2-tc1))##Mass flow rate of water in kg/s\n", + "psat=0.07375##The saturated pressure at 40 degree centigrade in bar\n", + "pair=(p-psat)*100##The pressure of air in kPa\n", + "vf=0.001008##Specific volume of saturated liquid in m**3/kg\n", + "vfg=19.544##Specific volume of vapour in m**3/kg\n", + "v2=vf+(x2*vfg)##Specific volume at 40 degree centigrade in m**3/kg\n", + "wair=(pair*((ws*1000)/3600)*v2)/(R*(tsat+273))##Mass flow rate of air in kg/s\n", + "t1m=(8-2)/log(8/2)##The mean temperature in degree centigrade\n", + "Ao=(((ws*1000)/3600)*h)/(U*t1m)##The area of the tubes in m**2\n", + "n=(wc*(4/pi)*(1/di**2)*(1/(P*V)))##The number of tubes\n", + "l=Ao/(pi*do*n)##Yhe length of tubes in m\n", + "\n", + "#Output\n", + "print \" (a) The rate of flow of cooling water = %3.1f kg/s \\n (b) The rate of air leakage into the condenser shell = %3.3f kg/s \\n (c) The length of tubes = %3.2f m \\n (d) The number of tubes = %3.0f \"%(wc,wair,l,n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 8.2 Pg: 592" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The rate of saving of condensate and the rate of saving in the heat supply in the boiler due to seperate air extraction pump = 1.105 kW \n", + " (b) The percentage reduction in air ejector load due to this seperate air extraction method = 80.8 percent \n", + " (c) The rate of cooling water flow = 533 kg/s \n" + ] + } + ], + "source": [ + "#Input data\n", + "ws=20##The amount of dry saturated steam received by a surface condencer in t/h\n", + "tsh=40##The temperature of dry saturated steam in degree centigrade \n", + "wa=(0.35/1000)##The mass flow rate of air per 1000 kg of steam in kg\n", + "tc=38##The temperature at which condensate leaves the temperature in degree centigrade\n", + "tm=10##The temperature at which makeup water is supplied in degree centigrade\n", + "tc1=32##The inlet temperature of cooling water in degree centigrade\n", + "tc2=38##The outlet temperature of cooling water in degree centigrade\n", + "tas=27##The temperature of air along with steam in degree centigrade\n", + "psat=0.07384##The saturated pressure at 40 degree C in bar\n", + "vg=19.52##The specific volume at 40 degree C in m**3/kg\n", + "R=0.287##Real gas constant in kJ/kgK\n", + "Cpc=4.187##The specific heat of water in kJ/kgK\n", + "Cp=1.005##The specific heat of air in kJ/kgK\n", + "\n", + "#Calculations\n", + "pair=((wa*R*(tsh+273)*1000)/vg)*10**-5##The pressure of air in bar\n", + "psat1=0.06624##The saturated pressure at 38 degree C in bar\n", + "vg1=21.63##The specific volume at 38 degree C in m**3/kg\n", + "pair1=psat-psat1##The pressure of air in bar\n", + "wa1=(ws*1000)*wa##Mass of air removed per hour in kg/h\n", + "V1=((wa1*R*(273+tc2)*1000))/(pair1*10**5)##Volume of air remove per hour\n", + "ws1=V1/vg1##The mass of steam accompanying air in kg/h\n", + "psat2=0.03564##The saturated pressure at 27 degree C in bar\n", + "vg2=38.81##The specific volume at 27 degree C in m**3/kg\n", + "pair2=psat-psat2##The pressure of air in bar\n", + "V2=(wa1*R*1000*(tas+273))/(pair2*10**5)##Volume of air removed per hr in m**3/hr\n", + "ws2=V2/vg2##The mass of steam accompanying air in kg/h\n", + "ws3=ws1-ws2##Saving mass of steam by using seperate extraction in kg/hr\n", + "Q3=(ws3*Cpc*(tc-tm))/3600##Saving in heat supply in the boiler in kW\n", + "V=((V1-V2)/V1)*100##Percentage reduction in air ejector load in %\n", + "hc=159.3##Enthalpy at 38 degree C in kJ/kg\n", + "hs1=2574.3##Enthalpy at 40 degree C in kJ/kg\n", + "hs2=2550.3##Enthalpy at 27 degree C in kJ/kg\n", + "Q2=(((ws*1000)*(hs1-hc))-((wa1*(Cp*(tsh-tas))))-(ws3*hs2))/3600##The amount of heat in kW\n", + "wc=Q2/(Cpc*(tc2-tc1))##The mass flow rate of water in kg/s\n", + "\n", + "#Output\n", + "print \" (a) The rate of saving of condensate and the rate of saving in the heat supply in the boiler due to seperate air extraction pump = %3.3f kW \\n (b) The percentage reduction in air ejector load due to this seperate air extraction method = %3.1f percent \\n (c) The rate of cooling water flow = %3.0f kg/s \"%(Q3,V,wc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 8.3 Pg: 593" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The temperature of water leaving the tower = 21.8 degree centigrade \n", + " (b) The fraction of water evaporated = 0.0125 kg/kg dry air \n", + " (c) The approach of the cooling tower = 6.6 degree centigrade \n", + " The Range of the cooling tower = 8.2 degree centigrade \n" + ] + } + ], + "source": [ + "#Input data\n", + "tw3=30##The inlet temperature of water in degree centigrade\n", + "wc=1.15##Mass flow rate of cooling water in kg per kg air\n", + "tdb1=20##The dry bulb temperature of air in degree centigrade\n", + "R1=60##Relative humidity of air while entering in percentage\n", + "tdb2=28##The dry bulb temperature while leaving in degree centigrade\n", + "R2=90##Relative humidity of air while leaving in percentage\n", + "tm=20##The temperature of makeup water in degree centigrade\n", + "Cpc=4.187##The specific heat of water in kJ/kgK\n", + "G=1##Mass flow rate of dry air in kg/s\n", + "\n", + "#Calculations\n", + "twb1=15.2## from psychrometric chart The wet bulb temperature while entering in degree centigrade\n", + "twb2=26.7## from psychrometric chart The wet bulb temperature while leaving in degree centigrade\n", + "h1=43##The enthalpy from chart for dry air in kJ/kg dry air\n", + "h2=83.5##The enthalpy from chart in kJ/kg dry air\n", + "W1=0.0088##Humidity in kg water vapour/kg dry air\n", + "W2=0.0213##Humidity in kg water vapour/kg dry air\n", + "hw3=125.8##Enthalpy of water entering the tower in kJ/kg\n", + "hw=84##Enthalpy of makeup water in kJ/kg\n", + "hwc=((G/wc)*((h2-h1)-(W2-W1)*hw))##The change in enthalpy of water in kJ/kg\n", + "tw4=tw3-(hwc/Cpc)##The exit temperature of water in degree centigrade\n", + "ta=tw4-twb1##The approach temperature in degree centigrade\n", + "tr=tw3-tw4##The range temperature in degree centigrade\n", + "x=G*(W2-W1)##Fraction of water evaporated in kg/kg dry air\n", + "\n", + "#Output\n", + "print \" (a) The temperature of water leaving the tower = %3.1f degree centigrade \\n (b) The fraction of water evaporated = %3.4f kg/kg dry air \\n (c) The approach of the cooling tower = %3.1f degree centigrade \\n The Range of the cooling tower = %3.1f degree centigrade \"%(tw4,x,ta,tr)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 8.4 Pg: 593" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The final temperature of the water = 16.60 degree centigrade \n", + " (b)The amount of makeup water required per second = 0.2605 kg/s \n" + ] + } + ], + "source": [ + "#Input data\n", + "tw3=45##The temperature of warm water in degree centigrade\n", + "wc1=6##The cooling water inflow in kg/s\n", + "V=10##volume flow of ID fan in m**3/s\n", + "Ws=4.90##Heat absorbed by air in kW\n", + "ti=20##The temperature of air entering the tower in degree centigrade\n", + "R=60##The relative humidity in percentage\n", + "to=26##The temperature of air leaving the tower in degree centigrade\n", + "p=1.013##The constant pressure throughout the tower in bar\n", + "r=0.287##Real gas constant in kJ/kgK\n", + "Cpc=4.187##The specific heat of water in kJ/kgK\n", + "Cp=1.005##The specific heat of air in kJ/kgK\n", + "\n", + "#Calculations\n", + "ps=0.0234##The pressure at 20 degreec in bar\n", + "ps1=(R/100)*ps##The pressure of water vapour in bar\n", + "pa1=p-ps1##The pressure of air in bar\n", + "G1=(pa1)/(r*10**-3*(ti+273))##The mass flow rate of dry air in kg/s\n", + "w1=(ps1*10**5*V)/(0.4619*10**3*(ti+273))##Mass flow rate of vopour in kg/s\n", + "W1=w1/G1##Moisture flow in kg vap/kg dry air\n", + "ps2=0.0336##The pressure at 26 degree C at exit in bar\n", + "pw2=0.0336##The pressure of water vapour at 26 degree C at exit in bar\n", + "W2=(0.622)*(pw2/(1-pw2))##oisture flow in kg vap/kg dry air\n", + "G2=G1##The mass flow rate of dry air in kg/s\n", + "w2=W2*G2##Moisture flow at exit in kg/s\n", + "wm=w2-w1##Makeup water required in kg/s\n", + "wc2=wc1-wm##The cooling water outflow in kg/s\n", + "hw3=Cpc*tw3##The enthalpy of warm water in kJ/kg\n", + "hg=2538.1##The enthalpy of gas at 20 degree C in kJ/kg\n", + "tsat=12##The saturation temperature in degree centigrade\n", + "pw1=0.01404##The pressure at 12 degree C in bar\n", + "hw1=hg+(1.88*(ti-tsat))##The enthalpy of warm water in kJ/kg\n", + "hw2=2548.4##The enthalpy of evaporation at 26 degree C in kJ/kg\n", + "hw4=((G1*((Cp*(to-ti))+W2*hw2-W1*hw1)-Ws)-(wc1*hw3))/-wc2##The enthapy of warm water at point 4 in kJ/kg\n", + "E=hw4/Cpc##Exit water temperature in degree centigrade\n", + "\n", + "#Output\n", + "print \" (a) The final temperature of the water = %3.2f degree centigrade \\n (b)The amount of makeup water required per second = %3.4f kg/s \"%(E,wm)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/Ch9.ipynb b/Power_Plant_Engineering_by_P._K._Nag/Ch9.ipynb new file mode 100644 index 00000000..b7e3e772 --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/Ch9.ipynb @@ -0,0 +1,474 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 9 : Nuclear Power Plant" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.1 Pg: 648" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The mass defect = 0.13703 amu \n", + " The binding energy per nucleon = 7.97 MeV \n" + ] + } + ], + "source": [ + "#Input data\n", + "mp=1.007277##Atomic Mass of proton in amu\n", + "mn=1.008665##Atomic Mass of neutron in amu\n", + "me=0.00055##Atomic Mass of electron in amu\n", + "mo=15.99491##Atomic Mass of oxygen in amu\n", + "np=8##Number of protons in oxygen\n", + "ne=8##Number of electrons in oxygen\n", + "nn=8##Number of neutrons in oxygen\n", + "a=931##One amu in MeV\n", + "No=16##Number of nucleons in oxygen\n", + "\n", + "#Calculations\n", + "m=(np*mp)+(ne*me)+(nn*mn)-mo##The mass defect in amu\n", + "B=m*a##Binding energy in MeV\n", + "Bn=B/No##Binding energy per nucleon\n", + "\n", + "#Output\n", + "print \" The mass defect = %3.5f amu \\n The binding energy per nucleon = %3.2f MeV \"%(m,Bn)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.2 Pg: 649" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The decay constant = 1.36e-11 s**-1\n", + "The initial activity of 1 g of radium 226 = 3.61e+10 dis/s\n" + ] + } + ], + "source": [ + "#Input data\n", + "amr=226.095##Atomic mass of radium in amu\n", + "AC=6.023*10**23##Avogadro constant in molecules/g.mol\n", + "h=1620##Half life of radium in years\n", + "\n", + "#Calculations\n", + "D=(0.6931/(h*365*24*3600))##The decay constant in 1/s\n", + "Na=AC/amr##Number of atoms per gram of radium \n", + "Ao=D*Na##Initial activity in dis/s\n", + "\n", + "#Output\n", + "print \"The decay constant = %0.2e s**-1\"%D\n", + "print \"The initial activity of 1 g of radium 226 = %0.2e dis/s\"%Ao," + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.3 Pg: 649" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fuel consumed of U-235 per day = 3.9 g/day \n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Input data\n", + "F=190##Each fission of U-235 yeilds in MeV\n", + "a=85##Assuming the Neutrons absorbed by U-235 cause fission in percentage\n", + "b=15##Non fission capture to produce an isotope U-236 in percentage\n", + "Q=3000##The amount of thermal power produced in MW\n", + "\n", + "#Calculations\n", + "E=F*1.60*10**-13##Each fission yields a useful energy in J\n", + "N=1/E##Number of fissions required \n", + "B=((10**6)*(N*86400))/(a/100)##One day operation of a reactor the number of U-235 nuclei burned is in absorptions per day\n", + "M=(B*235)/(6.023*10**23)##Mass of U-235 consumed to produce one MW power in g/day\n", + "M1=M*3##Mass of U-235 consumed to produce 3000 MW power in g/day\n", + "\n", + "#Output\n", + "print \"The fuel consumed of U-235 per day = %3.1f g/day \"%(M1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.4 Pg: 650" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The cross section of neutrons = 5.02 barns \n" + ] + } + ], + "source": [ + "#Input data\n", + "sa1=10##Cross section of nucleus in barns\n", + "N=2200##Neutrons in m/s\n", + "En1=0.1##Kinetic energy of neutrons increases in eV\n", + "En2=0.02525##Kinetic energy of neutron in eV\n", + "\n", + "#Calculations\n", + "sa2=sa1/((En1/En2)**0.5)##The cross section of neutrons in barns\n", + "\n", + "#Output\n", + "print \"The cross section of neutrons = %3.2f barns \"%(sa2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.5 Pg: 650" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The microscopic absorption cross section of natural uranium = 7.6 barns \n" + ] + } + ], + "source": [ + "#Input data\n", + "U1=99.285##Uranium consists of U-238 in percentage \n", + "U2=0.715##Uranium consists of U-235 in Percentage\n", + "E=0.025##The energy of neutrons in eV\n", + "sc=2.72##Capture cross section for U-238 in barns\n", + "sf=0##fission cross section for U-238 in barns\n", + "sc1=101##Capture cross section for U-235 in barns\n", + "sf1=579##fission cross section for U-235 in barns\n", + "\n", + "#Calculations\n", + "sa=(U1/100)*(sc+sf)+(U2/100)*(sc1+sf1)##The microscopic absorption cross section of natural uranium in barns\n", + "\n", + "#Output\n", + "print \"The microscopic absorption cross section of natural uranium = %3.1f barns \"%(sa)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.6 Pg: 650" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The microscopic capture cross section of water = 0.0222 cm**-1 \n" + ] + } + ], + "source": [ + "#Input data\n", + "p=1##The density of water in g/cm**3\n", + "sch=0.332##The microscopic capture cross section of hydrogen in barn\n", + "sco=0.0002##The microscopic capture cross section of oxygen in barn\n", + "\n", + "#Calculations\n", + "N=(6.023*10**23)*p/18##Number of molecules of water per cm**3\n", + "scw=(2*N*sch*10**-24)+(N*sco*10**-24)##The microscopic capture cross section of water in cm**-1\n", + "\n", + "#output\n", + "print \"The microscopic capture cross section of water = %3.4f cm**-1 \"%(scw)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.7 Pg: 650" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The thermal neutron flux = 9.10e+11 cm**-2s**-1\n", + "The average distance that a neutron travels before it is absorbed = 0.010 cm\n" + ] + } + ], + "source": [ + "#Input data\n", + "m=230##The amount of boron piece in g\n", + "mw=10##The molecular weight of boron \n", + "R=9.57*10**13##Reaction rate in cm**-3s**-1\n", + "d=2.3##Density of boron in g/cm**3\n", + "sa=755##Absorbption cross section in barns\n", + "ss=4##Elastic scattering cross section in barns\n", + "\n", + "#Calculations\n", + "st=sa+ss##The total cross section in barns\n", + "N=(d*6.023*10**23)/mw##The number density of neutrons in cm**-3\n", + "S=N*st*10**-24##Number density of neutrons for total in cm**-1\n", + "F=R/S##Neutron flux in cm**-2s**-1\n", + "L=1/S##Average distance a neutron travels before it is absorbed in cm\n", + "\n", + "#Output\n", + "print \"The thermal neutron flux = %0.2e cm**-2s**-1\"%F\n", + "print \"The average distance that a neutron travels before it is absorbed = %0.3f cm\"%L" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.8 Pg: 651" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " The logarithmic energy decrement representing the neutron energy loss per elastic collision = 0.158 \n", + " The number of collisions necessary = 121 \n" + ] + } + ], + "source": [ + "from math import log\n", + "#Input data\n", + "Eni=4.8##The energy of the newly born electron in MeV\n", + "Enf=0.025##The energy of the electron after slow down in eV\n", + "A=12##The mass number of the graphite (carbon)\n", + "\n", + "#Calculations\n", + "L=1-(((A-1)**2/(2*A))*log((A+1)/(A-1)))##The logarithmic energy decrement\n", + "n=(log(Eni*10**6/Enf))/L##The number of collisions required to slowdown the neutron \n", + "\n", + "#Output\n", + "print \" The logarithmic energy decrement representing the neutron energy loss per elastic collision = %3.3f \\n The number of collisions necessary = %3.0f \"%(L,n)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.9 Pg: 651" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " (a) The rating of the reactor = 3.36 MW/tonne \n", + " (b)The rate of consumption of U-235 per day = 0.353 kg/day (or) 353 g/day \n" + ] + } + ], + "source": [ + "#Input data\n", + "f=100##The reactor is fuelled of natural uranium in tonnes\n", + "A=238.05##The atomic mass of natural uranium \n", + "F=10**13##The average thermal neutron flux in neutrons/cm**2s\n", + "A1=235.04##The atomic mass of U-235\n", + "sf=579##The fission cross section of U-235 in barns\n", + "sc=101##The capture cross section of U-235 in barns\n", + "E=200##The energy released per fission in MeV\n", + "P=0.715##U-235 in natural uranium in percentage\n", + "N=2200##The average thermal neutron in m/s\n", + "\n", + "#Calculations\n", + "n=((f*1000)*6.023*10**26*(P/100))/A##The number of U-235 atoms in the reactor in atoms\n", + "R=(sf*10**-24)*F*n##The rate of fission in the reactor in fissions/s\n", + "T=R*E*1.602*10**-19##Thermal power of the reactor in MW\n", + "Rr=T/f##Rating the reactor MW/tonne\n", + "Rc=(((R*A1*60*60*24))/(6.023*10**26))##The rate of consumption of U-235 by fission in kg/day\n", + "Rcc=Rc*1000##The rate of consumption of U-235 by fission in g/day\n", + "\n", + "#Output\n", + "print \" (a) The rating of the reactor = %3.2f MW/tonne \\n (b)The rate of consumption of U-235 per day = %3.3f kg/day (or) %3.0f g/day \"%(Rr,Rc,Rcc)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.10 Pg: 652" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The specific energy release rate for a light water uranium reactor = 138.13 W/cm**3\n" + ] + } + ], + "source": [ + "#Input data\n", + "f=3.5##Mass fraction of U-235 in the fuel in percentage\n", + "G=180##Energy per fission in Mev\n", + "F=10**13##The neutron flux in neutrons/cm**2s\n", + "sf=577##Fission cross section of U-235 in barns\n", + "M=1.602*10**-13##One MeV in J\n", + "\n", + "#Calculations\n", + "N=2.372*(f/100)*10**22##The fuel density for a uranium oxide fuel in nuclei/cm**3\n", + "q=G*N*sf*10**-24*F##The rate of energy release in MeV/cm**3s\n", + "qg=q*M##The rate of energy release in W/cm**3\n", + "\n", + "#Output\n", + "print \"The specific energy release rate for a light water uranium reactor = %3.2f W/cm**3\"%(qg)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex: 9.11 Pg: 653" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The reactor power level at the end of 1s is 3.196 MW\n" + ] + } + ], + "source": [ + "from math import exp\n", + "#Input data\n", + "P=1##The operating power of a reactor in W\n", + "K=1.0015##The effective multiplication factor of Reactor becomes suppercritical \n", + "t=0.0001##The average neutron life in s\n", + "t1=1.0001##Neutron life time in s\n", + "\n", + "#Calculations \n", + "d=(K-1)/K##The reactivity \n", + "Z=(d*P)/t##The number of neutrons\n", + "n=exp(Z)/10**6##Neutron density * 10**6\n", + "\n", + "#Output\n", + "print \"The reactor power level at the end of 1s is %3.3f MW\"%(n)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve.png b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve.png Binary files differnew file mode 100644 index 00000000..509068ba --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve.png diff --git a/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve2.png b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve2.png Binary files differnew file mode 100644 index 00000000..e9a1e64f --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadCurve2.png diff --git a/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadDurationCurve.png b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadDurationCurve.png Binary files differnew file mode 100644 index 00000000..adc15e4e --- /dev/null +++ b/Power_Plant_Engineering_by_P._K._Nag/screenshots/LoadDurationCurve.png diff --git a/sample_notebooks/FameethaAlamBasha/Chapter7_1.ipynb b/sample_notebooks/FameethaAlamBasha/Chapter7_1.ipynb new file mode 100644 index 00000000..b42dd470 --- /dev/null +++ b/sample_notebooks/FameethaAlamBasha/Chapter7_1.ipynb @@ -0,0 +1,78 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7:Starting And Speed Control Of Induction Motor " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7.1,Page 7-6" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "x= 0.7453\n", + "Thus 74.53 % tapping is required\n", + "Ist= 3.33 IF.L\n", + "Thus supply starting current is 3.33 times the full load current\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "#variable declaration\n", + "Isc=6 # starting current at rated voltage(amp)\n", + "Sf=5 # full load slip\n", + "Tfl=Tst=1 # full load torque=starting torque(N-m)\n", + "Ifl=1 # full load current(amp)\n", + "\n", + "#calculation\n", + "x=math.sqrt((Tst/Tfl) * ((float(Ifl)/float(Isc))**2)*(1/(float(Sf)/100)))\n", + "print 'x=',str (x)[:6]\n", + "print 'Thus',str(x*100)[:5],'% tapping is required'\n", + "\n", + "#Ist(supply)=x**2 * 6\n", + "Ist=(x**2) * Isc # supply starting current(IF.L)\n", + "Ist=round(Ist,2)\n", + "print 'Ist=',Ist,'IF.L'\n", + "print 'Thus supply starting current is',Ist,'times the full load current'" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/Hrituraj/Ch-6_1.ipynb b/sample_notebooks/Hrituraj/Ch-6_1.ipynb new file mode 100644 index 00000000..4aa9a7a3 --- /dev/null +++ b/sample_notebooks/Hrituraj/Ch-6_1.ipynb @@ -0,0 +1,547 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Ch-6 : Frequency response, bode plots and resonance" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.1 Page No: 476" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "peak value of Vout = 6.00 volts\n", + "phase angle of Vout = 70.00 degrees\n", + "with frequency equal to = 1000.00\n" + ] + } + ], + "source": [ + "from math import pi, cos, sin, atan, sqrt\n", + "# given V_in(t)=2*cos(2000*pi*t+A), A=40*pi/180\n", + "w=2000*pi# #omega\n", + "f=w/(2*pi)# #frequency\n", + "A=40*pi/180# #40 degrees = %0.2f radians\n", + "#equation of straight line of H_magnitude vs f is x+1000*y-4000=0\n", + "H_max=(4000-f)/1000# #magnitude of H(traansfer function)\n", + "#equation of straight line of H_phase angle vs f is 6000*y=pi*x (phase angle = %0.2f radians)\n", + "H_phi=pi*f/6000# #phase angle of H\n", + "H=H_max*complex(cos(H_phi),sin(H_phi))\n", + "V_in=2*complex(cos(A),sin(A))# #input voltage phasor\n", + "V_out=H*V_in# #output voltage phasor\n", + "V_out_R=(V_out.real)# #real part\n", + "V_out_I=(V_out.imag)# #imaginary part\n", + "V_out_max=sqrt((V_out_R**2)+(V_out_I**2))# #peak value\n", + "V_out_phi=atan(V_out_I/V_out_R)\n", + "print 'peak value of Vout = %0.2f volts'%V_out_max\n", + "print 'phase angle of Vout = %0.2f degrees'%(V_out_phi*180/pi)\n", + "print 'with frequency equal to = %0.2f'%f" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.2 Page No: 477" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is Vout1+Vout2+Vout3 where\n", + "\n", + "FOR Vout1:\n", + "peak value = 12.00 volts\n", + "phase angle = 0.00 degrees\n", + "with frequency = 0.00 hertz\n", + "\n", + "FOR Vout2:\n", + "peak value = 6.00 volts\n", + "phase angle = 30.00 degrees\n", + "with frequency = 1000.00 hertz\n", + "\n", + "FOR Vout3:\n", + "peak value = 2.00 volts\n", + "phase angle = -10.00 degrees\n", + "with frequency = 2000.00 hertz\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "from math import pi, cos, sin, atan, sqrt\n", + "\n", + "#given V_in(t)=3+2*cos(2000*pi*t)+cos(4000*pi*t-A), A=70*pi/180\n", + "#the three parts of V_in(t) are V_in_1=3, V_in_2=2*cos(2000*pi*t),V_in_3=cos(4000*pi*t-A)\n", + "\n", + "#first component V_1\n", + "V_in_1=3\n", + "f_1=0# #as omega is zero\n", + "#equation of straight line of H_magnitude vs f is x+1000*y-4000=0\n", + "H_1_max=(4000-f_1)/1000# #magnitude of H(traansfer function)\n", + "#equation of straight line of H_phase angle vs f is 6000*y=pi*x (phase angle = %0.2f radians)\n", + "H_1_phi=pi*f_1/6000# #phase angle of H\n", + "H_1=H_1_max*complex(cos(H_1_phi),sin(H_1_phi))\n", + "V_out_1=H_1*V_in_1\n", + "V_out_1_R=(V_out_1).real# #real part\n", + "V_out_1_I=(V_out_1).imag# #imaginary part\n", + "V_out_1_max=sqrt((V_out_1_R**2)+(V_out_1_I**2))# #peak value\n", + "V_out_1_phi=atan(V_out_1_I/V_out_1_R)# #phase angle\n", + "\n", + "#second component V_in_2\n", + "V_in_2=2*complex(cos(0),sin(0))# #V_in_2 phasor\n", + "w=2000*pi# #omega\n", + "f_2=w/(2*pi)# #frequency\n", + "#equation of straight line of H_magnitude vs f is x+1000*y-4000=0\n", + "H_2_max=(4000-f_2)/1000# #magnitude of H(traansfer function)\n", + "#equation of straight line of H_phase angle vs f is 6000*y=pi*x (phase angle = %0.2f radians)\n", + "H_2_phi=pi*f_2/6000# #phase angle of H\n", + "H_2=H_2_max*complex(cos(H_2_phi),sin(H_2_phi))\n", + "V_out_2=H_2*V_in_2\n", + "V_out_2_R=(V_out_2).real# #real part\n", + "V_out_2_I=(V_out_2).imag# #imaginary part\n", + "V_out_2_max=sqrt((V_out_2_R**2)+(V_out_2_I**2))# #peak value\n", + "V_out_2_phi=atan(V_out_2_I/V_out_2_R)# #phase angle\n", + "\n", + "#third component\n", + "A=-70*pi/180# #-70 degrees = %0.2f radians\n", + "V_in_3=complex(cos(A),sin(A))# #V_in_3 phasor\n", + "w=4000*pi# #omega\n", + "f_3=w/(2*pi)# #frequency\n", + "#equation of straight line of H_magnitude vs f is x+1000*y-4000=0\n", + "H_3_max=(4000-f_3)/1000# #magnitude of H(traansfer function)\n", + "#equation of straight line of H_phase angle vs f is 6000*y=pi*x (phase angle = %0.2f radians)\n", + "H_3_phi=pi*f_3/6000# #phase angle of H\n", + "H_3=H_3_max*complex(cos(H_3_phi),sin(H_3_phi))\n", + "V_out_3=H_3*V_in_3\n", + "V_out_3_R=(V_out_3).real# #real part\n", + "V_out_3_I=(V_out_3).imag# #imaginary part\n", + "V_out_3_max=sqrt((V_out_3_R**2)+(V_out_3_I**2))# #peak value\n", + "V_out_3_phi=atan(V_out_3_I/V_out_3_R)# #phase angle\n", + "\n", + "print 'Output voltage is Vout1+Vout2+Vout3 where'\n", + "print ''\n", + "print 'FOR Vout1:'\n", + "print 'peak value = %0.2f volts'%V_out_1_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_1_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_1\n", + "print ''\n", + "print 'FOR Vout2:'\n", + "print 'peak value = %0.2f volts'%V_out_2_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_2_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_2\n", + "print ''\n", + "print 'FOR Vout3:'\n", + "print 'peak value = %0.2f volts'%V_out_3_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_3_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_3" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.3 Page No: 477" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\n", + "Output voltage is Vout1+Vout2+Vout3 where\n", + "\n", + "FOR Vout1:\n", + "peak value = 4.98 volts\n", + "phase angle = -5.71 degrees\n", + "with frequency = 10.00 hertz\n", + "\n", + "FOR Vout2:\n", + "peak value = 3.54 volts\n", + "phase angle = -45.00 degrees\n", + "with frequency = 100.00 hertz\n", + "\n", + "FOR Vout3:\n", + "peak value = 0.50 volts\n", + "phase angle = -84.29 degrees\n", + "with frequency = 1000.00 hertz\n" + ] + } + ], + "source": [ + "from math import pi, cos, sin, atan, sqrt\n", + "\n", + "R=1000/(2*pi)# #resistance\n", + "C=10*10**-6# #capacitance\n", + "f_B=1/(2*pi*R*C)# #half-power frequency\n", + "#the three parts of V_in are V_1=5*cos(20*pi*t)+5*cos(200*pi*t)+5*cos(2000*pi*t)\n", + "\n", + "#first component V_in_1\n", + "V_in_1=5*complex(cos(0),sin(0))# #V_in_1 phasor\n", + "w_1=20*pi# #omega\n", + "f_1=w_1/(2*pi)# #frequency\n", + "H_1=1/(1+1J*(f_1/f_B))# #transfer function\n", + "V_out_1=H_1*V_in_1\n", + "V_out_1_R=(V_out_1).real# #real part\n", + "V_out_1_I=(V_out_1).imag# #imaginary part\n", + "V_out_1_max=sqrt((V_out_1_R**2)+(V_out_1_I**2))# #peak value\n", + "V_out_1_phi=atan(V_out_1_I/V_out_1_R)# #phase angle\n", + "\n", + "#second component V_in_2\n", + "V_in_2=5*complex(cos(0),sin(0))# #V_in_2 phasor\n", + "w_2=200*pi# #omega\n", + "f_2=w_2/(2*pi)# #frequency\n", + "H_2=1/(1+1J*(f_2/f_B))# #transfer function\n", + "V_out_2=H_2*V_in_2\n", + "V_out_2_R=(V_out_2).real #real part\n", + "V_out_2_I=(V_out_2).imag #imaginary part\n", + "V_out_2_max=sqrt((V_out_2_R**2)+(V_out_2_I**2))# #peak value\n", + "V_out_2_phi=atan(V_out_2_I/V_out_2_R)# #phase angle\n", + "\n", + "#third component V_in_3\n", + "V_in_3=5*complex(cos(0),sin(0))# #V_in_3 phasor\n", + "w_3=2000*pi# #omega\n", + "f_3=w_3/(2*pi)# #frequency\n", + "H_3=1/(1+1J*(f_3/f_B))# #transfer function\n", + "V_out_3=H_3*V_in_3\n", + "V_out_3_R=(V_out_3).real #real part\n", + "V_out_3_I=(V_out_3).imag #imaginary part\n", + "V_out_3_max=sqrt((V_out_3_R**2)+(V_out_3_I**2))# #peak value\n", + "V_out_3_phi=atan(V_out_3_I/V_out_3_R)# #phase angle\n", + "\n", + "print \" All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\"\n", + "print 'Output voltage is Vout1+Vout2+Vout3 where'\n", + "print ''\n", + "print 'FOR Vout1:'\n", + "print 'peak value = %0.2f volts'%V_out_1_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_1_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_1\n", + "print ''\n", + "print 'FOR Vout2:'\n", + "print 'peak value = %0.2f volts'%V_out_2_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_2_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_2\n", + "print ''\n", + "print 'FOR Vout3:'\n", + "print 'peak value = %0.2f volts'%V_out_3_max\n", + "print 'phase angle = %0.2f degrees'%(V_out_3_phi*180/pi)\n", + "print 'with frequency = %0.2f hertz'%f_3\n", + "#we can observe that there is a clear discrimination = %0.2f output signals based on frequencies i.e, lesser the frequency lesser the effect." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.4 Page No: 478" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\n", + "Break frequency = 1897.37 Hz\n" + ] + } + ], + "source": [ + "H_max=-30# #transfer function magnitude\n", + "f=60\n", + "m=20# #low-frequency asymptote slope rate = %0.2f db/decade\n", + "#f_B must be K higher than f where K is\n", + "K=abs(H_max)/m\n", + "#(base 10)log(f_B/60)=1.5 ==>\n", + "f_B=60*10**1.5\n", + "print \" All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\"\n", + "print 'Break frequency = %0.2f Hz'%f_B" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.5 Page No: 479" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Phasor voltage across Resistance\n", + "peak value = 1.00 volts\n", + "phase angle = 0.00 degrees\n", + "\n", + "Phasor voltage across Inductance\n", + "peak value = 10.00 volts\n", + "phase angle = 90.00 degrees\n", + "\n", + "Phasor voltage across Capacitance\n", + "peak value = 10.00 volts\n", + "phase angle = -90.00 degrees\n" + ] + } + ], + "source": [ + "from math import pi, cos, sin, atan, sqrt\n", + "V_s=1*complex(cos(0),sin(0))\n", + "L=159.2*10**-3\n", + "R=100\n", + "C=0.1592*10**-6\n", + "f_o=1/(2*pi*sqrt(L*C))# #resonant frequency\n", + "Q_s=2*pi*f_o*L/R# #quality factor\n", + "B=f_o/Q_s# #Bandwidth\n", + "#Approximate half-power frequencies are\n", + "f_H=f_o+(B/2)\n", + "f_L=f_o-(B/2)\n", + "#At resonance\n", + "Z_L=1J*2*pi*f_o*L# #impedance of inductance\n", + "Z_C=-1J/(2*pi*f_o*C)# #impedance of capacitance\n", + "Z_s=R+Z_L+Z_C\n", + "I=V_s/Z_s# #phasor current\n", + "#voltages across diffrent elements are\n", + "#for resistance\n", + "V_R=R*I\n", + "V_R_R=(V_R).real #real part\n", + "V_R_I=(V_R).imag #imaginary part\n", + "V_R_max=sqrt((V_R_R**2)+(V_R_I**2))# #peak value\n", + "V_R_phi=atan(V_R_I/V_R_R)# #phase angle\n", + "#for inductance\n", + "V_L=Z_L*I\n", + "V_L_R=(V_L).real #real part\n", + "V_L_I=(V_L).imag #imaginary part\n", + "V_L_max=sqrt((V_L_R**2)+(V_L_I**2))# #peak value\n", + "#Z_L is pure imaginary ==> V_L is pure imaginary which means V_L_phi can be +or- pi/2\n", + "if ((V_L/1J)==abs(V_L)):\n", + " V_L_phi=pi/2\n", + "elif ((V_L/1J)==-abs(V_L)):\n", + " V_L_phi=-pi/2\n", + "\n", + "\n", + "#for capacitance\n", + "V_C=Z_C*I\n", + "V_C_R=(V_C).real #real part\n", + "V_C_I=(V_C).imag #imaginary part\n", + "V_C_max=sqrt((V_C_R**2)+(V_C_I**2))# #peak value\n", + "#Z_C is pure imaginary ==> V_C is pure imaginary which means V_C_phi can be +or- pi/2\n", + "if ((V_C/1J)==abs(V_C)) :\n", + " V_C_phi=pi/2\n", + "elif ((V_C/1J)==-abs(V_C)) :\n", + " V_C_phi=-pi/2\n", + "\n", + " \n", + "print 'Phasor voltage across Resistance'\n", + "print 'peak value = %0.2f volts'%V_R_max\n", + "print 'phase angle = %0.2f degrees'%(V_R_phi*180/pi)\n", + "print ''\n", + "print 'Phasor voltage across Inductance'\n", + "print 'peak value = %0.2f volts'%V_L_max\n", + "print 'phase angle = %0.2f degrees'%(V_L_phi*180/pi)\n", + "print ''\n", + "print 'Phasor voltage across Capacitance'\n", + "print 'peak value = %0.2f volts'%V_C_max\n", + "print 'phase angle = %0.2f degrees'%(V_C_phi*180/pi)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.6 Page No: 480" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current phasor across Resistance\n", + "peak value = 0.001 amperes\n", + "phase angle = 0 degrees\n", + "\n", + "Current phasor across Inductance\n", + "peak value = 0.010 amperes\n", + "phase angle = -90.00 degrees\n", + "\n", + "current phasor across capacitance\n", + "peak value = 0.010 amperes\n", + "phase angle = 90.00 degrees\n" + ] + } + ], + "source": [ + "from math import pi, cos, sin, atan, sqrt\n", + "R=10*10**3\n", + "f_o=1*10**6\n", + "B=100*10**3\n", + "I=10**-3*complex(cos(0),sin(0))\n", + "Q_p=f_o/B# #quality factor\n", + "L=R/(2*pi*f_o*Q_p)\n", + "C=Q_p/(2*pi*f_o*R)\n", + "#At resonance\n", + "V_out=I*R\n", + "Z_L=1J*2*pi*f_o*L\n", + "Z_C=-1J/(2*pi*f_o*C)\n", + "\n", + "#across resistance\n", + "I_R=V_out/R\n", + "I_R_R=(I_R).real# #real part\n", + "I_R_I=(I_R).imag# #imaginary part\n", + "I_R_max=sqrt((I_R_R**2)+(I_R_I**2))# #peak value\n", + "I_R_phi=atan(I_R_I/I_R_R)# #phase angle\n", + "\n", + "#across inductance\n", + "I_L=V_out/Z_L\n", + "I_L_R=(I_L).real #real part\n", + "I_L_I=(I_L).imag# #imaginary part\n", + "I_L_max=sqrt((I_L_R**2)+(I_L_I**2))# #peak value\n", + "#Z_L is pure imaginary ==> V_L is pure imaginary which means V_L_phi can be +or- pi/2\n", + "if ((I_L/1J)==abs(I_L)):\n", + " I_L_phi=pi/2\n", + "elif ((I_L/1J)==-abs(I_L)) :\n", + " I_L_phi=-pi/2\n", + "\n", + "\n", + "#across capacitor\n", + "I_C=V_out/Z_C\n", + "I_C_R=(I_C).real# #real part\n", + "I_C_I=(I_C).imag# #imaginary part\n", + "I_C_max=sqrt((I_C_R**2)+(I_C_I**2))# #peak value\n", + "#Z_C is pure imaginary ==> V_C is pure imaginary which means V_C_phi can be +or- pi/2\n", + "if ((I_C/1J)==abs(I_C)):\n", + " I_C_phi=pi/2\n", + "elif ((I_C/1J)==-abs(I_C)) :\n", + " I_C_phi=-pi/2\n", + "\n", + "\n", + "print 'Current phasor across Resistance'\n", + "print 'peak value = %0.3f amperes'%I_R_max\n", + "print 'phase angle = %0.f degrees'%(I_R_phi*180/pi)\n", + "print ''\n", + "print 'Current phasor across Inductance'\n", + "print 'peak value = %0.3f amperes'%I_L_max\n", + "print 'phase angle = %0.2f degrees'%(I_L_phi*180/pi)\n", + "print ''\n", + "print 'current phasor across capacitance'\n", + "print 'peak value = %0.3f amperes'%I_C_max\n", + "print 'phase angle = %0.2f degrees'%(I_C_phi*180/pi)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example: 6.7 Page No: 481" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\n", + "\n", + "The required second order circuit configuration is\n", + "Inductance = 50.00 KH\n", + "Capacitance = 0.51 mF(micro Farads)\n", + "Resistance = 314.16 ohms\n" + ] + } + ], + "source": [ + "from math import pi\n", + "#We need a high-pass filter\n", + "L=50*10**-3\n", + "#for the transfer function to be approximately constant = %0.2f passband area(from graph given = %0.2f the text), we choose\n", + "Q_s=1\n", + "f_o=1*10**3\n", + "C=1/(((2*pi)**2)*f_o**2*L)\n", + "R=2*pi*f_o*L/Q_s\n", + "print \" All the values in the textbook are approximated, hence the values in this code differ from those of Textbook\"\n", + "print ''\n", + "print 'The required second order circuit configuration is'\n", + "print 'Inductance = %0.2f KH'%(L*10**3)\n", + "print 'Capacitance = %0.2f mF(micro Farads)'%(C*10**6)\n", + "print 'Resistance = %0.2f ohms'%R\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/NishthaRani/Chapter12.ipynb b/sample_notebooks/NishthaRani/Chapter12.ipynb new file mode 100644 index 00000000..63660758 --- /dev/null +++ b/sample_notebooks/NishthaRani/Chapter12.ipynb @@ -0,0 +1,428 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 12 : Absorption of Gases" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 669 Example 12.1" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " For S02:\n", + "\n", + " kGa = 0.00012 ol/s m**3(kN/m**2)\n", + "\n", + " For NH3:\n", + "\n", + " kGa = 0.00016 kmol/s m**3(kN/m**2)\n", + "\n", + " For a very soluble gas such as NH3, kGa = KGa.\n", + "\n", + " For NH3 the liquid-film resistance will be small, and:\n", + "\n", + " kGa =KGa = 0.00016 mol/s m**3(kN/m**2)\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "#Overall liquid transfer coefficient KLa = 0.003 kmol/s.m**3(kmol/m**3)\n", + "\n", + "#(1/KLa)=(1/kLa)+(1/HkGa)\n", + "# let (KLa)=x\n", + "x = 0.003#\n", + "overall = 1/x#\n", + "\n", + "#For the absorption of a moderately soluble gas it is reasonable to assume that the liquid and gas phase resistances are of the same order ofmagnitude, assuming them to be equal.\n", + "#(1/KLa)=(1/kLa)+(1/HkGa)\n", + "#let 1/kLa = 1/HkGa = y\n", + "y = (1/(2*x))#\n", + "z = (1/y)# #z is in kmol/s m**3(kmol/m**3)\n", + "print \"\\n For S02:\"\n", + "print \"\\n kGa = %0.5f ol/s m**3(kN/m**2)\"%(z/50)#\n", + "print \"\\n For NH3:\"\n", + "d_SO2 = 0.103# #diffusivity at 273K for SO2 in cm**2/sec\n", + "d_NH3 = 0.170# #diffusivity at 273K for NH3 in cm**2/sec\n", + "print \"\\n kGa = %0.5f kmol/s m**3(kN/m**2)\"%((z/50)*(d_NH3/d_SO2)**0.56)#\n", + "print \"\\n For a very soluble gas such as NH3, kGa = KGa.\"\n", + "print \"\\n For NH3 the liquid-film resistance will be small, and:\"\n", + "print \"\\n kGa =KGa = %0.5f mol/s m**3(kN/m**2)\"%((z/50)*(d_NH3/d_SO2)**0.56)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 671 Example 12.2" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " d = 0.05 mm\n", + "\n", + " kG = 1.19*10**(-8) kmol/m**2sec(N/m**2)\n", + "\n", + " kG = 7.59*10**(-4)kg SO2/m**2sec(kN/m**2)\n" + ] + } + ], + "source": [ + "G = 2# #air flow rate in kg/m**2.sec\n", + "Re = 5160# #Re stands for Reynolds number\n", + "f = 0.02# #friction factor = R/pu**2\n", + "d_SO2 = 0.116*10**(-4)# #diffusion coefficient in m**2/sec\n", + "v = 1.8*10**(-5)# #viscosity in mNs/m**2\n", + "p = 1.154# #density is in kg/m**3\n", + "\n", + "#(hd/u)(Pm/P)(u/p/d_SO2)**0.56 = BRe**(-0.17)=jd\n", + "\n", + "def a1():\n", + " x = (v/(p*d_SO2))**(0.56)#\n", + " return x\n", + "\n", + "\n", + "#jd = f =R/pu**2\n", + "def a2():\n", + " y = f/a1()#\n", + " return y\n", + "#G = pu\n", + "u = (G/p)# #u is in m/sec\n", + "\n", + "def a3():\n", + " x1 = a2()*u#\n", + " return x1\n", + "\n", + "def d1():\n", + " d = Re*v/(G)#\n", + " return d\n", + "\n", + "print \"\\n d = %0.2f mm\"%(d1())#\n", + "R = 8314# #R is in m**3(N/m**2)/K kmol\n", + "T = 298# #T is in Kelvins\n", + "def kG1():\n", + " kG = a3()/(R*T)#\n", + " return kG\n", + "\n", + "print \"\\n kG = %.2f*10**(-8) kmol/m**2sec(N/m**2)\"%(kG1()*10**(8))#\n", + "print \"\\n kG = %.2f*10**(-4)kg SO2/m**2sec(kN/m**2)\"%(kG1()*10**(7)*64)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 694 Example 12.3" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " x1 = 0.01\n", + "\n", + " NoL =7.38\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve, log\n", + "#as the system are dilute mole fractions are approximately equal to mole ratios.\n", + "#At the bottom of the tower\n", + "y1 =0.015# #mole fraction\n", + "G = 1# #Gas flow rate is in kg/m**2sec\n", + "#At the top of the tower\n", + "y2 = 0.00015# #mole fraction\n", + "x2 = 0#\n", + "L = 1.6# #liquid flow rate is in kg/m**2.sec\n", + "Lm = 1.6/18# #liquid flow rate is in kmol/m**2.sec\n", + "Gm = 1.0/29# #gas flow rate is in kmol/m**2.sec\n", + "\n", + "x1 = symbols('x1')\n", + "x11 = solve(Gm*(y1-y2)-Lm*(x1),x1)[0]\n", + "\n", + "print \"\\n x1 = %0.2f\"%(x11)#\n", + "def henry_law(x):\n", + " ye1 = 1.75*x#\n", + " return ye1\n", + "bottom_driving_force = y1 - henry_law(x11)#\n", + "def log_mean():\n", + " lm = (bottom_driving_force-y2)/log(bottom_driving_force/y2)\n", + " return lm\n", + "NoG = (y1-y2)/log_mean()\n", + "NoL = NoG*1.75*(Gm/Lm)#\n", + "print \"\\n NoL =%.2f\"%(NoL)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 699 Example 12.4" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Z =7.80m\n", + "\n", + " Height of transfer unit HoG = 0.375m\n", + "\n", + " Number of transfer units NoG = 21.00\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve\n", + "from math import ceil\n", + "Gm = 0.015#\n", + "KGa = 0.04#\n", + "top = 0.0003#\n", + "Y1 = 0.03#\n", + "Ye = 0.026#\n", + "bottom = (Y1-Ye)#\n", + "log_mean_driving_force = (bottom-top)/log(bottom/top)#\n", + "Z = symbols('Z')\n", + "Z1 = solve(Gm*(Y1-top)-KGa*log_mean_driving_force*Z,Z)[0]\n", + "print \"\\n Z =%.2fm\"%(Z1)#\n", + "HoG = Gm/KGa#\n", + "print \"\\n Height of transfer unit HoG = %.3fm\"%(HoG)#\n", + "print \"\\n Number of transfer units NoG = %0.2f\"%(ceil(7.79/0.375))#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 700 Example 12.5" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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L+mwcFBNTWV1tVRo3NMkWoBfwBlAdmA+0V9V12Rw7J/1JIe/bo0ePfx7XqVOH\nOnXq5OC0xkTf339D165uRr++faFxY//WNQCS1ifRelxrCkpBxt8/nipnVIl3SCZMCQkJJCQkRPSY\nWbU8PsJNQVsUmOU9747rPnoPuCebY/8JBE8LVhrXgghFyPsGJw9j/EzVXUH15JNw883uDvGTTop3\nVJnbtm8bXSd1ZdSyUbxy3Ss0uaQJBSSUQSmM36T/w/r555/P8TGzSh4nq2o/cGNZqWovb3k/EWke\nwrHnAuW9K7XWAw2BRplsm/7vrnD2Ncb31qxxBfFVq+Czz+Cqq+IdUeaCu6gaVGxAcptku+HPHCar\n5BH8hT4s3bqC2R1YVVNE5HHc9LUFgcGqmiwirbz174vI6bgrqY4HAiLyJFBRVXdntG/Ir8oYn0hJ\ngbfegldegaeecl1VRYrEO6rMJa1Pos24NhSQAtZFZbKU6aW6IvIi0EdVd6VbXh54RVXrxyC+LNml\nusbPfv7ZFcRPPhneew/Kl493RJmzLqr8JaqX6qpqt/SJw1u+wg+Jwxi/2rkT2raF2293kzNNmODf\nxBHQAIPnDabiOxUpWKAgyW2SebjKw5Y4TLayutoq+I5v5b/dWKqqr0ctKmNyIVUYNcoljrp1XUH8\n5JPjHVXm0rqoRIRx94/j0jMujXdIJhfJquZRjH+TRitgQEwiMiYX+uMPVxBfsQKGD4err453RJnb\ntm8bz05+lpHJI62LyhyxLIcn+Wcjkfmq6rvKmdU8TLylpMDbb0PPnu4S3E6d4Kij4h1VxgIaYMj8\nIXSd3JV7LriHl659ya6iyqciUfMI+Q5zY8x/zZ0LLVtC8eKQmAjnnRfviDI3b8M8Wo9tbV1UJmIs\neRgTpp07oVs3+OILN1HTAw/49w7x4C6qntf1pOklTa2LykREpv+LRGRx2g9QIfi5iCyKYYzG+Mao\nUW4O8d27XUH8wQf9mTiCr6IShKVtltKsSjNLHCZismp53BazKIzxuT/+gCeegOXL4ZNPoHbteEeU\nubQuKoCxjcdStWTVOEdk8qJMk4eqro5hHMb4UkoK9OsHL7/sLsH98kv/FsSti8rEUlb3efyexX6q\nquWiEI8xvpGU5AriJ5wAM2ZAhQrxjihjAQ3w0YKPeGbSM9x9wd0sbbOUk47x8YiLJk/IqtuqetBj\nxdVHGgIdgHnRDMqYeNq1yxXEP/8c+vTxb10DrIvKxE9Ww5NsUdUtwDZc/SMBuByop6rZDcduTK70\n7beuIP6kxppGAAAZlUlEQVT33/DLL/DQQ/5MHNv3bafN2DbU+7QeLS5tQWLzREscJqay6rYqAjQD\n2uHmLr9DVVfGKjBjYmntWlfTWLoUPv4Y/DqnmHVRGb/IqtvqNyAFeAv4A7hYRC7GDVeiqjoyBvEZ\nE1WpqdC/P7z4orua6vPP/VsQn7dhHm3GtUFVrYvKxF1WyWOi9+/F3k96ljxMrpaU5IZML1oUpk+H\n88+Pd0QZ275vO89OfpYRySPoeW1PG/XW+EJWl+o2jWEcxsTM7t2uID58OPTuDU2a+LOuEdAAQxcM\npcukLtx9wd0kt0m2LirjGzY8iclXRo92o99ee627Q/yUU+IdUcasi8r4nSUPky+sW+cK4r/8AkOH\nwjXXxDuijG3ft51uU7rx1dKvrIvK+Jr9rzR5WmqqGzL9kkugUiVYtMifiSNtuPQL3rmAgAZIbpNM\n80ubW+IwvpXVpbr3cPgMgmnsaivje/PmuYL4scf6uyA+f8N82oxrQ6qm8l3j76hWslq8QzImW9kN\njJjVTEuWPIwv7d4N3bu7AQx79YKmTf1ZELcuKpOb2dVWJk8ZM8YVxOvUcfWNU0+Nd0SHC76K6q7z\n77KrqEyulG3BXEROBLoDabMyJwAvqOrfUYzLmLD8+acriC9eDEOGuKup/Mi6qExeEUob+UNgJ9AA\nuBfYBQyJZlDGhCo11Q2ZXrmyG5Nq0SJ/Jo7t+7bz+LjHqftpXZpVacbM5jMtcZhcLZRLdc9R1buD\nnvcQkYXRCsiYUC1Y4IZMP/po+OknuOCCeEd0uLQuqmcmP8OdFe5kaeulnHzsyfEOy5gcCyV57BOR\nq1T1JwARuRLYG92wjMnc7t3Qo4cbwDCtIF7Ah3XmtC6qlEAKYxqNsZaGyVNCSR6PAh+LyAne8+1A\nk+iFZEzmvvvOFcSvvtoVxE87Ld4RHS74KqqXr33Z5g43eVK2yUNVF+BG1D3ee74z6lEZk86ff8KT\nT8LChTBoEFx/fbwjOlzwVVR3nm9dVCZvC+Vqq+LAQ0BZoJC4C+ZVVdtGNzRjXEH8vfdcN9Vjj8Gw\nYXDMMfGO6nCTfptExwkdKVKwiF1FZfKFULqtxgEzgUVAAG8+j2gGZQy4gnirVlCkCEybBhUrxjui\nw/3y1y90mtCJ5VuX0+u6XtSvWB/x4x2JxkRYKMnjKFV9OuqRGOPZs8e1NIYOhVdegYcf9l9BfMOu\nDTw35Tm+Xf4tz1z1DKMajuKoQj6dRcqYKAjlIzlcRFqKyBkiclLaT9QjM/nS2LHufo0NG1xBvHlz\nfyWO3Qd3031Kdy567yKKH1Oc5Y8v56maT1niMPlOKC2P/cCrQFdctxW4bqty0QrK5D/r17uC+Pz5\n8MEHcMMN8Y7ov1ICKXw4/0N6JPTgmrOvIallEmVPLBvvsIyJm1CSR3vcjYJboh2MyX9SU+H9991A\nhq1auXs3/FQQV1XGrhhLpwmdKFG0BKMbjbZiuDGEljxWAPuiHYjJfxYudAmjUCFISHDdVX6StD6J\nDhM6sHH3Rvpc34dbz7vViuHGeEJJHnuBBSIyBTjgLbNLdc0R27MHnn8ePvoIevaEZs38VddYs2MN\nXSd3ZdLvk+hRuwfNL21OoQI26aYxwUL5RHzj/aRdnmuX6pojNm4ctGkDtWq5EXBLlIh3RP/asX8H\nPX/qyeD5g2lTvQ3v3fIexY4qFu+wjPGlUJLHL6o6N3iBiNwWpXhMHrVhgyuIJyW5GseNN8Y7on8d\nTD3Ie3Peo+f0ntx23m0sfmwxJYuVjHdYxvhaKJ0FA0WkUtoTEWkEdIteSCYvCQTcHeIXXwznnusu\nv/VL4lBVvlryFRXfqcj4VeOZ+OBEBt0+yBKHMSEIpeVRHxghIo2Bq3BDlfjsQkrjR4sWuYJ4gQIw\nZQpcdFG8I/pX4tpEOvzYgX0p+xhw6wCuL+fDwbKM8TFRzb58ISIVcHWPNcDdquqLIdlFREOJ38TW\n3r3wwgvw4Yfw0kvwyCP+KYiv2LqC/036H3P+nMNL177EAxc/YCPemnxHRFDVHF06mGnLQ0QWp1t0\nEq6ba7b3pX1xTk5s8qbvv3cF8Zo1Xcvj9NPjHZGzec9mXpj6Ap/98hntL2/PJ3d9wjGFfXRDiTG5\nTFbdVlYUNyHbsAHatYOff3Y1jptuindEzr5D+3hr9lv0TexLo4sakdwmmVOPOzXeYRmT62WaPFR1\ndQzjMLlUIAADB0K3bq576sMP4dhj4x2Vm1vjk0Wf8OzkZ6lWshqJzRM57+Tz4h2WMXmG3flkjtji\nxa4gDjB5MlSqlPX2sRI8t8bwe4Zz5VlXxjskY/IcSx4mbHv3wosvuhn9XnoJWrTwR0Hc5tYwJnZ8\n8JE3ucn48e6S29Wr/215xDtxbNi1gRajW3Dt0Gu58ZwbWdp6KQ0ubGCJw5goiurHXkTqisgyEVkh\nIp0z2eZtb/1CEakStHy1iCwSkfki8nM04zTZ27gRGjVyU8G+8w589ln8r6SyuTWMiZ+oJQ8RKQj0\nB+oCFYFGInJBum3qAeeqanmgJfBe0GoF6qhqFVWtEa04TdYCATecSKVKUKYMLFkCN98c35hSAikM\nTBrIef3OY+X2lSS1TKLPDX0ofkzx+AZmTD4SzZpHDWBl2lVbIvI5cAeQHLTN7cBQAFWdLSInikgJ\nVd3krbd+hzj65RfXLRUIwKRJboiReLK5NYzxj2gmj1LA2qDn64DLQtimFLAJ1/KYKCKpwPuq+kEU\nYzVB9u51hfAPPnCF8ZYt41/XsLk1jPGXaCaPUMcNyewb4EpVXS8ipwITRGSZqv6UfqMePXr887hO\nnTrUqVMn3DhNkB9+gNatoVo1d4f4GWfENx6bW8OYnEtISCAhISGixwxpbKsjOrBITaCHqtb1nncB\nAqraO2ibAUCCqn7uPV8G1A7qtkrbrjuwW1VfS7fcxraKkE2b3B3iM2e6gni9evGNJ/3cGh1rdbS5\nNYyJkEiMbRXNzoi5QHkRKSsiRYCGwOh024zGjdKblmx2qOomETlWRIp5y48DbgTSj7VlIiAQcN1T\nlSpB6dKuzhHPxHEw9SBvzXqLCv0rsG3fNhY/tpgXrnnBEocxPhO19r+qpojI48APQEFgsKomi0gr\nb/37qjpOROqJyEpgD/Cwt/vpwEivT7sQ8Kmq/hitWPOrJUtcQTwlBSZMgMqV4xeLqjJi6Qi6TOpC\n+ZPLM/HBiVQq4ZNb1o0xh4lat1UsWLfVkdm3zxXEBw50Q6e3bAkFC8YvnuC5NV694VWbW8OYKIvq\nkOwmb5owwd3od+mlsHAhlIzjpHk2t4YxuZclj3zir79cQXzGDFcQv+WWOMay5y9emvYSwxcPt7k1\njMml7M+8PC4QcAMYXnSRa2UsWRK/xPH79t9pM7YNFfpXQFVJbpNMl6u6WOIwJheylkcetnSpK4gf\nPAg//giXXBKfOBZtWkTvGb0Zv3I8raq2IrlNMqcX9ckUg8aYI2Itjzxo3z549lm4+mq47z5ITIx9\n4lBVpq2ZRr1P61H3k7pULlGZ39r+Rs/relriMCYPsJZHHjNxIjz6KFSp4u4Qj3VBPKABvvv1O3pN\n78XmvZvpWKsjIxuO5OhCR8c2EGNMVFnyyCP++gvat4effoL+/eHWW2N7/kOphxi+eDh9EvtwdKGj\n+d8V/+PuC+6mYIE4XgNsjIkaSx65XCAAQ4ZAly7w0EPuDvGiRWN3/j0H9zBo3iBem/ka5U8uz5s3\nvcn15a63QQuNyeMseeRiS5e6Lqr9+92AhlWqZL9PpGzdu5X+P/fnnTnvcHWZq/n63q+pXqp67AIw\nxsSVFcxzof37oVs3VxC/9143mGGsEscff//BU+Ofony/8qzduZafHv6JEfeOsMRhTD5jLY9cZtIk\n19qoXNndIV6qVGzOu3TzUvrM6MPo5aNpVqUZix9bTKnjY3RyY4zvWPLIJTZvdgXxqVNdQfy222Jz\n3plrZ9JrRi9mrZtF2xptWdV2lU33aoyx5OF3qq4g/r//wYMPujvEo10QD2iA8SvH03tGb/74+w86\nXN6Bz+75jGMLHxvdExtjcg1LHj6WnOy6qPbuhfHj3WCG0XQg5QCfLv6U12a+RpGCRehweQcaXtTQ\nZu4zxhzGvhV8aP9+6NkT3n0Xund308JGc8j0rXu3MmDuAPrP6c8lp1/C23Xf5tqzr7XLbY0xmbLk\n4TOTJ7vWxkUXwYIFcOaZ0TvXb9t/442Zb/DJ4k+48/w7+fGBH20CJmNMSCx5+MTmzdChA0yZAv36\nwR13RO9cs9fNpu/Mvkz5fQotLm3BktZLKFksjhN7GGNyHUsecaYKQ4dC585w//2uIF4sCtN1BzTA\nmOVj6DuzL+t2rqNdzXYMuWMIRYvE8HZ0Y0yeYckjjpYtc11Uu3fD999HpyC+79A+Pl74Ma/NfI0T\njj6BjrU6cvcFd1sR3BiTI/YNEgf790OvXu5+jeeegzZtIl8Q37xnM+/OeZd3577LZaUuY9Dtg7jq\nrKusCG6MiQhLHjE2ZYprbVSsCPPnQ+nSkT3+r1t/5fWZr/PFki9oULEBU5tO5fxTzo/sSYwx+Z4l\njxjZssUVxCdPhrffhjvvjNyxVZUZa2fQN7EviWsTebTaoyxrs4wSRUtE7iTGGBPEkkeUqcLHH0On\nTtC4cWQL4qmBVEYtG0XfxL5s3beVp2s+zfB7htud4MaYqLPkEUXLl7suqp07Ydw4qFo1Msfdc3AP\nQxYM4Y1Zb1DiuBJ0vqIzt1e43SZeMsbEjCWPKDhwwBXE+/Vzc4k//jgUisA7vXH3Rvr/3J/3k97n\n6jJXM+yuYdQqXSvnBzbGmDBZ8oiwhATX2qhQIXIF8aWbl/L6zNf5OvlrGl/UmJnNZ3LuSefm/MDG\nGHOELHlEyNat0LEjTJjgWhw5LYirKlPXTKVvYl/mrp9Lm+ptWPHECk459pTIBGyMMTlgySOHVGHY\nMFcQv+8+NzVsTgriKYEURiwdQd/Evuw+uJv2l7dnxL0jOLrQ0ZEL2hhjcsiSRw78+qvrotqxA777\nDqpVO/Jj7Tqwi8HzB/PmrDcpc2IZutfuzi3n3UIBsZmCjTH+Y8njCBw4AL17u/s1unaFJ5448oL4\nnzv/5O3ZbzNo/iCuL3c9Xzb4khqlakQ2YGOMiTBLHmGaNg1atYLy5WHePDjrrCM7zqJNi3ht5muM\nWT6Ghyo/xNwWczm7+NmRDdYYY6JEVDXeMRwxEdFYxb91q6tr/PCDa3HcdReEO0yUqjLxt4n0ndmX\nxZsW0/aytrSq2srmBDfGxJSIoKo5GujOWh7ZUIVPPnFXUt17ryuIH398+MeZtW4Wrb5rRWoglQ61\nOjD6vtEcVeioyAdsjDExYC2PLKxY4Qri27bBwIFQvfqRH+v37b+zfOtybjrnJhvZ1hgTV5FoeVjy\nyMCBA9CnD7z1FjzzDLRtG5k7xI0xxg+s2yoKfvrJFcTPOQeSkqBMmXhHZIwx/mPJw7NtmyuIjx/v\nWhx33x1+QdwYY/KLfH8HWlpBvGJFOOYYN2T6PfdY4jDGmKzk65bHihXw2GNuoqbRo6GG3ZtnjDEh\nyZctj4MH4eWX4fLL4eabYe5cSxzGGBOOfNfymD4dWraEs892SaNs2XhHZIwxuU++SR7btkHnzm5G\nv7fesrqGMcbkRJ7vtlKFTz+FCy+Eo45yd4jXr2+JwxhjciJPtzxWrXIF8U2b4Jtv4LLL4h2RMcbk\nDXmy5XHwIPTs6ZLFDTe42oYlDmOMiZw81/KYPt3dIV6mjBXEjTEmWqLa8hCRuiKyTERWiEjnTLZ5\n21u/UESqhLNvsO3b3VVUDRtCjx4wdqwlDmOMiZaoJQ8RKQj0B+oCFYFGInJBum3qAeeqanmgJfBe\nqPumUYXPPnN3iBcu7O4Qb9DAXwXxhISEeIcQEoszsnJDnLkhRrA4/SiaLY8awEpVXa2qh4DPgTvS\nbXM7MBRAVWcDJ4rI6SHuC0DduvDKKzBqFLzzDpx4YrRezpHLLf+hLM7Iyg1x5oYYweL0o2gmj1LA\n2qDn67xloWxTMoR9AbjuOjf6bc2aOY7XGGNMiKJZMA91oo0cdTB16pSTvY0xxhyJqE0GJSI1gR6q\nWtd73gUIqGrvoG0GAAmq+rn3fBlQGzg7u3295bl3JitjjIkjP08GNRcoLyJlgfVAQ6BRum1GA48D\nn3vJZoeqbhKRrSHsm+MXb4wx5shELXmoaoqIPA78ABQEBqtqsoi08ta/r6rjRKSeiKwE9gAPZ7Vv\ntGI1xhgTnlw9h7kxxpj48O3wJLG8wTAecYpIaRGZIiJLROQXEWnrtxiD1hUUkfkiMiZaMeY0ThE5\nUURGiEiyiCz1ukH9GGcX73e+WESGi8hR8YpTRM4XkZkisl9E2oezrx/ijOVnKCdxBq2P+ucoh7/z\n8D5Dquq7H1xX1UqgLFAYWABckG6besA47/FlwKxQ9/VJnKcDl3iPiwLLoxFnTmIMWv808Ckw2o+/\nc+/5UKCZ97gQcILf4vT2+Q04ynv+BdAkjnGeClQDXgLah7OvT+KMyWcop3EGrY/q5yinMYb7GfJr\nyyMmNxjGMc4SqrpRVRd4y3cDybj7W3wTI4CInIn7MhxEDi+rjlacInICcJWqfuitS1HVv/0WJ7AT\nOAQcKyKFgGOBP+MVp6puVtW5Xkxh7euHOGP4GcpRnBCzz9ERx3gknyG/Jo+Y3GAYAUca55nBG3hX\nlVUBZkc8wpy9lwBvAB2BQBRiCzWGrLY5E3dp92YRGSIi80TkAxE51mdxllLVbcBrwB+4qwh3qOrE\nOMYZjX3DFZFzRfkzBDmPMxafo5zEGPZnyK/JIyY3GEbAkcb5z34iUhQYATzp/fUUaUcao4jIrcBf\nqjo/g/WRlpP3shBwKfCuql6Ku3LvfxGMLf35QnHY+yUi5wBP4boVSgJFReT+yIX2Hzm5EiaWV9Hk\n+Fwx+AxBDuKM4ecoJ+9l2J8hvyaPP4HSQc9L47JoVtuc6W0Tyr6RcqRx/gkgIoWBr4FPVPUbH8ZY\nC7hdRH4HPgOuFZGPfRjnOmCdqs7xlo/AfRD8Fmc1IFFVt6pqCjAS9x7HK85o7BuuHJ0rRp8hyFmc\nsfoc5STG8D9D0SjcRKDwUwhYhfsLrQjZFyVr8m9RMtt9fRKnAB8Db/j1vUy3TW1gjF/jBKYB53mP\newC9/RYncAnwC3CM9/sfCrSJV5xB2/bgv4VoX32GsogzJp+hnMaZbl3UPkc5jTHcz1BU3/AcvhE3\n466eWAl08Za1AloFbdPfW78QuDSrff0WJ3Alrv9zATDf+6nrpxjTHaM2UbzaKgK/88rAHG/5SKJ0\ntVUE4uwELAEW45JH4XjFibtaaS3wN7AdV4spmtm+foszlp+hnL6fQceI6ucoh7/zsD5DdpOgMcaY\nsPm15mGMMcbHLHkYY4wJmyUPY4wxYbPkYYwxJmyWPIwxxoTNkocxxpiwWfIwviIiZUVkcRSO+5GI\n3JPJ8nUiUsR7fop3J3CGsYhIj7ShrEXkRW/I9QUiMklESqfbdq6IFBGRiA+ZERxHdstFZLWInJTd\nEOYi8qyI/Coiy0VksohUjHTcJu+w5GHyCyXzsX9SgGZhHCdNH1WtrKqXAN8A3dNWiMjZuOEeDmZx\n3pzI7JgZvc6054eAdqp6Ie7O9zYicoEX7+PesotVtQLwCjA6mvONmNzNkofxLREp543wWU1EmorI\nNyLyo4j8LiKPi0gHb/1MESnu7XOJiMzyWgQjReTE4ENmcBoF3gLaiUgon4d/jqGqu4KWFwW2BD2v\nC4xP93pOEZFEEblZROqIyFTvNa0SkV4i8qCI/Cwii0SknLdPWa8VsFBEJqZv3WQX439eaNZDmHcC\nHlfV/d76CUAiEK2BG00uZ8nD+JKIVMANztZE3fwDABcCdwHVgZeBnepGAJ0JPORt8zHQUVUr44YA\n6U72/gCme8dI/1f7Od7sb/NFZD5uqIfgUZFfFpE/gCZAr6D9biIoeYjIacB3QDdV/d5bfLF3vAuA\nB4FzVLUGbs6HJ7xt+gFDvNfzKfB2Nq9FcIkwOObD5rgIHsJcRI4HjlPV1ek2m4t7z405jCUP40en\n4bqBGqtqWs1BgSmqukdVtwA7gLTpPBcDZb0vwRNU9Sdv+VDg6hDOp7humo4c/plYpapV0n6AAfy3\n9dFVVc8CPsLN2YBXPzkz6Mu4CDAJl9QmBR17jqpu8rq2VgI/eMt/wQ1uB64rabj3+BPceE7ZvZbX\n08W8PniDMIYwj/eUB8bHLHkYP9oBrAGuSrf8QNDjQNDzAG5E0fRC/vJT1ZW4AfYahh7mfwzHtYjA\nxT09aN0h3F/xddPtE+rrCfdLPNPtMxrCXFV3Anu8Ok2wqrhEZsxhLHkYPzoI3A08JCKNvGVZfYEK\n/PMluF1E0v46fxBICOF8acd+GegQapAiUj7o6R24UV3BJYlxQesUV5A/X0Q6hXp8TyJwn/f4ftyw\n2cExh0xEBBgMLFXVN9OtfhV4W0SO9ra9HriCf1s9xvxHRn+tGRNvqqp7vRnYJniXuqa/iij947Tn\nTYAB3hSaq4CHM9mH9MtVdamIJOFqAVntk7bsFa82k+qd6zFveW3g2XSvR71EOFpEdgFLs4knbd0T\nwBAR6Qj8FfR6srp6LLOrra4AHgAWebUQgGdU9XtV7edddLBYRFKBDcDtqnoAYzJgQ7IbE0Eicibw\nvqreEu9YjIkmSx7GGGPCZjUPY4wxYbPkYYwxJmyWPIwxxoTNkocxxpiwWfIwxhgTNksexhhjwmbJ\nwxhjTNj+D3ist4aa/CKkAAAAAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f4b195450d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The height of the tower is 4.91 meters\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib.pyplot import plot, show, xlabel, ylabel, title\n", + "from numpy import arange\n", + "y1 = 0.10#\n", + "Y1 = 0.10/(1-0.10)#\n", + "y2 = 0.001#\n", + "Y2 = y2#\n", + "mass_flowrate_gas = 0.95# #mass flow rate in kg/m**2.sec\n", + "mass_percent_air = (0.9*29/(0.1*17+0.9*29))*100#\n", + "mass_flowrate_air = (mass_percent_air*mass_flowrate_gas)##in kg/m**2.sec\n", + "Gm = (mass_flowrate_air/29)#\n", + "Lm = 0.65/18# #Lm is in kmol/m**2.sec\n", + "#A mass balance between a plane in the tower where the compositions are X and Y and the top of the tower gives:\n", + "\n", + "#Y = 1.173X+0.001\n", + "X = arange(0,0.159,0.001)\n", + "Y=[]\n", + "for x in X:\n", + " Y.append(1.173*x+0.001)\n", + "plot(X,Y)#\n", + "\n", + "x=[0.021,0.031,0.042,0.053,0.079,0.106,0.159]\n", + "\n", + "PG = [12,18.2,24.9, 31.7, 50.0, 69.6, 114.0]\n", + "i=0\n", + "Y1=[]\n", + "while i<7:\n", + " Y1.append(PG[(i)]/(760-PG[(i)]))\n", + " i=i+1\n", + "\n", + "plot(x,Y1)#\n", + "#xlabel(\"Area under the curve is 3.82m\"%(\"kmolNH3/kmolH2O\"%(\"kmolNH3/kmol air\"\n", + "\n", + " \n", + "title(\"Operating and equilibrium lines\")\n", + "xlabel(\"kmol NH3/kmol H2O\")\n", + "ylabel(\"kmol NH3/kmol air\")\n", + "show()\n", + "#If the equilibrium line is assumed to be straight, then:\n", + "#Gm(Y2 − Y1) = KGaZdeltaPlm\n", + "\n", + "#Top driving force\n", + "deltaY2 = 0.022#\n", + "#Bottom driving force\n", + "deltaY1 = 0.001#\n", + "deltaYlm = 0.0068#\n", + "Z = (0.0307*0.11)/(0.001*0.688)#\n", + "print \"The height of the tower is %.2f meters\"%(Z)#" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Page No 708 Example 12.6" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Number of theoretical plates = 9\n", + "\n", + " LHS = 0.98\n", + "\n", + " 1/A = 1.00\n", + "\n", + " Gm/Lm = 0.333\n", + "\n", + " Actual/minimum Gm/Lm = 1.39\n", + "\n", + " Actual Gm/Lm = 0.656\n", + "\n", + " 1/A = mGm/Lm = 1.968\n", + "\n", + " The actual number of plates are : 17.0\n" + ] + } + ], + "source": [ + "from sympy import symbols, solve, log\n", + "print \"\\n Number of theoretical plates = %d\"%((30*0.3))#\n", + "\n", + "#At the bottom of the tower:\n", + "#Flowrate of steam = Gm (kmol/m**22s)\n", + "#Mole ratio of pentane in steam = Y1, and\n", + "#Mole ratio of pentane in oil = X1\n", + "X1 = 0.001#\n", + "\n", + "#At the top of the tower:\n", + "#exit steam composition = Y2\n", + "#inlet oil composition = X2\n", + "X2 = 0.06#\n", + "#flowrate of oil = Lm (kmol/m**2.sec)\n", + "\n", + "#The minimum steam consumption occurs when the exit steam stream is in equilibrium with the inlet oil, that is when:\n", + "#The equilibrium relation for the system may be taken as Ye = 3.0X, where Ye and X are expressed in mole ratios of pentane in the gas and liquid phases respectively.\n", + "Ye2 = X2*3#\n", + "#Lmin(X2 − X1) = Gmin(Y2 − Y1)\n", + "#If Y1 = 0, that is the inlet steam is pentane-free, then:\n", + "ratio_min = (X2 - X1)/Ye2#\n", + "\n", + "\n", + "#The operating line may be fixed by trial and error as it passes through the point (0.001, 0), and 9 theoretical plates are required for the separation. Thus it is a matter of selecting the operating line which, with 9 steps, will give X2 = 0.001 when X1 = 0.06. This is tedious but possible, and the problem may be better solved analytically since the equilibrium line is straight.\n", + "\n", + "#let x = 1/A\n", + "#for a stripping operation\n", + "LHS = (X2-X1)/(X2)#\n", + "print \"\\n LHS = %0.2f\"%(LHS)#\n", + "x = symbols('x')\n", + "x1 = solve((x**10-x)-LHS*(x**(10)-1),x)[0]\n", + "print \"\\n 1/A = %.2f\"%(x1)#\n", + "#A = Lm/mGm\n", + "print \"\\n Gm/Lm = %.3f\"%(x1/3)#\n", + "print \"\\n Actual/minimum Gm/Lm = %.2f\"%(0.457/0.328)#\n", + "\n", + "#If (actual Gm/Lm)/(min Gm/Lm) = 2,\n", + "print \"\\n Actual Gm/Lm = %.3f\"%(.328*2)#\n", + "x2 = 3*(0.656)#\n", + "print \"\\n 1/A = mGm/Lm = %.3f\"%(3*0.656)#\n", + "\n", + "#y = (1.968)**(N+1)\n", + "y = symbols('y')\n", + "y1 = solve(0.983*(y-1)-(y-1.968),y)[0]\n", + "N = log(y1)/log(1.968)-1#\n", + "print \"\\n The actual number of plates are : %.1f\"%(ceil(N/0.3))#" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/SaleemAhmed/Chapter10_1.ipynb b/sample_notebooks/SaleemAhmed/Chapter10_1.ipynb new file mode 100644 index 00000000..70597c5b --- /dev/null +++ b/sample_notebooks/SaleemAhmed/Chapter10_1.ipynb @@ -0,0 +1,441 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter10 - Optical Fiber Systems" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.1 Page No: 349" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage required to have a pi radian phase change = 4.56 volt\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "r=30.8*10**-12## electro optice coefficient in m/V\n", + "L=3*10**-2## length in m\n", + "y=1.3*10**-6## wavelength in m\n", + "n=2.1#\n", + "d=30*10**-6## distance between the electrodes in m\n", + "V=(y*d)/((n)**3*r*L)## voltage required to have a pi radian phase change in volt\n", + "print \"The voltage required to have a pi radian phase change = %0.2f volt\"%( V)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.2 Page No: 350" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total channel loss =31 dB\n" + ] + } + ], + "source": [ + "a_fc=4## fider cable loss in dB/km\n", + "aj=0.7## splice loss in db/km\n", + "L=5## length in km\n", + "a_cr1=4## connector losses\n", + "a_cr2=3.5## connector losses\n", + "CL=(a_fc+aj)*L+(a_cr1+a_cr2)## total channel loss in dB\n", + "print \"The total channel loss =%d dB\"%( CL)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.3 Page No: 350" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "with mode coupling, the total rms broadening = 1.73 ns\n", + "\n", + " The dispersion equalization penalty = 1.84 dB\n", + "\n", + " without mode coupling, the total rms broadening = 6.00 dB\n", + "\n", + " without mode coupling, equalization penalty = 0.03 dB\n", + "\n", + " without mode coupling,dispersion equalization penalty with 125 Mb/s = 83.98 dB\n", + "\n", + " with mode coupling,dispersion equalization penalty with 125 Mb/s = 0.28 dB\n", + "\n", + " The answer is wrong in the textbook\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "p=0.5*10**-9## pulse broadening in s/km\n", + "L=12## length in km\n", + "Pt=p*sqrt(L)## with mode coupling, the total rms broadening in s\n", + "BT=20*10**6##\n", + "DL=2*(2*Pt*BT*sqrt(2))**4## dispersion equalization penalty in dB\n", + "Pt1=p*L## without mode coupling, the total rms broadening in s\n", + "DL1=2*(2*Pt1*BT*sqrt(2))**4## without mode coupling, equalization penalty in dB\n", + "DL2=2*(2*Pt1*150*10**6*sqrt(2))**4## without mode coupling,dispersion equalization penalty with 125 Mb/s\n", + "DL3=2*(2*Pt*125*10**6*sqrt(2))**4## with mode coupling,dispersion equalization penalty with 125 Mb/s\n", + "print \"with mode coupling, the total rms broadening = %0.2f ns\"%( Pt*10**9)#\n", + "print \"\\n The dispersion equalization penalty = %0.2f dB\"%( DL*10**4)#\n", + "print \"\\n without mode coupling, the total rms broadening = %0.2f dB\"%( Pt1*10**9)#\n", + "print \"\\n without mode coupling, equalization penalty = %0.2f dB\"%( DL1)#\n", + "print \"\\n without mode coupling,dispersion equalization penalty with 125 Mb/s = %0.2f dB\"%( DL2)#\n", + "print \"\\n with mode coupling,dispersion equalization penalty with 125 Mb/s = %0.2f dB\"%( DL3)#\n", + "print \"\\n The answer is wrong in the textbook\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.4 Page No: 351" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The length when system operating at 25 Mbps = 78.89 km\n", + "\n", + " The length when system operating at 350 Mbps = 56.67 km\n" + ] + } + ], + "source": [ + "Pi=-2.5## mean optical power launched into the fiber in dBm\n", + "Po=-45## mean output optical power available at the receiver in dBm\n", + "a_fc=0.35## fider cable loss in dB/km\n", + "aj=0.1## splice loss in db/km\n", + "a_cr=1## connector losses\n", + "Ma=6## safety margin in dB\n", + "L=(Pi-Po-a_cr-Ma)/(a_fc+aj)## length in km when system operating at 25 Mbps\n", + "Po1=-35## mean output optical power available at the receiver in dBm\n", + "L1=(Pi-Po1-a_cr-Ma)/(a_fc+aj)## length in km when system operating at 350 Mbps\n", + "print \"The length when system operating at 25 Mbps = %0.2f km\"%( L)#\n", + "print \"\\n The length when system operating at 350 Mbps = %0.2f km\"%( L1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.5 Page No: 351" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fider loss = 20.00 dB\n", + "\n", + " The total splicing loss = 3.20 dB\n", + "\n", + " The fangle effects & future splice = 5.00 dB\n", + "\n", + " The total attenuation = 29.20 dB\n", + "\n", + " The excess power margin = 2.80 dB\n", + "\n", + " hence the system can operate with small excess power margin\n" + ] + } + ], + "source": [ + "Tx=-80## transmitter output in dBm\n", + "Rx=-40## receiver sensitivity in dBm\n", + "sm=32## system margin in dB\n", + "L=10## in km\n", + "fl=2*L## fider loss in dB\n", + "cl=1## detector coupling loss in dB\n", + "tl=0.4*8## total splicing loss in dB\n", + "ae=5## angle effects & future splice in dB\n", + "ta=29.2## total attenuation in dB\n", + "Ep=2.8## excess power margin in dB\n", + "print \"The fider loss = %0.2f dB\"%( fl)#\n", + "print \"\\n The total splicing loss = %0.2f dB\"%( tl)#\n", + "print \"\\n The fangle effects & future splice = %0.2f dB\"%( ae)#\n", + "print \"\\n The total attenuation = %0.2f dB\"%( ta)#\n", + "print \"\\n The excess power margin = %0.2f dB\"%( Ep)#\n", + "print \"\\n hence the system can operate with small excess power margin\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.6 Page No: 352" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The max transmission length when transmission star coupler is used = 1.99 km\n", + "\n", + " The max transmission length when reflection star coupler is used = 1.24 km\n" + ] + } + ], + "source": [ + "from math import log\n", + "Lc=1## connector loss in db\n", + "Ls=5## star coupler insertion loss in dB\n", + "af=2## fider loss in dB\n", + "Ps=-14## transmitted power in dBm\n", + "Pr=-49## receiver sensitivity in dBm\n", + "sm=6## system margin in dB\n", + "N=16#\n", + "L=(Ps-Pr-Ls-4*Lc-(10*log(N))/log(10)-sm)/(2*af)## max transmission length in km when transmission star coupler is used\n", + "N1=32#\n", + "L1=(Ps-Pr-Ls-4*Lc-(10*log(N1))/log(10)-sm)/(2*af)## max transmission length in km when reflection star coupler is used\n", + "print \"The max transmission length when transmission star coupler is used = %0.2f km\"%( L)#\n", + "print \"\\n The max transmission length when reflection star coupler is used = %0.2f km\"%( L1)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.7 Page No: 353" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The system rise time = 18.51 ns\n", + "\n", + " The max bit rate for NRZ coding = 37.81 Mbit/s\n", + "\n", + " The max bit rate for RZ coding = 18.90 Mbit/s\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "y=860*10**-9## wavelength in m\n", + "L=5000## length in m\n", + "X=0.024#\n", + "dy=20*10**-9## spectral width in m\n", + "dts=6*10**-9## silica optical link rise time in s\n", + "dtr=8*10**-9## detector rise in s\n", + "c=3*10**8## speed of light in m/s\n", + "dtm=-(L*dy*X)/(c*y)## material dispersion delay time in s\n", + "id=2.5*10**-12## intermodel dispersion in s/m\n", + "dti=id*L## intermodel dispersion delay time\n", + "dtsy=sqrt((dts**2)+(dtr**2)+(dtm**2)+(dti**2))## system rise time in s\n", + "Br_max=0.7/dtsy## max bit rate for NRZ coding in bit/s\n", + "Br_max1=0.35/dtsy## max bit rate for RZ coding in bit/s\n", + "print \"The system rise time = %0.2f ns\"%( dtsy*10**9)#\n", + "print \"\\n The max bit rate for NRZ coding = %0.2f Mbit/s\"%( Br_max/10**6)#\n", + "print \"\\n The max bit rate for RZ coding = %0.2f Mbit/s\"%( Br_max1/10**6)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.8 Page No: 353" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The transmission distance for Si fiber = 71.43 m\n", + "\n", + " The transmission distance for GRIN fiber = 420.00 m\n" + ] + } + ], + "source": [ + "Br=50*10**6## data rate in b/s\n", + "c=3*10**8## speed of light in m/s\n", + "n1=1.47## \n", + "dl=0.02## \n", + "n12=n1*dl## the difference b/w n1 and n2\n", + "L_si=(0.35*c)/(n12*Br)## transmission distance for Si fiber\n", + "L_GI=(2.8*c*n1**2)/(2*n1*n12*Br)## transmission distance for GRIN fiber\n", + "print \"The transmission distance for Si fiber = %0.2f m\"%( L_si)#\n", + "print \"\\n The transmission distance for GRIN fiber = %0.2f m\"%( L_GI)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.9 Page No: 353" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The material dispersion limited transmission distance =627 m\n" + ] + } + ], + "source": [ + "Br=20.0*10**6## data rate in b/s\n", + "c=3*10**8## speed of light in m/s\n", + "y=86*10**-9## wavelength in m\n", + "dy=30*10**-9## spectral width in m\n", + "X=0.024#\n", + "Tb=1/Br#\n", + "Lmax=(0.35*Tb*c*y)/(dy*X)## material dispersion limited transmission distance for RZ coding in m\n", + "print \"The material dispersion limited transmission distance =%d m\"%( Lmax)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex:10.10 Page No: 354" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The material dispersion limited distance = 12.54*10**10*1/Br m\n", + "\n", + " The model dispersion limited distance = 2.10*10**10*1/Br m\n", + "\n", + " The attenuation limited distance =60-20log(Br) km\n" + ] + } + ], + "source": [ + "from __future__ import division\n", + "y=860*10**-9## wavelength in m\n", + "c=3*10**8## speed of light in m/s\n", + "n1=1.47## \n", + "dl=0.02## \n", + "n12=n1*dl## the difference b/w n1 and n2\n", + "La=1/1000## loss a in dB/m\n", + "Pr=-65## receiver power in dB\n", + "Pt=-5## transmitted power in dB\n", + "dy=30*10**-9## line width in m\n", + "X=0.024#\n", + "Lmax=(0.35*c*y)/(dy*X)## material dispersion limited distance for RZ coding in m\n", + "L_GI=(1.4*c*n1)/(n12)## model dispersion limited distance for RZ coding in m\n", + "L_At=(Pt-Pr)/(La)## attenuation limited distance for RZ coding in m \n", + "print \"The material dispersion limited distance = %0.2f*10**10*1/Br m\"%( Lmax/10**10)#\n", + "print \"\\n The model dispersion limited distance = %0.2f*10**10*1/Br m\"%( L_GI/10**10)#\n", + "print \"\\n The attenuation limited distance =%d-20log(Br) km\"%( L_At/10**3)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/sample_notebooks/VivekMaindola/chap3.ipynb b/sample_notebooks/VivekMaindola/chap3.ipynb new file mode 100644 index 00000000..299b3f5a --- /dev/null +++ b/sample_notebooks/VivekMaindola/chap3.ipynb @@ -0,0 +1,617 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter No.3 : The cellular concept system design fundamentals" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.1 Page No.61" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " The number of channels available per cell for 4-cell reuse system = 165 channels\n", + "\n", + " One control channel and 160 voice channels would be assigned to each cell.\n", + "\n", + " \n", + " The number of channels available per cell for 7-cell reuse system = 95 channels\n", + "\n", + " Each cell would have one control channel, four cells would have 90 voice channels and three cells would have 91 voice channels.\n", + "\n", + " \n", + " The number of channels available per cell for 12-cell reuse system = 55 channels\n", + "\n", + " Each cell would have one control channel, eight cells would have 53 voice channels and four cells would have 54 voice channels.\n" + ] + } + ], + "source": [ + "from math import ceil\n", + "# To compute the number of channels available per cell for a)four-cell reuse system a)seven-cell reuse system a)12-cell reuse system\n", + "\n", + "# Given data\n", + "B=33*10**6# # Total bandwidth allocated to particular FDD system in Hz\n", + "Bc=25*10**3# # Bandwidth per channel in Hz\n", + "Nc=2# # Number of simplex channels\n", + "Bc=Bc*Nc# # Channel bandwidth in Hz\n", + "\n", + "Ntotal=B/Bc# # Total number of channels\n", + "\n", + "#a) To compute the number of channels available per cell for four-cell reuse system\n", + "N=4# # frequency reuse factor\n", + "chpercell=Ntotal/N# # number of channels available per cell for four-cell reuse system\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n The number of channels available per cell for 4-cell reuse system = %0.0f channels\"%(chpercell)\n", + "print \"\\n One control channel and 160 voice channels would be assigned to each cell.\"\n", + "\n", + "# b) To compute the number of channels available per cell for seven-cell reuse system\n", + "N=7# # frequency reuse factor\n", + "chpercell=ceil(Ntotal/N)# # number of channels available per cell for seven-cell reuse system\n", + "\n", + "# Answer is varrying due to round-off error\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n The number of channels available per cell for 7-cell reuse system = %0.0f channels\"%(chpercell)\n", + "print \"\\n Each cell would have one control channel, four cells would have 90 voice channels and three cells would have 91 voice channels.\"\n", + "\n", + "# c) To compute the number of channels available per cell for 12-cell reuse system\n", + "N=12# # frequency reuse factor\n", + "chpercell=Ntotal/N# # number of channels available per cell for seven-cell reuse system\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n The number of channels available per cell for 12-cell reuse system = %0.0f channels\"%(chpercell)\n", + "print \"\\n Each cell would have one control channel, eight cells would have 53 voice channels and four cells would have 54 voice channels.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.2 Page No.72" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Signal to noise ratio for n=4 with frequency reuse factor N=7 = 18.66 dB\n", + "\n", + " Signal to noise ratio for n=3 with frequency reuse factor N=7 = 12.05 dB\n", + "\n", + " Signal to noise ratio for n=3 with frequency reuse factor N=12 = 15.56 dB\n", + "\n", + " Since SIR is for n=3 with frequency reuse factor N=7 greater than the minimum required, so N=12 is used.\n" + ] + } + ], + "source": [ + "from math import sqrt, log10\n", + "from __future__ import division\n", + "# To find frequency reuse factor for path loss exponent (n) a)n=4 b)n=3\n", + "\n", + "# Given data\n", + "SIdB=15# # Signal to interference(dB)\n", + "io=6# # Number of cochannel cell\n", + "\n", + "# For n=4\n", + "n1=4# # Path loss exponent\n", + "N1=7# # First consideration: frequency reuse factor N=7\n", + "DR1=sqrt(3*N1)# # Co-channel reuse ratio\n", + "si1=(1/io)*(DR1)**n1# # Signal to interference\n", + "sidB1=10*log10(si1)# # Signal to interference(dB)\n", + "\n", + "# For n=3\n", + "n2=3# # Path loss exmponent\n", + "si=(1/io)*(DR1)**n2# # Signal to interference for first consideration: frequency reuse factor N=7\n", + "sidB=10*log10(si)# # Signal to interference(dB)\n", + "\n", + "N2=12# # second consideration : frequency reuse factor N=12 since sidB<SIdB \n", + "DR2=sqrt(3*N2)# # Co-channel reuse ratio\n", + "si2=(1/io)*(DR2)**n2# # Signal to interference\n", + "sidB2=10*log10(si2)# # Signal to interference(dB)\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Signal to noise ratio for n=4 with frequency reuse factor N=7 = %0.2f dB\"%(sidB1)\n", + "print \"\\n Signal to noise ratio for n=3 with frequency reuse factor N=7 = %0.2f dB\"%(sidB)\n", + "print \"\\n Signal to noise ratio for n=3 with frequency reuse factor N=12 = %0.2f dB\"%(sidB2)\n", + "print \"\\n Since SIR is for n=3 with frequency reuse factor N=7 greater than the minimum required, so N=12 is used.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.4 Page No.80" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Total number of users for 1 channel = 1\n", + "\n", + " Total number of users for 5 channel = 11\n", + "\n", + " Total number of users for 10 channel = 40\n", + "\n", + " Total number of users for 20 channel = 111\n", + "\n", + " Total number of users for 100 channel = 809\n" + ] + } + ], + "source": [ + "# To find number of users for Number of channels (C) a)C=1 b)C=5 c)C=10 d)C=20 e)C=100\n", + "\n", + "# Given data\n", + "GOS=0.005# #G rade of Service\n", + "Au=0.1# # Traffic intensity per user\n", + "\n", + "# a)To find number of users for C=1\n", + "C1=1# # Number of channels\n", + "A1=0.005# # Total traffic intensity from Erlangs B chart\n", + "U1=(A1/Au)# # Number of users\n", + "U1=1# # Since one user could be supported on one channel\n", + "\n", + "# b)To find number of users for C=5\n", + "C2=5# # Number of channels\n", + "A2=1.13# # Total traffic intensity from Erlangs B chart\n", + "U2=round(A2/Au)# # Number of users\n", + "\n", + "# c)To find number of users for C=10\n", + "C3=10# # Number of channels\n", + "A3=3.96# # Total traffic intensity from Erlangs B chart\n", + "U3=round(A3/Au)# # Number of users\n", + "\n", + "# Answer is varrying due to round off error\n", + "\n", + "# d)To find number of users for C=20\n", + "C4=20# # Number of channels\n", + "A4=11.10# # Total traffic intensity from Erlangs B chart\n", + "U4=round(A4/Au)# # Number of users\n", + "\n", + "# Answer is varrying due to round off error\n", + "\n", + "# e)To find number of users for C=100\n", + "C5=100# # Number of channels\n", + "A5=80.9# # Total traffic intensity from Erlangs B chart\n", + "U5=round(A5/Au)# # Number of users\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Total number of users for 1 channel = %0.0f\"%(U1)\n", + "print \"\\n Total number of users for 5 channel = %0.0f\"%(U2)\n", + "print \"\\n Total number of users for 10 channel = %0.0f\"%(U3)\n", + "print \"\\n Total number of users for 20 channel = %0.0f\"%(U4)\n", + "print \"\\n Total number of users for 100 channel = %0.0f\"%(U5)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.5 Page No.83" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Total number of users in system A = 47280\n", + "\n", + " The percentage market penetration of system A = 2.36\n", + "\n", + " \n", + " Total number of users in system B = 44100\n", + "\n", + " The percentage market penetration of system B = 2.205\n", + "\n", + " \n", + " Total number of users in system C = 43120\n", + "\n", + " The percentage market penetration of system C = 2.156\n", + "\n", + " \n", + " Total number of users in all 3 systems = 134500\n", + "\n", + " The combined Market penetration percentage of all systems = 6.725\n" + ] + } + ], + "source": [ + "# To find number of users for a)system A b)system B c)system C\n", + "\n", + "# Given data\n", + "GOS=0.02# # Grade of Service (Probability of bloacking)\n", + "lamda=2# # Average calls per hour\n", + "H=(3/60)# # Call duration in seconds\n", + "\n", + "Au=lamda*H# # Traffic intensity per user\n", + "\n", + "# a)To find number of users for System A\n", + "C1=19# # Number of channels used\n", + "A1=12# # Traffic intensity from Erlang B chart\n", + "U1=round(A1/Au)# # Number of users per cell\n", + "cells1=394\n", + "TU1=U1*cells1# # Total number of users\n", + "MP1=TU1/(2*10**6)*100# # Market penetration percentage\n", + "\n", + "# b)To find number of users for System B\n", + "C2=57# # No. of channels used\n", + "A2=45# # Traffic intensity from Erlang B chart\n", + "U2=round(A2/Au)# # Number of users per cell\n", + "cells2=98\n", + "TU2=U2*cells2# # Total no. of users\n", + "MP2=TU2/(2*10**6)*100# # Market penetration percentage\n", + "\n", + "# c)To find number of users for System C\n", + "C3=100# # Number of channels used\n", + "A3=88# # traffic intensity from Erlang B chart\n", + "U3=round(A3/Au)# # Number of users per cell\n", + "cells3=49\n", + "TU3=U3*cells3# # Total no. of users\n", + "MP3=TU3/(2*10**6)*100# # Market penetration percentage\n", + "\n", + "TU=TU1+TU2+TU3# # Total number of users in all 3 systems\n", + "MP=TU/(2*10**6)*100# # Combined Market penetration percentage\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Total number of users in system A = %0.0f\"%(TU1)\n", + "print \"\\n The percentage market penetration of system A = %0.2f\"%(MP1)\n", + "print \"\\n \\n Total number of users in system B = %0.0f\"%(TU2)\n", + "print \"\\n The percentage market penetration of system B = %0.3f\"%(MP2)\n", + "print \"\\n \\n Total number of users in system C = %0.0f\"%(TU3)\n", + "print \"\\n The percentage market penetration of system C = %0.3f\"%(MP3)\n", + "print \"\\n \\n Total number of users in all 3 systems = %0.0f\"%(TU)\n", + "print \"\\n The combined Market penetration percentage of all systems = %0.3f\"%(MP)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.6 Page No.84" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Number of cells in given system = 31 cells\n", + "\n", + " \n", + " Number of channels per cell in given system = 95 channels/cell\n", + "\n", + " \n", + " Traffic intensity in given system = 84 Erlangs/cell\n", + "\n", + " \n", + " Maximum carried traffic in given system = 2604 Erlangs\n", + "\n", + " \n", + " Total number of users = 86800 users\n", + "\n", + " \n", + " Number of mobiles per unique channel = 130 mobiles/channel\n", + "\n", + " \n", + " Theoretically maximum number of served mobiles is the number of available channels in the system.\n", + "\n", + " Theoretical Maximum number of users could be served at one time = 2945 users\n", + "It is 3.4% of customer base.\n" + ] + } + ], + "source": [ + "# To find a)Number of cells in given area b)Number of channels/cell c)Traffic intensity per cell d)Maximum carried traffic e)Total number of users for 2% GOS f) Number of mobiles per unique channel g)Maximum number of users could be served at one time\n", + "\n", + "# Given data\n", + "Area=1300# # Total coverage area in m**2\n", + "R=4# # Radius of cell in m\n", + "N=7# # Frequecy reuse factor\n", + "S=40*10**6# # Allocated spectrum in Hz\n", + "Ch=60*10**3# # Channel width in Hz\n", + "\n", + "# a)Number of cells\n", + "CA=2.5981*R**2# # Area of hexagonal cell in m**2\n", + "Nc=round(Area/CA)# # Number of cells\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Number of cells in given system = %0.0f cells\"%(Nc)\n", + "\n", + "# b)Number of channels/cell\n", + "C1=round(S/(Ch*N))# # Number of channels\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Number of channels per cell in given system = %0.0f channels/cell\"%(C1)\n", + "\n", + "# c) Traffic intensity per cell\n", + "C1=95# # Number of channels from b)\n", + "GOS=0.02# # Grade of service\n", + "A=84# # Traffic intensity from Erlang B chart\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Traffic intensity in given system = %0.0f Erlangs/cell\"%(A)\n", + "\n", + "# d)Maximum carried traffic\n", + "traffic=Nc*A# # Maximum carried traffic\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Maximum carried traffic in given system = %0.0f Erlangs\"%(traffic)\n", + "\n", + "# e)Total number of users for 2% GOS \n", + "trafficperuser=0.03# # Given traffic per user\n", + "U=traffic/trafficperuser# # Total number of users\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Total number of users = %0.0f users\"%(U)\n", + "\n", + "# f) Number of mobiles per unique channel\n", + "C=666# # Number of channels\n", + "mobilesperchannel=round(U/C)# # Number of mobiles per unique channel\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Number of mobiles per unique channel = %0.0f mobiles/channel\"%(mobilesperchannel)\n", + "\n", + "# g)Maximum number of users could be served at one time\n", + "print \"\\n \\n Theoretically maximum number of served mobiles is the number of available channels in the system.\"\n", + "C=C1*Nc# # Maximum number of users could be served at one time\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Theoretical Maximum number of users could be served at one time = %0.0f users\"%(C)\n", + "print \"It is 3.4% of customer base.\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.7 Page 85" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Number of users per square km in given system = 62 users/sq km\n", + "\n", + " \n", + " Percentage of probability that delayed call have to wait longer than t=10 sec = 56.29 percent\n", + "\n", + " \n", + " Percentage of probability that call is delayed more than 10 sec = 2.81 percent\n" + ] + } + ], + "source": [ + "from math import exp\n", + "# To find a)number of users per square km b)probability that delayed call have to wait longer than t=10sec c)probability that call is delayed more than 10 sec\n", + "\n", + "# Given data\n", + "R=1.387# # Radius of cell in m\n", + "Area=2.598*R**2# # Area of hexagonal cell in m**2\n", + "cellpercluster=4# # Number of cells/cluster\n", + "channels=60# # Number of channels\n", + "\n", + "channelspercell=channels/cellpercluster# # Number of channels per cell\n", + "\n", + "# a)To find number of users per square km\n", + "A=0.029# # Traffic intensity per user\n", + "delayprob=0.05# # Grade of service\n", + "traffic=9# # Traffic intensity from Erlang chart C\n", + "U1=traffic/A# # Total number of users in 5sq.km.\n", + "U=round(U1/Area)# # Number of users per square km\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Number of users per square km in given system = %0.0f users/sq km\"%(U)\n", + "\n", + "# b)To find the probability that delayed call have to wait longer than t=10sec\n", + "lamda=1# # Holding time\n", + "H1=A/lamda# # Duration of call\n", + "H=H1*3600# # Duration of call in second\n", + "t=10\n", + "Pr=exp(-(channelspercell-traffic)*t/H)*100# # probability that delayed call have to wait longer than t=10sec.\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Percentage of probability that delayed call have to wait longer than t=10 sec = %0.2f percent\"%(Pr)\n", + "\n", + "# c)To find the probability that call is delayed more than 10 sec\n", + "Pr10=delayprob*Pr# # probability that call is delayed more than 10 sec\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n \\n Percentage of probability that call is delayed more than 10 sec = %0.2f percent\"%(Pr10)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.8 Page 89" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Number of channels without use of microcell = 300 channels\n", + "\n", + " \n", + " Number of channels with the use of lettered microcells = 660 channels\n", + "\n", + " \n", + " Number of channels if all base stations are replaced by microcells = 1020 channels\n" + ] + } + ], + "source": [ + "# To find number of channels in 3 km by 3 km square centered around A in Figure 3.9 for a)without use of microcell b)with the use of lettered microcells c)all base stations are replaced by microcells\n", + "\n", + "# Given data\n", + "R=1# # Cell radius in km\n", + "r=0.5# # Micro-cell radius in km\n", + "Nc=60# # Number of channels in base station\n", + "\n", + "# a)To find number of channels without use of microcell\n", + "Nb1=5# # Number of base stations in given area\n", + "N1=Nb1*Nc# # Number of channels without use of microcell\n", + "\n", + "# b)To find number of channels with the use of lettered microcells\n", + "Nb2=6# # Number of lettered microcells\n", + "Nb2=Nb1+Nb2# # Total number of base stations in given area\n", + "N2=Nb2*Nc# # Number of channels with the use of lettered microcells\n", + "\n", + "# c)To find number of channels if all base stations are replaced by microcells\n", + "Nb3=12# # Number of all the microcells\n", + "Nb3=Nb1+Nb3# # Total number of base stations in given area\n", + "N3=Nb3*Nc# # Number of channels if all base stations are replaced by microcells\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Number of channels without use of microcell = %0.0f channels\"%(N1)\n", + "print \"\\n \\n Number of channels with the use of lettered microcells = %0.0f channels\"%(N2)\n", + "print \"\\n \\n Number of channels if all base stations are replaced by microcells = %0.0f channels\"%(N3)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3.9 Page 92" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Cell capacity of unsectored system = 1326 calls/hour\n", + "\n", + " \n", + " Cell capacity of 120 degree sectored system = 1008 calls/hour\n", + "\n", + " \n", + " Decrease in cell capacity in 120 degree sectored system = 24 percent\n" + ] + } + ], + "source": [ + "# To analyze trunking efficiency capacity of sectoring and unsectoring\n", + "\n", + "# Given data\n", + "H=2/60# # Average call duration in hour\n", + "GOS=0.01# # Probability of blocking\n", + "\n", + "# Unsectored system\n", + "C1=57# # Number of traffic channels per cell in unsectored system\n", + "A=44.2# # Carried traffic in unsectored system\n", + "calls1=1326# # Number of calls per hour in unsectored system from Erlangs B table\n", + "\n", + "# 120 degree sectored system\n", + "C2=C1/3# # Number of traffic channels per antenna sector in 120 degree sectored system\n", + "calls2=336# # Number of calls per hour in 120 degree sectored system from Erlangs B table\n", + "Ns1=3# # Number of sectors\n", + "capacity=Ns1*calls2# # Cell capacity or number of calls handled by system per hour\n", + "\n", + "dif=calls1-capacity# # decrease in cell capacity in 120 degree sectored system\n", + "percentdif=(dif/calls1)*100# # decrease in cell capacity in 120 degree sectored system in percentage\n", + "\n", + "# Displaying the result in command window\n", + "print \"\\n Cell capacity of unsectored system = %0.0f calls/hour\"%(calls1)\n", + "print \"\\n \\n Cell capacity of 120 degree sectored system = %0.0f calls/hour\"%(capacity)\n", + "print \"\\n \\n Decrease in cell capacity in 120 degree sectored system = %0.0f percent\"%(percentdif)" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |
