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diff --git a/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter15_1.ipynb b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter15_1.ipynb new file mode 100644 index 00000000..31dadb42 --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter15_1.ipynb @@ -0,0 +1,1424 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 15 - Ideal gas power cycles" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 1: pg 436" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.1\n", + " The thermal efficiency of the cycle is (percent) = 49.0\n" + ] + } + ], + "source": [ + "#pg 436\n", + "print('Example 15.1');\n", + "\n", + "# aim : To determine \n", + "# the thermal efficiency of the cycle\n", + "\n", + "# given values\n", + "T1 = 273.+400;# temperature limit, [K]\n", + "T3 = 273.+70;# temperature limit, [K]\n", + "\n", + "# solution\n", + "# using equation [15] of section 15.3\n", + "n_the = (T1-T3)/T1*100;# thermal efficiency \n", + "#results\n", + "print ' The thermal efficiency of the cycle is (percent) = ',round(n_the)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2: pg 437" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.2\n", + " (a) The volume ratio of the adiabatic process is = 4.425\n", + " The volume ratio of the isothermal process is = 3.39\n", + " (b) The thermal efficiency of the cycle is (percent) = 44.8\n" + ] + } + ], + "source": [ + "#pg 437\n", + "print('Example 15.2');\n", + "\n", + "# aim : To determine\n", + "# (a) the volume ratios of the isothermal and adiabatic processes\n", + "# (b) the thermal efficiency of the cycle\n", + "\n", + "# given values\n", + "T1 = 273.+260;# temperature, [K]\n", + "T3 = 273.+21;# temperature, [K]\n", + "er = 15.;# expansion ratio\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# (a)\n", + "T2 = T1;\n", + "T4 = T3;\n", + "# for adiabatic process\n", + "rva = (T1/T4)**(1/(Gama-1));# volume ratio of adiabatic\n", + "rvi = er/rva;# volume ratio of isothermal\n", + "print ' (a) The volume ratio of the adiabatic process is = ',round(rva,3)\n", + "print ' The volume ratio of the isothermal process is = ',round(rvi,2)\n", + "\n", + "# (b)\n", + "n_the = (T1-T4)/T1*100;# thermal efficiency\n", + "print ' (b) The thermal efficiency of the cycle is (percent) = ',round(n_the,1)\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 3: pg 438" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.3\n", + "(a) At line 1\n", + " V1 = 0.096 m^3, t1 = 300.0 C, P1 = 1730.0 kN/m^2 \n", + "At line 2\n", + " V2 = 0.288 m^3, t2 = 300.0 C, P2 = 576.7 kN/m^2\n", + "At line 3\n", + " V3 = 0.576 m^3, t3 = 161.0 C, P3 = 218.5 kN/m^2\n", + "At line 4\n", + " V4 = 0.192 m^3, t4 = 161.3 C, P4 = 655.5 kN/m^2\n", + " (b) The thermal efficiency of the cycle (percent) = 24.0\n", + " (c) The work done per cycle is (kJ) = 44.2\n", + " (d) The work ratio is = 0.156\n", + "there is calculation mistake in the book so answer is not matching\n" + ] + } + ], + "source": [ + "#pg 438\n", + "print('Example 15.3');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure, volume and temperature at each corner of the cycle\n", + "# (b) the thermal efficiency of the cycle\n", + "# (c) the work done per cycle\n", + "# (d) the work ratio\n", + "from math import log\n", + "# given values\n", + "m = 1;# mass of air, [kg]\n", + "P1 = 1730.;# initial pressure of carnot engine, [kN/m^2]\n", + "T1 = 273.+300;# initial temperature, [K]\n", + "R = .29;# [kJ/kg K]\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.15\n", + "# (a)\n", + "# for the isothermal process 1-2\n", + "# using ideal gas law\n", + "V1 = m*R*T1/P1;# initial volume, [m^3]\n", + "T2 = T1;\n", + "V2 = 3.*V1;# given condition\n", + "# for isothermal process, P1*V1=P2*V2, so\n", + "P2 = P1*(V1/V2);# [MN/m^2]\n", + "# for the adiabatic process 2-3\n", + "V3 = 6.*V1;# given condition\n", + "T3 = T2*(V2/V3)**(Gama-1);\n", + "# also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so\n", + "P3 = P2*(V2/V3)**Gama;\n", + "# for the isothermal process 3-4\n", + "T4 = T3;\n", + "# for both adiabatic processes, the temperataure ratio is same, \n", + "# T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so\n", + "V4 = 2*V1;\n", + "# for isothermal process, 3-4, P3*V3=P4*V4, so\n", + "P4 = P3*(V3/V4);\n", + "print '(a) At line 1'\n", + "print ' V1 = ',round(V1,3),' m^3, t1 = ',T1-273,' C, P1 = ',P1,' kN/m^2 '\n", + "\n", + "print 'At line 2'\n", + "print ' V2 = ',round(V2,3),' m^3, t2 = ',T2-273,' C, P2 = ',round(P2,1),' kN/m^2'\n", + "\n", + "print 'At line 3'\n", + "print ' V3 = ',round(V3,3),' m^3, t3 = ',round(T3-273),' C, P3 = ',round(P3,1),' kN/m^2'\n", + "\n", + "\n", + "print 'At line 4'\n", + "print ' V4 = ',round(V4,3),' m^3, t4 = ',round(T4-273,1),' C, P4 = ',round(P4,1),' kN/m^2'\n", + "\n", + "\n", + "# (b)\n", + "n_the = (T1-T3)/T1;# thermal efficiency\n", + "print ' (b) The thermal efficiency of the cycle (percent) = ',round(n_the*100)\n", + "\n", + "# (c)\n", + "W = m*R*T1*log(V2/V1)*n_the;# work done, [J]\n", + "print ' (c) The work done per cycle is (kJ) = ',round(W,1)\n", + "\n", + "# (d)\n", + "wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));# work ratio\n", + "print ' (d) The work ratio is = ',round(wr,3)\n", + "\n", + "print 'there is calculation mistake in the book so answer is not matching'\n", + "\n", + "# End\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 4: pg 446" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.4\n", + " (a) P1 = 100.0 kN/m^2, V1 = 0.084 m^3, t1 = 28.0 C,\n", + " P2 = 1229.0 kN/m^2, V2 = 0.014 m^3, t2 = 343.0 C,\n", + " P3 = 1229.0 kN/m^2, V3 = 0.019 m^3, t3 = 549.0 C,\n", + " P4 = 100.0 kN/m^2, V4 = 0.112 m^3, t4 = 128.0 C\n", + " (b) The heat received is (kJ) = 20.07\n", + " (c) The work done is (kJ) = 10.27\n", + " (d) The thermal efficiency is (percent) = 51.2\n", + " (e) The carnot efficiency is (percent) = 63.4\n", + " (f) The work ratio is = 0.293\n", + " (g) The mean effective pressure is (kN/m^2) = 104.77\n", + " there is minor variation in answer reported in the book due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 446\n", + "print('Example 15.4');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure, volume and temperature at cycle state points\n", + "# (b) the heat received\n", + "# (c) the work done \n", + "# (d) the thermal efficiency\n", + "# (e) the carnot efficiency\n", + "# (f) the work ration\n", + "# (g) the mean effective pressure\n", + "\n", + "# given values\n", + "ro = 8.;# overall volume ratio;\n", + "rv = 6.;# volume ratio of adiabatic compression\n", + "P1 = 100.;# initial pressure , [kN/m^2]\n", + "V1 = .084;# initial volume, [m^3]\n", + "T1 = 273.+28;# initial temperature, [K]\n", + "Gama = 1.4;# heat capacity ratio\n", + "cp = 1.006;# specific heat capacity, [kJ/kg K]\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.18\n", + "# (a)\n", + "V2 = V1/rv;# volume at stage2, [m^3] \n", + "V4 = ro*V2;# volume at stage 4;[m^3]\n", + "# using PV^(Gama)=constant for process 1-2\n", + "P2 = P1*(V1/V2)**(Gama);# pressure at stage2,. [kN/m^2]\n", + "T2 = T1*(V1/V2)**(Gama-1);# [K]\n", + "\n", + "P3 = P2;# pressure at stage 3, [kN/m^2]\n", + "V3 = V4/rv;# volume at stage 3, [m^3]\n", + "# since pressure is constant in process 2-3 , so using V/T=constant, so\n", + "T3 = T2*(V3/V2);# temperature at stage 3, [K]\n", + "\n", + "# for process 1-4\n", + "T4 = T1*(V4/V1);# temperature at stage4, [K\n", + "P4 = P1;# pressure at stage4, [kN/m^2]\n", + "\n", + "print ' (a) P1 = ',P1,' kN/m^2, V1 = ',V1,' m^3, t1 = ',T1-273,' C,\\n P2 = ',round(P2),' kN/m^2, V2 = ',V2,' m^3, t2 = ',round(T2-273),' C,\\n P3 = ',round(P3),' kN/m^2, V3 = ',round(V3,3),' m^3, t3 = ',round(T3-273),' C,\\n P4 = ',P4,' kN/m^2, V4 = ',V4,' m^3, t4 = ',round(T4-273),' C'\n", + "\n", + "# (b)\n", + "R = cp*(Gama-1)/Gama;# gas constant, [kJ/kg K]\n", + "m = P1*V1/(R*T1);# mass of gas, [kg]\n", + "Q = m*cp*(T3-T2);# heat received, [kJ]\n", + "print ' (b) The heat received is (kJ) = ',round(Q,2)\n", + "\n", + "# (c) \n", + "W = P2*(V3-V2)-P1*(V4-V1)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);# work done, [kJ]\n", + "print ' (c) The work done is (kJ) = ',round(W,2)\n", + "\n", + "# (d)\n", + "TE = 1-T1/T2;# thermal efficiency\n", + "print ' (d) The thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# (e)\n", + "CE = (T3-T1)/T3;# carnot efficiency\n", + "print ' (e) The carnot efficiency is (percent) = ',round(CE*100,1)\n", + "\n", + "# (f)\n", + "PW = P2*(V3-V2)+(P3*V3-P4*V4)/(Gama-1);# positive work done, [kj]\n", + "WR = W/PW;# work ratio\n", + "print ' (f) The work ratio is = ',round(WR,3)\n", + "\n", + "# (g)\n", + "Pm = W/(V4-V2);# mean effective pressure, [kN/m^2]\n", + "print ' (g) The mean effective pressure is (kN/m^2) = ',round(Pm,2)\n", + "\n", + "print ' there is minor variation in answer reported in the book due to rounding off error'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 5: pg 450" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.5\n", + " (a) The actual thermal efficiency of the turbine is (percent) = 26.88\n", + " (b) The specific fuel consumption of the turbine is (kg/kWh) = 0.311\n" + ] + } + ], + "source": [ + "#pg 450\n", + "print('Example 15.5');\n", + "\n", + "# aim : To determine\n", + "# (a) the actual thermal efficiency of the turbine\n", + "# (b) the specific fuel consumption of the turbine in kg/kWh\n", + "\n", + "# given values\n", + "P2_by_P1 = 8;\n", + "n_tur = .6;# ideal turbine thermal efficiency\n", + "c = 43.*10**3;# calorific value of fuel, [kJ/kg]\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# (a)\n", + "rv = P2_by_P1;\n", + "n_tur_ide = 1-1/(P2_by_P1)**((Gama-1)/Gama);# ideal thermal efficiency\n", + "ate = n_tur_ide*n_tur;# actual thermal efficiency\n", + "print ' (a) The actual thermal efficiency of the turbine is (percent) = ',round(ate*100,2)\n", + "\n", + "# (b)\n", + "ewf = c*ate;# energy to work fuel, [kJ/kg]\n", + "kWh = 3600.;# energy equivalent ,[kJ]\n", + "sfc = kWh/ewf;# specific fuel consumption, [kg/kWh]\n", + "print ' (b) The specific fuel consumption of the turbine is (kg/kWh) = ',round(sfc,3)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 6: pg 456" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.6\n", + " The relative efficiency of the engine is (percent) = 38.8\n" + ] + } + ], + "source": [ + "#pg 456\n", + "print('Example 15.6');\n", + "\n", + "# aim : To determine\n", + "# the relative efficiency of the engine\n", + "import math\n", + "# given values\n", + "d = 80;# bore, [mm]\n", + "l = 85;# stroke, [mm]\n", + "V1 = .06*10**6;# clearence volume, [mm^3]\n", + "ate = .22;# actual thermal efficiency of the engine\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "sv = math.pi*d**2/4*l;# stroke volume, [mm^3]\n", + "V2 = sv+V1;# [mm^3]\n", + "rv = V2/V1;\n", + "ite = 1-(1/rv)**(Gama-1);# ideal thermal efficiency\n", + "re = ate/ite;# relative thermal efficiency\n", + "#results\n", + "print ' The relative efficiency of the engine is (percent) = ',round(re*100,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 7: pg 457" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.7\n", + " (a) P1 = 103.0 kN/m^2, V1 = 1.039 m^3, t1 = 100.0 C,\n", + " P2 = 1265.0 kN/m^2, V2 = 0.173 m^3, t2 = 491.0 C,\n", + " P3 = 3450.0 kN/m^2, V3 = 0.173 m^3, t3 = 1809.0 C,\n", + " P4 = 280.8 kN/m^2, V4 = 1.039 m^3, t4 = 744.0 C\n", + " (b) The heat transferred to the air is (kJ) = 946.0\n", + " (c) The heat rejected by the air is (kJ) = 462.0\n", + " (d) The ideal thermal efficiency is (percent) = 51.2\n", + " (e) The work done is (kJ) = 484.0\n", + " (f) The mean effective pressure is (kN/m^2) = 558.85\n", + "The answers are a bit different due to rounding off error in textbook\n" + ] + } + ], + "source": [ + "#pg 457\n", + "print('Example 15.7');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure, volume and temperature at each cycle process change points\n", + "# (b) the heat transferred to air\n", + "# (c) the heat rejected by the air\n", + "# (d) the ideal thermal efficiency\n", + "# (e) the work done \n", + "# (f) the mean effective pressure\n", + "\n", + "# given values\n", + "m = 1.;# mass of air, [kg]\n", + "rv = 6.;# volume ratio of adiabatic compression\n", + "P1 = 103.;# initial pressure , [kN/m^2]\n", + "T1 = 273.+100;# initial temperature, [K]\n", + "P3 = 3450.;# maximum pressure, [kN/m^2]\n", + "Gama = 1.4;# heat capacity ratio\n", + "R = .287;# gas constant, [kJ/kg K]\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.20\n", + "# (a)\n", + "# for point 1\n", + "V1 = m*R*T1/P1;# initial volume, [m^3]\n", + "\n", + "# for point 2\n", + "V2 = V1/rv;# volume at point 2, [m^3] \n", + "# using PV^(Gama)=constant for process 1-2\n", + "P2 = P1*(V1/V2)**(Gama);# pressure at point 2,. [kN/m^2]\n", + "T2 = T1*(V1/V2)**(Gama-1);# temperature at point 2,[K]\n", + "\n", + "# for point 3\n", + "V3 = V2;# volume at point 3, [m^3]\n", + "# since volume is constant in process 2-3 , so using P/T=constant, so\n", + "T3 = T2*(P3/P2);# temperature at stage 3, [K]\n", + "\n", + "# for point 4\n", + "V4 = V1;# volume at point 4, [m^3]\n", + "P4 = P3*(V3/V4)**Gama;# pressure at point 4, [kN/m^2] \n", + "# again since volume is constant in process 4-1 , so using P/T=constant, so\n", + "T4 = T1*(P4/P1);# temperature at point 4, [K]\n", + "\n", + "print ' (a) P1 = ',P1,' kN/m^2, V1 = ',round(V1,3),' m^3, t1 = ',T1-273,' C,\\n P2 = ',round(P2),' kN/m^2, V2 = ',round(V2,3),' m^3, t2 = ',round(T2-273),' C,\\n P3 = ',round(P3),' kN/m^2, V3 = ',round(V3,3),' m^3, t3 = ',round(T3-273),' C,\\n P4 = ',round(P4,1),' kN/m^2, V4 = ',round(V4,3),' m^3, t4 = ',round(T4-273),' C'\n", + "\n", + "# (b)\n", + "cv = R/(Gama-1);# specific heat capacity, [kJ/kg K]\n", + "Q23 = m*cv*(T3-T2);# heat transferred, [kJ]\n", + "print ' (b) The heat transferred to the air is (kJ) = ',round(Q23)\n", + "\n", + "# (c) \n", + "Q34 = m*cv*(T4-T1);# heat rejected by air, [kJ]\n", + "print ' (c) The heat rejected by the air is (kJ) = ',round(Q34,1)\n", + "\n", + "# (d)\n", + "TE = 1-Q34/Q23;# ideal thermal efficiency\n", + "print ' (d) The ideal thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# (e)\n", + "W = Q23-Q34;# work done ,[kJ]\n", + "print ' (e) The work done is (kJ) = ',round(W,1)\n", + "# (f)\n", + "Pm = W/(V1-V2);# mean effective pressure, [kN/m^2]\n", + "print ' (f) The mean effective pressure is (kN/m^2) = ',round(Pm,2)\n", + "\n", + "print 'The answers are a bit different due to rounding off error in textbook'\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8: pg 460" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.8\n", + " (a) P1 = 101.0 kN/m^2, V1 = 0.003 m^3, t1 = 18.0 C,\n", + " P2 = 2213.0 kN/m^2, V2 = 0.0 m^3, t2 = 436.0 C,\n", + " P3 = 4500.0 kN/m^2, V3 = 0.0 m^3, t3 = 1168.0 C,\n", + " P4 = 205.3 kN/m^2, V4 = 0.003 m^3, t4 = 319.0 C\n", + " (b) The thermal efficiency is (percent) = 59.0\n", + " (c) The theoretical output is (kW) = 55.5\n", + " (g) The mean effefctive pressure is (kN/m^2) = 415.9\n", + " (e) The carnot efficiency is (percent) = 79.8\n" + ] + } + ], + "source": [ + "#pg 460\n", + "print('Example 15.8');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure, volume and temperature at cycle state points\n", + "# (b) the thermal efficiency\n", + "# (c) the theoretical output\n", + "# (d) the mean effective pressure\n", + "# (e) the carnot efficiency\n", + "\n", + "# given values\n", + "rv = 9.;# volume ratio\n", + "P1 = 101.;# initial pressure , [kN/m^2]\n", + "V1 = .003;# initial volume, [m^3]\n", + "T1 = 273.+18;# initial temperature, [K]\n", + "P3 = 4500.;# maximum pressure, [kN/m^2]\n", + "N = 3000.;\n", + "cp = 1.006;# specific heat capacity at constant pressure, [kJ/kg K]\n", + "cv = .716;# specific heat capacity at constant volume, [kJ/kg K]\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.20\n", + "# (a)\n", + "# for process 1-2\n", + "Gama = cp/cv;# heat capacity ratio\n", + "R = cp-cv;# gas constant, [kJ/kg K]\n", + "V2 = V1/rv;# volume at stage2, [m^3] \n", + "# using PV^(Gama)=constant for process 1-2\n", + "P2 = P1*(V1/V2)**(Gama);# pressure at stage2,. [kN/m^2]\n", + "T2 = T1*(V1/V2)**(Gama-1);# [K]\n", + "\n", + "# for process 2-3\n", + "V3 = V2;# volume at stage 3, [m^3]\n", + "# since volume is constant in process 2-3 , so using P/T=constant, so\n", + "T3 = T2*(P3/P2);# temperature at stage 3, [K]\n", + "\n", + "# for process 3-4\n", + "V4 = V1;# volume at stage 4\n", + "# using PV^(Gama)=constant for process 3-4\n", + "P4 = P3*(V3/V4)**(Gama);# pressure at stage2,. [kN/m^2]\n", + "T4 = T3*(V3/V4)**(Gama-1);# temperature at stage 4,[K]\n", + "\n", + "print ' (a) P1 = ',P1,' kN/m^2, V1 = ',round(V1,3),' m^3, t1 = ',T1-273,' C,\\n P2 = ',round(P2),' kN/m^2, V2 = ',round(V2,3),' m^3, t2 = ',round(T2-273),' C,\\n P3 = ',round(P3),' kN/m^2, V3 = ',round(V3,3),' m^3, t3 = ',round(T3-273),' C,\\n P4 = ',round(P4,1),' kN/m^2, V4 = ',round(V4,3),' m^3, t4 = ',round(T4-273),' C'\n", + "\n", + "# (b)\n", + "TE = 1-(T4-T1)/(T3-T2);# thermal efficiency\n", + "print ' (b) The thermal efficiency is (percent) = ',round(TE*100)\n", + "\n", + "# (c)\n", + "m = P1*V1/(R*T1);# mass os gas, [kg] \n", + "W = m*cv*((T3-T2)-(T4-T1));# work done, [kJ]\n", + "Wt = W*N/60;# workdone per minute, [kW]\n", + "print ' (c) The theoretical output is (kW) = ',round(Wt,1)\n", + "\n", + "# (d)\n", + "Pm = W/(V1-V2);# mean effective pressure, [kN/m^2]\n", + "print ' (g) The mean effefctive pressure is (kN/m^2) = ',round(Pm,1)\n", + "\n", + "# (e)\n", + "CE = (T3-T1)/T3;# carnot efficiency\n", + "print ' (e) The carnot efficiency is (percent) = ',CE*100\n", + "\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9: pg 467" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.9\n", + " (a) P1 = 90.0 kN/m^2, V1 = 0.998 m^3, t1 = 40.0 C,\n", + " P2 = 4365.0 kN/m^2, V2 = 0.062 m^3, t2 = 676.0 C,\n", + " P3 = 4365.0 kN/m^2, V3 = 0.11 m^3, t3 = 1400.0 C,\n", + " P4 = 199.1 kN/m^2, V4 = 0.998 m^3, t4 = 419.0 C\n", + " (b) The work done is (kJ) = 454.959974156\n", + " (c) The thermal efficiency is (percent) = 62.6\n", + " (d) The work ratio is = 0.318\n", + " (e) The mean effefctive pressure is (kN/m^2) = 486.4\n", + " (f) The carnot efficiency is (percent) = 81.3\n", + "value of t2 printed in the book is incorrect\n" + ] + } + ], + "source": [ + "#pg 467\n", + "print('Example 15.9');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure and temperature at cycle process change points\n", + "# (b) the work done \n", + "# (c) the thermal efficiency\n", + "# (d) the work ratio\n", + "# (e) the mean effective pressure\n", + "# (f) the carnot efficiency\n", + "\n", + "\n", + "# given values\n", + "rv = 16.;# volume ratio of compression\n", + "P1 = 90.;# initial pressure , [kN/m^2]\n", + "T1 = 273.+40;# initial temperature, [K]\n", + "T3 = 273.+1400;# maximum temperature, [K]\n", + "cp = 1.004;# specific heat capacity at constant pressure, [kJ/kg K]\n", + "Gama = 1.4;# heat capacoty ratio\n", + "\n", + "# solution\n", + "cv = cp/Gama;# specific heat capacity at constant volume, [kJ/kg K]\n", + "R = cp-cv;# gas constant, [kJ/kg K]\n", + "# for one kg of gas\n", + "V1 = R*T1/P1;# initial volume, [m^3]\n", + "# taking reference Fig. 15.22\n", + "# (a)\n", + "# for process 1-2\n", + "# using PV^(Gama)=constant for process 1-2\n", + "# also rv = V1/V2\n", + "P2 = P1*(rv)**(Gama);# pressure at stage2,. [kN/m^2]\n", + "T2 = T1*(rv)**(Gama-1);# temperature at stage 2, [K]\n", + "\n", + "# for process 2-3\n", + "P3 = P2;# pressure at stage 3, [kN/m^2]\n", + "V2 = V1/rv;#[m^3]\n", + "# since pressure is constant in process 2-3 , so using V/T=constant, so\n", + "V3 = V2*(T3/T2);# volume at stage 3, [m^3]\n", + "\n", + "# for process 1-4\n", + "V4 = V1;# [m^3]\n", + "P4 = P3*(V3/V4)**(Gama)\n", + "# since in stage 1-4 volume is constant, so P/T=constant, \n", + "T4 = T1*(P4/P1);# temperature at stage 4,[K]\n", + "\n", + "print ' (a) P1 = ',P1,' kN/m^2, V1 = ',round(V1,3),' m^3, t1 = ',T1-273,' C,\\n P2 = ',round(P2),' kN/m^2, V2 = ',round(V2,3),' m^3, t2 = ',round(T2-273),' C,\\n P3 = ',round(P3),' kN/m^2, V3 = ',round(V3,3),' m^3, t3 = ',round(T3-273),' C,\\n P4 = ',round(P4,1),' kN/m^2, V4 = ',round(V4,3),' m^3, t4 = ',round(T4-273),' C'\n", + "\n", + "# (b)\n", + "W = cp*(T3-T2)-cv*(T4-T1);# work done, [kJ]\n", + "print ' (b) The work done is (kJ) = ',W\n", + "\n", + "# (c) \n", + "TE = 1-(T4-T1)/((T3-T2)*Gama);# thermal efficiency\n", + "print ' (c) The thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# (d)\n", + "PW = cp*(T3-T2)+R*(T3-T4)/(Gama-1);# positive work done\n", + "WR = W/PW;# work ratio\n", + "print ' (d) The work ratio is = ',round(WR,3)\n", + "\n", + "# (e)\n", + "Pm = W/(V1-V2);# mean effective pressure, [kN/m^2]\n", + "print ' (e) The mean effefctive pressure is (kN/m^2) = ',round(Pm,1)\n", + "\n", + "# (f)\n", + "CE = (T3-T1)/T3;# carnot efficiency\n", + "print ' (f) The carnot efficiency is (percent) = ',round(CE*100,1)\n", + "\n", + "print 'value of t2 printed in the book is incorrect'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10: pg 470" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 10\n", + " (a) The maximum temperature during the cycle is (C) = 1595.4\n", + " (b) The thermal efficiency of the cycle is (percent) = 60.3\n" + ] + } + ], + "source": [ + "#pg 470\n", + "print('Example 10');\n", + "\n", + "# aim : To determine\n", + "# (a) the maximum temperature attained during the cycle\n", + "# (b) the thermal efficiency of the cycle\n", + "\n", + "# given value\n", + "rva =7.5;# volume ratio of adiabatic expansion\n", + "rvc =15.;# volume ratio of compression\n", + "P1 = 98.;# initial pressure, [kn/m^2]\n", + "T1 = 273.+44;# initial temperature, [K]\n", + "P4 = 258.;# pressure at the end of the adiabatic expansion, [kN/m^2]\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# by seeing diagram\n", + "# for process 4-1, P4/T4=P1/T1\n", + "T4 = T1*(P4/P1);# [K]\n", + "# for process 3-4\n", + "T3 = T4*(rva)**(Gama-1);\n", + "print ' (a) The maximum temperature during the cycle is (C) = ',round(T3-273,1)\n", + "\n", + "# (b)\n", + "\n", + "# for process 1-2,\n", + "T2 = T1*(rvc)**(Gama-1);# [K]\n", + "n_the = 1-(T4-T1)/((Gama)*(T3-T2));# thermal efficiency\n", + "print ' (b) The thermal efficiency of the cycle is (percent) = ',round(n_the*100,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 11: pg 471" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.11\n", + " (a) the thermal efficiency of the cycle is (percent) = 55.1\n", + " (c) The indicated power of the cycle is (kW) = 26.59\n" + ] + } + ], + "source": [ + "#pg 471\n", + "print('Example 15.11');\n", + "\n", + "# aim : To determine\n", + "# (a) the thermal efficiency of the cycle\n", + "# (b) the indicared power of the cycle\n", + "\n", + "# given values\n", + "# taking basis one second\n", + "rv = 11.;# volume ratio\n", + "P1 = 96.;# initial pressure , [kN/m^2]\n", + "T1 = 273.+18;# initial temperature, [K]\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.24\n", + "# (a)\n", + "Beta = 2;# ratio of V3 and V2\n", + "TE = 1-(Beta**(Gama)-1)/((rv**(Gama-1))*Gama*(Beta-1));# thermal efficiency\n", + "print ' (a) the thermal efficiency of the cycle is (percent) = ',round(TE*100,1)\n", + "\n", + "# (b) \n", + "# let V1-V2=.05, so\n", + "V2 = .05*.1;# [m^3]\n", + "# from this\n", + "V1 = rv*V2;# [m^3]\n", + "V3 = Beta*V2;# [m^3]\n", + "V4 = V1;# [m^3]\n", + "P2 = P1*(V1/V2)**(Gama);# [kN/m^2]\n", + "P3 = P2;# [kn/m^2]\n", + "P4=P3*(V3/V4)**(Gama);# [kN/m^2]\n", + "# indicated power\n", + "W = P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);# indicated power, [kW]\n", + "print ' (c) The indicated power of the cycle is (kW) = ',round(W,2)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 12: pg 477" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.12\n", + " (a) The pressure at the end of compression is (MN/m^2) = 5.0\n", + " The temperature at the end of compression is (C) = 621.0\n", + " (b) The pressure at the end of constant volume process is (MN/m^2) = 6.9\n", + " The temperature at the end of constant volume process is (C) = 962.0\n", + " (c) The temperature at the end of constant pressure process is (C) = 1517.0\n" + ] + } + ], + "source": [ + "#pg 477\n", + "print('Example 15.12');\n", + "# aim : To determine\n", + "# (a) the pressure and temperature at the end of compression\n", + "# (b) the pressure and temperature at the end of the constant volume process\n", + "# (c) the temperature at the end of constant pressure process\n", + "\n", + "# given values\n", + "P1 = 103.;# initial pressure, [kN/m^2]\n", + "T1 = 273.+22;# initial temperature, [K]\n", + "rv = 16.;# volume ratio of the compression\n", + "Q = 244.;#heat added, [kJ/kg]\n", + "Gama = 1.4;# heat capacity ratio\n", + "cv = .717;# heat capacity, [kJ/kg k]\n", + "\n", + "# solution\n", + "# taking reference as Fig.15.26\n", + "# (a)\n", + "# for compression\n", + "# rv = V1/V2\n", + "P2 = P1*(rv)**Gama;# pressure at end of compression, [kN/m^2]\n", + "T2 = T1*(rv)**(Gama-1);# temperature at end of compression, [K]\n", + "print ' (a) The pressure at the end of compression is (MN/m^2) = ',round(P2*10**-3)\n", + "print ' The temperature at the end of compression is (C) = ',round(T2-273)\n", + "\n", + "# (b)\n", + "# for constant volume process, \n", + "# Q = cv*(T3-T2), so\n", + "T3 = T2+Q/cv;# temperature at the end of constant volume, [K]\n", + "\n", + "# so for constant volume, P/T=constant, hence\n", + "P3 = P2*(T3/T2);# pressure at the end of constant volume process, [kN/m^2]\n", + "print ' (b) The pressure at the end of constant volume process is (MN/m^2) = ',round(P3*10**-3,1)\n", + "print ' The temperature at the end of constant volume process is (C) = ',round(T3-273)\n", + "\n", + "# (c)\n", + "S = rv-1;# stroke\n", + "# assuming \n", + "V3 = 1;# [volume]\n", + "#so\n", + "V4 = V3+S*.03;# [volume]\n", + "# also for constant process V/T=constant, hence\n", + "T4 = T3*(V4/V3);# temperature at the end of constant presure process, [k] \n", + "print ' (c) The temperature at the end of constant pressure process is (C) = ',round(T4-273)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13: pg 479" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.13\n", + " (a) P1 = 0.097 kN/m^2, V1 = 0.084 m^3, t1 = 28.0 C,\n", + " P2 = 4.0 kN/m^2, V2 = 0.0056 m^3, t2 = 620.0 C,\n", + " P3 = 6.0 kN/m^2, V3 = 0.0056 m^3, t3 = 1010.0 C,\n", + " P4 = 6.2 kN/m^2, V4 = 0.006955 m^3, t4 = 1320.0 C \n", + " P5 = 0.2 kN/m^2, V5 = 0.084 m^3, t5 = 313.0 C\n", + " (b) The net work done is (kJ) = 36.4\n", + " (c) The thermal efficiency is (percent) = 65.5\n", + " (d) The heat received is (kJ) = 55.6\n", + " (f) The work ratio is = 0.477\n", + " (e) The mean effective pressure is (kN/m^2) = 464.0\n", + " (f) The carnot efficiency is (percent) = 81.0\n" + ] + } + ], + "source": [ + "#pg 479\n", + "print('Example 15.13');\n", + "\n", + "# aim : To determine\n", + "# (a) the pressure, volume and temperature at cycle process change points\n", + "# (b) the net work done \n", + "# (c) the thermal efficiency\n", + "# (d) the heat received\n", + "# (e) the work ratio\n", + "# (f) the mean effective pressure\n", + "# (g) the carnot efficiency\n", + "\n", + "\n", + "# given values\n", + "rv = 15.;# volume ratio\n", + "P1 = 97.*10**-3;# initial pressure , [MN/m^2]\n", + "V1 = .084;# initial volume, [m^3]\n", + "T1 = 273.+28;# initial temperature, [K]\n", + "T4 = 273.+1320;# maximum temperature, [K]\n", + "P3 = 6.2;# maximum pressure, [MN/m^2]\n", + "cp = 1.005;# specific heat capacity at constant pressure, [kJ/kg K]\n", + "cv = .717;# specific heat capacity at constant volume, [kJ/kg K]\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.27\n", + "# (a)\n", + "R = cp-cv;# gas constant, [kJ/kg K]\n", + "Gama = cp/cv;# heat capacity ratio\n", + "# for process 1-2\n", + "V2 = V1/rv;# volume at stage2, [m^3] \n", + "# using PV^(Gama)=constant for process 1-2\n", + "P2 = P1*(V1/V2)**(Gama);# pressure at stage2,. [MN/m^2]\n", + "T2 = T1*(V1/V2)**(Gama-1);# temperature at stage 2, [K]\n", + "\n", + "# for process 2-3\n", + "# since volumee is constant in process 2-3 , so using P/T=constant, so\n", + "T3 = T2*(P3/P2);# volume at stage 3, [K]\n", + "V3 = V2;# volume at stage 3, [MN/m^2]\n", + "\n", + "# for process 3-4\n", + "P4 = P3;# pressure at stage 4, [m^3]\n", + "# since in stage 3-4 P is constant, so V/T=constant, \n", + "V4 = V3*(T4/T3);# temperature at stage 4,[K]\n", + "\n", + "# for process 4-5\n", + "V5 = V1;# volume at stage 5, [m^3]\n", + "P5 = P4*(V4/V5)**(Gama);# pressure at stage5,. [MN/m^2]\n", + "T5 = T4*(V4/V5)**(Gama-1);# temperature at stage 5, [K]\n", + "\n", + "print ' (a) P1 = ',P1,' kN/m^2, V1 = ',round(V1,3),' m^3, t1 = ',T1-273,' C,\\n P2 = ',round(P2),' kN/m^2, V2 = ',round(V2,6),' m^3, t2 = ',round(T2-273),' C,\\n P3 = ',round(P3),' kN/m^2, V3 = ',round(V3,6),' m^3, t3 = ',round(T3-273),' C,\\n P4 = ',round(P4,1),' kN/m^2, V4 = ',round(V4,6),' m^3, t4 = ',round(T4-273),' C \\n P5 = ',round(P5,1),' kN/m^2, V5 = ',round(V5,3),' m^3, t5 = ',round(T5-273),' C'\n", + "\n", + "\n", + "# (b)\n", + "W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10**3;# work done, [kJ]\n", + "print ' (b) The net work done is (kJ) = ',round(W,1)\n", + "\n", + "# (c) \n", + "TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));# thermal efficiency\n", + "print ' (c) The thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# (d)\n", + "Q = W/TE;# heat received, [kJ]\n", + "print ' (d) The heat received is (kJ) = ',round(Q,1)\n", + "\n", + "# (e)\n", + "PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1)\n", + "WR = W*10**-3/PW;# work ratio\n", + "print ' (f) The work ratio is = ',round(WR,3)\n", + "\n", + "# (e)\n", + "Pm = W/(V1-V2);# mean effective pressure, [kN/m^2]\n", + "print ' (e) The mean effective pressure is (kN/m^2) = ',round(Pm,1)\n", + "\n", + "# (f)\n", + "CE = (T4-T1)/T4;# carnot efficiency\n", + "print ' (f) The carnot efficiency is (percent) = ',round(CE*100)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 14: pg 487" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.14\n", + " (a) The thermal efficiency is (percent) = 52.6\n", + " (b) The heat received is (kJ/cycle) = 6.22\n", + " (c) The heat rejected is (kJ/cycle) = 2.95\n", + " (d) The net work is (kJ/cycle) = 3.27\n", + " (e) The work ratio is = 0.347\n", + " (f) The mean effefctive pressure is (kN/m^2) = 167.32\n", + " (g) The carnot efficiency is (percent) = 72.7\n", + "there is minor variation in answer reported in the book due to rounding off error\n" + ] + } + ], + "source": [ + "#pg 487\n", + "print('Example 15.14');\n", + "\n", + "# aim : To determine\n", + "# (a) the thermal efficiency\n", + "# (b) the heat received\n", + "# (c) the heat rejected\n", + "# (d) the net work \n", + "# (e) the work ratio\n", + "# (f) the mean effective pressure\n", + "# (g) the carnot efficiency\n", + "\n", + "\n", + "# given values\n", + "P1 = 101.;# initial pressure , [kN/m**2]\n", + "V1 = 14.*10**-3;# initial volume, [m**3]\n", + "T1 = 273.+15;# initial temperature, [K]\n", + "P3 = 1850.;# maximum pressure, [kN/m**2]\n", + "V2 = 2.8*10**-3;# compressed volume, [m**3]\n", + "Gama = 1.4;# heat capacity\n", + "R = .29;# gas constant, [kJ/kg k]\n", + "\n", + "# solution\n", + "# taking reference Fig. 15.29\n", + "# (a)\n", + "# for process 1-2\n", + "# using PV**(Gama)=constant for process 1-2\n", + "P2 = P1*(V1/V2)**(Gama);# pressure at stage2,. [MN/m**2]\n", + "T2 = T1*(V1/V2)**(Gama-1);# temperature at stage 2, [K]\n", + "\n", + "# for process 2-3\n", + "# since volumee is constant in process 2-3 , so using P/T=constant, so\n", + "T3 = T2*(P3/P2);# volume at stage 3, [K]\n", + "\n", + "# for process 3-4\n", + "P4 = P1;\n", + "T4 = T3*(P4/P3)**((Gama-1)/Gama);# temperature\n", + "\n", + "TE = 1-Gama*(T4-T1)/(T3-T2);# thermal efficiency\n", + "print ' (a) The thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# (b)\n", + "cv = R/(Gama-1);# heat capacity at copnstant volume, [kJ/kg k]\n", + "m = P1*V1/(R*T1);# mass of gas, [kg]\n", + "Q1 = m*cv*(T3-T2);# heat received, [kJ/cycle]\n", + "print ' (b) The heat received is (kJ/cycle) = ',round(Q1,2)\n", + "\n", + "# (c)\n", + "cp = Gama*cv;# heat capacity at constant at constant pressure, [kJ/kg K]\n", + "Q2 = m*cp*(T4-T1);# heat rejected, [kJ/cycle]\n", + "print ' (c) The heat rejected is (kJ/cycle) = ',round(Q2,2)\n", + "\n", + "# (d)\n", + "W = Q1-Q2;# net work , [kJ/cycle]\n", + "print ' (d) The net work is (kJ/cycle) = ',round(W,2)\n", + "\n", + "# (e)\n", + "# pressure is constant for process 1-4, so V/T=constant\n", + "V4 = V1*(T4/T1);# volume, [m**3]\n", + "V3 = V2;# for process 2-3\n", + "P4 = P1;# for process 1-4\n", + "PW = (P3*V3-P1*V1)/(Gama-1);# positive work done, [kJ/cycle]\n", + "WR = W/PW;# work ratio\n", + "print ' (e) The work ratio is = ',round(WR,3)\n", + "\n", + "# (f)\n", + "Pm = W/(V4-V2);# mean effective pressure, [kN/m**2]\n", + "print ' (f) The mean effefctive pressure is (kN/m^2) = ',round(Pm,2)\n", + "\n", + "# (g)\n", + "CE = (T3-T1)/T3;# carnot efficiency\n", + "print ' (g) The carnot efficiency is (percent) = ',round(CE*100,1)\n", + "\n", + "print 'there is minor variation in answer reported in the book due to rounding off error'\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 15: pg 492" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.15\n", + " (a) The net work done is (kJ) = 28.0\n", + " (b) The ideal thermal efficiency is (percent) = 68.9\n", + " (c) the thermal efficiency if the process of regeneration is not included is (percent) = 39.5\n" + ] + } + ], + "source": [ + "#pg 492\n", + "print('Example 15.15');\n", + "\n", + "# aim : To determine\n", + "# (a) the net work done\n", + "# (b) the ideal thermal efficiency\n", + "# (c) the thermal efficiency if the process of generation is not included\n", + "from math import log\n", + "# given values\n", + "P1 = 110.;# initial pressure, [kN/m^2)\n", + "T1 = 273.+30;# initial temperature, [K]\n", + "V1 = .05;# initial volume, [m^3]\n", + "V2 = .005;# volume, [m^3]\n", + "T3 = 273.+700;# temperature, [m^3]\n", + "R = .289;# gas constant, [kJ/kg K]\n", + "cv = .718;# heat capacity, [kJ/kg K]\n", + "\n", + "# solution\n", + "# (a)\n", + "m = P1*V1/(R*T1);# mass , [kg]\n", + "W = m*R*(T3-T1)*log(V1/V2);# work done, [kJ]\n", + "print ' (a) The net work done is (kJ) = ',round(W)\n", + "\n", + "# (b)\n", + "n_the = (T3-T1)/T3;# ideal thermal efficiency\n", + "print ' (b) The ideal thermal efficiency is (percent) = ',round(n_the*100,1)\n", + "\n", + "# (c)\n", + "V4 = V1;\n", + "V3 = V2;\n", + "T4 = T3;\n", + "T2 = T1;\n", + "\n", + "Q_rej = m*cv*(T4-T1)+m*R*T1*log(V1/V2);# heat rejected\n", + "Q_rec = m*cv*(T3-T2)+m*R*T3*log(V4/V3);# heat received\n", + "\n", + "n_th = (1-Q_rej/Q_rec);# thermal efficiency\n", + "print ' (c) the thermal efficiency if the process of regeneration is not included is (percent) = ',round(n_th*100,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 16: pg 493" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.16\n", + " (a) The maximum temperature is (C) = 313.0\n", + " (b) The net work done is (kJ) = 12.88\n", + " (c) The ideal thermal efficiency is (percent) = 50.0\n", + " (d) the thermal efficiency if the process of regeneration is not included is (percent) = 24.0\n" + ] + } + ], + "source": [ + "#pg 493\n", + "print('Example 15.16');\n", + "from math import log\n", + "# aim : To determine\n", + "# (a) the maximum temperature\n", + "# (b) the net work done\n", + "# (c) the ideal thermal efficiency\n", + "# (d) the thermal efficiency if the process of regeneration is not included\n", + "\n", + "# given values\n", + "P1 = 100.;# initial pressure, [kN/m^2)\n", + "T1 = 273.+20;# initial temperature, [K]\n", + "V1 = .08;# initial volume, [m^3]\n", + "rv = 5;# volume ratio\n", + "R = .287;# gas constant, [kJ/kg K]\n", + "cp = 1.006;# heat capacity, [kJ/kg K]\n", + "V3_by_V2 = 2;\n", + "\n", + "# solution\n", + "# (a)\n", + "# using Fig.15.33\n", + "# process 1-2 is isothermal\n", + "T2 = T1;\n", + "# since process 2-3 isisobaric, so V/T=constant\n", + "T3 = T2*(V3_by_V2);# maximumtemperature, [K]\n", + "print ' (a) The maximum temperature is (C) = ',T3-273\n", + "\n", + "# (b)\n", + "m = P1*V1/(R*T1);# mass , [kg]\n", + "W = m*R*(T3-T1)*log(rv);# work done, [kJ]\n", + "print ' (b) The net work done is (kJ) = ',round(W,2)\n", + "\n", + "# (c)\n", + "TE = (T3-T1)/T3;# ideal thermal efficiency\n", + "print ' (c) The ideal thermal efficiency is (percent) = ',TE*100\n", + "\n", + "# (d)\n", + "T4 = T3;\n", + "T2 = T1;\n", + "\n", + "Q_rej = m*cp*(T4-T1)+m*R*T1*log(rv);# heat rejected\n", + "Q_rec = m*cp*(T3-T2)+m*R*T3*log(rv);# heat received\n", + "\n", + "n_th = (1-Q_rej/Q_rec);# thermal efficiency\n", + "print ' (d) the thermal efficiency if the process of regeneration is not included is (percent) = ',round(n_th*100)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 17: pg 495" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.17\n", + " (a) The net work done is (kJ) = 44.7\n", + " (b) The thermal efficiency is (percent) = 56.7\n" + ] + } + ], + "source": [ + "#pg 495\n", + "print('Example 15.17');\n", + "\n", + "# aim : To determine \n", + "# (a) the net work done\n", + "# (b) thethermal efficiency\n", + "from math import log\n", + "# given values\n", + "m = 1.;# mass of air, [kg]\n", + "T1 = 273.+230;# initial temperature, [K]\n", + "P1 = 3450.;# initial pressure, [kN/m^2]\n", + "P2 = 2000.;# pressure, [kN/m^2]\n", + "P3 = 140.;# pressure, [kN/m^2]\n", + "P4 = P3;\n", + "Gama = 1.4; # heat capacity ratio\n", + "cp = 1.006;# heat capacity, [kJ/kg k]\n", + "\n", + "# solution\n", + "T2 =T1;# isothermal process 1-2\n", + "# process 2-3 and 1-4 are adiabatic so\n", + "T3 = T2*(P3/P2)**((Gama-1)/Gama);# temperature, [K] \n", + "T4 = T1*(P4/P1)**((Gama-1)/Gama);# [K]\n", + "R = cp*(Gama-1)/Gama;# gas constant, [kJ/kg K]\n", + "Q1 = m*R*T1*log(P1/P2);# heat received, [kJ]\n", + "Q2 = m*cp*(T3-T4);# heat rejected\n", + "\n", + "#hence\n", + "W = Q1-Q2;# work done\n", + "print ' (a) The net work done is (kJ) = ',round(W,1)\n", + "\n", + "# (b)\n", + "TE = 1-Q2/Q1;# thermal efficiency\n", + "print ' (b) The thermal efficiency is (percent) = ',round(TE*100,1)\n", + "\n", + "# End\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18: pg 497" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Example 15.18\n", + " The thermal efficiency is (percent) = 31.0\n", + " The carnot efficiency is = 73.7\n" + ] + } + ], + "source": [ + "#pg 497\n", + "print('Example 15.18');\n", + "\n", + "# aim : To determine \n", + "# thermal eficiency\n", + "# carnot efficiency\n", + "from math import log\n", + "# given values\n", + "rv = 5.;# volume ratio\n", + "Gama = 1.4;# heat capacity ratio\n", + "\n", + "# solution\n", + "# under given condition\n", + "\n", + "TE = 1-(1/Gama*(2-1/rv**(Gama-1)))/(1+2*((Gama-1)/Gama)*log(rv/2));# thermal efficiency\n", + "print ' The thermal efficiency is (percent) = ',round(TE*100)\n", + "\n", + "CE = 1-1/(2*rv**(Gama-1));# carnot efficiency\n", + "print ' The carnot efficiency is = ',round(CE*100,1)\n", + "\n", + "# End\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 +} |