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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 5 - Gases and Single Phase systems"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 1: pg 98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.1\n",
+ " The new pressure exerted on the air (mmHg) = 900.0\n",
+ " The difference in the two mercury column level (mm) = 135.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 98\n",
+ "#calculate the new pressure and difference in two levels\n",
+ "print('Example 5.1');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# new pressure exerted on the air and the difference in two mercury column level\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 765.;# atmospheric pressure, [mmHg]\n",
+ "V1 = 20000.;# [mm^3]\n",
+ "V2 = 17000.;# [mm^3]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# using boyle's law P*V=constant\n",
+ "# hence\n",
+ "P2 = P1*V1/V2;# [mmHg]\n",
+ "\n",
+ "del_h = P2-P1;# difference in Height of mercury column level\n",
+ "#results\n",
+ "print ' The new pressure exerted on the air (mmHg) = ',P2\n",
+ "print ' The difference in the two mercury column level (mm) = ',del_h\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 2: pg 99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.2\n",
+ " The new volume after expansion (m^3) = 0.7\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 99\n",
+ "#calculate the new volume\n",
+ "print('Example 5.2');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the new volume\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 300;# original pressure,[kN/m^2]\n",
+ "V1 = .14;# original volume,[m^3]\n",
+ "\n",
+ "P2 = 60.;# new pressure after expansion,[kn/m^2]\n",
+ "\n",
+ "# solution\n",
+ "# since temperature is constant so using boyle's law P*V=constant\n",
+ "V2 = V1*P1/P2;# [m^3]\n",
+ "\n",
+ "#results\n",
+ "print ' The new volume after expansion (m^3) = ',V2\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3: pg 101"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.3\n",
+ " The new volume of the gas trapped in the apparatus (mm^3) = 12302.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 101\n",
+ "#calculate the new volume\n",
+ "print('Example 5.3');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the new volume of the gas\n",
+ "\n",
+ "# Given values\n",
+ "V1 = 10000;# [mm^3]\n",
+ "T1 = 273.+18;# [K]\n",
+ "T2 = 273.+85;# [K]\n",
+ "\n",
+ "# solution\n",
+ "# since pressure exerted on the apparatus is constant so using charle's law V/T=constant\n",
+ "# hence\n",
+ "V2 = V1*T2/T1;# [mm^3]\n",
+ "\n",
+ "#results\n",
+ "print ' The new volume of the gas trapped in the apparatus (mm^3) = ',round(V2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4: pg 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.4\n",
+ " The final temperature of the gas (C) = 15.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 102\n",
+ "#calculate the final temperature\n",
+ "print('Example 5.4');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the final temperature\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .2;# original volume,[m^3]\n",
+ "T1 = 273+303;# original temperature, [K]\n",
+ "V2 = .1;# final volume, [m^3]\n",
+ "\n",
+ "# solution\n",
+ "# since pressure is constant, so using charle's law V/T=constant\n",
+ "# hence\n",
+ "T2 = T1*V2/V1;# [K]\n",
+ "t2 = T2-273;# [C]\n",
+ "#results\n",
+ "print ' The final temperature of the gas (C) = ',t2\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5: pg 106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.5\n",
+ "The new volume of the gas (m^3) = 0.0223\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 106\n",
+ "#calculate the new volume\n",
+ "print('Example 5.5');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the new volume of the gas\n",
+ "\n",
+ "# Given values\n",
+ "\n",
+ "# initial codition\n",
+ "P1 = 140;# [kN/m^2]\n",
+ "V1 = .1;# [m^3]\n",
+ "T1 = 273+25;# [K]\n",
+ "\n",
+ "# final condition\n",
+ "P2 = 700.;# [kN/m^2]\n",
+ "T2 = 273.+60;# [K]\n",
+ "\n",
+ "# by charasteristic equation, P1*V1/T1=P2*V2/T2\n",
+ "\n",
+ "V2=P1*V1*T2/(T1*P2);# final volume, [m^3]\n",
+ "\n",
+ "#results\n",
+ "print 'The new volume of the gas (m^3) = ',round(V2,4)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6: pg 106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.6\n",
+ " The mass of the gas present (kg) = 0.118\n",
+ " The new temperature of the gas (C) = 651\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 106\n",
+ "#calculate the new temperature and mass of gas\n",
+ "print('Example 5.6');\n",
+ "\n",
+ "# aim : To determine\n",
+ "# the mas of the gas and new temperature\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 350;# [kN/m^2]\n",
+ "V1 = .03;# [m^3]\n",
+ "T1 = 273+35;# [K]\n",
+ "R = .29;# Gas constant,[kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "# using charasteristic equation, P*V=m*R*T\n",
+ "m = P1*V1/(R*T1);# [Kg]\n",
+ "\n",
+ "# Now the gas is compressed\n",
+ "P2 = 1050;# [kN/m^2]\n",
+ "V2 = V1;\n",
+ "# since mass of the gas is constant so using, P*V/T=constant\n",
+ "# hence\n",
+ "T2 = T1*P2/P1# [K]\n",
+ "t2 = T2-273;# [C]\n",
+ "\n",
+ "#results\n",
+ "print ' The mass of the gas present (kg) = ',round(m,3)\n",
+ "print ' The new temperature of the gas (C) = ',t2\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7: pg 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.7\n",
+ " The heat transferred to the gas (kJ) = 172.8\n",
+ " The final pressure of the gas (kN/m^2) = 338.06\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 111\n",
+ "#calculate the final pressure and heat transferred\n",
+ "print('Example 5.7');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the heat transferred to the gas and its final pressure\n",
+ "\n",
+ "# Given values\n",
+ "m = 2;# masss of the gas, [kg]\n",
+ "V1 = .7;# volume,[m^3]\n",
+ "T1 = 273+15;# original temperature,[K]\n",
+ "T2 = 273+135;# final temperature,[K]\n",
+ "cv = .72;# specific heat capacity at constant volume,[kJ/kg K]\n",
+ "R = .29;# gas law constant,[kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "Q = m*cv*(T2-T1);# Heat transferred at constant volume,[kJ]\n",
+ "\n",
+ "# Now,using P1*V1=m*R*T1\n",
+ "P1 = m*R*T1/V1;# [kN/m^2]\n",
+ "\n",
+ "# since volume of the system is constant, so P1/T1=P2/T2\n",
+ "# hence\n",
+ "P2 = P1*T2/T1;# final pressure,[kN/m^2]\n",
+ "#results\n",
+ "print ' The heat transferred to the gas (kJ) = ',Q\n",
+ "print ' The final pressure of the gas (kN/m^2) = ',round(P2,2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 8: pg 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.8\n",
+ " The heat transferred to the gas (kJ) = -31.84\n",
+ " Work done on the gas during the process (kJ) = -9.19\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 114\n",
+ "#calculate the heat transferred and work done\n",
+ "print('Example 5.8');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the heat transferred from the gas and the work done on the gas\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 275;# pressure, [kN/m^2]\n",
+ "V1 = .09;# volume,[m^3]\n",
+ "T1 = 273+185;# initial temperature,[K]\n",
+ "T2 = 273+15;# final temperature,[K]\n",
+ "cp = 1.005;# specific heat capacity at constant pressure,[kJ/kg K]\n",
+ "R = .29;# gas law constant,[kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "# using P1*V1=m*R*T1\n",
+ "m = P1*V1/(R*T1);# mass of the gas\n",
+ "\n",
+ "# calculation of heat transfer\n",
+ "Q = m*cp*(T2-T1);# Heat transferred at constant pressure,[kJ]\n",
+ "\n",
+ "# calculation of work done\n",
+ "# Now,since pressure is constant so, V/T=constant\n",
+ "# hence\n",
+ "V2 = V1*T2/T1;# [m^3]\n",
+ "\n",
+ "W = P1*(V2-V1);# formula for work done at constant pressure,[kJ]\n",
+ "#results\n",
+ "print ' The heat transferred to the gas (kJ) = ',round(Q,2)\n",
+ "print ' Work done on the gas during the process (kJ) = ',round(W,2)\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 9: pg 117"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.9\n",
+ " The new pressure of the gas (kN/m^2) = 1299.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 117\n",
+ "#calculate the new pressure\n",
+ "print('Example 5.9');\n",
+ "\n",
+ "# aim : To determine\n",
+ "# the new pressure of the gas\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 300.;# original pressure,[kN/m**2]\n",
+ "T1 = 273.+25;# original temperature,[K]\n",
+ "T2 = 273.+180;# final temperature,[K]\n",
+ "\n",
+ "# solution\n",
+ "# since gas compressing according to the law,P*V**1.4=constant\n",
+ "# so,for polytropic process,T1/T2=(P1/P2)**((n-1)/n),here n=1.4\n",
+ "\n",
+ "# hence\n",
+ "P2 = P1*(T2/T1)**((1.4)/(1.4-1));# [kN/m**2]\n",
+ "\n",
+ "#results\n",
+ "print ' The new pressure of the gas (kN/m^2) = ',round(P2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 10: pg 118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.10\n",
+ " The new temperature of the gas (C) = 25.0\n",
+ " there is minor error in book answer due to rounding off error\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 118\n",
+ "#calculate the new temperature\n",
+ "print('Example 5.10');\n",
+ "\n",
+ "# aim : To determine\n",
+ "# the new temperature of the gas\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .015;# original volume,[m**3]\n",
+ "T1 = 273.+285;# original temperature,[K]\n",
+ "V2 = .09;# final volume,[m**3]\n",
+ "\n",
+ "# solution \n",
+ "# Given gas is following the law,P*V**1.35=constant\n",
+ "# so process is polytropic with\n",
+ "n = 1.35; # polytropic index\n",
+ "\n",
+ "# hence\n",
+ "T2 = T1*(V1/V2)**(n-1);# final temperature, [K]\n",
+ "\n",
+ "t2 = T2-273;# [C]\n",
+ "\n",
+ "#results\n",
+ "print ' The new temperature of the gas (C) = ',round(t2,1)\n",
+ "\n",
+ "print ' there is minor error in book answer due to rounding off error'\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 11: pg 119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.11\n",
+ " (a) The original volume of the gas (m^3) = 0.0765\n",
+ " and The final volume of the gas (m^3) = 0.306\n",
+ " (b) The final pressure of the gas (kN/m^2) = 231.0\n",
+ " (c) The final temperature of the gas (C) = 92.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 119\n",
+ "#calculate the final pressure, temperature and volume of gas\n",
+ "print('Example 5.11');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# (a) original and final volume of the gas\n",
+ "# (b) final pressure of the gas\n",
+ "# (c) final temperature of the gas\n",
+ "\n",
+ "# Given values\n",
+ "m = .675;# mass of the gas,[kg]\n",
+ "P1 = 1.4;# original pressure,[MN/m**2]\n",
+ "T1 = 273+280;# original temperature,[K]\n",
+ "R = .287;#gas constant,[kJ/kg K]\n",
+ "\n",
+ "# solution and results\n",
+ "\n",
+ "# (a)\n",
+ "# using characteristic equation, P1*V1=m*R*T1\n",
+ "V1 = m*R*T1*10**-3/P1;# [m**3]\n",
+ "# also Given \n",
+ "V2 = 4*V1;# [m**3]\n",
+ "print ' (a) The original volume of the gas (m^3) = ',round(V1,4)\n",
+ "print ' and The final volume of the gas (m^3) = ',round(V2,3)\n",
+ "\n",
+ "# (b)\n",
+ "# Given that gas is following the law P*V**1.3=constant\n",
+ "# hence process is polytropic with \n",
+ "n = 1.3; # polytropic index\n",
+ "P2 = P1*(V1/V2)**n;# formula for polytropic process,[MN/m**2]\n",
+ "print ' (b) The final pressure of the gas (kN/m^2) = ',round(P2*10**3)\n",
+ "\n",
+ "# (c)\n",
+ "# since mass is constant so,using P*V/T=constant\n",
+ "# hence\n",
+ "T2 = P2*V2*T1/(P1*V1);# [K]\n",
+ "t2 = T2-273;# [C]\n",
+ "print ' (c) The final temperature of the gas (C) = ',round(t2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 12: pg 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.12\n",
+ " (a) The change in internal energy of the air is del_U (kJ) = 30.73\n",
+ "since del_U>0, so it is gain of internal energy to the air\n",
+ " (b) The work done is (kJ) = -49.1\n",
+ "since W<0, so the work is done on the air\n",
+ " (c) The heat transfer is Q (kJ) = -18.4\n",
+ "since Q<0, so the heat is rejected by the air\n",
+ "The answer is a bit different from textbook due to rounding off error\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 120\n",
+ "#calculate the change in internal energy, work done and heat transfer\n",
+ "print('Example 5.12');\n",
+ "\n",
+ "# aim : T0 determine \n",
+ "# (a) change in internal energy of the air\n",
+ "# (b) work done\n",
+ "# (c) heat transfer\n",
+ "\n",
+ "# Given values\n",
+ "m = .25;# mass, [kg]\n",
+ "P1 = 140;# initial pressure, [kN/m**2]\n",
+ "V1 = .15;# initial volume, [m**3]\n",
+ "P2 = 1400;# final volume, [m**3]\n",
+ "cp = 1.005;# [kJ/kg K]\n",
+ "cv = .718;# [kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# assuming ideal gas\n",
+ "R = cp-cv;# [kJ/kg K]\n",
+ "# also, P1*V1=m*R*T1,hence\n",
+ "T1 = P1*V1/(m*R);# [K]\n",
+ "\n",
+ "# given that process is polytropic with \n",
+ "n = 1.25; # polytropic index\n",
+ "T2 = T1*(P2/P1)**((n-1)/n);# [K]\n",
+ "\n",
+ "# Hence, change in internal energy is,\n",
+ "del_U = m*cv*(T2-T1);# [kJ]\n",
+ "print ' (a) The change in internal energy of the air is del_U (kJ) = ',round(del_U,2)\n",
+ "if(del_U>0):\n",
+ " print('since del_U>0, so it is gain of internal energy to the air')\n",
+ "else:\n",
+ " print('since del_U<0, so it is gain of internal energy to the surrounding')\n",
+ "# (b)\n",
+ "W = m*R*(T1-T2)/(n-1);# formula of work done for polytropic process,[kJ]\n",
+ "print ' (b) The work done is (kJ) = ',round(W,1)\n",
+ "if(W>0):\n",
+ " print('since W>0, so the work is done by the air')\n",
+ "else:\n",
+ " print('since W<0, so the work is done on the air')\n",
+ "\n",
+ "\n",
+ "# (c)\n",
+ "Q = del_U+W;# using 1st law of thermodynamics,[kJ]\n",
+ "print ' (c) The heat transfer is Q (kJ) = ',round(Q,2)\n",
+ "if(Q>0):\n",
+ " print('since Q>0, so the heat is received by the air')\n",
+ "else:\n",
+ " print('since Q<0, so the heat is rejected by the air')\n",
+ "\n",
+ "print 'The answer is a bit different from textbook due to rounding off error'\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 13: pg 123"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.13\n",
+ "\n",
+ " The final volume of the gas is V2 (m^3) = 0.048\n",
+ "\n",
+ " The work done by the gas is (kJ) = 9.77\n",
+ "\n",
+ " The change of internal energy is (kJ) = -9.77\n",
+ "since del_U<0, so this is a loss of internal energy from the gas\n",
+ "The answer is a bit different from textbook due to rounding off error\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 123\n",
+ "#calculate the final volume, work done and the change in internal energy\n",
+ "print('Example 5.13');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# final volume, work done and the change in internal energy\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 700.;# initial pressure,[kN/m^2]\n",
+ "V1 = .015;# initial volume, [m^3]\n",
+ "P2 = 140.;# final pressure, [kN/m^2]\n",
+ "cp = 1.046;# [kJ/kg K]\n",
+ "cv = .752; # [kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "Gamma = cp/cv;\n",
+ "# for adiabatic expansion, P*V^gamma=constant, so\n",
+ "V2 = V1*(P1/P2)**(1/Gamma);# final volume, [m^3]\n",
+ "print '\\n The final volume of the gas is V2 (m^3) = ',round(V2,3)\n",
+ "\n",
+ "# work done\n",
+ "W = (P1*V1-P2*V2)/(Gamma-1);# [kJ]\n",
+ "print '\\n The work done by the gas is (kJ) = ',round(W,2)\n",
+ "\n",
+ "# for adiabatic process\n",
+ "del_U = -W;# [kJ]\n",
+ "print '\\n The change of internal energy is (kJ) = ',round(del_U,2)\n",
+ "if(del_U>0):\n",
+ " print 'since del_U>0, so the the gain in internal energy of the gas '\n",
+ "else:\n",
+ " print 'since del_U<0, so this is a loss of internal energy from the gas'\n",
+ "\n",
+ "print 'The answer is a bit different from textbook due to rounding off error'\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 14: pg 125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 14,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.14\n",
+ "\n",
+ " (a) The heat transferred during compression is Q (kJ) = -60.0\n",
+ "\n",
+ " (b) The change of the internal energy during the expansion is,del_U (kJ) = -45.2\n",
+ "\n",
+ " (c) The mass of the gas is,m (kg) = 0.478\n",
+ " There is calculation mistake in the book\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 125\n",
+ "#calculate the heat transfer, change of internal energy and mass of gas\n",
+ "print('Example 5.14');\n",
+ "import math\n",
+ "# aim : To determine the\n",
+ "# (a)heat transfer\n",
+ "# (b)change of internal energy\n",
+ "# (c)mass of gas\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .4;# initial volume, [m^3]\n",
+ "P1 = 100.;# initial pressure, [kN/m^2]\n",
+ "T1 = 273.+20;# temperature, [K]\n",
+ "P2 = 450.;# final pressure,[kN/m^2]\n",
+ "cp = 1.0;# [kJ/kg K]\n",
+ "Gamma = 1.4; # heat capacity ratio\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# for the isothermal compression,P*V=constant,so\n",
+ "V2 = V1*P1/P2;# [m^3]\n",
+ "W = P1*V1*math.log(P1/P2);# formula of workdone for isothermal process,[kJ]\n",
+ "\n",
+ "# for isothermal process, del_U=0;so\n",
+ "Q = W;\n",
+ "print '\\n (a) The heat transferred during compression is Q (kJ) = ',round(Q)\n",
+ "\n",
+ "\n",
+ "# (b)\n",
+ "V3 = V1;\n",
+ "# for adiabatic expansion\n",
+ "# also\n",
+ "\n",
+ "P3 = P2*(V2/V3)**Gamma;# [kN/m^2]\n",
+ "W = -(P3*V3-P2*V2)/(Gamma-1);# work done formula for adiabatic process,[kJ]\n",
+ "# also, Q=0,so using Q=del_U+W\n",
+ "del_U = -W;# [kJ]\n",
+ "print '\\n (b) The change of the internal energy during the expansion is,del_U (kJ) = ',round(del_U,1)\n",
+ "\n",
+ "# (c)\n",
+ "# for ideal gas\n",
+ "# cp-cv=R, and cp/cv=gamma, hence\n",
+ "R = cp*(1-1/Gamma);# [kj/kg K]\n",
+ "\n",
+ "# now using ideal gas equation\n",
+ "m = P1*V1/(R*T1);# mass of the gas,[kg]\n",
+ "print '\\n (c) The mass of the gas is,m (kg) = ',round(m,3)\n",
+ "\n",
+ "print' There is calculation mistake in the book'\n",
+ "\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 15: pg 128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 15,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.15\n",
+ " The heat received or rejected by the gas during this process is Q (kJ) = 1.03\n",
+ "since Q>0, so heat is received by the gas\n",
+ "\n",
+ " The polytropic specific heat capacity is cn (kJ/kg K) = 0.239\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 128\n",
+ "#calculate the the heat transferred and polytropic specific heat capacity\n",
+ "print('Example 5.15');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the heat transferred and polytropic specific heat capacity\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 1;# initial pressure, [MN/m^2]\n",
+ "V1 = .003;# initial volume, [m^3]\n",
+ "P2 = .1;# final pressure,[MN/m^2]\n",
+ "cv = .718;# [kJ/kg*K]\n",
+ "Gamma=1.4;# heat capacity ratio\n",
+ "\n",
+ "# solution\n",
+ "# Given process is polytropic with\n",
+ "n = 1.3;# polytropic index\n",
+ "# hence\n",
+ "V2 = V1*(P1/P2)**(1/n);# final volume,[m^3]\n",
+ "W = (P1*V1-P2*V2)*10**3/(n-1);# work done,[kJ]\n",
+ "# so\n",
+ "Q = (Gamma-n)*W/(Gamma-1);# heat transferred,[kJ]\n",
+ "\n",
+ "print ' The heat received or rejected by the gas during this process is Q (kJ) = ',round(Q,2)\n",
+ "if(Q>0):\n",
+ " print 'since Q>0, so heat is received by the gas'\n",
+ "else:\n",
+ " print 'since Q<0, so heat is rejected by the gas'\n",
+ "\n",
+ "# now\n",
+ "cn = cv*(Gamma-n)/(n-1);# polytropic specific heat capacity,[kJ/kg K]\n",
+ "print '\\n The polytropic specific heat capacity is cn (kJ/kg K) = ',round(cn,3)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 16: pg 129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 16,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.16\n",
+ "\n",
+ " (a) The initial partial pressure of the steam is (kN/m^2) = 7\n",
+ "\n",
+ " The initial partial pressure of the air is (kN/m^2) = 93.0\n",
+ " \n",
+ "(b) The final partial pressure of the steam is (kN/m^2) = 200.0\n",
+ "\n",
+ " The final partial pressure of the air is (kN/m^2) = 117.2\n",
+ "\n",
+ " (c) The total pressure after heating is (kN/m^2) = 317.2\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 129\n",
+ "\n",
+ "print('Example 5.16');\n",
+ "\n",
+ "# To determine the \n",
+ "# (a) initial partial pressure of the steam and air\n",
+ "# (b) final partial pressure of the steam and air\n",
+ "# (c) total pressure in the container after heating\n",
+ "\n",
+ "# Given values\n",
+ "T1 = 273.+39;# initial temperature,[K]\n",
+ "P1 = 100.;# pressure, [MN/m^2]\n",
+ "T2 = 273.+120.2;# final temperature,[K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# from the steam tables, the pressure of wet steam at 39 C is\n",
+ "Pw1 = 7;# partial pressure of wet steam,[kN/m^2]\n",
+ "# and by Dalton's law\n",
+ "Pa1 = P1-Pw1;# initial pressure of air, [kN/m^2]\n",
+ "\n",
+ "print '\\n (a) The initial partial pressure of the steam is (kN/m^2) = ',Pw1\n",
+ "print '\\n The initial partial pressure of the air is (kN/m^2) = ',Pa1\n",
+ "\n",
+ "# (b)\n",
+ "# again from steam table, at 120.2 C the pressure of wet steam is\n",
+ "Pw2 = 200.;# [kN/m^2]\n",
+ "\n",
+ "# now since volume is constant so assuming air to be ideal gas so for air P/T=contant, hence\n",
+ "Pa2 = Pa1*T2/T1 ;# [kN/m^2]\n",
+ "\n",
+ "print ' \\n(b) The final partial pressure of the steam is (kN/m^2) = ',Pw2\n",
+ "print '\\n The final partial pressure of the air is (kN/m^2) = ',round(Pa2,2)\n",
+ "\n",
+ "# (c)\n",
+ "Pt = Pa2+Pw2;# using dalton's law, total pressure,[kN/m^2]\n",
+ "print '\\n (c) The total pressure after heating is (kN/m^2) = ',round(Pt,2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 17: pg 130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 17,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.17\n",
+ "\n",
+ " The partial pressure of the air in the condenser is (kN/m^2) = 6.0\n",
+ "\n",
+ " The partial pressure of the steam in the condenser is (kN/m^2) = 8\n",
+ "\n",
+ " The mass of air which will associated with this steam is (kg) = 1430.5\n",
+ " There is misprint in book\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 130\n",
+ "print('Example 5.17');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the partial pressure of the air and steam, and the mass of the air\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 660.;# vaccum gauge pressure on condenser [mmHg]\n",
+ "P = 765.;# atmospheric pressure, [mmHg]\n",
+ "x = .8;# dryness fraction \n",
+ "T = 273.+41.5;# temperature,[K]\n",
+ "ms_dot = 1500.;# condense rate of steam,[kg/h]\n",
+ "R = .29;# [kJ/kg]\n",
+ "\n",
+ "# solution\n",
+ "Pa = (P-P1)*.1334;# absolute pressure,[kN/m^2]\n",
+ "# from steam table, at 41.5 C partial pressure of steam is\n",
+ "Ps = 8;# [kN/m^2]\n",
+ "# by dalton's law, partial pressure of air is\n",
+ "Pg = Pa-Ps;# [kN/m^2]\n",
+ "\n",
+ "print '\\n The partial pressure of the air in the condenser is (kN/m^2) = ',round(Pg)\n",
+ "print '\\n The partial pressure of the steam in the condenser is (kN/m^2) = ',Ps\n",
+ "\n",
+ "# also\n",
+ "vg = 18.1;# [m^3/kg]\n",
+ "# so\n",
+ "V = x*vg;# [m^3/kg]\n",
+ "# The air associated with 1 kg of the steam will occupiy this same volume\n",
+ "# for air, Pg*V=m*R*T,so\n",
+ "m = Pg*V/(R*T);# [kg/kg steam]\n",
+ "# hence\n",
+ "ma = m*ms_dot;# [kg/h]\n",
+ "\n",
+ "print '\\n The mass of air which will associated with this steam is (kg) = ',round(ma,1)\n",
+ "\n",
+ "print' There is misprint in book'\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 18: pg 130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 18,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.18\n",
+ " (a) The final pressure in the cylinder is (kN/m^2) = 707.1\n",
+ " (b) The final dryness fraction of the steam is = 0.83\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 130\n",
+ "print('Example 5.18');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# (a) final pressure\n",
+ "# (b) final dryness fraction of the steam\n",
+ "\n",
+ "# Given values\n",
+ "P1 = 130.;# initial pressure, [kN/m^2]\n",
+ "T1 = 273.+75.9;# initial temperature, [K]\n",
+ "x1 = .92;# initial dryness fraction\n",
+ "T2 = 273.+120.2;# final temperature, [K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# from steam table, at 75.9 C\n",
+ "Pws = 40.;# partial pressure of wet steam[kN/m^2]\n",
+ "Pa = P1-Pws;# partial pressure of air, [kN/m^2]\n",
+ "vg = 3.99# specific volume of the wet steam, [m^3/kg]\n",
+ "# hence\n",
+ "V1 = x1*vg;# [m^3/kg]\n",
+ "V2 = V1/5;# [m^3/kg]\n",
+ "# for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so\n",
+ "P2 = Pa*V1*T2/(V2*T1);# final pressure,[kN/m^2]\n",
+ "\n",
+ "# now for steam at 120.2 C\n",
+ "Ps = 200.;# final partial pressure of steam,[kN/m^2]\n",
+ "# so by dalton's law total pressure in cylindert is\n",
+ "Pt = P2+Ps;# [kN/m^2]\n",
+ "print ' (a) The final pressure in the cylinder is (kN/m^2) = ',round(Pt,1)\n",
+ "\n",
+ "# (b)\n",
+ "# from steam table at 200 kN/m^2 \n",
+ "vg = .885;# [m^3/kg]\n",
+ "# hence\n",
+ "x2 = V2/vg;# final dryness fraction of the steam\n",
+ "print ' (b) The final dryness fraction of the steam is = ',round(x2,2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 19: pg 131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 19,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.19\n",
+ " (a) The value of adiabatic index Gamma is = 1.426\n",
+ " (b) The change in internal energy during the adiabatic expansion is U2-U1 (This is loss of internal energy) (kJ) = -55.97\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 131\n",
+ "print('Example 5.19')\n",
+ "\n",
+ "# aim : To determine the \n",
+ "# (a) Gamma,\n",
+ "# (b) del_U\n",
+ "import math\n",
+ "# Given Values\n",
+ "P1 = 1400.;# [kN/m^2]\n",
+ "P2 = 100.;# [kN/m^2]\n",
+ "P3 = 220.;# [kN/m^2]\n",
+ "T1 = 273.+360;# [K]\n",
+ "m = .23;# [kg]\n",
+ "cp = 1.005;# [kJ/kg*K]\n",
+ "\n",
+ "# Solution\n",
+ "T3 = T1;# since process 1-3 is isothermal\n",
+ "\n",
+ "# (a)\n",
+ "# for process 1-3, P1*V1=P3*V3,so\n",
+ "V3_by_V1 = P1/P3;\n",
+ "# also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence\n",
+ "# and process process 2-3 is iso-choric so,V3=V2 and\n",
+ "V2_by_V1 = V3_by_V1;\n",
+ "# hence,\n",
+ "Gamma = math.log(P1/P2)/math.log(P1/P3); # heat capacity ratio\n",
+ "\n",
+ "print ' (a) The value of adiabatic index Gamma is = ',round(Gamma,3)\n",
+ "\n",
+ "# (b)\n",
+ "cv = cp/Gamma;# [kJ/kg K]\n",
+ "# for process 2-3,P3/T3=P2/T2,so\n",
+ "T2 = P2*T3/P3;# [K]\n",
+ "\n",
+ "# now\n",
+ "del_U = m*cv*(T2-T1);# [kJ]\n",
+ "print ' (b) The change in internal energy during the adiabatic expansion is U2-U1 (This is loss of internal energy) (kJ) = ',round(del_U,2)\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 20: pg 133"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 20,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.20\n",
+ " The mass of oxygen used (kg) = 5.5\n",
+ " The amount of heat transferred through the cylinder wall is (kJ) = 13.28\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 133\n",
+ "print('Example 5.20');\n",
+ "\n",
+ "# aim : To determine \n",
+ "# the mass of oxygen and heat transferred\n",
+ "\n",
+ "# Given values\n",
+ "V1 = 300.;# [L]\n",
+ "P1 = 3.1;# [MN/m^2]\n",
+ "T1 = 273.+18;# [K]\n",
+ "P2 = 1.7;# [MN/m^2]\n",
+ "T2 = 273.+15;# [K]\n",
+ "Gamma = 1.4; # heat capacity ratio\n",
+ "# density condition\n",
+ "P = .101325;# [MN/m^2]\n",
+ "T = 273.;# [K]\n",
+ "V = 1.;# [m^3]\n",
+ "m = 1.429;# [kg]\n",
+ "\n",
+ "# hence\n",
+ "R = P*V*10**3/(m*T);# [kJ/kg*K]\n",
+ "# since volume is constant\n",
+ "V2 = V1;# [L]\n",
+ "# for the initial conditions in the cylinder,P1*V1=m1*R*T1\n",
+ "m1 = P1*V1/(R*T1);# [kg]\n",
+ "\n",
+ "# after some of the gas is used\n",
+ "m2 = P2*V2/(R*T2);# [kg]\n",
+ "# The mass of oxygen remaining in cylinder is m2 kg,so\n",
+ "# Mass of oxygen used is\n",
+ "m_used = m1-m2;# [kg]\n",
+ "print ' The mass of oxygen used (kg) = ',round(m_used,1)\n",
+ "\n",
+ "# for non-flow process,Q=del_U+W\n",
+ "# volume is constant so no external work is done so,Q=del_U\n",
+ "cv = R/(Gamma-1);# [kJ/kg*K]\n",
+ "\n",
+ "# heat transfer is\n",
+ "Q = m2*cv*(T1-T2);# (kJ)\n",
+ "print ' The amount of heat transferred through the cylinder wall is (kJ) = ',round(Q,2)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 21: pg 134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 21,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.21\n",
+ " (a) The work transferred during the compression is (kJ) = -28.1\n",
+ " (b) The change in internal energy is (kJ) = 14.2\n",
+ " (c) The heat transferred during the compression is (kJ) = -14.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 134\n",
+ "print('Example 5.21');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# (a) work transferred during the compression\n",
+ "# (b) change in internal energy\n",
+ "# (c) heat transferred during the compression\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .1;# initial volume, [m^3]\n",
+ "P1 = 120.;# initial pressure, [kN/m^2]\n",
+ "P2 = 1200.; # final pressure, [kN/m^2]\n",
+ "T1 = 273.+25;# initial temperature, [K]\n",
+ "cv = .72;# [kJ/kg*K]\n",
+ "R = .285;# [kJ/kg*K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# given process is polytropic with\n",
+ "n = 1.2; # polytropic index\n",
+ "# hence\n",
+ "V2 = V1*(P1/P2)**(1/n);# [m^3]\n",
+ "W = (P1*V1-P2*V2)/(n-1);# workdone formula, [kJ]\n",
+ "print ' (a) The work transferred during the compression is (kJ) = ',round(W,1)\n",
+ "\n",
+ "# (b)\n",
+ "# now mass is constant so,\n",
+ "T2 = P2*V2*T1/(P1*V1);# [K]\n",
+ "# using, P*V=m*R*T\n",
+ "m = P1*V1/(R*T1);# [kg]\n",
+ "\n",
+ "# change in internal energy is\n",
+ "del_U = m*cv*(T2-T1);# [kJ]\n",
+ "print ' (b) The change in internal energy is (kJ) = ',round(del_U,1)\n",
+ "\n",
+ "# (c)\n",
+ "Q = del_U+W;# [kJ]\n",
+ "print ' (c) The heat transferred during the compression is (kJ) = ',round(Q,0)\n",
+ "\n",
+ "# End\n",
+ "\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 22: pg 135"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 22,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.22\n",
+ " (a) The new pressure of the air in the receiver is (kN/m^2) = 442.0\n",
+ " (b) The specific enthalpy of the air at 15 C is (kJ/kg) = 15.075\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 135\n",
+ "print('Example 5.22');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# (a) new pressure of the air in the receiver\n",
+ "# (b) specific enthalpy of air at 15 C\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .85;# [m^3]\n",
+ "T1 = 15.+273;# [K]\n",
+ "P1 = 275.;# pressure,[kN/m^2]\n",
+ "m = 1.7;# [kg]\n",
+ "cp = 1.005;# [kJ/kg*K]\n",
+ "cv = .715;# [kJ/kg*K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "\n",
+ "R = cp-cv;# [kJ/kg*K]\n",
+ "# assuming m1 is original mass of the air, using P*V=m*R*T\n",
+ "m1 = P1*V1/(R*T1);# [kg]\n",
+ "m2 = m1+m;# [kg]\n",
+ "# again using P*V=m*R*T\n",
+ "# P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so\n",
+ "P2 = P1*m2/m1;# [kN/m^2]\n",
+ "print ' (a) The new pressure of the air in the receiver is (kN/m^2) = ',round(P2)\n",
+ "\n",
+ "# (b)\n",
+ "# for 1 kg of air, h2-h1=cp*(T1-T0)\n",
+ "# and if 0 is chosen as the zero enthalpy, then\n",
+ "h = cp*(T1-273);# [kJ/kg]\n",
+ "print ' (b) The specific enthalpy of the air at 15 C is (kJ/kg) = ',h\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 23: pg 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 23,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.23\n",
+ " (a) The characteristic gas constant of the gas is R (kJ/kg K) = 0.185\n",
+ " (b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) = 0.827\n",
+ " (c) The specific heat capacity of the gas at constant volume cv (kJ/kg K) = 0.642\n",
+ " (d) The change in internal energy is (kJ) = 136.0\n",
+ " (e) The work transfer is W (kJ) = 39.0\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 136\n",
+ "print('Example 5.23');\n",
+ "\n",
+ "# aim : T determine the\n",
+ "# (a) characteristic gas constant of the gas\n",
+ "# (b) cp,\n",
+ "# (c) cv,\n",
+ "# (d) del_u \n",
+ "# (e) work transfer\n",
+ "\n",
+ "# Given values\n",
+ "P = 1.;# [bar] \n",
+ "T1 = 273.+15;# [K]\n",
+ "m = .9;# [kg]\n",
+ "T2 = 273.+250;# [K]\n",
+ "Q = 175.;# heat transfer,[kJ]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# using, P*V=m*R*T, given,\n",
+ "m_by_V = 1.875;\n",
+ "# hence\n",
+ "R = P*100/(T1*m_by_V);# [kJ/kg*K]\n",
+ "print ' (a) The characteristic gas constant of the gas is R (kJ/kg K) = ',round(R,3)\n",
+ "\n",
+ "# (b)\n",
+ "# using, Q=m*cp*(T2-T1)\n",
+ "cp = Q/(m*(T2-T1));# [kJ/kg K]\n",
+ "print ' (b) The specific heat capacity of the gas at constant pressure cp (kJ/kg K) = ',round(cp,3)\n",
+ "\n",
+ "# (c)\n",
+ "# we have, cp-cv=R,so\n",
+ "cv = cp-R;# [kJ/kg*K]\n",
+ "print ' (c) The specific heat capacity of the gas at constant volume cv (kJ/kg K) = ',round(cv,3)\n",
+ "\n",
+ "# (d)\n",
+ "del_U = m*cv*(T2-T1);# [kJ]\n",
+ "print ' (d) The change in internal energy is (kJ) = ',round(del_U)\n",
+ "\n",
+ "# (e)\n",
+ "# using, Q=del_U+W\n",
+ "W = Q-del_U;# [kJ]\n",
+ "print ' (e) The work transfer is W (kJ) = ',round(W)\n",
+ "\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 24: pg 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 24,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.24\n",
+ " (a) The work transfer is W (kJ) = 198.2\n",
+ " (b) The change of internal energy is del_U (kJ) = -157.3\n",
+ " (c) The heat transfer Q (kJ) = 40.8\n",
+ "The answer is a bit different due to rounding off error in textbook\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 136\n",
+ "print('Example 5.24');\n",
+ "\n",
+ "# aim : To determine the\n",
+ "# (a) work transfer,\n",
+ "# (b)del_U and,\n",
+ "# (c)heat transfer\n",
+ "\n",
+ "# Given values\n",
+ "V1 = .15;# [m^3]\n",
+ "P1 = 1200.;# [kN/m^2]\n",
+ "T1 = 273.+120;# [K]\n",
+ "P2 = 200.;# [kN/m^2]\n",
+ "cp = 1.006;#[kJ/kg K]\n",
+ "cv = .717;# [kJ/kg K]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# (a)\n",
+ "# Given, PV^1.32=constant, so it is polytropic process with\n",
+ "n = 1.32;# polytropic index\n",
+ "# hence\n",
+ "V2 = V1*(P1/P2)**(1./n);# [m^3]\n",
+ "# now, W\n",
+ "W = (P1*V1-P2*V2)/(n-1);# [kJ]\n",
+ "print ' (a) The work transfer is W (kJ) = ',round(W,1)\n",
+ "\n",
+ "# (b)\n",
+ "R = cp-cv;# [kJ/kg K]\n",
+ "m = P1*V1/(R*T1);# gas law,[kg]\n",
+ "# also for polytropic process\n",
+ "T2 = T1*(P2/P1)**((n-1)/n);# [K]\n",
+ "# now for gas,\n",
+ "del_U = m*cv*(T2-T1);# [kJ]\n",
+ "print ' (b) The change of internal energy is del_U (kJ) = ',round(del_U,1)\n",
+ "\n",
+ "# (c)\n",
+ "Q = del_U+W;# first law of thermodynamics,[kJ]\n",
+ "print ' (c) The heat transfer Q (kJ) = ',round(Q,1)\n",
+ "\n",
+ "print 'The answer is a bit different due to rounding off error in textbook'\n",
+ "# End\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 26: pg 141"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 25,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 5.26\n",
+ "heat transfer from the gas (kJ) = -248.2\n",
+ " The volume of gas before transfer is (m^3) = 0.145\n",
+ " The volume of pressure vessel is (m^3) = 0.27\n"
+ ]
+ }
+ ],
+ "source": [
+ "#pg 141\n",
+ "print('Example 5.26');\n",
+ "\n",
+ "# aim : To determine\n",
+ "# the volume of the pressure vessel and the volume of the gas before transfer\n",
+ "\n",
+ "# Given values\n",
+ "\n",
+ "P1 = 1400.;# initial pressure,[kN/m^2]\n",
+ "T1 = 273.+85;# initial temperature,[K]\n",
+ "\n",
+ "P2 = 700.;# final pressure,[kN/m^2]\n",
+ "T2 = 273.+60;# final temperature,[K]\n",
+ "\n",
+ "m = 2.7;# mass of the gas passes,[kg]\n",
+ "cp = .88;# [kJ/kg]\n",
+ "cv = .67;# [kJ/kg]\n",
+ "\n",
+ "# solution\n",
+ "\n",
+ "# steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W [1], \n",
+ "# given, there is no kinetic energy change and neglecting potential energy term\n",
+ "W = 0;# no external work done\n",
+ "# so final equation is,u1+P1*v1+Q=u2 [2]\n",
+ "# also u2-u1=cv*(T2-T1)\n",
+ "# hence Q=cv*(T2-T1)-P1*v1 [3]\n",
+ "# and for unit mass P1*v1=R*T1=(cp-cv)*T1 [4]\n",
+ "# so finally\n",
+ "Q = cv*(T2-T1)-(cp-cv)*T1;# [kJ/kg]\n",
+ "# so total heat transferred is\n",
+ "Q2 = m*Q;# [kJ] \n",
+ "\n",
+ "# using eqn [4]\n",
+ "v1 = (cp-cv)*T1/P1;# [m^3/kg]\n",
+ "# Total volume is\n",
+ "V1 = m*v1;# [m^3]\n",
+ "\n",
+ "# using ideal gas equation P1*V1/T1=P2*V2/T2\n",
+ "V2 = P1*T2*V1/(P2*T1);# final volume,[m^3]\n",
+ "\n",
+ "print 'heat transfer from the gas (kJ) = ',round(Q2,1)\n",
+ "print ' The volume of gas before transfer is (m^3) = ',round(V1,3)\n",
+ "print ' The volume of pressure vessel is (m^3) = ',round(V2,2)\n",
+ " \n",
+ "# End\n"
+ ]
+ }
+ ],
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generated by cgit (git 2.34.1) at 2024-12-21 06:40:09 +0000