{ "metadata": { "name": "", "signature": "sha256:7f13a6535713e6dd307350b25e55b55427f77dbef3f6f6422a5890eea0412625" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 5 : First Law of Thermodynamics for Control Volumes" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.1 Page No : 119" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Variables\n", "Q = -24.; \t\t\t#KJ/kg\n", "\n", "p1 = 5e5; \t\t\t#N/m**2\n", "t1 = 227.; \t\t\t#\u00b0C\n", "V1 = 50.; \t\t\t#m/s\n", "v1 = 0.78; \t\t\t#m**3/kg\n", "\n", "p2 = 1e5; \t\t\t#N/m**2\n", "t2 = 57.; \t\t\t#\u00b0C\n", "V2 = 100.; \t\t\t#m/s\n", "v2 = 0.97; \t\t\t#m**3/kg\n", "\n", "g = 9.81; \t\t\t#m/s**2 \t\t\t#acceleration due to gravity\n", "\n", "delta_z = -5; \t\t\t#m \n", "Cv = 0.7; \t\t\t#KJ/kg \n", "delta_u = Cv*(t2 - t1); \t\t\t#KJ/kg \t\t\t#change in internal energy \t\t\t#u2-u1\n", "\n", "\t\t\t\n", "# Calculations and Results\n", "delta_h = delta_u + (p2*v2 - p1*v1)*.001; \t\t\t#KJ/kg \t\t\t#change in enthalpy \t\t\t#h2-h1\n", "\n", "W_x = Q - (delta_h + (V2**2 - V1**2)/2*.001 + g*delta_z*.001); \t\t\t#kJ/kg \t\t\t#Work Output\n", "\n", "print \"Work output = %.1f KJ/kg\"%(W_x);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work output = 384.3 KJ/kg\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.2 Page No : 120" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\t\t\t\n", "# Variables\n", "m = 5000./3600 \t\t\t# kg/s \t\t\t# flow rate\n", "W_x = 550. \t\t\t# KJ/s \t\t\t#power developed by turbine\n", "Q = 0. \t \t\t#Heat loss is negligible\n", "\n", "\t\t\t\n", "# Calculations and Results\n", "#Part (a)\n", "print \"Parta\"\n", "V1 = 0. \t\t\t# m/s \t\t\t#inlet velocity\n", "V2 = 360. \t\t\t# m/s \t\t\t#exit velocity\n", "g = 9.81 \t\t\t# m/s**2\n", "delta_z = 0. \t\t\t#m \t\t\t#z2-z1\n", "\n", "delta_h = ((Q-W_x)/m)-g*delta_z*.001-((V2**2-V1**2)/2000) \t\t\t#KJ/Kg \t\t\t#change in enthalpy\n", "print \"Change in enthalpy = %.2f KJ/kg\"%(delta_h)\n", "\n", "#Part (a)\n", "print \"Partb\"\n", "V1 = 60. \t\t\t# m/s \t\t\t#inlet velocity\n", "V2 = 360. \t\t\t# m/s \t\t\t#exit velocity\n", "g = 9.81 \t\t\t# m/s**2\n", "delta_z = 0. \t\t\t#m \t\t\t#z2-z1\n", "\n", "delta_h = ((Q-W_x)/m)-g*delta_z*.001-((V2**2-V1**2)/2000) \t\t\t#KJ/Kg \t\t\t#change in enthalpy\n", "print \"Change in enthalpy = %.2f KJ/kg\"%(delta_h)\n", "\n", "# note : rounding off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Parta\n", "Change in enthalpy = -460.80 KJ/kg\n", "Partb\n", "Change in enthalpy = -459.00 KJ/kg\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.3 Page No : 122" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\t\t\n", "# Variables\n", "mA = 0.8 \t\t\t# kg/s \t\t\t#flow rate of stream A\n", "pA = 15e5 \t\t\t# N/m**2 \t\t\t#Pressure of stream A\n", "tA = 250. \t\t\t#\u00b0C \t\t\t#temperature of stream A\n", "\n", "mB = 0.5 \t\t\t# kg/s \t\t\t#flow rate of stream B\n", "pB = 15e5 \t\t\t# N/m**2 \t\t\t#Pressure of stream B\n", "tB = 200. \t\t\t#\u00b0C \t\t\t#temperature of stream B\n", "\n", "Q = 0. \t\t\t#No heat loss\n", "\n", "p1 = 10e5 \t\t\t# N/m**2 \t\t\t#pressure supply of chamber\n", "t2 = 30. \t\t\t#\u00b0C \t\t\t#exhaust air temperature from turbine\n", "\n", "Cv = 0.718 \t\t\t# KJ/kgK \t\t\t#heat capacity at constant volume\n", "Cp = 1. \t\t\t# KJ/kgK \t\t\t#heat capacity at constant pressure\n", "\n", "\t\t\t\n", "# Calculations and Results\n", "#Part(a)\n", "print \"Part a\"\n", "t1 = ((mA*Cp*tA)+(mB*Cp*tB))/((mA*Cp)+(mB*Cp)) \t\t\t# \u00b0C \t\t\t#the temperature of air at inlet to the turbine\n", "print \"The temperature of air at inlet to the turbine = %.2f \u00b0C\"%(t1);\n", "\n", "#Part(b)\n", "print \"Part b\"\n", "WT = -1*(mA+mB)*Cp*(t2-t1) \t\t\t# \u00b0kW \t\t\t#power developed by turbine\n", "print \"Power developed by turbine = %.0f kW\"%(WT);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Part a\n", "The temperature of air at inlet to the turbine = 230.77 \u00b0C\n", "Part b\n", "Power developed by turbine = 261 kW\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.4 Page No : 123" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "# Variables\n", "d1 = 0.15 \t\t\t#m \t\t\t#inlet diameter\n", "m = 4000./3600 \t\t\t# kg/s \t\t\t#flow rate\n", "v1 = 0.285 \t\t\t#m**3/kg \t\t\t#specific volume at entry\n", "d2 = 0.25 \t\t\t#m \t\t\t#exit diameter\n", "v2 = 15. \t\t\t# m**3/kg \t\t\t#specific volume at exit\n", "\n", "\t\t\t\n", "# Calculations and Results\n", "\n", "A1 = math.pi*d1**2/4 \t\t\t#m**2 \t\t\t#inlet cross sectional area\n", "A2 = math.pi*d2**2/4 \t\t\t# m**2 \t\t\t# exit cross sectional area\n", "print \"Inlet cross sectional area A1)= %.5f m**2\"%(A1);\n", "print \"Exit cross sectional area A2)= %.4f m**2\"%(A2);\n", "\n", "V1 = m*v1/A1 \t\t\t#m/s \t\t\t#inlet velocity\n", "V2 = m*v2/A2 \t\t\t#m/s \t\t\t#exit velocity\n", "\n", "print \"Inlet velocity = %.1f m/s\"%(V1);\n", "print \"Exit velocity = %.1f m/s\"%(int(V2));\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inlet cross sectional area A1)= 0.01767 m**2\n", "Exit cross sectional area A2)= 0.0491 m**2\n", "Inlet velocity = 17.9 m/s\n", "Exit velocity = 339.0 m/s\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.5 Page No : 125" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\t\t\n", "# Variables\n", "p1 = 10.\t\t\t#bar \t\t\t#inlet pressure\n", "t1 = 300. \t\t\t#\u00b0C \t\t\t#inlet temperature\n", "\n", "p2 = 0.1 \t\t\t#bar \t\t\t#exit pressure\n", "Cp = 1. \t\t\t#kJ/kgK \t\t\t# heat capacity at constant pressure\n", "\t\t\t\n", "# Calculations and Results\n", "#Adiabatic process\n", "delta_h = 0 \t\t\t#change in enthalpy\n", "t2 = delta_h/Cp + t1\n", "print \"Temperature of air after throttling = %.0f \u00b0C\"%(t1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Temperature of air after throttling = 300 \u00b0C\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.6 Page No : 126" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\t\t\n", "# Variables\n", "p1 = 1e5 \t\t\t# N/m**2 \t\t\t#inlet pressure\n", "v1 = 0.08 \t\t\t#m**3/kg \t\t\t# specific volume at inlet\n", "p2 = 7e5 \t\t\t# N/m**2 \t\t\t#exit pressure\n", "v2 = 0.016 \t\t\t# m**3/kg \t\t\t#specific volume at exit\n", "u1 = 48. \t\t\t# kJ/kg \t\t\t# internal energy at inlet\n", "u2 = 200. \t\t\t# kJ/kg \t\t\t# internal energy at exit\n", "Q = -120. \t\t\t# kJ/kg \t\t\t# heat loss\n", "\t\t\t\n", "# Calculations and Results\n", "Wc = ((u2 - u1) + (p2*v2 - p1*v1)*.001 - Q)*-1 \t\t\t# kJ/kg \t\t\t# work input to compressor\n", "print \"Work input to compressor Wc) = %.1f kJ/kg\"%(Wc)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Work input to compressor Wc) = -275.2 kJ/kg\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.7 Page No : 128" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# variables\n", "mh = 9.45 \t\t\t# kg/s \t\t\t# flow rate of steam\n", "h_h2 = 140. \t\t\t# kJ/kg \t\t\t# enthalpy of condensate\n", "h_h1 = 2570. \t\t\t# kJ/kg \t\t\t# inlet enthalpy of steam\n", "t1 = 25. \t\t\t# \u00b0C \t\t\t#inlet temperature of cooling water\n", "t2 = 36. \t\t\t# \u00b0C \t\t\t#exit temperature of cooling water\n", "c = 4.189 \t\t\t# kJ/kg deg \t\t\t# specific heat of water\n", "\t\t\t\n", "# Calculations and Results\n", "mc = -1*(mh*(h_h2-h_h1))/(c*(t2-t1)) \t\t\t# kg/s \t\t\t#mass flow rate of cooling water\n", "print \"Mass flow rate of cooling water = %.2f kg/s\"%(mc)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass flow rate of cooling water = 498.35 kg/s\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.8 Page No : 129" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# variables\n", "mh = 9.45 \t\t\t# kg/s \t\t\t# flow rate of steam\n", "h_h2 = 140. \t\t\t# kJ/kg \t\t\t# enthalpy of condensate\n", "h_h1 = 2570. \t\t\t# kJ/kg \t\t\t# inlet enthalpy of steam\n", "t1 = 25. \t\t\t# \u00b0C \t\t\t#inlet temperature of cooling water\n", "t2 = 36. \t\t\t# \u00b0C \t\t\t#exit temperature of cooling water\n", "c = 4.189 \t\t\t# kJ/kg deg \t\t\t# specific heat of water\n", "fractionalheatloss = 0.1 \t\t\t# fractional heat loss\n", "\t\t\t\n", "# Calculations and Results\n", "mc = -1*((1-fractionalheatloss)*mh*(h_h2-h_h1))/(c*(t2-t1)) \t\t\t# kg/s \t\t\t#mass flow rate of cooling water\n", "print \"Mass flow rate of cooling water = %.1f kg/s\"%(mc)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass flow rate of cooling water = 448.5 kg/s\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.9 Page No : 130" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# variables\n", "V1 = 300 \t\t\t#m/s \t\t\t#inlet air velocity\n", "t2 = 100 \t\t\t#\u00b0C \t\t\t#exit air temperature\n", "V2 = 15 \t\t\t#m/s \t\t\t#exit air velocity\n", "\t\t\t\n", "# Calculations and Results\n", "t1 = t2 + .001*(V2**2 - V1**2)/2 \t\t\t# \u00b0C \t\t\t#inlet air temperature\n", "print \"Inlet air temperature = %.1f \u00b0C\"%(t1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Inlet air temperature = 55.1 \u00b0C\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 5.10 Page No : 131" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\t\t\t\n", "# Variables\n", "m1 = 0.8 \t\t\t#kg \t\t\t#initial mass of air\n", "p1 = 150. \t\t\t#kPa \t\t\t#initial pressure of air\n", "T1 = 300. \t\t\t#K \t\t\t#initial temperature of air\n", "p_p = 600. \t\t\t#kPa \t\t\t#pressure of air in pipe\n", "T_p = 330. \t\t\t#K \t\t\t# temperature of air in pipe\n", "\n", "\t\t\t\n", "# Calculations and Results\n", "m2T2 = (p_p/p1)*T1*m1\n", "m2 = ((0.718*(m2T2/m1-T1))/(331.65)*m1)+m1 \t\t\t#kg \t\t\t#final mass of air\n", "print \"Mass of air entering in vessel = %.4f kg\"%(m2-m1)\n", "T2 = m2T2/m2 \t\t\t#K \t\t\t#Temperature of air in vessel\n", "print \"Temperature of air in vessel = %.0f K\"%(T2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mass of air entering in vessel = 1.5588 kg\n", "Temperature of air in vessel = 407 K\n" ] } ], "prompt_number": 7 } ], "metadata": {} } ] }