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diff --git a/Industrial_Instrumentation/ch6.ipynb b/Industrial_Instrumentation/ch6.ipynb new file mode 100644 index 00000000..a3874b6a --- /dev/null +++ b/Industrial_Instrumentation/ch6.ipynb @@ -0,0 +1,1286 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 6 : Availability and Irreversibility" + ] + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.1 Page no : 313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the irreversibility if the sink temperature is 293 K. \n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "T0 = 293.; \t\t\t#K\n", + "T1 = 300.; \t\t\t#K\n", + "T2 = 370.; \t\t\t#K\n", + "cv = 0.716;\n", + "cp = 1.005;\n", + "R = 0.287;\n", + "p1 = 1.; \t\t\t#bar\n", + "p2 = 6.8; \t\t\t#bar\n", + "m = 1.; \t\t\t#kg\n", + "\n", + "# Calculations\n", + "Wmax = -(cv*(T2-T1) - T0*(cp*math.log(T2/T1)-R*math.log(p2/p1)));\n", + "n = 1/(1-(math.log(T2/T1)/math.log(p2/p1)));\n", + "Wact = m*R*(T1-T2)/(n-1);\n", + "I = Wmax - Wact;\n", + "\n", + "# Results\n", + "print (\"Irreversibility = %.3f\")%(I),(\"kJ/kg\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Irreversibility = 13.979 kJ/kg\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.2 Page no : 313" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find\n", + "(i) The entropy produced during heat transfer ;\n", + "(ii) The decrease in available energy after heat transfer.\n", + "'''\n", + "\n", + "# Variables\n", + "T1 = 1000.; \t\t\t#K\n", + "T2 = 500.; \t\t\t#K\n", + "T0 = 300.; \t\t\t#K\n", + "Q = 7200.; \t\t\t#kJ/min\n", + "\n", + "# Calculations and Results\n", + "print (\"(i) Net change of entropy :\")\n", + "dS_source = -Q/T1;\n", + "dS_system = Q/T2;\n", + "dS_net = dS_source+dS_system;\n", + "print (\"dS_net = \"), (dS_net), (\"kJ/min.K\")\n", + "\n", + "\n", + "print (\"(ii) Decrease in available energy :\")\n", + "AE_source = (T1-T0)*(-dS_source); \t\t\t#Available energy with the source\n", + "AE_system = (T2-T0)*dS_system; \t\t\t#Available energy with the system\n", + "dAE = AE_source - AE_system; \t\t\t#Decrease in available energy\n", + "print (\"dAE = \"), (dAE), (\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Net change of entropy :\n", + "dS_net = 7.2 kJ/min.K\n", + "(ii) Decrease in available energy :\n", + "dAE = 2160.0 kJ\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.3 Page no : 315" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine :\n", + "(i) The availability if the system goes through the ideal work producing process.\n", + "(ii) The availability and effectiveness if the air is cooled at constant pressure to atmos-\n", + "pheric temperature without bringing it to complete dead state. Take c v = 0.718 kJ/kg K ;\n", + "c p = 1.005 kJ/kg K\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m = 8.; \t\t\t#kg\n", + "T1 = 650.; \t\t\t#K\n", + "p1 = 5.5*10**5; \t\t\t#Pa\n", + "p0 = 1*10.**5; \t\t\t#Pa\n", + "T0 = 300.; \t\t\t#K\n", + "cp = 1.005; \t\t\t#kJ/kg.K\n", + "cv = 0.718;\n", + "R = 0.287;\n", + "#p1*v1/T1 = p0*v0/T0\n", + "#Let r = v1/v0 = 1/2.54\n", + "r = 1/2.54;\n", + "\n", + "# Calculations and Results\n", + "print (\"(i) Change in available energy(for bringing the system to dead state) = \")\n", + "ds = cv*math.log(T1/T0) + R*math.log(r);\n", + "dAE = m*(cv*(T1-T0) - T0*ds);\n", + "#dAE is the change in available energy in kJ\n", + "V1 = m*R*10**3*T1/p1;\n", + "V0 = V1/r;\n", + "L = p0*(V0 - V1)/10**3;\n", + "print (\"Loss of availability, L = \"), (L), (\"kJ\")\n", + "\n", + "print (\"(ii) Available Energy and Effectiveness\")\n", + "Q = m*cp*(T1-T0);\n", + "ds = m*cp*math.log(T1/T0);\n", + "Unavailable_energy = T0*ds;\n", + "Available_energy = Q - Unavailable_energy;\n", + "print (\"Available energy = %.3f\")% (Available_energy), (\"kJ\")\n", + "\n", + "Effectiveness = Available_energy/dAE;\n", + "print (\"Effectiveness = %.3f\")% (Effectiveness)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Change in available energy(for bringing the system to dead state) = \n", + "Loss of availability, L = 417.872 kJ\n", + "(ii) Available Energy and Effectiveness\n", + "Available energy = 949.066 kJ\n", + "Effectiveness = 0.719\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.4 Page no : 316" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Obtain the results on the basis of 1 kg of water.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "c_pg = 1.; \t\t\t#kJ/kgK\n", + "h_fg = 1940.7; \t\t\t#kJ/kg\n", + "Ts = 473.; \t \t\t#K ; Temperature of saturation of steam\n", + "T1 = 1273.; \t\t\t#K ; Initial temperature of gases\n", + "T2 = 773.; \t\t \t#K ; Final temperature of gases\n", + "T0 = 293.; \t\t\t #K ; atmospheric temperature\n", + "\n", + "# Calculations\n", + "#Heat lost by gases = Heat gained by 1 kg saturated water when it is converted to steam at 200 0C\n", + "m_g = h_fg/c_pg/(T1-T2);\n", + "dS_g = m_g*c_pg*math.log(T2/T1);\n", + "dS_w = h_fg/Ts;\n", + "dS_net = dS_g + dS_w;\n", + "\n", + "# Results\n", + "print (\"Net change in entropy = %.3f\")% (dS_net), (\"kJ/K\")\n", + "\n", + "E = T0*dS_net; \t\t\t#Increase in unavailable energy due to hea transfer\n", + "print (\"Increase in unavailable energy = %.3f\")%(E), (\"kJ per kg of steam formed\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net change in entropy = 2.167 kJ/K\n", + "Increase in unavailable energy = 634.847 kJ per kg of steam formed\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.5 Page no : 317" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find the loss in available energy due to above heat transfer.\n", + "'''\n", + "\n", + "import math\n", + "\n", + "# Variables\n", + "m_g = 3.; \t\t\t#kg\n", + "p1 = 2.5; \t\t\t#bar\n", + "T1 = 1200.; \t\t\t#K; Temperature of infinite source\n", + "T1a = 400.; \t\t\t#K; Initial temperature\n", + "Q = 600.; \t\t\t#kJ\n", + "cv = 0.81; \t\t\t#kJ/kg.K\n", + "T0 = 290.; \t\t\t#K; Surrounding Temperature\n", + "\n", + "# Calculations\n", + "#final temperature = T2a\n", + "T2a = Q/m_g/cv + T1a;\n", + "AE = (T1-T0)*Q/T1; \t\t\t#Available energy with the source\n", + "dS = m_g*cv*math.log(T2a/T1a); \t\t\t#Change in entropy of the gas\n", + "UAE = T0*dS; \t\t\t#Unavailability of the gas \n", + "A = Q-UAE; \t\t\t#Available energy with the gas\n", + "loss = AE-A;\n", + "\n", + "# Results\n", + "print (\"Loss in available energy due to heat transfer = %.3f\")%(loss),(\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in available energy due to heat transfer = 193.783 kJ\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.6 Page no : 318" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate the unavailable energy in 60 kg of water at 60\u00b0C \n", + "'''\n", + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables\n", + "m = 60.; \t\t\t#kg\n", + "T1 = 333.; \t\t\t#K\n", + "T0 = 279.; \t\t\t#K\n", + "p = 1.; \t\t\t#atm\n", + "cp = 4.187;\n", + "\n", + "# Calculations\n", + "def f16( T): \n", + "\t return m*cp*(1-T0/T)\n", + "\n", + "Wmax = quad(f16, T0, T1)[0]\n", + "Q1 = m*cp*(T1-T0);\n", + "#Let unavailable energy = E\n", + "E = Q1-Wmax;\n", + "\n", + "# Results\n", + "print (\"unavailable energy = %.3f\")%(E), (\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "unavailable energy = 12401.141 kJ\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.7 Page no : 319" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "find the loss in availability for the process.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m = 15.; \t\t\t#kg\n", + "T1 = 340.; \t\t\t#K\n", + "T0 = 300.; \t\t\t#K\n", + "cp = 4.187; \t\t\t#kJ/kgK\n", + "\n", + "# Calculations\n", + "#Work added during churning = Increase in enthalpy of water\n", + "W = m*cp*(T1-T0);\n", + "ds = cp*math.log(T1/T0);\n", + "AE = m*(cp*(T1-T0)-T0*ds);\n", + "AE_loss = W-AE; \t\t\t#Loss in availability\n", + "\n", + "# Results\n", + "print (\"Loss in availability %.3f\")% (AE_loss), (\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in availability 2358.261 kJ\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.8 Page no : 320" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "(i) Determine the availability of the system if the surrounding pressure and temperature\n", + "are 1 bar and 290 K respectively.\n", + "(ii) determine the availability and effectiveness\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m = 5.; \t\t\t#kg\n", + "T1 = 550.; \t\t\t#K\n", + "p1 = 4*10.**5; \t\t\t#Pa\n", + "T2 = 290.; \t\t\t#K\n", + "T0 = T2;\n", + "p2 = 1.*10**5; \t\t\t#Pa\n", + "p0 = p2;\n", + "cp = 1.005; \t\t\t#kJ/kg K\n", + "cv = 0.718; \t\t\t#kJ/kg K\n", + "R = 0.287; \t\t\t#kJ/kg K\n", + "\n", + "# Calculations and Results\n", + "print (\"(i) Availability of the system :\")\n", + "ds = cp*math.log(T1/T0) - R*math.log(p1/p0);\n", + "Availability = m*(cv*(T1-T0) - T0*ds);\n", + "print (\"Availability of the system = %.3f\")% (Availability), (\"kJ\")\n", + "\n", + "print (\"(ii) Available energy and Effectiveness\")\n", + "Q = m*cp*(T1-T0);\n", + "dS = m*cp*math.log(T1/T0);\n", + "E = T0*dS; \t\t\t#Unavailable energy\n", + "AE = Q-E;\n", + "print (\"Available Energy = %.3f\")%(AE), (\"kJ\")\n", + "\n", + "Effectiveness = AE/Availability;\n", + "print (\"Effectiveness = %.3f\")%(Effectiveness)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Availability of the system :\n", + "Availability of the system = 577.612 kJ\n", + "(ii) Available energy and Effectiveness\n", + "Available Energy = 373.806 kJ\n", + "Effectiveness = 0.647\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.9 Page no : 321" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine actual and minimum power required to run the compressor. \n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "R = 0.287; \t\t\t#kJ/kgK\n", + "cp = 1.005; \t\t\t#kJ/kgK\n", + "m = 25./60; \t\t\t#kg/s\n", + "p1 = 1.; \t\t\t#bar\n", + "p2 = 2.; \t\t\t#bar\n", + "T1 = 288.; \t\t\t#K\n", + "T0 = T1;\n", + "T2 = 373.; \t\t\t#K\n", + "\n", + "# Calculations and Results\n", + "W_act = cp*(T2-T1); \t\t\t#W_actual\n", + "W_total = m*W_act;\n", + "\n", + "print (\"Total actual power required = %.3f\")%(W_total), (\"kW\")\n", + "\n", + "ds = cp*math.log(T2/T1) - R*math.log(p2/p1);\n", + "Wmin = cp*(T2-T1) - T0*(ds);\n", + "\n", + "\n", + "W = m*Wmin;\n", + "print (\"Minimuumm work required = %.3f\")%(W), (\"kW\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Total actual power required = 35.594 kW\n", + "Minimuumm work required = 28.276 kW\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.10 Page no : 322" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the loss in availability if the surrounding temperature is 290 K.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m_O2 = 1.; \t\t\t#kg\n", + "m_H2 = 1.; \t\t\t#kg\n", + "p = 1*10.**5; \t\t\t#Pa\n", + "T_O2 = 450.; \t\t\t#K\n", + "T_H2 = 450.; \t\t\t#K\n", + "T0 = 290.; \t\t\t#K\n", + "R0 = 8.314;\n", + "M_O2 = 32.;\n", + "M_H2 = 2.;\n", + "\n", + "# Calculations\n", + "R_O2 = R0/M_O2;\n", + "v_O2 = m_O2*R_O2*T_O2/p;\n", + "R_H2 = R0/M_H2;\n", + "v_H2 = m_H2*R_H2*T_H2/p;\n", + "v_f = v_O2 + v_H2; \t\t\t#total volume after mixing\n", + "dS_O2 = R_O2*math.log(v_f/v_O2);\n", + "dS_H2 = R_H2*math.log(v_f/v_H2);\n", + "dS_net = dS_O2 + dS_H2;\n", + "#Let E be the loss in availability \n", + "E = T0*dS_net;\n", + "\n", + "# Results\n", + "print (\"Loss in availability = %.3f\")% (E), (\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in availability = 286.555 kJ\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.11 Page no : 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate the decrease in available energy \n", + "'''\n", + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables\n", + "T0 = 283.; \t\t\t#K\n", + "cp = 4.18; \t\t\t#kJ/kgK\n", + "m1 = 20.; \t\t\t#kg\n", + "T1 = 363.; \t\t\t#K\n", + "m2 = 30.; \t\t\t#kg\n", + "T2 = 303.; \t\t\t#K\n", + "T3 = 327.; \t\t\t#K\n", + "\n", + "\n", + "# Calculations\n", + "def f13( T): \n", + "\t return m1*cp*(1-T0/T)\n", + "\n", + "AE1 = quad(f13, T0, T1)[0]\n", + "\n", + "def f14( T): \n", + "\t return m2*cp*(1-T0/T)\n", + "\n", + "AE2 = quad(f14, T0, T2)[0]\n", + "AE_total = AE1 + AE2; \t\t\t#before mixing\n", + "#If T K is the final temperature after mixing\n", + "T = (m1*T1+m2*T2)/(m1+m2);\n", + "m_total = m1+m2;\n", + "\n", + "#Available energy of 50kg of water at 54 0C\n", + "AE3 = m_total*cp*((T3-T0) - T0*math.log(T3/T0));\n", + "\n", + "#Decrease in available energy due to mixing dAE\n", + "dAE = AE_total - AE3;\n", + "\n", + "# Results\n", + "print (\"Decrease in avialble energy = %.3f\")% (dAE), (\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Decrease in avialble energy = 234.184 kJ\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.12 Page no : 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine the loss in availability on the basis of one kg of oil per second.\n", + "'''\n", + "\n", + "import math\n", + "\n", + "# Variables\n", + "T_w1 = 323.; \t\t\t#K\n", + "T_w2 = 343.; \t\t\t#K\n", + "T_o1 = 513.; \t\t\t#K\n", + "T_o2 = 363.; \t\t\t#K\n", + "SG_oil = 0.82;\n", + "c_po = 2.6; \t\t\t#kJ/kg K\n", + "c_pw = 4.18; \t\t\t#kJ/kg K\n", + "T0 = 300.; \t\t\t#K\n", + "m_o = 1.; \t\t\t#kg\n", + "\n", + "# Calculations\n", + "#Heat lost by oil = Heat gained by water\n", + "m_w = (m_o*c_po*(T_o1-T_o2))/(c_pw*(T_w2-T_w1));\n", + "dS_w = m_w*c_pw*math.log(T_w2/T_w1);\n", + "dS_o = m_o*c_po*math.log(T_o2/T_o1);\n", + "dAE_w = m_w*(c_pw*(T_w2-T_w1))-T0*dS_w;\n", + "dAE_o = m_o*(c_po*(T_o2-T_o1))-T0*dS_o;\n", + "\n", + "# Loss in availability E = \n", + "E = dAE_w+dAE_o;\n", + "\n", + "# Results\n", + "print (\"Loss in availability = %.3f\")%(E),(\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Loss in availability = -81.676 kJ\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.13 Page no : 325" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "calculate the net increase in entropy and unavailable energy when the system reaches common temperature\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m_i = 1.; \t\t\t#kg\n", + "T_i = 273.; \t\t\t#K\n", + "m_w = 12.; \t\t\t#kg\n", + "T_w = 300.; \t\t\t#K\n", + "T0 = 288.; \t\t\t#K\n", + "c_pw = 4.18; \t\t\t#kJ/kg K\n", + "c_pi = 2.1; \t\t\t#kJ/kg K\n", + "L_i = 333.5; \t\t\t#kJ/kg\n", + "\n", + "# Calculations\n", + "Tc = (m_w*c_pw*T_w + m_i*c_pw*T_i - L_i)/(m_w*c_pw + m_i*c_pw);\n", + "dS_w = m_w*c_pw*math.log(Tc/T_w);\n", + "dS_i = m_i*c_pw*math.log(Tc/T_i) + L_i/T_i;\n", + "dS_net = dS_w+dS_i;\n", + "\n", + "# Results\n", + "print (\"Increase in entropy = %.3f\")% (dS_net), (\"kJ/K\")\n", + "\n", + "dAE = T0*dS_net;\n", + "print (\"Increase in unavailable energy = %.3f\")% (dAE),(\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Increase in entropy = 0.107 kJ/K\n", + "Increase in unavailable energy = 30.878 kJ\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.14 Page no : 326" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "What is the fraction of the available energy in the heat transferred from the process vapour\n", + "at 400\u00b0C that is lost due to the irreversible heat transfer at 200\u00b0C ?\n", + "'''\n", + "\n", + "# Variables\n", + "T1 = 673.; \t\t\t#K\n", + "T2 = 473.; \t\t\t#K\n", + "T0 = 303.; \t\t\t#K\n", + "T1a = T2;\n", + "\n", + "# Calculations\n", + "UAE = T0*(T1-T1a)/T1a/(T1-T0);\n", + "\n", + "# Results\n", + "print (\"the fraction of energy that becomes unavailable = %.3f\")%(UAE)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the fraction of energy that becomes unavailable = 0.346\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.15 Page no : 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate the effectiveness of the heating process when the atmospheric temperature is 15\u00b0C.\n", + "'''\n", + "import math \n", + "\n", + "# Variables\n", + "T1 = 293.; \t\t\t#K\n", + "T2 = 353.; \t\t\t#K\n", + "Tf = 1773.; \t\t\t#K\n", + "T0 = 288.; \t\t\t#K\n", + "c_pl = 6.3; \t\t\t#kJ/kg K\n", + "\n", + "# Calculations\n", + "dAE = c_pl*(T2-T1) - T0*c_pl*math.log(T2/T1);\n", + "n = (1-T0/Tf); \t\t\t#efficiency\n", + "E = c_pl*(T2-T1)*n;\n", + "Effectiveness = dAE/E;\n", + "\n", + "# Results\n", + "print (\"Effectiveness of the heating process = %.3f\")%(Effectiveness)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Effectiveness of the heating process = 0.126\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.16 Page no : 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "calculate :\n", + "(i) The ratio of mass flow of air initially at 100\u00b0C to that initially at 20\u00b0C.\n", + "(ii) The effectiveness of heating process, if the atmospheric temperature is 20\u00b0C.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "T0 = 293.; \t\t\t#K\n", + "T1 = 293.; \t\t\t#K\n", + "T2 = 373.; \t\t\t#K\n", + "T3 = 323.; \t\t\t#K\n", + "cp = 1.005;\n", + "\n", + "# Calculations and Results\n", + "print (\"(i) The ratio of mass flow\")\n", + "x = (T3-T1)/(T2-T3);\n", + "print (\"x = \"), (x)\n", + "\n", + "ds_13 = cp*math.log(T3/T1);\n", + "ds_32 = cp*math.log(T2/T3);\n", + "A = cp*(T3-T1) - T1*ds_13; \t \t\t#Increase of availability of system\n", + "B = x*(cp*(T2-T3)-T0*(ds_32));\t\t\t# Loss of availability of surroundings\n", + "Effectiveness = A/B;\n", + "print (\"Effectiveness of heating process = %.3f\")%(Effectiveness)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The ratio of mass flow\n", + "x = 0.6\n", + "Effectiveness of heating process = 0.306\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.17 Page no : 329" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "determine :\n", + "(i) The maximum work ;\n", + "(ii) The change in availability ;\n", + "(iii) The irreversibility.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m = 2.5; \t\t\t#kg\n", + "p1 = 6.*10**5; \t\t\t#Pa\n", + "r = 2.; \t\t\t#r = V2/V1\n", + "cv = 0.718; \t\t\t#kJ/kg K\n", + "R = 0.287; \t\t\t#kJ/kg K\n", + "T1 = 363.; \t\t\t#K\n", + "p2 = 1.*10**5; \t\t\t#Pa\n", + "T2 = 278.; \t\t\t#K\n", + "V1 = m*R*T1/p1;\n", + "V2 = 2*V1;\n", + "T0 = 278.; \t\t\t#K\n", + "p0 = 1.*10**5; \t\t\t#Pa\n", + "Q = 0.; \t\t\t#adiabatic process\n", + "\n", + "# Calculations and Results\n", + "dS = m*cv*math.log(T2/T1) + m*R*math.log(V2/V1);\n", + "Wmax = m*(cv*(T1-T2)) + T0*(cv*math.log(T2/T1) + R*math.log(V2/V1));\n", + "print (\"(i)The maximum work\"),(\"Wmax = %.3f\")% (Wmax), (\"kJ\")\n", + "\n", + "dA = Wmax+p0*(V1-V2);\n", + "print (\"(ii)Change in availability = %.3f\")%(dA), (\"kJ\")\n", + "\n", + "I = T0*m*(cv*math.log(T2/T1)+R*math.log(V2/V1));\n", + "print (\"(iii)Irreversibility = %.3f\")% (I),(\"kJ\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The maximum work Wmax = 154.628 kJ\n", + "(ii)Change in availability = 111.219 kJ\n", + "(iii)Irreversibility = 5.132 kJ\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.18 Page no : 330" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "determine per kg of air :\n", + "(i) The decrease in availability ;\n", + "(ii) The maximum work ;\n", + "(iii) The irreversibility.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "m = 1.; \t\t\t#kg\n", + "p1 = 7.*10**5; \t\t\t#Pa\n", + "T1 = 873.; \t\t\t#K\n", + "p2 = 1.*10**5; \t\t\t#Pa\n", + "T2 = 523.; \t\t\t#K\n", + "T0 = 288.; \t\t\t#K\n", + "Q = -9.; \t\t\t#kJ/kg\n", + "cp = 1.005; \t\t\t#kJ/kg K\n", + "R = 0.287; \t\t\t#kJ/kg K\n", + "\n", + "# Calculations and Results\n", + "print (\"(i) The decrease in availability \")\n", + "dA = cp*(T1-T2) - T0*(R*math.log(p2/p1) - cp*math.log(T2/T1));\n", + "print (\"dA = %.3f\")%(dA), (\"kJ/kg\")\n", + "\n", + "print (\"(ii) The maximum work\")\n", + "Wmax = dA; \t\t\t#change in availability\n", + "print (\"Wmax %.3f\")% (Wmax), (\"kJ/kg\")\n", + "\n", + "\n", + "W = cp*(T1-T2) + Q;\n", + "I = Wmax - W;\n", + "print (\"(iii)Irreversibility = %.3f\")%(I), (\"kJ/kg\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The decrease in availability \n", + "dA = 364.295 kJ/kg\n", + "(ii) The maximum work\n", + "Wmax 364.295 kJ/kg\n", + "(iii)Irreversibility = 21.545 kJ/kg\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.19 Page no : 331" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "determine\n", + "(i) The irreversibility ;\n", + "(ii) The effectiveness.\n", + "'''\n", + "\n", + "import math \n", + "\n", + "# Variables\n", + "cp = 1.005; \t\t\t#kJ/kg K\n", + "cv = 0.718; \t\t\t#kJ/kg K\n", + "R = 0.287; \t\t\t#kJ/kg K\n", + "m = 1.; \t\t\t#kg\n", + "T1 = 290.; \t\t\t#K\n", + "T0 = 290.; \t\t\t#K\n", + "T2 = 400.; \t\t\t#K\n", + "p1 = 1.; \t\t\t#bar\n", + "p0 = 1.; \t\t\t#bar\n", + "p2 = 6.; \t\t\t#bar\n", + "\n", + "# Calculations and Results\n", + "#Wrev = change in internal energy - T0*change in entropy\n", + "Wrev = -(cv*(T2-T1) - T0*(cp*math.log(T2/T1) - R*math.log(p2/p1)));\n", + "n = (1./(1-math.log(T2/T1)/math.log(p2/p1)));\n", + "Wact = m*R*(T1-T2)/(n-1);\n", + "\n", + "I = Wrev-Wact;\n", + "print (\"(i)Irreversibility = %.3f\")% (I), (\"kJ\")\n", + "\n", + "effectiveness = Wrev/Wact*100;\n", + "print (\"(ii)The effectiveness = %.3f\")%(effectiveness), (\"%\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Irreversibility = 9.945 kJ\n", + "(ii)The effectiveness = 93.109 %\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.20 Page no : 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "(i) find the rise in the temperature of the bearings when the flywheel has come to rest.\n", + "(ii) Calculate the greatest possible amount of the above heat which may be returned to the\n", + "flywheel? What would be the final r.p.m. of the flywheel, if it is set in motion with this available energy ?\n", + "'''\n", + "\n", + "import math \n", + "from scipy.integrate import quad \n", + "\n", + "# Variables\n", + "I = 0.62; \t\t\t#kg/m**2\n", + "N1 = 2500.; \t\t\t#rpm\n", + "w1 = 2*math.pi*N1/60; \t\t\t#rad/s\n", + "m = 1.9; \t\t\t#kg; Water equivalent of shaft bearings\n", + "cp = 4.18;\n", + "T0 = 293.; \t\t\t#K\n", + "t0 = 20.; \t\t\t#0C\n", + "\n", + "# Calculations and Results\n", + "print (\"(i)Rise in temperature of bearings\")\n", + "KE = 1./2*I*w1**2/1000; \t\t\t#kJ\n", + "dT = KE/(m*cp); \t\t\t#rise in temperature of bearings\n", + "print (\"dT = %.3f\")% (dT), (\"0C\")\n", + "\n", + "t2 = t0+dT;\n", + "print (\"Final temperature of the bearings = %.3f\")% (t2), (\"0C\")\n", + "\n", + "T2 = t2+273;\n", + "print (\"(ii)Final r.p.m. of the flywheel\")\n", + "def f15( T): \n", + "\t return m*cp*(1-T0/T)\n", + "\n", + "AE = quad(f15, T0, T2)[0]\n", + "UE = KE - AE;\n", + "\n", + "print (\"Available energy = %.3f\")% (AE), (\"kJ\")\n", + "\n", + "UAE = KE-AE;\n", + "print (\"Unavailable energy = %.3f\")%(UAE), (\"kJ\")\n", + "\n", + "w2 = math.sqrt(AE*10**3*2/I);\n", + "N2 = w2*60/2/math.pi;\n", + "print (\"Final rpm of the flywheel = %.3f\")%(N2), (\"rpm\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Rise in temperature of bearings\n", + "dT = 2.675 0C\n", + "Final temperature of the bearings = 22.675 0C\n", + "(ii)Final r.p.m. of the flywheel\n", + "Available energy = 0.096 kJ\n", + "Unavailable energy = 21.151 kJ\n", + "Final rpm of the flywheel = 168.407 rpm\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + "Example 6.21 Page no : 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Determine :\n", + "(i) Reversible work and actual work assuming the process to be adiabatic ;\n", + "(ii) Irreversibility and effectiveness of the system on the basis of 1 kg of air flow.\n", + "'''\n", + "\n", + "import math\n", + "\n", + "# Variables\n", + "p1 = 8.; \t\t\t#bar\n", + "T1 = 453.; \t\t\t#K\n", + "p2 = 1.4; \t\t\t#bar\n", + "T2 = 293.; \t\t\t#K\n", + "T0 = T2;\n", + "p0 = 1.; \t\t\t#bar\n", + "m = 1.; \t\t\t#kg\n", + "C1 = 80.; \t\t\t#m/s\n", + "C2 = 40.; \t\t\t#m/s\n", + "cp = 1.005; \t\t\t#kJ/kg K\n", + "R = 0.287; \t\t\t#kJ/kg K \n", + "\n", + "# Calculations and Results\n", + "print (\"(i) Reversible work and actual work \")\n", + "A1 = cp*(T1-T0)-T0*(cp*math.log(T1/T0)-R*math.log(p1/p0))+C1**2/2/10**3; \t\t\t#Availability at the inlet\n", + "A2 = cp*(T2-T0)-T0*(cp*math.log(T2/T0)-R*math.log(p2/p0))+C2**2/2/10**3; \t\t\t#Availability at the exit\n", + "\n", + "W_rev = A1-A2;\n", + "print (\"W_rev = %.3f\")%(W_rev), (\"kJ/kg\")\n", + "\n", + "W_act = cp*(T1-T2) + (C1**2-C2**2)/2/10**3;\n", + "print (\"W_act = %.3f\")%(W_act),(\"kJ/kg\")\n", + "\n", + "print (\"(ii) Irreversibilty and effectiveness = \")\n", + "\n", + "I = W_rev-W_act;\n", + "print (\"Irreversibilty = %.3f\")% (I), (\"kJ/kg\")\n", + "\n", + "Effectiveness = W_act/W_rev*100;\n", + "print (\"Effectiveness = %.3f\")%(Effectiveness),(\"%\")\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Reversible work and actual work \n", + "W_rev = 181.464 kJ/kg\n", + "W_act = 163.200 kJ/kg\n", + "(ii) Irreversibilty and effectiveness = \n", + "Irreversibilty = 18.264 kJ/kg\n", + "Effectiveness = 89.935 %\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 3, + "metadata": {}, + "source": [ + " Example 6.22 Page no : 335 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "'''\n", + "Calculate :\n", + "(i) The isentropic efficiency of the process ;\n", + "(ii) The loss of availability of the system assuming an atmospheric temperature of 20\u00b0C ;\n", + "(iii) The effectiveness of the process ;\n", + "'''\n", + "\n", + "# Variables\n", + "p1 = 20.; \t\t\t#bar\n", + "t1 = 400.; \t\t\t#0C\n", + "p2 = 4.; \t\t\t#bar\n", + "t2 = 250.; \t\t\t#0C\n", + "t0 = 20.; \t\t\t#0C\n", + "T0 = t0+273;\n", + "h1 = 3247.6; \t\t\t#kJ/kg\n", + "s1 = 7.127; \t\t\t#kJ/kg K\n", + "#let h2' = h2a and s2' = s2a\n", + "h2a = 2964.3; \t\t\t#kJ/kg\n", + "s2a = 7.379; \t\t\t#kJ/kg K\n", + "s2 = s1;\n", + "s1a = s1;\n", + "#By interpolation, we get\n", + "h2 = 2840.8; \t\t\t#kJ/kg\n", + "\n", + "# Calculations and Results\n", + "n_isen = (h1-h2a)/(h1-h2);\n", + "print (\"(i)Isentropic efficiency = %.3f\")%(n_isen)\n", + "\n", + "A = h1-h2a + T0*(s2a-s1a);\n", + "print (\"(ii)Loss of availability = %.3f\")%(A), (\"kJ/kg\")\n", + "\n", + "Effectiveness = (h1-h2a)/A;\n", + "print (\"(iii)Effectiveness = %.3f\")% (Effectiveness)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)Isentropic efficiency = 0.696\n", + "(ii)Loss of availability = 357.136 kJ/kg\n", + "(iii)Effectiveness = 0.793\n" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +}
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