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| author | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
|---|---|---|
| committer | Thomas Stephen Lee | 2015-09-04 22:04:10 +0530 |
| commit | 64419e47f762802600b3a2b6d8c433a16ccd3d55 (patch) | |
| tree | 14ad7c37c9547cd516f141494f3fa375621edbaa /Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb | |
| parent | 10f6fb8cd1d840a3042651dfaa6fd5af4924b94a (diff) | |
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diff --git a/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb new file mode 100755 index 00000000..13ddc7dd --- /dev/null +++ b/Chemical_Engineering_Thermodynamics_by_P_Ahuja/ch1_2.ipynb @@ -0,0 +1,770 @@ +{
+ "metadata": {
+ "name": "ch1_2",
+ "signature": "sha256:05ef5db2afef7232054abda0278780d1df8c92b807546fcd7648e8885fa4539f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1 : Introduction"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.1 Page number - 6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# \n",
+ "# Variables\n",
+ "#(a)\n",
+ " \n",
+ "m = 50.;\t\t\t#[kg] - Mass of piston\n",
+ "A = 0.05;\t\t\t#[m**(2)] - Area of piston\n",
+ "g = 9.81;\t\t\t#[m/s**(2)] - Acceleration due to gravity\n",
+ "Po = 101325;\t\t\t#[N/m**(2)] - Atmospheric pressure\n",
+ "\n",
+ "# Calculations and Results\n",
+ "P = (m*g/A)+Po;\t\t\t#[N/m**(2)]\n",
+ "P = P/100000.;\t\t\t#[bar]\n",
+ "print \" (a).Pressure = %f bar\"%P\n",
+ "\n",
+ "#(b)\n",
+ "print \" (b).Since the piston weight and surroundings pressure are the same, \\\n",
+ "the gas pressure in the piston-cylinder assembly remains %f bar\"%P\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a).Pressure = 1.111350 bar\n",
+ " (b).Since the piston weight and surroundings pressure are the same, the gas pressure in the piston-cylinder assembly remains 1.111350 bar\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.2 Page number - 8\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "\n",
+ "# Variables\n",
+ "P = 1.;\t\t\t#[atm] - Atmospheric pressure\n",
+ "P = 101325.;\t\t\t#[N/m**(2)]\n",
+ "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+ "T = 30.;\t\t\t#[C] - Temperature of air\n",
+ "T = 30.+273.15;\t\t\t#[K]\n",
+ "V = 5.*5*5;\t\t\t#[m**(3)] - Volume of the room\n",
+ "\n",
+ "# Calculations\n",
+ "#The number of moles of air is given by\n",
+ "n = (P*V)/(R*T)\t\t\t#[mol]\n",
+ "\n",
+ "#Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)= 28.84 g/mol\n",
+ "m = n*28.84;\t\t\t#[g]\n",
+ "m = m/1000.;\t\t\t#[kg]\n",
+ "\n",
+ "# Results\n",
+ "print \"The mass of air is, m = %f kg\"%m\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mass of air is, m = 144.928664 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.3 Page number - 13\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "\n",
+ "# Variables\n",
+ "P1 = 3.;\t\t\t# [bar] - initial pressure\n",
+ "V1 = 0.5;\t\t\t# [m**(3)] - initial volume\n",
+ "V2 = 1.0;\t\t\t# [m**(3)] - final volume\n",
+ "import math\n",
+ "#(a)\n",
+ "n = 1.5;\n",
+ "\n",
+ "# Calculations and Results\n",
+ "#Let P*V**(n)=C \t\t\t# Variables relation\n",
+ "#W (work done per mole)= (integrate('P'%'V'%V1%V2))\n",
+ "#and thus W=((P2*V2)-(P1*V1))/(1-n)\n",
+ "#The above math.expression is valid for all values of n%except n=1.0\n",
+ "P2 = (P1*((V1/V2)**(n)))\t\t\t#[bar] \t\t\t#pressure at state 2\n",
+ "\n",
+ "#we have%(V1/V2)=(V1t/(V2t)%since the number of moles are constant.Thus\n",
+ "W = ((P2*V2)-(P1*V1))/(1-n)*10**(5)\t\t\t#[J]\n",
+ "W = W/1000.;\t\t\t#[kJ]\n",
+ "print \" (a).The work done (for n=1.5) is %f kJ\"%W\n",
+ "\n",
+ "#(b)\n",
+ "#For n=1.0%we have% PV=C.\n",
+ "# w(wok done per mol)= (integrate('P'%'V'%V1%V2)) = (integrate('C/V'%'V'%V1%V2)) = C*ln(V2/V1)=P1*V1*ln(V2/V1)\n",
+ "W1 = P1*V1*math.log(V2/V1)*10**(5)\t\t\t#[J]\n",
+ "W1 = W1/1000.;\t\t\t#[kJ]\n",
+ "print \" (b).The work done (for n=1.0) is %f kJ\"%W1\n",
+ "\n",
+ "#(c)\n",
+ "#For n=0%we get P=Constant and thus\n",
+ "P = P1;\t\t\t#[bar]\n",
+ "# w =(integrate('P'%'V'%V1%V2)) = P*(V2-V1)\n",
+ "W2 = P*(V2-V1)*10**(5)\t\t\t#[J]\n",
+ "W2 = W2/1000.;\t\t\t#[kJ]\n",
+ "print \" (c).The work done (for n=0) is %f kJ\"%W2\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a).The work done (for n=1.5) is 87.867966 kJ\n",
+ " (b).The work done (for n=1.0) is 103.972077 kJ\n",
+ " (c).The work done (for n=0) is 150.000000 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.4 Page number - 17\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "\n",
+ "# Variables\n",
+ "#(a)\n",
+ "# Variables\n",
+ "V = 9.;\t\t\t# [m/s] - velocity\n",
+ "d = 1.;\t\t\t#[m] - diameter\n",
+ "A = 3.14*(d/2)**(2)\t\t\t#[m**(2)] - area\n",
+ "P = 1.;\t\t\t#[atm] - pressure\n",
+ "P = 101325.;\t\t\t# [N/m**(2)]\n",
+ "T = 300.;\t\t\t#[K] - Temperature\n",
+ "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+ "\n",
+ "# Calculations and Results\n",
+ "E = (V**(2))/2.;\t\t\t#[J/kg]\n",
+ "print \" (a).The wind energy per unit mass of air is %f J/kg\"%E\n",
+ "\n",
+ "#(b)\n",
+ "# Molecular weight of air(21 vol% O2 and 79 vol% N2)=(0.21*32)+(0.79*28)= 28.84 g/mol\n",
+ "M = 28.84*10**(-3)\t\t\t#[kg/mol]\n",
+ "r = (P*M)/(R*T)\t\t\t#[kg/m**(3)] - density\n",
+ "m = r*V*A;\t\t\t# [kg/s] - mass flow rate of air\n",
+ "pi = m*E;\t\t\t#[Watt] - power input\n",
+ "print \" (b).The wind power input to the turbine is %f Watt\"%pi\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a).The wind energy per unit mass of air is 40.500000 J/kg\n",
+ " (b).The wind power input to the turbine is 335.233787 Watt\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.5 Page number - 23\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Given\n",
+ "P = 1.;\t\t\t # [bar] - atospheric pressure\n",
+ "P1guz = 0.75;\t\t\t# [bar] - gauze pressure in 1st evaporator\n",
+ "P2Vguz = 0.25;\t\t\t# [bar] - vaccum gauze pressure in 2nd evaporator\n",
+ "\n",
+ "# Calculations\n",
+ "P1abs = P + P1guz;\t\t\t# [bar] - absolute pressure in 1st evaporator\n",
+ "P2abs = P - P2Vguz;\t\t\t# [bar] -absolute pressure in 2nd evaporator\n",
+ "\n",
+ "# Results\n",
+ "#From saturated steam table as reported in the book\n",
+ "print \" For P1abs (absolute pressure) = %f bar\"%P1abs\n",
+ "print \" The saturation temperature in first evaporator is 116.04 C\"\n",
+ "print \" For P2abs (absolute pressure) = %f bar\"%P2abs\n",
+ "print \" The saturation temperature in second evaporator is 91.76 C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " For P1abs (absolute pressure) = 1.750000 bar\n",
+ " The saturation temperature in first evaporator is 116.04 C\n",
+ " For P2abs (absolute pressure) = 0.750000 bar\n",
+ " The saturation temperature in second evaporator is 91.76 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.6 Page number - 23\n"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "V = 1.;\t\t\t # [kg] - volume of tank\n",
+ "P = 10.;\t\t\t# [bar] - pressure\n",
+ "\n",
+ "#Here degree of freedom =1(C=1%P=2%threfore F=1)\n",
+ "#From steam table at 10 bar as reported in the book\n",
+ "V_liq = 0.001127;\t\t\t# [m**(3)/kg] - volume in liquid phase\n",
+ "V_vap = 0.19444;\t\t\t# [m**(3)/kg] - volume in vapour phase\n",
+ "\n",
+ "# Calculations\n",
+ "#x*Vv=(1-x)*Vl \t\t\t# since two volumes are equal\n",
+ "x = (V_liq/(V_liq+V_vap))\t\t\t# [kg]\n",
+ "y = (1-x)\t\t\t#[kg]\n",
+ "\n",
+ "# Results\n",
+ "print \" Mass of saturated vapour is %f kg\"%x\n",
+ "print \" Mass of saturated liquid is %f kg\"%y\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Mass of saturated vapour is 0.005763 kg\n",
+ " Mass of saturated liquid is 0.994237 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.7 Page number - 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "V = 1.;\t\t\t# [m**(3)] - volume of tank\n",
+ "M = 10.;\t\t\t# [m**(3)] - total mass\n",
+ "T = (90+273.15)\t\t\t#[K] - temperature\n",
+ "\n",
+ "#From steam table at 90 C as reported in the book\n",
+ "#vapour pressure(pressure of rigid tank) = 70.14[kPa] = 0.7014[bar]\n",
+ "print \" Pressure of tank = 0.7014 bar\"\n",
+ "\n",
+ "# Calculations and Results\n",
+ "V_liq_sat=0.001036;\t\t\t# [m**(3)/kg] - saturated liquid specific volume\n",
+ "V_vap_sat=2.36056;\t\t\t# [m**(3)/kg] - saturated vapour specific volume\n",
+ "\n",
+ "#1=(V_liq_sat*(10-x))+(V_vap_sat*x)\n",
+ "x = (1-(10*V_liq_sat))/(V_vap_sat-V_liq_sat)\t\t\t#[kg]\n",
+ "y = (10-x)\t\t\t#[kg]\n",
+ "\n",
+ "print \" The amount of saturated liquid is %f kg\"%y\n",
+ "print \" The amount of saturated vapour is %f kg \"%x\n",
+ "\n",
+ "z = y*V_liq_sat;\t\t\t#[m**(3)] - Volume of saturated liquid \n",
+ "w = x*V_vap_sat;\t\t\t#[m**(3)] - Volume of saturated vapour\n",
+ "\n",
+ "print \" Total volume of saturated liquid is %f m**(3)\"%z\n",
+ "print \" Total volume of saturated vapour is %f m**(3)\"%w\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Pressure of tank = 0.7014 bar\n",
+ " The amount of saturated liquid is 9.580576 kg\n",
+ " The amount of saturated vapour is 0.419424 kg \n",
+ " Total volume of saturated liquid is 0.009925 m**(3)\n",
+ " Total volume of saturated vapour is 0.990075 m**(3)\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.8 Page number - 24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "V = 10.;\t\t\t# [m**(3)] - volume of vessel\n",
+ "P_1 = 1.;\t\t\t# [bar] - initial pressure\n",
+ "V_liq_sat = 0.05;\t\t\t# [m**(3)] - saturated liquid volume\n",
+ "V_gas_sat = 9.95;\t\t\t# [m**(3)] - saturated vapour volume\n",
+ "\n",
+ "#At 1 bar pressure\n",
+ "V_liq_1 = 0.001043;\t\t\t# [m**(3/kg)] - specific saturated liquid volume\n",
+ "U_liq_1 = 417.33;\t\t\t# [kJ/kg] - specific internal energy\n",
+ "V_gas_1 = 1.69400;\t\t\t# [m**(3/kg)] - specific saturated vapour volume\n",
+ "U_gas_1 = 2506.06;\t\t\t# [kJ/kg]\n",
+ "\n",
+ "# Calculations\n",
+ "M_liq_1 = V_liq_sat/V_liq_1;\t\t\t# [kg] - mass of saturated liqid\n",
+ "M_gas_1 = V_gas_sat/V_gas_1;\t\t\t# [kg] - mass of saturated vapour\n",
+ "M = (M_liq_1+M_gas_1)\t\t\t# [kg] - total mass\n",
+ "U_1t = (M_liq_1*U_liq_1)+(M_gas_1*U_gas_1)\t\t\t# [kJ] - initial internal energy\n",
+ "V_gas_2 = (V/M)\t\t\t#[m**(3/kg)]\n",
+ "\n",
+ "#from steam table at 10 bar pressure as reported in the book\n",
+ "V_vap_2 = 0.19444;\t\t\t# [m**(3/kg)]\n",
+ "U_vap_2 = 2583.64;\t\t\t# [kJ/kg]\n",
+ "\n",
+ "#from steam table at 11 bar pressure as reported in the book\n",
+ "V_vap_3 = 0.17753;\t\t\t#[m**(3/kg)]\n",
+ "U_vap_3 = 2586.40;\t\t\t#[kJ/kg]\n",
+ "\n",
+ "#Now computing pressure when molar volume of saturated vapour=Vg_2\n",
+ "#By interpolation (P2-10)/(11-10)=(Vg_2-Vv_2)/(Vv_3-Vv_2)\n",
+ "P_2 = (((V_gas_2 - V_vap_2)/(V_vap_3 - V_vap_2)*1)+10)\t\t\t# [bar] - final pressure\n",
+ "\n",
+ "#By interpolation calculating internal energy at state 2\n",
+ "#(P2-10)/(11-10)=(U2-Uv_2)/(Uv_3-Uv_2)\n",
+ "U_2 = (((P_2-10)/(11-10))*(U_vap_3 - U_vap_2))+U_vap_2;\t\t\t#[kJ/kg]\n",
+ "U_2t = U_2*M;\t\t\t#[kJ]\n",
+ "H = U_2t - U_1t;\t\t\t#[kJ] - Heat supplied\n",
+ "H = H/1000;\t\t\t#[MJ]\n",
+ "\n",
+ "# Results\n",
+ "print \" Total heat supplied is %f MJ\"%H\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Total heat supplied is 104.381244 MJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.9 Page number - 26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "#Antoine equation for water ln(Psat)=16.262-(3799.89/(T_sat + 226.35))\n",
+ "P = 2.;\t\t\t#[atm] - Pressure\n",
+ "P = (2.*101325)/1000;\t\t\t#[kPa]\n",
+ "\n",
+ "# Calculations\n",
+ "P_sat = P;\t\t\t# Saturation pressure\n",
+ "T_sat = (3799.89/(16.262-math.log(P_sat)))-226.35;\t\t\t#[C] - Saturation temperature\n",
+ "#Thus boiling at 2 atm occurs at Tsat = 120.66 C.\n",
+ "\n",
+ "#From steam tables%at 2 bar%Tsat = 120.23 C and at 2.25 bar%Tsat = 124.0 C\n",
+ "#From interpolation for T_sat = 120.66 C%P = 2.0265 bar\n",
+ "#For P_= 2.0265 bar%T_sat% from steam table by interpolation is given by\n",
+ "#((2.0265-2)/(2.25-2))=((Tsat-120.23)/(124.0-120.23))\n",
+ "T_sat_0 = (((2.0265-2)/(2.25-2))*(124.0-120.23))+120.23;\t\t\t#[C]\n",
+ "\n",
+ "# Results\n",
+ "print \" Saturation temperature (Tsat) = %f C which is close \\\n",
+ "to %f C as determined from Antoine equation\"%(T_sat_0,T_sat)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Saturation temperature (Tsat) = 120.629620 C which is close to 120.655450 C as determined from Antoine equation\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.10 Page number - 27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "# math.log(P)=-(1640/T)+10.56 (solid)\n",
+ "# math.log(P)=-(1159/T)+7.769 (liquid)%where T is in K\n",
+ "# F+P=C+2% at triple point F+3=1+2 or%F=0 i.e%vapour pressure of liquid and solid at triple point are same%we get\n",
+ "# -(1640/T)+10.56 = -(1159/T)+7.769\n",
+ "\n",
+ "# Calculations\n",
+ "T = (1640-1159)/(10.56-7.769)\t\t\t#[K]\n",
+ "P = 10**((-1640/T)+10.56)\t\t\t#[torr]\n",
+ "\n",
+ "# Results\n",
+ "print \" The temperature is %f K\"%T\n",
+ "print \" The pressure is %f torr (or mm Hg)\"%P\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The temperature is 172.339663 K\n",
+ " The pressure is 11.063907 torr (or mm Hg)\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.11 Page number - 29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "M_O2 = 31.999;\t\t\t#molecular weight of oxygen\n",
+ "M_N2 = 28.014;\t\t\t#molecular weight of nitrogen\n",
+ "Y = 1.4;\t\t\t#molar heat capacities ratio for air\n",
+ "\n",
+ "# Calculations and Results\n",
+ "#Molecular weight of air(21 vol% O2 and 79 vol% N2)is given by\n",
+ "M_air = (0.21*M_O2)+(0.79*M_N2)\t\t\t#(vol% = mol%)\n",
+ "\n",
+ "R = 8.314;\t\t\t#[J/mol*K] - Universal gas constant\n",
+ "R = (R*1/M_air)\t\t\t#[kJ/kg*K]\n",
+ "\n",
+ "print \" The value of universal gas constant (R) = %f kJ/kg-K \"%R\n",
+ "\n",
+ "#Y=Cp0/Cv0 and Cp0-Cv0=R\n",
+ "Cv_0 = R/(Y-1)\t\t\t#[kJ/kg*K] \n",
+ "Cp_0 = Y*Cv_0;\t\t\t#[kJ/kg*K]\n",
+ "print \" The value of Cp_0 for air is %f kJ/kg-K\"%Cp_0\n",
+ "print \" The value of Cv_0 for air is %f kJ/kg-K\"%Cv_0\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The value of universal gas constant (R) = 0.288172 kJ/kg-K \n",
+ " The value of Cp_0 for air is 1.008601 kJ/kg-K\n",
+ " The value of Cv_0 for air is 0.720429 kJ/kg-K\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.12 Page number - 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "Y = 1.4;\t\t\t#molar heat capacities ratio for air\n",
+ "R = 8.314;\t\t\t# [J/mol*K] - Universal gas constant\n",
+ "\n",
+ "# Calculations\n",
+ "Cv_0 = R/(Y-1)\t\t\t# [J/mol*K]\n",
+ "Cp_0 = Y*Cv_0;\t\t\t# [J/mol*K]\n",
+ "\n",
+ "# Results\n",
+ "print \" The molar heat capacity at constant volume (Cv_0) is %f J/mol-K\"%Cv_0\n",
+ "print \" The molar heat capacity at constant pressure (Cp_0) is %f J/mol-K\"%Cp_0\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The molar heat capacity at constant volume (Cv_0) is 20.785000 J/mol-K\n",
+ " The molar heat capacity at constant pressure (Cp_0) is 29.099000 J/mol-K\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.13 Page number - 30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "# Variables\n",
+ "# Cp0=7.7+(0.04594*10**(-2)*T)+(0.2521*10**(-5)*T**(2))-(0.8587*10**(-9)*T**(3))\n",
+ "T_1 = 400.;\t\t\t#[K]\n",
+ "T_2 = 500.;\t\t\t#[K]\n",
+ "\n",
+ "# Calculations\n",
+ "def f(T):\n",
+ " return 7.7+(0.04594*10**(-2)*T)+(0.2521*10**(-5)*T**(2))-(0.8587*10**(-9)*T**(3))\n",
+ "#(C)avg = q/(T_2 - T_1) = 1/(T_2 - T_1)*{(integrate('C'%'T'%T_1%T_2))}\n",
+ "#(Cp0)avg = 1/(T_2 - T_1)*{(integrate('Cp0'%'T'%T_1%T_2))}\n",
+ "Cp0_avg = (1/(T_2 - T_1))*quad(f,T_1,T_2)[0]\n",
+ "\n",
+ "# Results\n",
+ "print \" The mean heat capacity (Cp0_avg) for temerature range of 400 to 500 K is %f cal/mol-K\"%Cp0_avg\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The mean heat capacity (Cp0_avg) for temerature range of 400 to 500 K is 8.340118 cal/mol-K\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example - 1.14 Page number - 31"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " \n",
+ "# Variables\n",
+ "#(a)\n",
+ "P_1 = 0.2;\t\t\t# [MPa] - pressure\n",
+ "x_1 = 0.59;\t\t\t# mole fraction\n",
+ "\n",
+ "# Calculations and Results\n",
+ "#From saturated steam tables at 0.2 MPa\n",
+ "H_liq_1 = 504.7;\t\t\t# [kJ/kg] - Enthalpy of saturated liquid\n",
+ "H_vap_1 = 2706.7;\t\t\t# [kJ/kg]- Enthalpy of saturated vapour\n",
+ "H_1 = (H_liq_1*(1-x_1))+(x_1*H_vap_1)\t\t\t# [kJ/kg]\n",
+ "print \" (a).Enthalpy of 1 kg of water in tank is %f kJ/kg\"%H_1\n",
+ "\n",
+ "#(b)\n",
+ "T_2 = 120.23;\t\t\t# [C] - temperature\n",
+ "V_2 = 0.6;\t\t\t# [m**(3)/kg] - specific volume\n",
+ "\n",
+ "#From saturated steam tables at 120.23 C% as reported in the book\n",
+ "V_liq_2=0.001061;\t\t\t# [m**(3)/kg]\n",
+ "V_vap_2=0.8857;\t\t\t# [m**(3)/kg]\n",
+ "#since V_2 < Vv_2%dryness factor will be given by% V = ((1-x)*V_liq)+(x*V_vap)\n",
+ "x_2 = (V_2- V_liq_2)/(V_vap_2 - V_liq_2)\n",
+ "\n",
+ "#From steam table%at 120.2 C%the vapour pressure of water is 0.2 MPa.So%enthalpy is given by\n",
+ "H_2 = (H_liq_1*(1-x_2))+(H_vap_1*x_2)\t\t\t#kJ/kg]\n",
+ "print \" (b).Enthalpy of saturated steam is %f kJ/kg\"%H_2\n",
+ "\n",
+ "#(c)\n",
+ "P_3 = 2.5;\t\t\t#[MPa]\n",
+ "T_3 = 350;\t\t\t#[C]\n",
+ "#From steam tables at 2.5 MPa%T_sat = 223.99 C%as reported in the book\n",
+ "#since%T_3 > Tsat% steam is superheated\n",
+ "print \" (c).As steam is superheated%from steam table%enthalpy (H) is 3126.3 kJ/kg\"\n",
+ "\n",
+ "#(d)\n",
+ "T_4 = 350;\t\t\t#[C]\n",
+ "V_4 = 0.13857;\t\t\t#[m**(3)/kg]\n",
+ "#From steam table%at 350 C% V_liq = 0.001740 m**(3)/kg and V_vap = 0.008813 m**(3)/kg.Since%V > V_vap%therefore it is superheated.\n",
+ "#From steam table at 350 C and 1.6 MPa% V = 0.17456 m**(3)/kg\n",
+ "#At 350 C and 2.0 MPa% V = 0.13857 m**(3)/kg. So%\n",
+ "print \" (d).The enthalpy of superheated steam (H) is 3137.0 kJ/kg\"\n",
+ "\n",
+ "#(e)\n",
+ "P_4 = 2.0;\t\t\t#[MPa]\n",
+ "U_4 = 2900;\t\t\t# [kJ/kg] - internal energy\n",
+ "#From saturated table at 2.0 MPa% U_liq = 906.44kJ and U_vap = 2600.3 kJ/kg\n",
+ "#scince%U_4 > Uv% it is saturated.\n",
+ "#From superheated steam table at 2.0 MPa and 350 C% as reported in the book\n",
+ "U_1 = 2859.8;\t\t\t#[kJ/kg]\n",
+ "H_1 = 3137.0;\t\t\t#[kJ/kg]\n",
+ "#At 2.0 MPa and 400 C%\n",
+ "U_2 = 2945.2;\t\t\t#[kJ/kg]\n",
+ "H_2 = 3247.6;\t\t\t#[kJ/kg]\n",
+ "T = (((U_4 - U_1)/(U_2 - U_1))*(400 - 350)) + 350;\t\t\t#[C] - By interpolation\n",
+ "H = (((T - 350)/(400 - 350))*(H_2 - H_1)) + H_1;\t\t\t#[kJ/kg]\n",
+ "print \" (e).The enthalpy value (of superheated steam) obtained after interpolation is %f kJ/kg\"%H\n",
+ "\n",
+ "#(f)\n",
+ "P_5 = 2.5;\t\t\t#[MPa]\n",
+ "T_5 = 100;\t\t\t#[C]\n",
+ "#At 100 C%P_sat=101350 N/m**(2). Since P_5 > P_sat%it is compressed liquid\n",
+ "P_sat = 0.101350;\t\t\t#[MPa]\n",
+ "H_liq = 419.04;\t\t\t#[kJ/kg] - At 100 C and 0.10135 MPa\n",
+ "V_liq = 0.001044;\t\t\t#[m**(3)/kg] - At 100 C and 0.10135 MPa\n",
+ "H_0 = H_liq + (V_liq*(P_5 - P_sat))*1000;\t\t\t#kJ/kg]\n",
+ "print \" (f).The enthalpy of compressed liquid is %f kJ/kg\"%H_0\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " (a).Enthalpy of 1 kg of water in tank is 1803.880000 kJ/kg\n",
+ " (b).Enthalpy of saturated steam is 1995.549576 kJ/kg\n",
+ " (c).As steam is superheated%from steam table%enthalpy (H) is 3126.3 kJ/kg\n",
+ " (d).The enthalpy of superheated steam (H) is 3137.0 kJ/kg\n",
+ " (e).The enthalpy value (of superheated steam) obtained after interpolation is 3189.062295 kJ/kg\n",
+ " (f).The enthalpy of compressed liquid is 421.544191 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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