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diff --git a/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter9.ipynb b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter9.ipynb new file mode 100644 index 00000000..91d540ce --- /dev/null +++ b/Basic_And_Applied_Thermodynamics_by_P._K._Nag/Chapter9.ipynb @@ -0,0 +1,929 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 09:Properties of pure substances" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.1:pg-302" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.1\n", + "\n", + " At 1 MPa, \n", + " saturation temperature is 179.91 degree celcius\n", + "\n", + " Changes in specific volume is 0.193313 m**3/kg\n", + "\n", + " Change in entropy during evaporation is 4.4478 kJ/kg K\n", + "\n", + " The latent heat of vaporization is 2015.3 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "# At 1 MPa\n", + "tsat = 179.91 # Saturation temperature in degree Celsius\n", + "vf = 0.001127 # Specific volume of fluid in m**3/kg\n", + "vg = 0.19444 # Specific volume of gas in m**3/kg \n", + "sf = 2.1387 # Specific entropy of fluid in kJ/kgK\n", + "sg = 6.5865# Specific entropy of gas in kJ/kgK\n", + "print \"\\n Example 9.1\"\n", + "vfg = vg-vf # Change in specific volume due to evaporation\n", + "sfg = sg-sf# Change in specific entropy due to evaporation\n", + "hfg = 2015.3\n", + "print \"\\n At 1 MPa, \\n saturation temperature is \",tsat ,\" degree celcius\"\n", + "print \"\\n Changes in specific volume is \",vfg ,\" m**3/kg\"\n", + "print \"\\n Change in entropy during evaporation is \",sfg ,\" kJ/kg K\"\n", + "print \"\\n The latent heat of vaporization is \",hfg ,\" kJ/kg\"\n", + "# Data is given in the table A.1(b) in Appendix in the book\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.2:pg-302" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.2\n", + "\n", + " pressure = 0.6 Mpa\n", + " Temperature = 158.85 degree centigrade\n", + " Specific volume = 0.3156 m**3/kg\n", + " enthalpy = 2756.8 kJ/kg\n" + ] + } + ], + "source": [ + "# Given that\n", + "s = 6.76 # Entropy of saturated steam in kJ/kgK\n", + "print \"\\n Example 9.2\"\n", + "# From the table A.1(b) given in the book at s= 6.76 kJ/kgK\n", + "p = 0.6\n", + "t=158.85\n", + "v_g=0.3156\n", + "h_g=2756.8\n", + "print \"\\n pressure = \",p ,\" Mpa\\n Temperature = \",t ,\" degree centigrade\\n Specific volume = \",v_g ,\" m**3/kg\\n enthalpy = \",h_g ,\" kJ/kg\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.3:pg-302" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.3\n", + "\n", + " The enthalpy and entropy of the system are\n", + " 2614.55463998 kW and 5.96006442363 kJ/kg and kJ/kg K respectively.\n" + ] + } + ], + "source": [ + "\n", + "v = 0.09 # Specific volume of substance at a point in m**3/kg\n", + "vf = 0.001177 # Specific volume of fluid in m**3/kg\n", + "vg = 0.09963 # Specific volume of gas in m**3/kg\n", + "hf = 908.79 # Specific enthalpy of fluid in kJ/kg\n", + "hfg = 1890.7 # Latent heat of substance in kJ/kg\n", + "sf = 2.4474 # Specific entropy of fluid in kJ/kgK\n", + "sfg = 3.8935 # Entropy change due to vaporization\n", + "\n", + "print \"\\n Example 9.3\"\n", + "x = (v-vf)/(vg-vf) # steam quality\n", + "h = hf+(x*hfg) # Specific enthalpy of substance at a point in kJ/kg\n", + "s = sf+(x*sfg) # Specific entropy of substance at a point in kJ/kgK\n", + "\n", + "print \"\\n The enthalpy and entropy of the system are\\n \",h ,\" kW and \",s ,\" kJ/kg and kJ/kg K respectively.\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.5:pg-303" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.5\n", + "\n", + " The pressure is 3.973 MPa\n", + "\n", + " The total mass of mixture is 9.57329343706 kg\n", + "\n", + " Specific volume is 0.00417829039327 m3/kg\n", + "\n", + " Enthalpy is is 1188.13405609 kJ/kg\n", + "\n", + " The entropy is 2.9891336667 kJ/kg K\n", + "\n", + " The internal energy is 1171.53370836 kJ/kg\n", + "\n", + " At 250 degree Celsius, internal energy is 1171.53445483 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "Psat = 3.973 # Saturation pressure in MPa\n", + "vf = 0.0012512 # specific volume of fluid in m**3/kg\n", + "vg = 0.05013 # Specific volume of gas in m**3/kg\n", + "hf = 1085.36 # Specific enthalpy of fluid in kJ/kg\n", + "hfg = 1716.2 # Latent heat of vaporization in kJ/kg\n", + "sf = 2.7927 # Specific entropy of fluid in kJ/kgK\n", + "sfg = 3.2802 # Entropy change due to vaporization in kJ/kgK\n", + "mf = 9.0 # Mass of liquid in kg\n", + "V = 0.04 # Volume of vessel in m**3\n", + "# at T = 250\n", + "uf = 1080.39 #Specific internal energy in kJ/kg \n", + "ufg = 1522.0# Change in internal energy due to vaporization in kJ/kg\n", + "\n", + "print \"\\n Example 9.5\"\n", + "Vf = mf*vf # volume of fluid\n", + "Vg = V-Vf # volume of gas\n", + "mg = Vg/vg # mass of gas\n", + "m = mf+mg # mass if mixture\n", + "x = mg/m # quality of steam\n", + "v = vf+x*(vg-vf) # specific volume of mixture\n", + "h = hf+x*hfg # enthalpy of mixture\n", + "s = sf+(x*sfg) # entropy of mixture\n", + "u = h-Psat*1e6*v*1e-03 # Internal energy of mixture\n", + "u_ = uf+x*ufg # Internal energy at 250 degree Celsius\n", + "print \"\\n The pressure is \",Psat ,\" MPa\"\n", + "print \"\\n The total mass of mixture is \",m ,\" kg\"\n", + "print \"\\n Specific volume is \",v ,\" m3/kg\"\n", + "print \"\\n Enthalpy is is \",h ,\" kJ/kg\"\n", + "print \"\\n The entropy is \",s ,\" kJ/kg K\"\n", + "print \"\\n The internal energy is \",u ,\" kJ/kg\"\n", + "print \"\\n At 250 degree Celsius, internal energy is \",u_ ,\"kJ/kg\" #The answer provided in the textbook is wrong\n", + "\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.7:pg-305" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.7\n", + "\n", + " The ideal work output of the turbine is 882.40804932 kJ/Kg\n" + ] + } + ], + "source": [ + "\n", + "# At T = 40 degree\n", + "Psat = 7.384 # Saturation pressure in kPa\n", + "sf = 0.5725 # Entropy of fluid in kJ/kgK\n", + "sfg = 7.6845 # Entropy change due to vaporization in kJ/kgK\n", + "hf = 167.57 # Enthalpy of fluid in kJ/kg\n", + "hfg = 2406.7 # Latent heat of vaporization in kJ/kg\n", + "s1 = 6.9189 # Entropy at turbine inlet in kJ/kgK\n", + "h1 = 3037.6 # Enthalpy at turbine inlet in kJ/kg\n", + "print \"\\n Example 9.7\"\n", + "x2 = (s1-sf)/sfg # Steam quality\n", + "h2 = hf+(x2*hfg) # Enthalpy at turbine exit\n", + "W = h1-h2 # Net work done\n", + "print \"\\n The ideal work output of the turbine is \",W ,\" kJ/Kg\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.9:pg-308" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.9\n", + "\n", + " The quality of steam in pipe line is 0.96097673702\n", + "\n", + " Maximum moisture content that can be determined is 5.47886817645 percent\n" + ] + } + ], + "source": [ + "\n", + "h2 = 2716.2 # Enthalpy at turbine inlet in kJ/kg\n", + "hf = 844.89 # Enthalpy of fluid in kJ/kg\n", + "hfg = 1947.3 # Latent heat of vaporization in kJ/kg\n", + "h3 = 2685.5 # Enthalpy at turbine exit in kJ/kg\n", + "print \"\\n Example 9.9\"\n", + "x1 = (h2-hf)/hfg\n", + "x4 = (h3-hf)/hfg\n", + "print \"\\n The quality of steam in pipe line is \",x1 #The answers vary due to round off error\n", + "print \"\\n Maximum moisture content that can be determined is \",100-(x4*100) ,\" percent\"#The answer provided in the textbook is wrong\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.10:pg-309" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.10\n", + "\n", + " The quality of the steam in the pipe line is 0.909544295341\n" + ] + } + ], + "source": [ + "\n", + "# At 0.1Mpa, 110 degree\n", + "h2 = 2696.2 # Enthalpy at turbine inlet in kJ/kg\n", + "hf = 844.89 # Enthalpy of fluid in kJ/kg\n", + "hfg = 1947.3 # Latent heat of vaporization in kJ/kg\n", + "vf = 0.001023 # at T = 70 degree\n", + "V = 0.000150 # In m3\n", + "m2 = 3.24 # mass of condensed steam in kg\n", + "\n", + "print \"\\n Example 9.10\"\n", + "x2 = (h2-hf)/hfg # Quality of steam at turbine inlet\n", + "m1 = V/vf # mass of moisture collected in separator\n", + "x1 = (x2*m2)/(m1+m2) # quality of the steam\n", + "print \"\\n The quality of the steam in the pipe line is \",x1 \n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.11:pg-310" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.11\n", + "\n", + " The heat transfer during the process is 1788.19203218 MJ\n" + ] + } + ], + "source": [ + "\n", + "# P = 1MPa\n", + "vf = 0.001127 # specific volume of fluid in m**3/kg\n", + "vg = 0.1944# specific volume of gas in m**3/kg\n", + "hg = 2778.1 # specific enthalpy of gas in kJ/kg\n", + "uf = 761.68 # Specific internal energy of fluid in kJ/kg\n", + "ug = 2583.6 # Specific internal energy of gas in kJ/kg\n", + "ufg = 1822 # Change in specific internal energy due to phase change in kJ/kg \n", + "# Initial anf final mass\n", + "Vif = 5 # Initial volume of water in m**3 \n", + "Viw = 5# Initial volume of gas in m**3 \n", + "Vff = 6 # Final volume of gas in m**3 \n", + "Vfw = 4 # Final volume of water in m**3 \n", + "\n", + "\n", + "print \"\\n Example 9.11\"\n", + "ms = ((Viw/vf)+(Vif/vg)) - ((Vfw/vf)+(Vff/vg)) \n", + "U1 = ((Viw*uf/vf)+(Vif*ug/vg))\n", + "Uf = ((Vfw*uf/vf)+(Vff*ug/vg))\n", + "Q = Uf-U1+(ms*hg)\n", + "print \"\\n The heat transfer during the process is \",Q/1e3 ,\" MJ\"\n", + "#The answer provided in the textbook is wrong\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.12:pg-311" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.12\n" + ] + }, + { + "ename": "NameError", + "evalue": "name 'math' is not defined", + "output_type": "error", + "traceback": [ + "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m", + "\u001b[1;31mNameError\u001b[0m Traceback (most recent call last)", + "\u001b[1;32m<ipython-input-5-4824fe7bf0c8>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[0;32m 14\u001b[0m \u001b[1;32mprint\u001b[0m \u001b[1;34m\"\\n Example 9.12\"\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 15\u001b[0m \u001b[0mV1\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mm\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0mv1\u001b[0m \u001b[1;31m# total volume at point 1\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[1;32m---> 16\u001b[1;33m \u001b[0mVd\u001b[0m \u001b[1;33m=\u001b[0m \u001b[1;33m(\u001b[0m\u001b[0mmath\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mpi\u001b[0m\u001b[1;33m/\u001b[0m\u001b[1;36m4\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0md\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m1e-3\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m**\u001b[0m\u001b[1;36m2\u001b[0m\u001b[1;33m*\u001b[0m\u001b[0ml\u001b[0m\u001b[1;33m*\u001b[0m\u001b[1;36m1e-3\u001b[0m \u001b[1;31m# displaced volume\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m\u001b[0;32m 17\u001b[0m \u001b[0mV2\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mV1\u001b[0m\u001b[1;33m+\u001b[0m\u001b[0mVd\u001b[0m \u001b[1;31m# Total volume at point 2\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0;32m 18\u001b[0m \u001b[0mn\u001b[0m \u001b[1;33m=\u001b[0m \u001b[0mlog\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mP1\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mP2\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mlog\u001b[0m\u001b[1;33m(\u001b[0m\u001b[0mV2\u001b[0m\u001b[1;33m/\u001b[0m\u001b[0mV1\u001b[0m\u001b[1;33m)\u001b[0m \u001b[1;31m# polytropic index\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n", + "\u001b[1;31mNameError\u001b[0m: name 'math' is not defined" + ] + } + ], + "source": [ + "\n", + "m = 0.02 # Mass of steam in Kg\n", + "d = 280 # diameter of piston in mm\n", + "l = 305 # Stroke length in mm\n", + "P1 = 0.6 # Initial pressure in MPa\n", + "P2 = 0.12 # Final pressure in MPa\n", + "# At 0.6MPa, t = 200 degree\n", + "v1 = 0.352 # Specific volume in m**3/kg\n", + "h1 = 2850.1 # Specific enthalpy in kJ/kg\n", + "vf = 0.0010476 # specific volume of fluid in m**3/kg\n", + "vfg = 1.4271 # Specific volume change due to vaporization in m**3/kg\n", + "uf = 439.3 # specific enthalpy of fluid\n", + "ug = 2512.0 # Specific enthalpy of gas\n", + "print \"\\n Example 9.12\"\n", + "V1 = m*v1 # total volume at point 1\n", + "Vd = (math.pi/4)*(d*1e-3)**2*l*1e-3 # displaced volume\n", + "V2 = V1+Vd # Total volume at point 2\n", + "n = log(P1/P2)/log(V2/V1) # polytropic index\n", + "W12 = ((P1*V1)-(P2*V2))*1e6/(n-1) # work done\n", + "print \"\\n The value of n is \",n\n", + "print \"\\n The work done by the steam is \",W12/1e3 ,\"kJ \"\n", + "#The answers vary due to round off error\n", + "v2 = V2/m # specific volume\n", + "x2 = (v2-vf)/vfg # Steam quality\n", + "# At 0.12MPa\n", + "u2 = uf + (x2*(ug-uf)) # Internal energy \n", + "u1 = h1-(P1*1e6*v1*1e-03) # Internal energy\n", + "Q12 = m*(u2-u1)+ (W12/1e3) # Heat transfer\n", + "print \"\\n The heat transfer is \",Q12 ,\"kJ \"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.13:pg-312" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.13\n", + "\n", + " Final pressure is 3.5 bar\n", + "\n", + " Steam quality is 0.87 \n", + " Entropy change during the process is 0.4227 kJ/K\n" + ] + } + ], + "source": [ + "\n", + "x1 = 1 # Steam quality in first vessel\n", + "x2 = 0.8 # Steam quality in second vessel\n", + "# at 0.2MPa\n", + "vg = 0.8857 # Specific volume of gas in m**3/kg\n", + "h1 = 2706.7 # Enthalpy in first vessel in kJ/kg\n", + "v1 = vg # Specific volume of gas in first vessel in m**3/kg\n", + "hg = h1 # Enthalpy in first vessel 1 in kJ/kg\n", + "m1 = 5 # mass in first vessel in kg\n", + "V1 = m1*v1 # Volume of first vessel in m**3\n", + "# at 0.5MPa\n", + "m2 = 10 # mass in second vessel in kg\n", + "hf = 640.23 # Enthalpy in second vessel in kJ/kg\n", + "hfg = 2108.5 # Latent heat of vaporization in kJ/kg\n", + "vf = 0.001093 # Specific volume of fluid in second vessel in m**3/kg\n", + "vfg = 0.3749 # Change in specific volume in second vessel due to evaporation of gas in m**3/kg\n", + "v2 = vf+(x2*vfg) # Specific volume of gas in second vessel\n", + "V2 = m2*v2 # Volume of second vessel in m**3\n", + "#\n", + "Vm = V1+V2 # Total volume \n", + "m = m1+m2 # Total mass\n", + "vm = Vm/m # net specific volume\n", + "u1 = h1 # Internal energy\n", + "h2 = hf+(x2*hfg) # Enthalpy calculation\n", + "u2 = h2 # Internal energy calculation\n", + "m3 = m # Net mass calculation\n", + "h3 = ((m1*u1)+(m2*u2))/m3 # Resultant enthalpy calculation\n", + "u3 = h3 # Resultant internal energy calculation\n", + "v3 = vm # resultant specific volume calculation\n", + "# From Mollier diagram\n", + "x3 = 0.870 # Steam quality \n", + "p3 = 3.5 # Pressure in MPa\n", + "s3 = 6.29 # Entropy at state 3 in kJ/kgK\n", + "s1 = 7.1271 # Entropy at state 1 in kJ/kgK\n", + "sf = 1.8607 # Entropy in liquid state in kJ/kgK\n", + "sfg = 4.9606 # Entropy change due to vaporization in kJ/kgK\n", + "s2 = sf+(x2*sfg) # Entropy calculation\n", + "E = m3*s3-((m1*s1)+(m2*s2)) # Entropy change during process\n", + "\n", + "print \"\\n Example 9.13\"\n", + "print \"\\n Final pressure is \",p3 ,\" bar\"\n", + "print \"\\n Steam quality is \",x3 ,\n", + "print \"\\n Entropy change during the process is \",E ,\" kJ/K\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.14:pg-314" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.14\n", + "\n", + " The availability of the steam before the throttle valve 1263.6894 kJ/kg\n", + "\n", + " The availability of the steam after the throttle valve 1237.5538 kJ/kg\n", + "\n", + " The availability of the steam at the turbine exhaust 601.851036792 kJ/kg\n", + "\n", + " The specific work output from the turbine is 546.253422512 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "# At 6 MPa, 400 degree\n", + "h1 = 3177.2 # Enthalpy in kJ/kg\n", + "s1 = 6.5408 #Entropy in kJ/kgK\n", + "# At 20 degree\n", + "h0= 83.96 # Enthalpy in kJ/kg \n", + "s0 = 0.2966#Entropy in kJ/kgK\n", + "T0 = 20 # Surrounding temperature in degree Celsius \n", + "f1 = (h1-h0)-(T0+273)*(s1-s0) # Availability before throttling\n", + "# By interpolation at P= 5MPa, h= 3177.2\n", + "s2 = 6.63 #Entropy in kJ/kgK\n", + "h2 = h1 # Throttling\n", + "f2 = (h2-h0)-(T0+273)*(s2-s0) # Availability after throttling\n", + "df = f1-f2 # Change in availability\n", + "x3s = (s2-1.5301)/(7.1271-1.5301) #Entropy at state 3 in kJ/kgK\n", + "h3s = 504.7+(x3s*2201.9) #Enthalpy at state 3 in kJ/kg\n", + "eis = 0.82 # isentropic efficiency\n", + "h3 = h2-eis*(h1-h3s) # Enthalpy at state 3 in kJ/kgK\n", + "x3 = (h3-504.7)/2201.7 # Steam quality at state 3\n", + "s3 = 1.5301+(x3*5.597) # Entropy at state 3\n", + "f3 = (h3-h0)-(T0+273)*(s3-s0) # Availability at state 3\n", + "\n", + "print \"\\n Example 9.14\"\n", + "print \"\\n The availability of the steam before the throttle valve \",f1 ,\" kJ/kg\"\n", + "print \"\\n The availability of the steam after the throttle valve \",f2 ,\" kJ/kg\"\n", + "print \"\\n The availability of the steam at the turbine exhaust \",f3 ,\" kJ/kg\"\n", + "print \"\\n The specific work output from the turbine is \",h2-h3 ,\" kJ/kg\"\n", + "#The answers vary due to round off error\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.15:pg-316" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.15\n", + "\n", + " Availability of steam entering is 1057.4864 kJ/kg\n", + "\n", + " Availability of steam leaving the turbine is 656.7062 kJ/kg\n", + "\n", + " Maximum work is 741.14568 kJ/kg\n", + "\n", + " Irreversibility is 21.36505104 kJ/kg\n" + ] + } + ], + "source": [ + "\n", + "# At 25 bar, 350 degree\n", + "h1 = 3125.87 # Enthalpy in kJ/kg\n", + "s1 = 6.8481# Entropy in kJ/kgK\n", + "# 30 degree\n", + "h0 = 125.79 # Enthalpy in kJ/kg\n", + "s0 = 0.4369# Entropy in kJ/kgK\n", + "# At 3 bar, 200 degree\n", + "h2 = 2865.5 # Enthalpy in kJ/kg\n", + "s2 = 7.3115 #Entropy in kJ/kgK\n", + "# At 0.2 bar 0.95 dry\n", + "hf = 251.4 # Enthalpy of liquid in kJ/kg\n", + "hfg = 2358.3 # Latent heat of vaporization in kJ/kg\n", + "sf = 0.8320 # Entropy of liquid in kJ/kgK\n", + "sg = 7.0765# Entropy of liquid in kJ/kgK\n", + "h3 = hf+0.92*hfg # Enthalpy at state 3 in kJ/kg\n", + "s3 = sf+(0.92*sg) # Entropy at state 3 in kJ/kgK\n", + "# Part (a)\n", + "T0 = 30 # Atmospheric temperature in degree Celsius\n", + "f1 = (h1-h0)-((T0+273)*(s1-s0)) # Availability at steam entering turbine\n", + "f2 = (h2-h0)-((T0+273)*(s2-s0)) # Availability at state 2\n", + "f3 = (h3-h0)-((T0+273)*(s3-s0))# Availability at state 3\n", + "\n", + "print \"\\n Example 9.15\"\n", + "print \"\\n Availability of steam entering is \",f1 ,\" kJ/kg\"\n", + "print \"\\n Availability of steam leaving the turbine is \",f2 ,\" kJ/kg\"\n", + "\n", + "# Part (b)\n", + "m2m1 = 0.25 # mass ratio\n", + "m3m1 = 0.75 # mass ratio\n", + "Wrev = f1-(m2m1*f2)-(m3m1*f3) # Maximum work\n", + "print \"\\n Maximum work is \",Wrev ,\" kJ/kg\"\n", + "\n", + "# Part (c)\n", + "w1 = 600 # mass flow at inlet of turbine in kg/h\n", + "w2 = 150 # mass flow at state 2 in turbine in kg/h\n", + "w3 = 450# mass flow at state 2 in turbine in kg/h\n", + "Q = -10 # Heat loss rate kJ/s\n", + "I = ((T0+273)*(w2*s2+w3*s3-w1*s1)-Q*3600)*103/600\n", + "print \"\\n Irreversibility is \",I/1e3 ,\" kJ/kg\"\n", + "#The answer provided in the textbook is wrong\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.16:pg-317" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.16\n", + "\n", + " Energy of system in Part (a) is 73.2 kJ\n", + "\n", + " Energy of system in Part (b) is 197.3474 kJ\n", + "\n", + " Energy of system in Part (c) is 498.2624 kJ\n", + "\n", + " Energy of system in Part (d) is 121.8 kJ\n" + ] + } + ], + "source": [ + "\n", + "# At dead state of 1 bar, 300K\n", + "u0 = 113.1 # Internal energy in kJ/kg\n", + "h0 = 113.2 # Enthalpy in kJ/kg\n", + "v0 = 0.001005 # Specific volume in m**3/kg\n", + "s0 = 0.395 # Entropy in kJ/kg\n", + "T0 = 300 # Atmospheric temperature in K\n", + "P0 = 1 # Atmospheric pressure in bar \n", + "K = h0-T0*s0\n", + "# Part (a)\n", + "# At 1bar and 90 degree Celsius \n", + "u = 376.9 # Internal energy in kJ/kg\n", + "h = 377 # Enthalpy in kJ/kg\n", + "v = 0.001035 # specific volume in m**3/kg\n", + "s = 1.193 # Entropy in kJ/kgK\n", + "m = 3 # Mass of water in kg\n", + "fi = m*(h-(T0*s)-K) #Energy of system\n", + "\n", + "print \"\\n Example 9.16\"\n", + "print \"\\n Energy of system in Part (a) is \",fi ,\" kJ\"\n", + "#The answers vary due to round off error\n", + "\n", + "# Part (b)\n", + "# At P = 4 Mpa, t = 500 degree\n", + "u = 3099.8# Internal energy in kJ/kg \n", + "h = 3446.3 # Enthalpy in kJ/kg \n", + "v = 0.08637 # specific volume in m**3/kg \n", + "s = 7.090 # Entropy in kJ/kgK\n", + "m = 0.2 # Mass of steam in kg \n", + "fib = m*(u+P0*100*v-T0*s-K) # Energy of system\n", + "print \"\\n Energy of system in Part (b) is \",fib ,\" kJ\"\n", + "\n", + "# Part (c) # P = 0.1 bar\n", + "m = 0.4 # Mass of wet steam in kg \n", + "x = 0.85 # Quality\n", + "u = 192+x*2245 # Internal energy \n", + "h = 192+x*2392# Enthalpy\n", + "s = 0.649+x*7.499 # Entropy\n", + "v = 0.001010+x*14.67 # specific volume\n", + "fic = m*(u+P0*100*v-T0*s-K) # Energy of system\n", + "print \"\\n Energy of system in Part (c) is \",fic ,\" kJ\"\n", + "\n", + "# Part (d) \n", + "# P = 1 Bar, t = -10 degree Celsius\n", + "m = 3 # Mass of ice in kg \n", + "h = -354.1 # Enthalpy in kJ/kg \n", + "s = -1.298 # at 1000kPa, -10 degree\n", + "fid = m*((h-h0)-T0*(s-s0)) # Energy of system\n", + "\n", + "print \"\\n Energy of system in Part (d) is \",fid ,\" kJ\" #The answer provided in the textbook is wrong\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.17:pg-318" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.17\n", + "\n", + " In parallel flow\n", + "\n", + " The rate of irreversibility is 10.98 kW\n", + "\n", + " The Second law efficiency is 24.275862069 percent\n", + "\n", + "\n", + " In counter flow\n", + "\n", + " The rate of irreversibility is 10.9454545455 kW\n", + "\n", + " The Second law efficiency is 32.1594684385 percent\n" + ] + } + ], + "source": [ + "\n", + "# Given\n", + "th1 = 90.0 # Inlet temperature of hot water in degree Celsius\n", + "tc1 = 25.0# Inlet temperature of cold water in degree Celsius\n", + "tc2 = 50.0# Exit temperature of cold water in degree Celsius\n", + "mc = 1.0 # mass flow rate of cold water in kg/s\n", + "T0 = 300.0 # Atmospheric temperature in K\n", + "th2p = 60.0 # Temperature limit in degree Celsius for parallel flow\n", + "th2c = 35.0 # Temperature limit in degree Celsius for counter flow\n", + "mhp = (tc2-tc1)/(th1-th2p) # mass flow rate of hot water in kg/s for parallel flow\n", + "mhc = (tc2-tc1)/(th1-th2c) # mass flow rate of hot water in kg/s for counter flow\n", + "# At 300 K\n", + "h0 = 113.2 # ENthalpy in kJ/kg\n", + "s0 = 0.395 # ENtropy in kJ/kgK\n", + "T0 = 300.0 # temperature in K\n", + "# At 90 degree celsius\n", + "h1 = 376.92 # Enthalpy in kJ/kg \n", + "s1 = 1.1925 # Entropy in kJ/kgK\n", + "af1 = mhp*((h1-h0)-T0*(s1-s0)) # Availability\n", + "# Parallel Flow\n", + "# At 60 degree\n", + "h2 = 251.13 # Enthalpy in kJ/kg \n", + "s2 =0.8312 # Entropy in kJ/kgK\n", + " # At 25 degree\n", + "h3 = 104.89 # Enthalpy in kJ/kg \n", + "s3 = 0.3674 # Entropy in kJ/kgK\n", + "# At 50 degree\n", + "h4 = 209.33 # Enthalpy in kJ/kg \n", + "s4 = 0.7038 # Entropy in kJ/kgK\n", + "REG = mc*((h4-h3)-T0*(s4-s3)) # Rate of energy gain\n", + "REL = mhp*((h1-h2)-T0*(s1-s2)) # Rate of energy loss\n", + "Ia = REL-REG # Energy destruction\n", + "n2a = REG/REL # Second law efficiency\n", + "\n", + "print \"\\n Example 9.17\"\n", + "print \"\\n In parallel flow\"\n", + "print \"\\n The rate of irreversibility is \",Ia ,\" kW\"\n", + "print \"\\n The Second law efficiency is \",n2a*100 ,\" percent\"\n", + "#The answers vary due to round off error\n", + "\n", + "\n", + "# Counter flow\n", + "h2_ = 146.68 \n", + "sp = 0.5053 # At 35 degree\n", + "REG_b = REG # Rate of energy gain by hot water is same in both flows\n", + "REL_b = mhc*((h1-h2_)-T0*(s1-sp))\n", + "Ib = mhc*((h1-h2_)-(T0*(s1-sp))) # Energy destruction\n", + "n2b = REG_b/Ib # Second law efficiency\n", + "print \"\\n\\n In counter flow\"\n", + "print \"\\n The rate of irreversibility is \",Ib ,\" kW\"\n", + "print \"\\n The Second law efficiency is \",n2b*100 ,\" percent\"\n", + "#The answers vary due to round off error\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex9.18:pg-320" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Example 9.18\n", + "\n", + " The maximum cooling rate is 106.207042424 kW\n" + ] + } + ], + "source": [ + "\n", + "m = 50.0# mass flow rate in kg/h\n", + "Th = 23.0 # Home temperature in degree Celsius\n", + "# State 1\n", + "T1 = 150.0 # Saturated vapor temperature in degree Celsius\n", + "h1 = 2746.4 # Saturated vapor enthalpy in kJ/kg\n", + "s1 = 6.8387 #Saturated vapor entropy in kJ/kgK\n", + "# State 2\n", + "h2 = 419.0 # Saturated liquid enthalpy in kJ/kg\n", + "s2 = 1.3071 #Saturated liquid entropy in kJ/kg \n", + "T0 = 45.0 # Atmospheric temperature in degree Celsius\n", + "#\n", + "b1 = h1-((T0+273)*s1) # Availability at point 1\n", + "b2 = h2-((T0+273)*s2) # Availability at point 2\n", + "Q_max = m*(b1-b2)/((T0+273)/(Th+273)-1) # maximum cooling rate\n", + "\n", + "print \"\\n Example 9.18\"\n", + "print \"\\n The maximum cooling rate is \",Q_max/3600 ,\" kW\"\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |