From e5af489d177d5304b9fbcb803e7611577903ef9e Mon Sep 17 00:00:00 2001 From: Trupti Kini Date: Thu, 3 Nov 2016 23:30:59 +0600 Subject: Added(A)/Deleted(D) following books A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter1.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter10_qYi9AAs.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter11_EZtJ7kK.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter12_KsTKwv5.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter13_DZJQwFk.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter14_GYUnehZ.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter15_teB3fFs.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter16_fpjEDzx.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter17_szOwhWr.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter18_TVmT3rf.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter19_5sx3l6T.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter2.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter20_6AjJCXE.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter21_iYkzq89.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter22_OEH4UuY.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter3.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter4.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter6_s6H0KKG.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter7_dQdnyuw.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter8_oFmkmxA.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter9_8b0ahS6.ipynb A Basic_And_Applied_Thermodynamics_by_P._K._Nag/screenshots/16.11_yr7v4ev.png A Basic_And_Applied_Thermodynamics_by_P._K._Nag/screenshots/3.3_3S1AvbP.png A Basic_And_Applied_Thermodynamics_by_P._K._Nag/screenshots/7.10_BFuA7JB.png A Digital_Communications_by_S._Haykin/README.txt A Modern_Digital_And_Analog_Communication_System_by_B._P._Lathi/README.txt A sample_notebooks/SPANDANAARROJU/Chapter4_J3M7PEz.ipynb A "sample_notebooks/Sadananda CharyArroju/Chapter2.ipynb" --- .../Chapter5.ipynb | 465 +++++++++++++++++++++ 1 file changed, 465 insertions(+) create mode 100644 Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb (limited to 'Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb') diff --git a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb new file mode 100644 index 00000000..ce4f45c9 --- /dev/null +++ b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter5.ipynb @@ -0,0 +1,465 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 05:First law applied to Flow Processes" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.1:pg-97" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.1\n", + "\n", + " The rate of work input is 116.0 kW\n", + "\n", + " The ratio of the inlet pipe diameter and outet pipe diameter is 0.0 \n" + ] + } + ], + "source": [ + "# Part(a)\n", + "import math\n", + "V1 = 0.95 # Inlet volume flow rate in m**3/kg\n", + "\n", + "P1 = 100 # Pressure at inlet in kPa\n", + "\n", + "v1 = 7 # velocity of flow at inlet in m/s\n", + "\n", + "V2 = 0.19 # Exit volume flow rate in m**3/kg\n", + "\n", + "P2 = 700 # Pressure at exit in kPa \n", + "\n", + "v2 = 5 # velocity of flow at exit in m/s\n", + "\n", + "w = 0.5 # mass flow rate in kg/s\n", + "\n", + "u21 = 90 # change in internal energy in kJ/kg\n", + "\n", + "Q = -58 # Heat transfer in kW\n", + "\n", + "W = - w*( u21 + (P2*V2-P1*V1) + ((v2**2-v1**2)/2) ) + Q # W = dW/dt \n", + "\n", + "print \"\\n Example 5.1\"\n", + "\n", + "print \"\\n The rate of work input is \",abs(W) ,\" kW\"\n", + "\n", + "#The answers given in textbook is wrong\n", + "\n", + "# Part (b)\n", + "\n", + "A = (v2/v1)*(V1/V2) # A = A1/A2\n", + "\n", + "d_ratio = math.sqrt(A) # d = d1/d2\n", + "\n", + "print \"\\n The ratio of the inlet pipe diameter and outet pipe diameter is \",d_ratio ,\" \"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.2:pg-98" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.2\n", + "\n", + " The internal energy decreases by 20.0 kJ\n" + ] + } + ], + "source": [ + "import math\n", + "V1 = 0.37 # volume flow rate at inlet in m**3/kg\n", + "\n", + "P1 = 600# Inlet pressure in kPa\n", + "\n", + "v1 = 16 # Inlet velocity of flow in m/s\n", + "\n", + "V2 = 0.62 # volume flow rate at exit in m**3/kg \n", + "\n", + "P2 = 100# Exit pressure in kPa\n", + "\n", + "v2 = 270 # Exit velocity of flow in m/s\n", + "\n", + "Z1 = 32 # Height of inlet port from datum in m\n", + "\n", + "Z2 = 0 #Height of exit port from datum in m\n", + "\n", + "g = 9.81 # Acceleration due to gravity\n", + "\n", + "Q = -9 # Heat transfer in kJ/kg\n", + "\n", + "W = 135 # Work transfer in kJ/kg\n", + "\n", + "U12 = (P2*V2-P1*V1) + ((v2**2-v1**2)/2000) + (Z2-Z1)*g*1e-3 + W - Q # Change in internal energy in kJ\n", + "\n", + "\n", + "\n", + "print \"\\n Example 5.2\"\n", + "\n", + "print \"\\n The internal energy decreases by \",round(U12) ,\" kJ\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.3:pg-99" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.3\n", + "\n", + " The steam flow rate is 53.5854836932 Kg/s\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "P1 = 4 # Boiler pressure in MPa\n", + "\n", + "t1 = 400 # Exit temperature at boiler in degree Celsius\n", + "\n", + "h1 = 3213 # Enthalpy at boiler exit in kJ/kg\n", + "\n", + "V1 = 0.073 # specific volume at boiler exit in m**3/kg\n", + "\n", + "P2 = 3.5 # Pressure at turbine end in MPa\n", + "\n", + "t2 = 392 # Turbine exit temperature in degree Celsius\n", + "\n", + "h2 = 3202 # Enthalpy at turbine exit in kJ/kg\n", + "\n", + "V2 = 0.084 # specific volume at turbine exit in m**3/kg\n", + "\n", + "Q = -8.5 # Heat loss from pipeline in kJ/kg\n", + "\n", + "v1 = math.sqrt((2*(h1-h2+Q)*1e3)/(1.15**2-1)) # velocity of flow in m/s\n", + "\n", + "A1 = (math.pi/4)*0.2**2 # Area of pipe in m**2\n", + "\n", + "w = (A1*v1)/V1 # steam flow rate in Kg/s\n", + "\n", + "\n", + "\n", + "print \"\\n Example 5.3\"\n", + "\n", + "print \"\\n The steam flow rate is \",w ,\" Kg/s\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.4:pg-100" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.4\n", + "\n", + " The amount of heat that should be supplied is 703.880549402 Kg/h\n" + ] + } + ], + "source": [ + "import math\n", + "h1 = 313.93 # Enthalpy of water at heater inlet in kJ/kg\n", + "\n", + "h2 = 2676 # Enthalpy of hot water at temperature 100.2 degree Celsius\n", + "\n", + "h3 = 419 #Enthalpy of water at heater inlet in kJ/kg\n", + "\n", + "w1 = 4.2 # mass flow rate in kg/s\n", + "\n", + "\n", + "\n", + "print \"\\n Example 5.4\"\n", + "\n", + "w2 = w1*(h3-h1)/(h2-h3)# Steam rate \n", + "\n", + "print \"\\n The amount of heat that should be supplied is \",w2*3600 ,\" Kg/h\"\n", + "\n", + "\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.5:pg-100" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.5\n", + "\n", + " The rate of heat transfer to the air in the heat exchanger is 1577.85 kJ/s\n", + "\n", + " The power output from the turbine assuming no heat loss is 298 kW\n", + "\n", + " The velocity at the exit of the nozzle is 552.358579186 m/s\n" + ] + } + ], + "source": [ + "import math\n", + "t1 = 15 # Heat exchanger inlet temperature in degree Celsius\n", + "\n", + "t2 = 800 # Heat exchanger exit temperature in degree Celsius\n", + "\n", + "t3 = 650 # Turbine exit temperature in degree Celsius\n", + "\n", + "t4 = 500 # Nozzle exit temperature in degree Celsius\n", + "\n", + "v1 = 30 # Velocity of steam at heat exchanger inlet in m/s\n", + "\n", + "v2 = 30# Velocity of steam at turbine inlet in m/s\n", + "\n", + "v3 = 60 # Velocity of steam at nozzle inlet in m/s\n", + "\n", + "w = 2 # mass flow rate in kg/s\n", + "\n", + "cp = 1005 # Specific heat capacity of air in kJ/kgK\n", + "\n", + "\n", + "\n", + "print \"\\n Example 5.5\"\n", + "\n", + "Q1_2 = w*cp*(t2-t1) # rate of heat transfer\n", + "\n", + "print \"\\n The rate of heat transfer to the air in the heat exchanger is \",Q1_2/1e3 ,\" kJ/s\"\n", + "\n", + "\n", + "\n", + "W_T = w*( ((v2**2-v3**2)/2) + cp*(t2-t3)) # power output from the turbine\n", + "\n", + "print \"\\n The power output from the turbine assuming no heat loss is \",W_T/1000 ,\" kW\"\n", + "\n", + "v4 = math.sqrt( (v3**2) + (2*cp*(t3-t4)) ) # velocity at the exit of the nozzle\n", + "\n", + "print \"\\n The velocity at the exit of the nozzle is \",v4 ,\" m/s\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.6:pg-102" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.6\n", + "\n", + " Velocity of exhaust gas is 541.409855832 m/s\n" + ] + } + ], + "source": [ + "import math\n", + "\n", + "ha = 260 # Enthalpy of air in kJ/kg\n", + "\n", + "hg = 912 # Enthalpy of gas in kJ/kg\n", + "\n", + "Va = 270 # Velocity of air in m/s\n", + "\n", + "wf = 0.0190 # mass of fuel in Kg\n", + "\n", + "wa = 1 # mass of air in Kg\n", + "\n", + "Ef = 44500 # Chemical energy of fuel in kJ/kg\n", + "\n", + "Q = 21 # Heat loss from the engine in kJ/kg\n", + "\n", + "\n", + "\n", + "print \"\\n Example 5.6\"\n", + "\n", + "Eg = 0.05*wf*Ef/(1+wf) # As 5% of chemical energy is not released in reaction\n", + "\n", + "wg = wa+wf # mass of flue gas\n", + "\n", + "Vg = math.sqrt(2000*(((ha+(Va**2*0.001)/2+(wf*Ef)-Q)/(1+wf))-hg-Eg)) \n", + "\n", + "\n", + "\n", + "print \"\\n Velocity of exhaust gas is \",Vg ,\" m/s\"\n", + "\n", + "#Answer given in textbook is wrong\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.8:pg-103" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 5.8\n", + "\n", + " The rate at which air flows out of the tank is 0.85 kg/h\n" + ] + } + ], + "source": [ + "import math\n", + "# Given that\n", + "\n", + "V = 0.12 # Volume of tank in m**3\n", + "\n", + "p = 1 # Pressure in MPa\n", + "\n", + "T = 150 # Temperature in degree centigrade\n", + "\n", + "P = 0.1 # Power to peddle wheel in kW\n", + "\n", + "print \"\\n Example 5.8\"\n", + "\n", + "u0 = 0.718*273 # Internal energy at 0 degree Celsius\n", + "\n", + "# Function for internal energy of gas\n", + "\n", + "def f1(t):\n", + " u = u0+(0.718*t)\n", + " pv = 0.287*(273+t)\n", + " return (u,pv)\n", + " \n", + "U,PV=f1(T)\n", + " \n", + " \n", + "hp = U+PV # At 150 degree centigrade\n", + "m_a = P/hp\n", + " \n", + "print \"\\n The rate at which air flows out of the tank is \",round(m_a*3600,2) ,\" kg/h\"\n", + "\n", + "#The answers vary due to round off error\n", + "\n", + "\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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} -- cgit