{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Chapter 5 : Second Law of Thermodynamics and Entropy" ] }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.1 Page no : 237" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The thermal efficiency \n", "(ii) The rate of heat rejection.\n", "'''\n", "\n", "# Variables\n", "Q1 = 1500./60; \t\t#kJ/s\n", "W = 8.2; \t\t\t#kW\n", "\n", "# Calculations and Results\n", "print (\"(i) Thermal efficiency\")\n", "n = W/Q1;\n", "print (\"n = \"),(n)\n", "\n", "print (\"(ii) Rate of heat rejection\")\n", "Q2 = Q1-W; \n", "print (\"Q2 = \"),(Q2), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Thermal efficiency\n", "n = 0.328\n", "(ii) Rate of heat rejection\n", "Q2 = 16.8 kW\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.2 Page no : 238" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Is it possible to reach initial state by an adiabatic process ?\n", "'''\n", "\n", "# Variables\n", "Q_12 = 30.; \t\t#kJ\n", "W_12 = 60; \t\t\t#kJ\n", "\n", "# Calculations\n", "dU_12 = Q_12-W_12;\n", "Q_21 = 0;\n", "W_21 = Q_21+dU_12;\n", "\n", "# Results\n", "print (\"W_21 = \"),(W_21)\n", "print (\"Thus 30 kJ work has to be done on the system to restore it to original state, by adiabatic process.\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "W_21 = -30.0\n", "Thus 30 kJ work has to be done on the system to restore it to original state, by adiabatic process.\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.3 Page no : 239" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Find the co-efficient of performance and heat transfer rate in the condenser of a refrigerator \n", "'''\n", "# Variables\n", "Q2 = 12000.; \t\t\t#kJ/h\n", "W = 0.75*60*60; \t\t#kJ/h\n", "\n", "# Calculations and Results\n", "COP = Q2/W;\n", "print (\"Coefficient of performance %.3f\")%(COP)\n", "\n", "Q1 = Q2+W;\n", "print (\"heat transfer rate = %.3f\")%(Q1), (\"kJ/h\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Coefficient of performance 4.444\n", "heat transfer rate = 14700.000 kJ/h\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.4 Page no : 239" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the least power necessary to pump this heat out continuously\n", "'''\n", "# Variables\n", "T2 = 261.; \t\t\t#K\n", "T1 = 308.; \t\t\t#K\n", "Q2 = 2.; \t\t\t#kJ/s\n", "\n", "# Calculations\n", "Q1 = Q2*(T1/T2);\n", "W = Q1-Q2;\n", "\n", "# Results\n", "print (\"Least power required to pump the heat continuosly %.3f\")%(W),(\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Least power required to pump the heat continuosly 0.360 kW\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.5 Page no :239" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) Heat abstracted from outside ;\n", "(ii) Co-efficient of performance.\n", "'''\n", "\n", "import math \n", "\n", "\n", "# Variables\n", "Q1 = 2*10**5; \t\t\t#kJ/h\n", "W = 3*10**4; \t\t\t#kJ/h\n", "\n", "# Calculations and Results\n", "Q2 = Q1-W;\n", "print (\"Heat abstracted from outside = \"),(Q2), (\"kJ/h\")\n", "\n", "\n", "COP_hp = Q1/(Q1-Q2);\n", "print (\"Co-efficient of performance = %.2f\")%(COP_hp)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat abstracted from outside = 170000 kJ/h\n", "Co-efficient of performance = 6.00\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.6 Page no : 240" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "What is the highest possible theoretical efficiency of a heat engine operating\n", "with a hot reservoir of furnace gases \n", "'''\n", "\n", "# Variables\n", "T1 = 2373; \t\t\t#K\n", "T2 = 288.; \t\t\t#K\n", "\n", "# Calculations\n", "n_max = 1-T2/T1;\n", "\n", "# Results\n", "print (\"Highest possible theoritical efficiency = %.3f\")% (n_max*100), (\"%\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Highest possible theoritical efficiency = 87.863 %\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.7 Page no : 240" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "find :\n", "(i) Efficiency of the system ;\n", "(ii) The net work transfer ;\n", "(iii) Heat rejected to sink.\n", "'''\n", "\n", "# Variables\n", "T1 = 523.; \t\t\t#K\n", "T2 = 258.; \t\t\t#K\n", "Q1 = 90.; \t\t\t#kJ\n", "\n", "# Calculations and Results\n", "n = 1-T2/T1;\n", "print (\"(i) Efficiency of the system %.3f\")%(n*100), (\"%\")\n", "\n", "W = n*Q1;\n", "print (\"(ii) The net work transfer\"),(\"W = %.3f\")%(W),(\"kJ\")\n", " \n", "Q2 = Q1-W;\n", "print (\"(iii) Heat rejected to the math.sink\"),(\"Q2 = %.3f\")%(Q2),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Efficiency of the system 50.669 %\n", "(ii) The net work transfer W = 45.602 kJ\n", "(iii) Heat rejected to the math.sink Q2 = 44.398 kJ\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.8 Page no : 241" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "An inventor claims that his engine has few specifications :\n", "State whether his claim is valid or not.\n", "'''\n", "\n", "# Variables\n", "T1 = 1023.; \t\t#K\n", "T2 = 298.; \t\t\t#K\n", "\n", "# Calculations\n", "n_carnot = 1-T2/T1;\n", "W = 75*1000*60*60;\n", "Q = 3.9*74500*1000;\n", "n_thermal = W/Q;\n", "\n", "# Results\n", "print (\"n_carnot = %.3f\")%(n_carnot)\n", "\n", "print (\"n_thermal = %.3f\")%(n_thermal)\n", "\n", "print (\"Since \u03b7thermal > \u03b7carnot, therefore claim of the inventor is not valid (or possible)\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "n_carnot = 0.709\n", "n_thermal = 0.929\n", "Since \u03b7thermal > \u03b7carnot, therefore claim of the inventor is not valid (or possible)\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.9 Page no : 241" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Find the least rate of heat rejection per kW net output of the engine ?\n", "'''\n", "\n", "# Variables\n", "T1 = 1273.; \t\t#K\n", "T2 = 313.; \t\t\t#K\n", "n_max = 1-T2/T1;\n", "Wnet = 1.;\n", "\n", "# Calculations\n", "Q1 = Wnet/n_max;\n", "Q2 = Q1-Wnet;\n", "\n", "# Results\n", "print (\"the least rate of heat rejection = %.3f\")%(Q2), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "the least rate of heat rejection = 0.326 kW\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.10 Page no : 242" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate the power required\n", "'''\n", "\n", "# Variables\n", "one_ton_of_refrigeration = 210.; \t\t\t#kJ/min\n", "Cooling_required = 40*(one_ton_of_refrigeration); \t\t\t#kJ/min\n", "T1 = 303.; \t\t\t#K\n", "T2 = 238.; \t\t\t#K\n", "\n", "# Calculations\n", "COP_refrigerator = T2/(T1-T2);\n", "COP_actual = 0.20*COP_refrigerator;\n", "W = Cooling_required/COP_actual/60;\n", "\n", "# Results\n", "print (\"power required = %.1f\")% (W), (\"kW\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "power required = 191.2 kW\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.11 Page no : 242" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Which source (1 or 2) would you choose to supply energy to an ideal reversible heat engine that is to produce large amount of\n", "power if the temperature of the surroundings is 35\u00b0C ?\n", "'''\n", "\n", "# Variables\n", "E = 12000.; \t\t#kJ/min\n", "T2 = 308.; \t\t\t#K\n", "# Source 1\n", "T1 = 593.; \t\t\t#K\n", "\n", "# Calculations\n", "n1 = 1-T2/T1;\n", "# Source 2\n", "T1 = 343.; \t\t\t#K\n", "n2 = 1-T2/T1;\n", "W1 = E*n1;\n", "\n", "# Results\n", "print (\"W1 = %.3f\")% (W1),(\"kJ/min\")\n", "\n", "W2 = E*n2;\n", "print (\"W2 = %.3f\")% (W2),(\"kJ/min\")\n", "\n", "print (\"Thus, choose source 2.\")\n", "print (\"The source 2 is selected even though efficiency in this case is lower, because the criterion for selection is the larger output.\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "W1 = 5767.285 kJ/min\n", "W2 = 1224.490 kJ/min\n", "Thus, choose source 2.\n", "The source 2 is selected even though efficiency in this case is lower, because the criterion for selection is the larger output.\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ " Example 5.12 Page no : 243" ] }, { "cell_type": "code", "collapsed": false, "input": [ ".'''\n", "(i) Determine the heat transfer to the refrigerant and the net heat transfer to the reservoir\n", "at 50\u00b0C ;\n", "(ii) Reconsider (i) given that the efficiency of the heat engine and the C.O.P. of the refrig-\n", "erator are each 45 per cent of their maximum possible values.\n", "'''\n", "\n", "\n", "# Variables\n", "T1 = 973.; \t\t\t#K\n", "T2 = 323.; \t\t\t#K\n", "T3 = 248.; \t\t\t#K\n", "\n", "Q1 = 2500.; \t\t\t#kJ\n", "W = 400.; \t\t\t#kJ\n", "\n", "# Calculations and Results\n", "n_max = 1-T2/T1;\n", "W1 = n_max*Q1;\n", "COP_max = T3/(T2-T3);\n", "W2 = W1-W;\n", "Q4 = COP_max*W2;\n", "COP1 = round(Q4/W2,3);\n", "Q3 = Q4+W2;\n", "Q2 = Q1-W1;\n", "print (\"Heat rejection to the 50\u00b0C reservoir = %.3f\")%(Q2+Q3), (\"kJ\")\n", "\n", "\n", "n = 0.45*n_max;\n", "W1 = n*Q1;\n", "W2 = W1-W;\n", "COP2 = 0.45*COP1;\n", "Q4 = W2*COP2;\n", "Q3 = Q4+W2;\n", "Q2 = Q1-W1;\n", "\n", "print (\"Heat rejected to 50\u00b0C reservoir = %.3f\")% (Q2+Q3), (\"kJ\")\n", "\n", "# Note : Answers are slightly different then book because of Rounding Error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Heat rejection to the 50\u00b0C reservoir = 6299.773 kJ\n", "Heat rejected to 50\u00b0C reservoir = 2623.147 kJ\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.13 Page no : 244" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) determine the C.O.P. of the machine and work input required.\n", "(ii) determine the overall C.O.P. of the system.\n", "'''\n", "\n", "# Variables\n", "T1 = 298.; \t\t\t#K\n", "T2 = 273.; \t\t\t#K\n", "Q1 = 24.; \t\t\t#kJ/s\n", "T3 = 653.; \t\t\t#K\n", "\n", "# Calculations and Results\n", "COP = T1/(T1-T2);\n", "print (\"(i) determine COP and work input required\")\n", "\n", "print (\"Coefficient of performance = \"),(COP)\n", "\n", "COP_ref = T2/(T1-T2);\n", "W = Q1/COP_ref;\n", "print (\"Work input required = %.3f\")%(W),(\"kW\")\n", "\n", " \n", "Q4 = T1*W/(T3-T1);\n", "Q3 = Q4+W;\n", "Q2 = Q1+W;\n", "COP = Q1/Q3;\n", "print (\"(ii)Determine overall COP of the system \"),(\"COP = %.3f\")%(COP)\n", "\n", "COP_overall = (Q2+Q4)/Q3;\n", "print (\"Overall COP = %.3f\")%(COP_overall)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) determine COP and work input required\n", "Coefficient of performance = 11.92\n", "Work input required = 2.198 kW\n", "(ii)Determine overall COP of the system COP = 5.937\n", "Overall COP = 6.937\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.14 Page no : 245" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) Power developed by the engine ;\n", "(ii) Fuel consumed per hour.\n", "Take enthalpy of fusion of ice = 334.5 kJ/kg.\n", "'''\n", "\n", "# Variables\n", "T_e1 = 493.; \t\t\t#K\n", "T_e2 = 298.; \t\t\t#K\n", "T_p1 = 298.; \t\t\t#K\n", "T_p2 = 273.; \t\t\t#K\n", "Amt = 15.; \t\t \t#tonnes produced per day\n", "h = 334.5; \t\t\t #kJ/kg\n", "Q_abs = 44500.; \t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "Q_p2 = Amt*10**3*h/24/60;\n", "COP_hp = T_p2/(T_p1-T_p2);\n", "W = Q_p2/COP_hp/60;\n", "print (\"(i)Power developed by the engine = %.3f\")%(W),(\"kW\")\n", "\n", "print (\"(ii) Fuel consumed per hour\")\n", "n_carnot = 1-(T_e2/T_e1);\n", "Q_e1 = W/n_carnot*3600; \t\t\t#kJ/h\n", "fuel_consumed = Q_e1/Q_abs;\n", "print (\"Quantity of fuel consumed/hour = %.3f\")%(fuel_consumed),(\"kg/h\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i)Power developed by the engine = 5.318 kW\n", "(ii) Fuel consumed per hour\n", "Quantity of fuel consumed/hour = 1.088 kg/h\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.15 Page no : 247" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the intermediate temperature.\n", "'''\n", "\n", "# Variables\n", "T1 = 550.; \t\t\t#K\n", "T3 = 350.; \t\t\t#K\n", "\n", "# Calculations\n", "T2 = (T1+T3)/2;\n", "\n", "# Results\n", "print (\"Intermediate temperature = \"), (T2),(\"K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Intermediate temperature = 450.0 K\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.16 Page no : 247" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine :\n", "(i) The temperature T 3 such that heat supplied to engine Q 1 is equal to the heat absorbed\n", "by refrigerator Q 2 .\n", "(ii) The efficiency of Carnot engine and C.O.P. of Carnot refrigerator.\n", "'''\n", "\n", "# Variables\n", "T1 = 600.; \t\t\t#K\n", "T2 = 300.; \t\t\t#K\n", "\n", "# Calculations and Results\n", "T3 = 2*T1/(T1/T2+1);\n", "print (\"(i) When Q1 = Q2\"),(\"T3 = \"),(T3),(\"K\")\n", "\n", "\n", "print (\"(ii) Efficiency of Carnot engine and COP of carnot refrigerator\")\n", "n = (T1-T3)/T1; \t\t\t#carnot engine\n", "COP = T2/(T3-T2); \t\t\t#refrigerator\n", "\n", "print (\"Efficiency of carnot engine = %.3f\")% (n)\n", "\n", "print (\"COP of carnot refrigerator = \"), (COP)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) When Q1 = Q2 T3 = 400.0 K\n", "(ii) Efficiency of Carnot engine and COP of carnot refrigerator\n", "Efficiency of carnot engine = 0.333\n", "COP of carnot refrigerator = 3.0\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.17 Page no : 249" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "estimate the energy taken from the reservoir at 1077\u00b0C.\n", "'''\n", "\n", "# Variables\n", "T3 = 278.; \t\t\t#K\n", "T2 = 350.; \t\t\t#K\n", "T4 = T2;\n", "T1 = 1350.; \t\t\t#K\n", "\n", "# Calculations\n", "Q1 = 100/(((T4/T1)*(T1-T2)/(T4-T3))+T2/T1) \t#Q4+Q2 = 100; Q4 = Q1*((T4/T1)*(T1-T2)/(T4-T3)); Q2 = T2/T1*Q1;\n", "\n", "# Results\n", "print (\"Q1 = %.3f\")%(Q1),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Q1 = 25.906 kJ\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.18 Page no : 250" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Classify which of the result report a reversible cycle or irreversible cycle or impossible\n", "results.\n", "'''\n", "# Variables\n", "Q1 = 300.; \t\t\t#kJ/s\n", "T1 = 290.; \t\t\t#0C\n", "T2 = 8.5; \t\t\t#0C\n", "\n", "# Calculations and Results\n", "print (\"let \u03a3dQ/T = A\")\n", "\n", "print (\"(i) 215 kJ/s are rejected\")\n", "Q2 = 215.; \t\t\t#kJ/s\n", "A = Q1/(T1+273) - Q2/(T2+273)\n", "print (\"Since, A<0, Cycle is irreversible.\")\n", "\n", "\n", "print (\"(ii) 150 kJ/s are rejected\")\n", "Q2 = 150; \t\t\t#kJ/s\n", "A = Q1/(T1+273) - Q2/(T2+273)\n", "print (\"Since A = 0, cycle is reversible\")\n", "\n", "\n", "print (\"(iii) 75 kJ/s are rejected.\")\n", "Q2 = 75; \t\t\t#kJ/s\n", "A = Q1/(T1+273) - Q2/(T2+273)\n", "print (\"Since A>0, cycle is impossible\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "let \u03a3dQ/T = A\n", "(i) 215 kJ/s are rejected\n", "Since, A<0, Cycle is irreversible.\n", "(ii) 150 kJ/s are rejected\n", "Since A = 0, cycle is reversible\n", "(iii) 75 kJ/s are rejected.\n", "Since A>0, cycle is impossible\n" ] } ], "prompt_number": 65 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.19 Page no : 251" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Verify the Clausius inequality for the cycle.\n", "'''\n", "\n", "# Variables\n", "P1 = 0.124*10**5; \t\t\t#N/m**2\n", "T1 = 433; \t\t\t#K\n", "T2 = 323; \t\t\t#K\n", "h_f1 = 687; \t\t\t#kJ/kg\n", "h2 = 2760; \t\t\t#kJ/kg\n", "h3 = 2160; \t\t\t#kJ/kg\n", "h_f4 = 209; \t\t\t#kJ/kg\n", "\n", "# Calculations and Results\n", "Q1 = h2-h_f1;\n", "Q2 = h_f4-h3;\n", "print (\"Let A = \u03a3dQ/T\")\n", "A = Q1/T1+Q2/T2;\n", "print (A)\n", "print (\"A<0. Hence classius inequality is verified\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Let A = \u03a3dQ/T\n", "-3\n", "A<0. Hence classius inequality is verified\n" ] } ], "prompt_number": 66 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.20 Page no :251" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''Verify the Clausius Inequality'''\n", "\n", "# Variables\n", "T1 = 437.; \t\t\t#K\n", "T2 = 324.; \t \t\t#K\n", "h2 = 2760.; \t\t\t#kJ/kg\n", "h1 = 690.; \t\t \t#kJ/kg\n", "h3 = 2360.; \t\t\t#kJ/kg\n", "h4 = 450.; \t\t\t #kJkg\n", "\n", "# Calculations\n", "Q1 = h2-h1;\n", "Q2 = h4-h3;\n", "\n", "# Results\n", "print (\"Let A = \u03a3dQ/T\")\n", "A = Q1/T1 + Q2/T2;\n", "print \"%.3f\"%(A)\n", "print (\"Since A<0, Classius inequality is verified\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Let A = \u03a3dQ/T\n", "-1.158\n", "Since A<0, Classius inequality is verified\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.21 Page no : 266" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Find the entropy changes for the iron cube and the water. Is the process reversible ? If so why ?\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "T0 = 273.; \t\t\t#K\n", "T1 = 673.; \t\t\t#K\n", "T2 = 298.; \t\t\t#K\n", "m_w = 10.; \t\t\t#kg\n", "T3 = 323.; \t\t\t#K\n", "c_pw = 4186.; \t\t#kJ/kg.K\n", "\n", "# Calculations and Results\n", "print (\"Let C = mi*cpi\")\n", "C = m_w*c_pw*(T3-T2)/(T1-T3);\n", "S_iT1 = C*math.log(T1/T0); \t\t\t# Entropy of iron at 673 K\n", "S_wT2 = m_w*c_pw*math.log(T2/T0); \t#Entropy of water at 298 K\n", "S_iT3 = C*math.log(T3/T0); \t\t\t#Entropy of iron at 323 K\n", "S_wT3 = m_w*c_pw*math.log(T3/T0); \t#Entropy of water at 323 K\n", "\n", "dS_i = S_iT3 - S_iT1;\n", "dS_w = S_wT3 - S_wT2; \n", "dS_net = dS_i + dS_w\n", "\n", "print (\"Since dS>0, process is irreversible\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Let C = mi*cpi\n", "Since dS>0, process is irreversible\n" ] } ], "prompt_number": 68 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.23 Page no : 267" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The net heat flow from the air.\n", "(ii) The net entropy change.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "T1 = 293.; \t\t\t#K\n", "V1 = 0.025; \t\t\t#m**3\n", "V3 = V1;\n", "p1 = 1.05*10**5; \t\t\t#N/m**2\n", "p2 = 4.5*10**5; \t\t\t#N/m**2\n", "R = 0.287*10**3; \n", "cv = 0.718;\n", "cp = 1.005;\n", "T3 = 293.; \t\t\t#K\n", "\n", "# Calculations and Results\n", "m = p1*V1/R/T1;\n", "T2 = p2/p1*T1;\n", "Q_12 = m*cv*(T2-T1);\n", "Q_23 = m*cp*(T3-T2)\n", "\n", "Q_net = Q_12+Q_23;\n", "print (\"Net heat flow = \"),(Q_net), (\"kJ\")\n", "\n", "\n", "dS_32 = m*cp*math.log(T2/T1);\n", "dS_12 = m*cv*math.log(T2/T1);\n", "dS_31 = dS_32 - dS_12;\n", "print (\"Decrease in entropy = %.3f\")% (dS_31), (\"kJ/K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Net heat flow = -8.625 kJ\n", "Decrease in entropy = 0.013 kJ/K\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.24 Page no : 269" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) The change of entropy,\n", "(ii) The heat flow, and\n", "(iii) The work done.\n", "'''\n", "\n", "%pylab inline\n", "\n", "from matplotlib.pyplot import *\n", "from numpy import *\n", "import math \n", "\n", "# Variables\n", "p1 = 1.05*10**5; \t#N/m**2\n", "V1 = 0.04; \t\t\t#m**3\n", "T1 = 288.; \t\t\t#K\n", "p2 = 4.8*10**5;\n", "T2 = T1;\n", "R0 = 8314.;\n", "M = 28.;\n", "\n", "# Calculations and Results\n", "R = R0/M;\n", "m = p1*V1/R/T1;\n", "dS = m*R*math.log(p1/p2)\n", "print (\"Decrease in entropy = %.3f\")% (-dS), (\"J/K\")\n", "\n", "\n", "print \n", "Q = T1*(-dS);\n", "print (\"(ii)Heat rejected = \"),(\"Q = %.3f\")%(Q),(\"J\")\n", "\n", "\n", "W = Q;\n", "print (\"Work done = %.3f\")% (W), (\"J\")\n", "\n", "V2 = p1*V1/p2;\n", "v1 = V1/m; \t\t\t#specific volume\n", "v2 = V2/m; \t\t\t#specific volume\n", "\n", "v = linspace(v2,0.8081571,64);\n", "\n", "\n", "def f(v):\n", " return p1*v1/v\n", "plot(v,f(v))\n", "\n", "p = []\n", "for i in range(len(v)):\n", " p.append(p1)\n", "plot(v,p,'--')\n", "\n", "p = [0 ,p2]\n", "v = [v2 ,v2]\n", "plot(v,p,'--')\n", "\n", "p = [0 ,p1]\n", "v = [v1 ,v1]\n", "plot(v,p,'--')\n", "\n", "T = [288, 288]\n", "s = [10 ,(10-dS)]\n", "plot(s,T)\n", "\n", "s = [10 ,10]\n", "T = [0 ,288]\n", "plot(s,T,'--')\n", "\n", "s = [(10-dS), (10-dS)]\n", "T = [0 ,288]\n", "plot(s,T,'--')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", "Decrease in entropy = 22.164 J/K\n", "\n", "(ii)Heat rejected = Q = 6383.268 J\n", "Work done = 6383.268 J\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 13, "text": [ "[]" ] }, { "metadata": {}, 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"text": [ "" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.25 Page no : 270" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the change in entropy.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "R = 287.; \t\t\t#kJ/kg.K\n", "dU = 0;\n", "W = 0;\n", "Q = dU+W;\n", "\n", "# Calculations\n", "dS = R*math.log(2); \t\t\t#v2/v1 = 2\n", "\n", "# Results\n", "print (\"Change in entropy = %.3f\")%(dS),(\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Change in entropy = 198.933 kJ/kg.K\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.26 Page no : 271" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the change of entropy\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "m = 0.04; \t\t\t#kg\n", "p1 = 1*10.**5; \t\t\t#N/m**2\n", "T1 = 293.; \t\t\t#K\n", "p2 = 9*10.**5; \t\t\t#N/m**2\n", "V2 = 0.003; \t\t\t#m**3\n", "cp = 0.88; \t\t\t#kJ/kg.K\n", "R0 = 8314.;\n", "M = 44.;\n", "\n", "# Calculations\n", "R = R0/M;\n", "T2 = p2*V2/m/R;\n", "ds_2A = R/10**3*math.log(p2/p1);\n", "ds_1A = cp*math.log(T2/T1);\n", "ds_21 = ds_2A - ds_1A;\n", "dS_21 = m*ds_21;\n", "\n", "# Results\n", "print (\"Decrease in entropy = %.3f\")% (dS_21),(\"kJ/K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Decrease in entropy = 0.010 kJ/K\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.27 Page no : 272" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the change of entropy \n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "p1 = 7.*10**5; \t\t\t#N/m**2\n", "T1 = 873.; \t\t\t#K\n", "p2 = 1.05*10**5; \t\t\t#N/M62\n", "n = 1.25;\n", "m = 1.; \t\t\t#kg\n", "R = 0.287;\n", "cp = 1.005;\n", "\n", "# Calculations\n", "T2 = T1*(p2/p1)**((n-1)/n);\n", "\n", "# At constant temperature from 1 to A\n", "ds_1A = R*math.log(p1/p2);\n", "# At constant pressure from A to 2\n", "ds_2A = cp*math.log(T1/T2);\n", "ds_12 = ds_1A - ds_2A;\n", "\n", "# Results\n", "print (\"Increase in entropy = %.3f\")% (ds_12), (\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Increase in entropy = 0.163 kJ/kg.K\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.28 Page no : 274" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Show that the process is irreversible ;\n", "(ii) Calculate the change of entropy per kg of air.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "p1 = 7*10**5; \t\t#Pa\n", "T1 = 733.; \t\t\t#K\n", "p2 = 1.012*10**5; \t#Pa\n", "T2a = 433.; \t\t#K\n", "y = 1.4;\n", "cp = 1.005;\n", "\n", "# Calculations and Results\n", "print (\"(i) To prove that the process is irreversible\")\n", "T2 = T1*(p2/p1)**((y-1)/y);\n", "print (\"T2 = %.3f\")% (T2)\n", "print (\"But the actual temperature is 433K at th epressure of 1.012 bar, Hence the process is irreversible. Proved.\")\n", "\n", "\n", "print (\"(ii) Change of entropy per kg of air\")\n", "ds = cp*math.log(T2a/T2);\n", "print (\"Increase of entropy = %.3f\")% (ds), (\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) To prove that the process is irreversible\n", "T2 = 421.820\n", "But the actual temperature is 433K at th epressure of 1.012 bar, Hence the process is irreversible. Proved.\n", "(ii) Change of entropy per kg of air\n", "Increase of entropy = 0.026 kJ/kg.K\n" ] } ], "prompt_number": 25 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.29 Page no : 275" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine : \n", "(i) Change in enthalpy ;\n", "(ii) Change in internal energy ;\n", "(iii) Change in entropy ;\n", "(iv) Heat transfer ;\n", "(v) Work transfer.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "V1 = 0.3; \t\t\t#m**3\n", "p1 = 4*10**5; \t\t#N/m**2\n", "V2 = 0.08; \t\t\t#m**3\n", "n = 1.25; \n", "\n", "# Calculations and Results\n", "p2 = p1*(V1/V2)**n;\n", "\n", "dH = n*(p2*V2-p1*V1)/(n-1)/10**3;\n", "print (\"(i) Change in enthalpy\"), (\"dH = %.3f\")% (dH), (\"kJ\")\n", "\n", "dU = dH-(p2*V2 - p1*V1)/10**3;\n", "print (\"(ii) Change in internal energy\"),(\"dU = %.3f\")% (dU), (\"kJ\")\n", "\n", "dS = 0;\n", "print (\"(iii) Change in entropy\"),(\"dS\"), (dS)\n", "\n", "Q = 0;\n", "print (\"(iv)Heat transfer\"),(\"Q = \"), (Q)\n", "\n", "W = Q-dU;\n", "print (\"(v) Work transfer\"),(\"W = %.3f\")%(W),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Change in enthalpy dH = 234.947 kJ\n", "(ii) Change in internal energy dU = 187.958 kJ\n", "(iii) Change in entropy dS 0\n", "(iv)Heat transfer Q = 0\n", "(v) Work transfer W = -187.958 kJ\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.30 Page no : 277" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) Change in internal energy,\n", "(ii) Work done,\n", "(iii) Heat transferred, and\n", "(iv) Change in entropy\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "m = 20.; \t\t\t#kg\n", "p1 = 4.*10**5; \t\t\t#Pa\n", "p2 = 8.*10**5; \t\t\t#Pa\n", "V1 = 4.; \t\t\t#m**3\n", "V2 = V1;\n", "cp = 1.04; \t\t\t#kJ/kg.K\n", "cv = 0.7432; \t\t\t#kJ/kg.K\n", "R = cp-cv;\n", "T1 = p1*V1/R/1000.; \t\t\t#kg.K; T = mass*temperature\n", "T2 = p2*V2/R/1000.; \t\t\t#kg.K\n", "\n", "# Calculations and Results\n", "dU = cv*(T2-T1);\n", "print (\"(i) Change in internal energy\"),(\"dU = %.3f\")% (dU), (\"kJ\")\n", "\n", "Q = 0;\n", "W = Q-dU;\n", "print (\"(ii) Work done\"),(\"W %.3f\")% (W), (\"kJ\")\n", "\n", "print (\"(iii) Heat transferred = \"), (Q)\n", "\n", "dS = m*cv*math.log(T2/T1);\n", "print (\"(iv) Change in entropy = %.3f\")%(dS), (\"kJ/K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Change in internal energy dU = 4006.469 kJ\n", "(ii) Work done W -4006.469 kJ\n", "(iii) Heat transferred = 0\n", "(iv) Change in entropy = 10.303 kJ/K\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.31 Page no : 278" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "determine the net work. Also plot the processes on T-S diagram\n", "'''\n", "\n", "import math \n", "from matplotlib.pyplot import *\n", "%pylab inline\n", "\n", "# Variables\n", "V1 = 5.; \t\t\t#m**3\n", "p1 = 2.*10**5; \t\t\t#Pa\n", "T1 = 300.; \t\t\t#K\n", "p2 = 6.*10**5; \t\t\t#Pa\n", "p3 = 2.*10**5; \t\t\t#Pa\n", "R = 287.;\n", "n = 1.3;\n", "y = 1.4;\n", "\n", "# Calculations and Results\n", "m = p1*V1/R/T1;\n", "T2 = T1*(p2/p1)**((n-1)/n);\n", "T3 = T2*(p3/p2)**((y-1)/y);\n", "W_12 = m*R*(T1-T2)/(n-1)/1000; \t\t\t#polytropic compression\n", "W_23 = m*R*(T2-T3)/(y-1)/1000; \t\t\t#Adiabatic expansion\n", "\n", "\n", "W_net = W_12+W_23;\n", "print (\"Net work done on the air = %.3f\")%(-W_net), (\"kJ\")\n", "\n", "T = [T1, 310, 320, 330, 340, 350, 360, 370, 380, T2];\n", "def f(T):\n", " return (y-n)/(y-1)/(1-n)*R/10**3*math.log(T);\n", "\n", "s = [f(T1), f(310), f(320), f(330), f(340), f(350), f(360), f(370), f(380), f(T2)]\n", "\n", "plot(s,T)\n", "\n", "T = [T2, T3];\n", "s = [f(T2), f(T2)];\n", "plot(s,T,'r')\n", "\n", "# Answers are slightly diffferent because of rounding error." ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", "Net work done on the air = 94.023 kJ\n" ] }, { "output_type": "stream", "stream": "stderr", "text": [ "WARNING: pylab import has clobbered these variables: ['f', 'draw_if_interactive']\n", "`%pylab --no-import-all` prevents importing * from pylab and numpy\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 9, "text": [ "[]" ] }, { "metadata": {}, "output_type": "display_data", "png": 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TExPD+vXrAXC73WRmZuJ2uwkNDSU/P7/FpiIREXFWq0NE/XJRDREVEWk3R4aI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\nkSCmIiAiEsT+P3S/G55r8UZJAAAAAElFTkSuQmCC\n", "text": [ "" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.32 Page no : 279" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) The heat supplied.\n", "(ii) The entropy change.\n", "'''\n", "\n", "# Variables\n", "V1 = 0.004; \t\t\t#m**3\n", "p1 = 1.*10**5; \t\t\t#Pa\n", "T1 = 300.; \t\t\t#K\n", "T2 = 400.; \t\t\t#K\n", "y = 1.4;\n", "M = 28.;\n", "R0 = 8.314;\n", "R = R0/M;\n", "\n", "# Calculations and Results\n", "print (\"(i) The heat supplied\")\n", "m = p1*V1/R/1000/T1; \t\t\t#kg\n", "cv = R/(y-1);\n", "Q = m*cv*(T2-T1);\n", "print (\"Q %.3f\")%(Q), (\"kJ\")\n", "\n", "print (\"(ii) The entropy change\")\n", "dS = m*cv*math.log(T2/T1);\n", "print (\"dS = %.8f\")%(dS), (\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The heat supplied\n", "Q 0.333 kJ\n", "(ii) The entropy change\n", "dS = 0.00095894 kJ/kg.K\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.33 Page no : 279" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine :\n", "(i) Change in entropy.\n", "(ii) Work done\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "V1 = 0.05; \t\t\t#m**3\n", "p1 = 1.*10**5; \t\t\t#Pa\n", "T1 = 280.; \t\t\t#K\n", "p2 = 5.*10**5; \t\t\t#Pa\n", "\n", "# Calculations\n", "print (\"(i) Change in entropy\")\n", "R0 = 8.314;\n", "M = 28.;\n", "R = R0/M;\n", "m = p1*V1/R/T1/1000;\n", "dS = m*R*math.log(p1/p2);\n", "\n", "# Results\n", "print (\"dS = %.3f\")%(dS), (\"kJ/K\")\n", "\n", "print (\"(ii)Work done\")\n", "Q = T1*dS;\n", "print (\"Q = %.3f\")%(Q),(\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Change in entropy\n", "dS = -0.029 kJ/K\n", "(ii)Work done\n", "Q = -8.047 kJ\n" ] } ], "prompt_number": 30 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.34 Page no : 280" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) Final specific volume and temperature.\n", "(ii) Change of internal energy, work done and heat interaction.\n", "(iii) Change in entropy.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "R = 0.287; \t\t\t#kJ/kg.K\n", "m = 1.; \t\t\t#kg\n", "p1 = 8.*10**5; \t\t\t#Pa\n", "p2 = 1.6*10**5; \t\t\t#Pa\n", "T1 = 380.; \t\t\t#K\n", "n = 1.2;\n", "y = 1.4;\n", "\n", "\n", "# Calculations and Results\n", "print (\"(i) Final specific volume and temperature\")\n", "v1 = R*T1/p1*10**3; \t\t\t#m**3/kg\n", "v2 = v1*(p1/p2)**(1/n);\n", "T2 = T1*(p2/p1)**((n-1)/n);\n", "\n", "print (\"v2 = %.3f\")%(v2), (\"m**3/kg\")\n", "print (\"T2 = %.3f\")% (T2),(\"K\")\n", "\n", "print (\"(ii) Change of internal energy, work done and heat interaction\")\n", "dU = R/(y-1)*(T2-T1);\n", "print (\"dU = %.3f\")%(dU), (\"kJ/kg\")\n", "\n", "W = R*(T1-T2)/(n-1);\n", "print (\"W = %.3f\")%(W), (\"kJ/kg\")\n", "\n", "Q = dU + W;\n", "print (\"Q = %.3f\")%(Q),(\"kJ/kg\")\n", "\n", "print (\"(iii) Change in entropy\")\n", "dS = R/(y-1)*math.log(T2/T1) + R*math.log(v2/v1)\n", "print (\"dS = %.3f\")%(dS),(\"kJ/kg.K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Final specific volume and temperature\n", "v2 = 0.521 m**3/kg\n", "T2 = 290.595 K\n", "(ii) Change of internal energy, work done and heat interaction\n", "dU = -64.148 kJ/kg\n", "W = 128.296 kJ/kg\n", "Q = 64.148 kJ/kg\n", "(iii) Change in entropy\n", "dS = 0.192 kJ/kg.K\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.35 Page no : 281" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(b)What would be the percentage error if the entropy change is calculated by dividing the\n", "quantity of heat exchanged by the mean absolute temperature during the process ?\n", "\n", "'''\n", "%pylab inline\n", "import math\n", "from matplotlib.pyplot import *\n", "\n", "# Variables\n", "y = 1.4;\n", "cv = 0.718; \t\t#kJ/kg.K\n", "m = 1.; \t\t\t #kg\n", "T1 = 290.; \t\t\t#K\n", "n = 1.3;\n", "r = 16.;\n", "y = 1.4;\n", "\n", "# Calculations and Results\n", "T2 = T1*(r)**(n-1);\n", "\n", "print (\"(a)\")\n", "\n", "T = [T1, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530, 540, 550, 560, 570, 580, 590, 600, 610, 620, 630, 640, 650, 660, T2];\n", "def f(T):\n", " return (y-n)*cv/(1-n)/10**3*math.log(T);\n", "\n", "s = [f(T1),f(300),f(310),f(320),f(330),f(340),f(350),f(360),f(370),f(380),f(390),f(400),f(410),f(420),f(430),f(440),f(450),f(460),f(470),f(480),f(490),f(500),f(510),f(520),f(530),f(540),f(550),f(560),f(570),f(580),f(590),f(600),f(610),f(620),f(630),f(640),f(650),f(660), f(T2)];\n", "plot(s,T)\n", "\n", "T = [0, T2];\n", "s = [f(T2), f(T2)];\n", "plot(s,T,'r--')\n", "\n", "T = [0 ,T1];\n", "s = [f(T1),f(T1)];\n", "plot(s,T,'r--')\n", "\n", "T = [T1 ,T2];\n", "s = [f(T1), f(T2)];\n", "plot(s,T,'r--' )\n", "suptitle(\"T-S diagram\")\n", "xlabel(\"S\")\n", "ylabel(\"T\")\n", "text(-0.00150,400,\"pV**n = C\")\n", "\n", "print (\"(b)Entropy change\")\n", "dS = cv*((n-y)/(n-1))*math.log(T2/T1);\n", "print (\"dS = %.3f\")%(dS), (\"kJ/kg.K\")\n", "print (\"There is decrease in entropy\")\n", "\n", "Q = cv*((y-n)/(n-1))*(T1-T2);\n", "Tmean = (T1+T2)/2;\n", "dS_app = Q/Tmean;\n", "\n", "error = ((-dS) - (-dS_app))/(-dS) * 100;\n", "print (\"age error = %.3f\")%(error), (\"%\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Populating the interactive namespace from numpy and matplotlib\n", "(a)\n", "(b)Entropy change" ] }, { "output_type": "stream", "stream": "stdout", "text": [ "\n", "dS = -0.199 kJ/kg.K\n", "There is decrease in entropy\n", "age error = 5.393 %\n" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.36 Page no : 282" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "calculate :\n", "(i) The net heat flow.\n", "(ii) The overall change in entropy.\n", "'''\n", "\n", "from matplotlib.pyplot import *\n", "from numpy import *\n", "# Variables\n", "cp = 1.005; \t\t\t#kJ/kg.K\n", "R = 0.287; \t\t\t#kJ/kg.K\n", "V1 = 1.2; \t\t\t#m**3\n", "p1 = 1.*10**5; \t\t\t#Pa\n", "p2 = p1;\n", "T1 = 300.; \t\t\t#K\n", "T2 = 600.; \t\t\t#K\n", "T3 = T1;\n", "p1 = 1.*10**5; \t\t\t#Pa\n", "cv = cp-R;\n", "\n", "# Calculations and Results\n", "print (\"(i) The net heat flow\")\n", "m = p1*V1/R/1000/T1; \t\t\t#kg\n", "Q = m*R*(T2-T1);\n", "print (\"Q = \"), (Q), (\"kJ\")\n", "\n", "print (\"(ii) The overall change in entropy\")\n", "dS_12 = m*cp*math.log(T2/T1);\n", "dS_23 = m*(cp-R)*math.log(T3/T2); \t\t\t#cv = cp-R\n", "dS_overall = dS_12+dS_23;\n", "print (\"Overall change in entropy = %.3f\")%(dS_overall),(\"kJ/K\")\n", "\n", "s = linspace(math.sqrt(300),math.sqrt(600),100);\n", "T = s**2;\n", "plot(s,T)\n", "\n", "s = linspace(22.18,math.sqrt(600),100)\n", "T = 10*(s-16.725)**2;\n", "plot(s,T,'r')\n", "\n", "s = [17, 25];\n", "T = [600, 600];\n", "plot(s,T,'--')\n", "\n", "s = [17 ,25];\n", "T = [300 ,300];\n", "plot(s,T,'--')\n", "\n", "suptitle(\"T-s diagram \")\n", "xlabel(\"S\")\n", "ylabel(\"T\")\n", "text(24,400,\"v = C\")\n", "text(20,450,\"p = C\")" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) The net heat flow\n", "Q = 120.0 kJ\n", "(ii) The overall change in entropy\n", "Overall change in entropy = 0.277 kJ/K\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 2, "text": [ "" ] }, { "metadata": {}, "output_type": "display_data", "png": 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"text": [ "" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.37 Page no : 283" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Constant volume heat addition till pressure becomes 5 bar, \n", "(ii) Constant pressure cooling, and \n", "(iii) Isothermal heating to initial state.\n", "'''\n", "\n", "import math \n", "from matplotlib.pyplot import *\n", "\n", "# Variables\n", "cv = 0.718; \t\t\t#kJ/kg.K\n", "R = 0.287 \t\t\t#kJ/kg.K\n", "p1 = 1.*10**5; \t\t\t#Pa\n", "T1 = 300. \t\t\t#K\n", "V1 = 0.018; \t\t\t#m**3\n", "p2 = 5.*10**5 \t\t\t#Pa\n", "T3 = T1;\n", "cp = cv+R;\n", "p3 = p2;\n", "\n", "# Calculations and Results\n", "m = p1*V1/R/T1/1000; \t\t\t#kg\n", "T2 = T1*p2/p1;\n", "\n", "print (\"(i) constant volume process\")\n", "dS_12 = m*cv*math.log(T2/T1);\n", "print (\"dS = %.3f\")%(dS_12), (\"kJ/K\")\n", "\n", "print (\"(ii) Constant prssure process \")\n", "dS_23 = m*cp*math.log(T3/T2);\n", "print (\"dS = %.3f\")%(dS_23), (\"kJ/K\")\n", "\n", "print (\"(iii) Isothermal process\")\n", "dS_31 = m*R*math.log(p3/p1);\n", "print (\"dS = %.5f\")%(dS_31),(\"kJ/K\")\n", "\n", "print (\"T-s diagram\")\n", "s = linspace(math.sqrt(300),math.sqrt(600),72);\n", "T = s**2;\n", "#plot(s,T)\n", "\n", "s = linspace(22.18,math.sqrt(600),24)\n", "T = 10*(s-16.725)**2;\n", "#plot(s,T,'r')\n", "\n", "s = [math.sqrt(300), 22.18];\n", "T = [300 ,300];\n", "#plot(s,T,'g')\n", "\n", "print (\"p-V diagram\")\n", "\n", "V = [0.018, 0.018];\n", "p = [1 ,5];\n", "#plot(V,p)\n", "\n", "p = [5 ,5];\n", "V = [0.0036, 0.018];\n", "#plot(V,p,'r')\n", "\n", "V = linspace(0.0036,0.018,145)\n", "\n", "def f():\n", " return 1*0.018/V;\n", "f1 = f()\n", "\n", "plot(V,f1,'g')\n", "suptitle(\"p-V diagram\")\n", "xlabel(\"V\")\n", "ylabel(\"p\")\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) constant volume process\n", "dS = 0.024 kJ/K\n", "(ii) Constant prssure process \n", "dS = -0.034 kJ/K\n", "(iii) Isothermal process\n", "dS = 0.00966 kJ/K\n", "T-s diagram\n", "p-V diagram\n" ] }, { "metadata": {}, "output_type": "pyout", "prompt_number": 3, "text": [ "" ] }, { "metadata": {}, "output_type": "display_data", "png": 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pJBaFhuCogog0FYtCQ3BUQUSaikWhQTiqICJNxKLQIBxVEJEmYlFomD9HFSdunJA6ChER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\n6sSiICKiOrEoiIioTiwKIiKq0/8HJIEAIIPDm7IAAAAASUVORK5CYII=\n", "text": [ "" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.39 Page no : 285" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the entropy change of 4 kg of a perfect gas \n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "m = 4; \t\t\t #kg\n", "T1 = 400; \t\t\t#K\n", "T2 = 500; \t\t\t#K\n", "\n", "# Calculations\n", "def f12( T): \n", "\t return m*(0.48+0.0096*T)/T\n", "\n", "dS = quad(f12, T1,T2)[0]\n", "\n", "# Results\n", "print (\"dS = %.3f\")%(dS), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "dS = 4.268 kJ\n" ] } ], "prompt_number": 34 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.40 Page no : 286" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "find the entropy of the gas at 25 bar and 750 K temperature.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "p1 = 1*10**5; \t\t\t#Pa\n", "T1 = 273; \t\t\t#K\n", "p2 = 25*10**5; \t\t\t#Pa\n", "T2 = 750; \t\t\t#K\n", "R = 0.29; \t\t\t#kJ/kg.K ; cp = 0.85+0.00025*T; cv = 0.56+0.00025*T; R = cp-cv;\n", "\n", "# Calculations\n", "v2 = R*T2/p2;\n", "v1 = R*T1/p1;\n", "\n", "def f8( T): \n", "\t return (0.56+0.00025*T)/T\n", "\n", "def f9(v):\n", " return R/v\n", "ds = quad(f8, T1, T2)[0] + quad(f9,v1,v2)[0]\n", "\n", "# Results\n", "print (\"ds = %.3f\")%(ds),(\"kJ/kg K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "ds = 0.045 kJ/kg K\n" ] } ], "prompt_number": 35 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.41 Page no : 287" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate :\n", "(i) Final equilibrium temperature,\n", "(ii) Final pressure on each side of the diaphragm, and\n", "(iii) Entropy change of system.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "cv = 0.715; \t\t\t#kJ/kg K\n", "R = 0.287; \t\t\t#kJ/kg K\n", "V_A = 0.25; \t\t\t#m**3\n", "p_Ai = 1.4; \t\t\t#bar\n", "T_Ai = 290; \t\t\t#K\n", "V_B = 0.25; \t\t\t#m**3\n", "p_Bi = 4.2; \t\t\t#bar\n", "T_Bi = 440; \t\t\t#K\n", "\n", "# Calculations and Results\n", "print (\"(i) Final equilibrium temperature\")\n", "m_A = p_Ai * 10**5 * V_A / R / 1000/ T_Ai; \t\t\t#kg\n", "m_B = p_Bi * 10**5 * V_B / R / 1000/ T_Bi; \t\t\t#kg\n", "\n", "T_f = (m_B * T_Bi + m_A * T_Ai)/(m_A + m_B);\n", "print (\"T_f = %.3f\")% (T_f), (\"K\")\n", "\n", "\n", "print (\"(ii) Final pressure on each side of the diaphragm\")\n", "p_Af = p_Ai*T_f/T_Ai;\n", "print (\"p_Af = %.3f\")%(p_Af),(\"bar\")\n", "\n", "p_Bf = p_Bi*T_f/T_Bi;\n", "print (\"p_Bf = %.3f\")%(p_Bf),(\"bar\")\n", "\n", "\n", "print (\"(iii) Entropy change of the system\")\n", "dS_A = m_A*cv*math.log(T_f/T_Ai);\n", "dS_B = m_B*cv*math.log(T_f/T_Bi);\n", "dS_net = dS_A+dS_B;\n", "print (\"Net change of entropy = %.3f\")%(dS_net), (\"kJ/K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Final equilibrium temperature\n", "T_f = 389.618 K\n", "(ii) Final pressure on each side of the diaphragm\n", "p_Af = 1.881 bar\n", "p_Bf = 3.719 bar\n", "(iii) Entropy change of the system\n", "Net change of entropy = 0.016 kJ/K\n" ] } ], "prompt_number": 36 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.42 Page no : 287" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Find the change in entropy in each of the adiabatic processes.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "cv = 1.25; \t\t\t#kJ/kg.K\n", "T1 = 530.; \t\t\t#K\n", "v1 = 0.0624; \t\t\t#m**3/kg\n", "v2 = 0.186; \t\t\t#m**3/kg\n", "dT_31 = 25.; \t\t\t#K\n", "T3 = T1-dT_31; \t\t\t#K\n", "dT_21 = 165.; \t\t\t#K\n", "\n", "# Calculations\n", "T2 = T1-dT_21; \t\t\t#K\n", "# Path 1-2 : Reversible adiabatic process\n", "ds_12 = 0;\n", "v3 = 0.186; \t\t\t#m**3/kg\n", "v3 = v2;\n", "ds_13 = cv*math.log(T3/T2);\n", "\n", "# Results\n", "print (\"Chang in entropy = %.4f\")%(ds_13), (\"kJ/kgK\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Chang in entropy = 0.4058 kJ/kgK\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.44 Page no : 289" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Determine the heat interactions with the other two sources of heat.\n", "(ii) Evaluate the entropy change due to each heat interaction with the engine.\n", "(iii) Total entropy change during the cycle.\n", "'''\n", "\n", "import math \n", "from numpy import *\n", "\n", "# Variables\n", "T1 = 500.; \t\t\t#K\n", "T2 = 400.; \t\t\t#K\n", "T3 = 300.; \t\t\t#K\n", "Q1 = 1500.; \t\t\t#kJ/min\n", "W = 200.; \t\t\t#kJ/min\n", "\n", "# Calculations and Results\n", "A = [[1,-1],[(1./400),(-1./300)]];\n", "B = [(-1300),(-3)];\n", "X = linalg.solve(A,B)\n", "\n", "Q2 = X[0];\n", "print (\"Q2 = \"), (Q2), (\"kJ/min\")\n", "\n", "Q3 = X[1];\n", "print (\"Q3 = \"), (Q3), (\"kJ/min\")\n", "\n", "print (\"(ii) Entropy change \")\n", "dS1 = (-Q1)/T1;\n", "print (\"Entropy change of source 1 = \"), (dS1), (\"kJ/K\")\n", "\n", "dS2 = (-Q2)/T2;\n", "print (\"Entropy change of math.sink 2 = \"), (dS2), (\"kJ/K\")\n", "\n", "dS3 = Q3/T3;\n", "print (\"Entropy change of source 3 = \"),(dS3), (\"kJ/K\")\n", "\n", "\n", "print (\"(iii) Net change of the entropy\")\n", "dSnet = dS1 + dS2 + dS3;\n", "print (\"dSnet = %d\")% (dSnet)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Q2 = -1600.0 kJ/min\n", "Q3 = -300.0 kJ/min\n", "(ii) Entropy change \n", "Entropy change of source 1 = -3.0 kJ/K\n", "Entropy change of math.sink 2 = 4.0 kJ/K\n", "Entropy change of source 3 = -1.0 kJ/K\n", "(iii) Net change of the entropy\n", "dSnet = 0\n" ] } ], "prompt_number": 13 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.45 Page no : 291" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Determine the maximum amount of work that can be recovered as the system is cooled down to the temperature of the reservoir.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "T1 = 250; \t\t\t#K\n", "T2 = 125; \t\t\t#K\n", "\n", "# Calculations\n", "#cv = 0.0045*T**2\n", "def f10( T): \n", "\t return 0.045*T**2\n", "\n", "Q1 = quad(f10, T1, T2)[0]\n", "\n", "def f11( T): \n", "\t return 0.045*T\n", "\n", "dS_system = quad(f11, T1, T2)[0]\n", "\n", "dS_universe = 0;\n", "\n", "W_max = ((-Q1) -T2*(dS_universe-dS_system))/1000;\n", "\n", "# Results\n", "print (\"W_max = %.3f\")%(W_max), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "W_max = 73.242 kJ\n" ] } ], "prompt_number": 39 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.46 Page no : 292" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Establish the direction of the flow of the air in the duct.\n", "'''\n", "\n", "import math \n", "from scipy.integrate import quad \n", "\n", "# Variables\n", "cp = 1.005; \t\t\t#kJ/kg K\n", "T_A = 333.; \t\t\t#K\n", "T_B = 288.; \t\t\t#K\n", "p_A = 140.; \t\t\t#kPa\n", "p_B = 110.; \t\t\t#kPa\n", "#h = cp*T\n", "#v/T = 0.287/p\n", "\n", "# Calculations\n", "def f9( T): \n", "\t return cp/T\n", "\n", "def f10(p):\n", " return 0.287/p\n", " \n", "ds_system = quad(f9, T_B, T_A)[0] + quad(f10,p_A,p_B)[0]\n", "ds_surr = 0;\n", "ds_universe = ds_system+ds_surr;\n", "\n", "# Results\n", "print (\"change in entropy of universe = -%.4f\")% (ds_universe), (\"kJ/kgK\")\n", "print (\"Since change in entropy of universe from A to B is -ve\")\n", "print (\"The flow is from B to A\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "change in entropy of universe = -0.0767 kJ/kgK\n", "Since change in entropy of universe from A to B is -ve\n", "The flow is from B to A\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.47 Page no : 292" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "Calculate the change of entropy due to mixing process.\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "m1 = 3.; \t\t\t#kg\n", "m2 = 4.; \t\t\t#kg\n", "T0 = 273.; \t\t\t#K\n", "T1 = 80.+273; \t\t\t#K\n", "T2 = 15.+273; \t\t\t#K\n", "c_pw = 4.187; \t\t\t#kJ/kgK\n", "\n", "# Calculations\n", "tm = (m1*T1 + m2*T2)/(m1+m2);\n", "Si = m1*c_pw*math.log(T1/T0) + m2*c_pw*math.log(T2/T0);\n", "Sf = (m1+m2)*c_pw*math.log(tm/T0);\n", "dS = Sf-Si;\n", "\n", "# Results\n", "print (\"Net change in entropy = %.3f\")%(dS),(\"kJ/K\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Net change in entropy = 0.150 kJ/K\n" ] } ], "prompt_number": 41 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.49 Page no : 294" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(a)find :\n", "(i) Entropy change of water ;\n", "(ii) Entropy change of the heat reservoir ;\n", "(iii) Entropy change of the universe.\n", "(b)what will the entropy change of the universe be ?\n", "(c) Explain how water might be heated from 0\u00b0C to 90\u00b0C \n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "m = 1.; \t\t\t#kg\n", "T1 = 273.; \t\t\t#K\n", "T2 = 363.; \t\t\t#K\n", "c = 4.187;\n", "\n", "# Calculations and Results\n", "print (\"(a)\")\n", "ds_water = m*c*math.log(T2/T1);\n", "print (\"(i) Entropy of water = %.3f\")%(ds_water), (\"kJ/kgK\")\n", "\n", "print (\"(ii) Entropy change of the reservoir \")\n", "Q = m*c*(T2-T1);\n", "ds_reservoir = -Q/T2;\n", "print (\"ds_reservoir = %.3f\")% (ds_reservoir), (\"kJ/K\")\n", "\n", "ds_universe = ds_water+ds_reservoir;\n", "print (\"(iii) Entropy change of universe = %.3f\")% (ds_universe), (\"kJ/K\")\n", "\n", "print (\"(b)\")\n", "T3 = 313; \t\t\t#K\n", "ds_water = m*c*(math.log(T3/T1) + math.log(T2/T3));\n", "ds_res1 = -m*c*(T3-T1)/T3;\n", "ds_res2 = -m*c*(T2-T3)/T2;\n", "\n", "ds_universe = ds_water+ds_res1+ds_res2;\n", "print (\"(iii) Entropy change of universe = %.3f\")%(ds_universe), (\"kJ/K\")\n", "\n", "print (\"(c) The entropy change of universe would be less and less, if the water is heated in more and more stages, by bringing\")\n", "print (\"the water in contact successively with more and more heat reservoirs, each succeeding reservoir being at a higher temperature\") \n", "print (\"than the preceding one.\")\n", "\n", "print (\"When water is heated in infinite steps, by bringing in contact with an infinite number of reservoirs in succession, so that\") \n", "print (\"at any insmath.tant the temperature difference between the water and the reservoir in contact is infinitesimally small, then\") \n", "print (\"the entropy change of the universe would be zero and the water would be reversibly heated.\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a)\n", "(i) Entropy of water = 1.193 kJ/kgK\n", "(ii) Entropy change of the reservoir \n", "ds_reservoir = -1.038 kJ/K\n", "(iii) Entropy change of universe = 0.155 kJ/K\n", "(b)\n", "(iii) Entropy change of universe = 0.081 kJ/K\n", "(c) The entropy change of universe would be less and less, if the water is heated in more and more stages, by bringing\n", "the water in contact successively with more and more heat reservoirs, each succeeding reservoir being at a higher temperature\n", "than the preceding one.\n", "When water is heated in infinite steps, by bringing in contact with an infinite number of reservoirs in succession, so that\n", "at any insmath.tant the temperature difference between the water and the reservoir in contact is infinitesimally small, then\n", "the entropy change of the universe would be zero and the water would be reversibly heated.\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 3, "metadata": {}, "source": [ "Example 5.50 Page no : 295" ] }, { "cell_type": "code", "collapsed": false, "input": [ "'''\n", "(i) Determine the entropy increase of the universe.\n", "(ii) What is the minimum amount of work necessary to convert the water back into ice at \u2013 5\u00b0C ?\n", "'''\n", "\n", "import math \n", "\n", "# Variables\n", "cp = 2.093; \t\t\t#kJ/kg0C\n", "c = 4.187;\n", "Lf = 333.33; \t\t\t#kJ/kg\n", "m = 1.; \t\t\t#kg\n", "T0 = 273.; \t\t\t#K\n", "T1 = 268.; \t\t\t#K\n", "T2 = 298.; \t\t\t#K\n", "\n", "# Calculations and Results\n", "Q_s = m*cp*(T0-T1);\n", "Q_f = m*Lf;\n", "Q_l = m*c*(T2-T0);\n", "Q = Q_s+Q_f+Q_l;\n", "\n", "print (\"(i) Entropy increase of the universe\")\n", "ds_atm = -Q/T2;\n", "ds_sys1 = m*cp*math.log(T0/T1);\n", "ds_sys2 = Lf/T0;\n", "ds_sys3 = m*c*math.log(T2/T0);\n", "ds_total = ds_sys1+ds_sys2+ds_sys3;\n", "ds_universe = ds_total+ds_atm;\n", "\n", "print (\"Entropy increase of universe = %.3f\")%(ds_universe), (\"kJ/K\")\n", "\n", "\n", "print (\"(ii) Minimum amount of work necessary to convert the water back into ice at \u2013 5\u00b0C, Wmin.\")\n", "dS_refrigerator = 0;\n", "\n", "dS_system = -1.6263; \t\t\t#kJ/kg K\n", "T = 298; \t\t\t#K\n", "#For minimum work \n", "W_min = T*(-dS_system)-Q;\n", "\n", "print (\"Minimum work done = %.3f\")% (W_min), (\"kJ\")\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Entropy increase of the universe\n", "Entropy increase of universe = 0.122 kJ/K\n", "(ii) Minimum amount of work necessary to convert the water back into ice at \u2013 5\u00b0C, Wmin.\n", "Minimum work done = 36.167 kJ\n" ] } ], "prompt_number": 43 } ], "metadata": {} } ] }