{ "cells": [ { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "# Chapter 10:Principles of heat exchangers" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.1:pg-415" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\n", "The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4) in W\n", "Q= 11286.0\n", "(a) For a parallel flow arrangement\n", "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n", "LMTD= 35.4699026617\n", "Area(A)=Q/(U*LMTD) in m**2\n", "A= 0.397731567932\n", "(b)For counterflow arrangement\n", "LMTD=(deltaT2-deltaT1)/(ln(deltaT2/deltaT1))in°C \n", "LMTD= 36.3143018164\n", "Area(A)=Q/(U*LMTD) in m**2\n", "A= 0.38848330532\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 1\"\n", "#A brine solution is heated from temprature ,T1=8°C to temprature,T2=14°C in a double pipe heat exchanger.\n", "T1=8.0;\n", "T2=14.0;\n", "#Water entering at temprature T3=55°C and leaving at temprature,T4=40°C at the mass flow rate of (mdot)=0.18kg/s\n", "mdot=0.18;\n", "T3=55.0;\n", "T4=40.0;\n", "#Specific heat (cp) of water =4.18kJ/(kg*K)\n", "cp=4.18*10**3;\n", "#overall heat transfer coefficient(U)=800 W/(m**2*K)\n", "U=800.0;\n", "#The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4)\n", "print\"The rate of heat transfer from water is given by Q=mdot*cp*(T3-T4) in W\"\n", "Q=mdot*cp*(T3-T4)\n", "print\"Q=\",Q\n", "print\"(a) For a parallel flow arrangement\"\n", "#For a parallel flow arrangement deltaT1=T3-T1 and deltaT2=T4-T2. \n", "deltaT1=T3-T1;#deltaT1 is temprature difference \n", "deltaT2=T4-T2;#deltaT2 is temprature difference \n", "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for both parallel and counter flow.\n", "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n", "#let X=math.log10((deltaT2/deltaT1)) and Y=math.log10(2.718281)\n", "X=math.log10((deltaT2/deltaT1));\n", "Y=math.log10(2.718281);\n", "#ln=(ln(deltaT2/deltaT1))\n", "ln=X/Y;\n", "LMTD=(deltaT2-deltaT1)/ln\n", "print\"LMTD=\",LMTD\n", "#Area(A)=Q/(U*LMTD) in m**2\n", "print\"Area(A)=Q/(U*LMTD) in m**2\"\n", "A=Q/(U*LMTD)\n", "print\"A=\",A\n", "print\"(b)For counterflow arrangement\"\n", "deltaT1=T3-T2;\n", "deltaT2=T4-T1;\n", "print\"LMTD=(deltaT2-deltaT1)/(ln(deltaT2/deltaT1))in°C \"\n", "X=math.log10((deltaT2/deltaT1));\n", "Y=math.log10(2.718281);\n", "ln=X/Y;\n", "LMTD=(deltaT2-deltaT1)/ln\n", "print\"LMTD=\",LMTD\n", "print\"Area(A)=Q/(U*LMTD) in m**2\"\n", "A=Q/(U*LMTD)\n", "print\"A=\",A" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.2:pg-416" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\n", "The inlet temprature(Tc1) of cold oil in °C \n", "Tc1= 55.0948103792\n", "The rate of heat transfer Q=mdoth*ch*(Th1-Th2) in W\n", "Q= 190190.0\n", "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n", "LMTD= 28.996452611\n", "Area(A)=Q/(U*LMTD) in m**2\n", "A= 10.0908895279\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 2\"\n", "#Hot oil(specific heat,ch=2.09kJ/(kg*K)) flows through counter flow heat excahnger at the mass flow rate of mdoth=(0.7kg/s)\n", "ch=2.09*10**3;\n", "mdoth=0.7;\n", "#overall heat transfer coefficient(U)=650 W/(m**2*K)\n", "U=650;\n", "#It enters at temprature,Th1=200°C and leaves at temprature,Th2=70°C \n", "Th1=200;\n", "Th2=70;\n", "#Cold oil(specific heat,cc=1.67kJ/(kg*K) exits at temprature,Tc2=150°C at the mass flow rate of mdotc=(1.2kg/s)\n", "mdotc=1.2;\n", "cc=1.67*10**3;\n", "Tc2=150;\n", "#The unknown inlet temprature(Tc1) of cold oil may be found from energy balance mdotc*(Tc2-Tc1)=mdoth*(Th2-Th1)\n", "print\"The inlet temprature(Tc1) of cold oil in °C \"\n", "Tc1=Tc2-((mdoth*ch)/(mdotc*cc))*(Th1-Th2)\n", "print\"Tc1=\",Tc1\n", "#The rate of heat transfer can be calculate as Q=mdoth*ch*(Th1-Th2)\n", "print\"The rate of heat transfer Q=mdoth*ch*(Th1-Th2) in W\"\n", "Q=mdoth*ch*(Th1-Th2)\n", "deltaT1=Th1-Tc2;\n", "print\"Q=\",Q\n", "#deltaT1 is temprature difference between hot oil inlet temprature and cold oil exit temprature\n", "deltaT2=Th2-Tc1;\n", "#deltaT2 is temprature difference between hot oil exit temprature and cold oil inlet temprature\n", "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n", "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n", "#let X=log10((deltaT2/deltaT1)) and Y=log10(2.718281)\n", "X=math.log10((deltaT2/deltaT1));\n", "Y=math.log10(2.718281);\n", "#ln=(ln(deltaT2/deltaT1))\n", "ln=X/Y;\n", "LMTD=(deltaT2-deltaT1)/ln\n", "print\"LMTD=\",LMTD\n", "#Area(A)=Q/(U*LMTD) in m**2\n", "print\"Area(A)=Q/(U*LMTD) in m**2\"\n", "A=Q/(U*LMTD)\n", "print\"A=\",A" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.3:pg-417" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\n", "The outlet temprature(Tho) of oil in °C \n", "Tho= 70.6666666667\n", "For a counterflow heat exchanger\n", "deltaT1= 10\n", "deltaT2= 20.6666666667\n", "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n", "LMTD= 14.6936488511\n", "dimensionless parameters P and R are\n", "P= 0\n", "R= 0.733333333333\n", "correction factor(F) for the cross flow arrangement as obtained from graph of F vs Single Pass flow with fluids unmixed\n", "Q= 167200.0\n", "overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\n", "U= 758.604399737\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 3\"\n", "#A cross flow heat exchanger with both fluids unmixed is used to heat water((specific heat,cc=4.18kJ/(kg*K)) from temprature Tci=50°C to Tco=90°C \n", "#flowing at the mass flow rate of (mdotc)=1kg/s\n", "Tci=50;\n", "Tco=90;\n", "cc=4.18*10**3;\n", "mdotc=1;\n", "#The hot engine oil has (specific heat,ch=1.9kJ/(kg*K)) flowing at the mass flow rate of mdoth=3kg/s enters at temprature Thi=100°C\n", "mdoth=3;\n", "Thi=100;\n", "ch=1.9*10**3;\n", "#The unknown outlet temprature(Tho) of oil may be found from energy balance mdotc*(Tco-Tci)=mdoth*(Tho-Thi)\n", "print\"The outlet temprature(Tho) of oil in °C \"\n", "Tho=Thi-((mdotc*cc)/(mdoth*ch))*(Tco-Tci)\n", "print\"Tho=\",Tho\n", "print\"For a counterflow heat exchanger\"\n", "deltaT1=Thi-Tco;#deltaT1 is temprature difference \n", "deltaT2=Tho-Tci;#deltaT2 is temprature difference \n", "print\"deltaT1=\",deltaT1\n", "print\"deltaT2=\",deltaT2\n", "#LMTD(math.log mean temprature difference) is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n", "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n", "#let X=log10((deltaT2/deltaT1)) and Y=log10(2.718281)\n", "X=math.log10((deltaT2/deltaT1));\n", "Y=math.log10(2.718281);\n", "ln=X/Y;\n", "LMTD=(deltaT2-deltaT1)/ln\n", "print\"LMTD=\",LMTD\n", "#Area(A)=20m**2\n", "A=20;\n", "#We have to employ correction factor(F) for the cross flow arrangement.\n", "#We evaluate dimensionless parameters P=(Tco-Tci)/(Thi-Tco) and R=(Thi-Tho)/(Tco-Tci).\n", "print\"dimensionless parameters P and R are\"\n", "P=(Tco-Tci)/(Thi-Tci)\n", "R=(Thi-Tho)/(Tco-Tci)\n", "print\"P=\",P\n", "print\"R=\",R\n", "print\"correction factor(F) for the cross flow arrangement as obtained from graph of F vs Single Pass flow with fluids unmixed\"\n", "F=0.75\n", "#The rate of heat transfer can be calculate as Q=mdoth*ch*(Th1-Th2)\n", "Q=mdotc*cc*(Tco-Tci);\n", "print\"Q=\",Q\n", "#overall heat transfer coefficient(U)=Q/(A*F*LMTD)\n", "print\"overall heat transfer coefficient(U)=Q/(A*F*LMTD)in W/(m**2*K)\"\n", "U=Q/(A*F*LMTD)\n", "print\"U=\",U" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.5:pg-419" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\n", "(a)Applying LMTD method\n", "The rate of heat transfer Q=mdotw*cpw*(Tout-Tin) in W\n", "Q= 300960.0\n", "The unknown outlet temprature(Thout) of geothermal fluid in °C \n", "Thout= 125.085846868\n", "LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \n", "LMTD= 81.903612671\n", "Area(A)=Q/(U*LMTD) in m**2\n", "A= 6.12427197827\n", "To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\n", "L= 129.961087757\n", "(b)Applying NTU method\n", "The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cpw in KW/°C\n", "Ch= 8.62\n", "Cc= 5.016\n", "C=Cmin/Cmax\n", "C= 0.581902552204\n", "Heat transfer effectiveness is defined as eff=Q/(Cmin*(Thin-Tin))\n", "eff= 0.461538461538\n", "NTU is determined by NTU=(1/(C-1))*ln((eff-1)/(eff*C-1))\n", "NTU= 0.732568418453\n", "Area(A)=(NTU*Cmin)/U in m**2\n", "A= 6.12427197827\n", "To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\n", "Hence same result is obtained for both methods\n", "L= 129.961087757\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 5\"\n", "#Water is heated from temprature ,Tin=30°C to Tout=90°C in a counter flow double pipe heat exchanger.\n", "Tin=30;\n", "Tout=90;\n", "#Water flows at a mass flow rate of mdotw=1.2kg/s\n", "mdotw=1.2;\n", "#The heating is accomplished by a geothermal fluid which enters the heat exchanger at temprature ,Thin=160°C at the mass flow rate of mdoth=2kg/s\n", "mdoth=2;\n", "Thin=160;\n", "#The inner tube is thin walled having diameter(D)=15mm or 0.015m\n", "D=0.015;\n", "#overall heat transfer coefficient(U)=600 W/(m**2*K)\n", "U=600;\n", "#The specific heat of water and geothermal fluid is (cpw=4.18kJ/(kg*K))and(cph=4.31kJ/(kg*K)) respectively\n", "cpw=4.18*10**3;\n", "cph=4.31*10**3;\n", "#The rate of heat transfer in heat exchanger can be calculate as Q=mdotw*cpw*(Tout-Tin)\n", "print\"(a)Applying LMTD method\"\n", "print\"The rate of heat transfer Q=mdotw*cpw*(Tout-Tin) in W\"\n", "Q=mdotw*cpw*(Tout-Tin)\n", "print\"Q=\",Q\n", "#The unknown outlet temprature(Thout) of geothermal fluid may be found from energy balance mdotw*cpw*(Tout-Tin)=mdoth*cph*(Thin-Thout)\n", "print\"The unknown outlet temprature(Thout) of geothermal fluid in °C \"\n", "Thout=Thin-Q/(mdoth*cph)\n", "print\"Thout=\",Thout\n", "deltaT1=Thin-Tout;#Temprature difference between inlet temprature of hot fluid and outlet temprature of cold fluid\n", "deltaT2=Thout-Tin;#Temprature difference between outlet temprature of hot fluid and inlet temprature of cold fluid\n", "#LMTD is defined as (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) for counter flow.\n", "print\"LMTD is given by (deltaT2-deltaT1)/(ln(deltaT2/deltaT1)) in °C \"\n", "#let X=math.log10((deltaT2/deltaT1))and Y=math.log10(2.718281)\n", "X=math.log10((deltaT2/deltaT1));\n", "Y=math.log10(2.718281);\n", "ln=X/Y;\n", "LMTD=(deltaT2-deltaT1)/ln\n", "print\"LMTD=\",LMTD\n", "#Area(A)=Q/(U*LMTD) in m**2\n", "print\"Area(A)=Q/(U*LMTD) in m**2\"\n", "A=Q/(U*LMTD)\n", "print\"A=\",A\n", "print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n", "L=A/(math.pi*D)\n", "print\"L=\",L\n", "print\"(b)Applying NTU method\"\n", "#The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cw in KW/°C\n", "print\"The heat capacity rates are defined as Ch=mdoth*cph and Cc=mdotw*cpw in KW/°C\"\n", "Ch=(mdoth*cph)/1000\n", "Cc=(mdotw*cpw)/1000\n", "print\"Ch=\",Ch\n", "print\"Cc=\",Cc\n", "#So Cmin=Cc and Cmax=Ch\n", "Cmin=Cc;\n", "Cmax=Ch;\n", "#C is defined as Cmin/Cmax\n", "print\"C=Cmin/Cmax\"\n", "C=Cmin/Cmax\n", "print\"C=\",C\n", "#Heat transfer effectiveness is (eff)\n", "print\"Heat transfer effectiveness is defined as eff=Q/(Cmin*(Thin-Tin))\"\n", "eff=(Q/1000)/(Cmin*(Thin-Tin))\n", "print\"eff=\",eff\n", "print\"NTU is determined by NTU=(1/(C-1))*ln((eff-1)/(eff*C-1))\"\n", "#let X=math.log10((eff-1)/(eff*C-1)) and Y=math.log10(2.718281)\n", "X=math.log10((eff-1)/(eff*C-1));\n", "Y=math.log10(2.718281);\n", "#ln=ln((eff-1)/(eff*C-1))\n", "ln=X/Y;\n", "#NTU is Number of transfer units\n", "NTU=(1/(C-1))*ln\n", "print\"NTU=\",NTU\n", "#NTU =U*A/Cmin\n", "print\"Area(A)=(NTU*Cmin)/U in m**2\"\n", "A=(NTU*Cmin*1000)/U\n", "print\"A=\",A\n", "print\"To provide this surface area ,The length(L) of the tube required is given by L=A/(pi*D) in m\"\n", "L=A/(math.pi*D)\n", "print\"Hence same result is obtained for both methods\"\n", "print\"L=\",L" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.6:pg-422" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\n", "NTU is defined as (U*A)/Cmin \n", "NTU= 1.86170212766\n", "Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\n", "eff= 0.683715054322\n", "The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\n", "Q= 163.886498521\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 6\"\n", "#Water having specific heat,cw=4.18kJ/(kg*K) enters a counterflow double pipe heat exchanger at temprature,Tci=35°C flowing at the mass flow rate of mdotw=0.8 kg/s.\n", "cw=4.18;\n", "mdotw=0.8;\n", "Tci=35;\n", "#It is heated by oil having specific heat,co=1.88kJ/(kg*K) flowing at the mass flow rate of mdoto=1.5 kg/s from an inlet temprature(Thi) of 120°C.\n", "co=1.88;\n", "mdoto=1.5;\n", "Thi=120;\n", "#For an area(A) of 15m**2 and an overall heat transfer coefficient(U) of 350W/(m**2*K).\n", "A=15;\n", "U=350;\n", "#Cwater and Co are heat capacities for water and oil respectively\n", "#Cwater=mdotw*cw and Co=mdoto*co\n", "Cwater=mdotw*cw;\n", "Co=mdoto*co;\n", "#C=Cmin/Cmax\n", "Cmin=min(Cwater,Co);\n", "Cmax=max(Cwater,Co);\n", "C=Cmin/Cmax;\n", "#NTU is number of transfer units\n", "#NTU=(U*A)/Cmin\n", "print\"NTU is defined as (U*A)/Cmin \"\n", "NTU=(U*A)/(Cmin*1000)\n", "print\"NTU=\",NTU\n", "#Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\n", "print\"Heat transfer effectiveness(eff) is defined as (1-e**(-NTU*(1-C))/(1-C*e**(-NTU*(1-C))\"\n", "eff=(1-math.e**(-NTU*(1-C)))/(1-C*math.e**(-NTU*(1-C)))\n", "print\"eff=\",eff\n", "#Hence The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci)in kW.\n", "print\"The total heat transfer rate (Q)=eff*Cmin*(Thi-Tci) in kW\" \n", "Q=eff*Cmin*(Thi-Tci)\n", "print\"Q=\",Q" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.7:pg-424 " ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\n", "NTU is defined as (U*A)/Cmin \n", "NTU= 5.44554455446\n", "The effectiveness of heat exchanger is\n", "eff= 9.98333889769\n", "The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci) in W\n", "Q= 10587330.901\n", "The exit temprature of air in °C \n", "Tho= -923.250584257\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 7\"\n", "#Water enters a cross flow heat exchanger (both fluids unmixed) at temprature(Tci)=20°C amd flows at a mass flow rate of mdotw=7kg/s\n", "Tci=20;\n", "mdotw=7;\n", "#The air flows at a mass flow rate of mdota=10kg/s from Temprature(Thi)=125°C \n", "mdota=10;\n", "Thi=125;\n", "#The overall heat transfer coefficient(U)=220W/(m**2*K)and Area(A)=250m**2.\n", "U=220;\n", "A=250;\n", "#The specific heat of air (cpa=1.01kJ/(kg*K)) and water is (cpw=4.18kJ/(kg*K))\n", "cpa=1.01;\n", "cpw=4.18;\n", "#Cair and Cwater are heat capacities of air and water respectively\n", "Cair=mdota*cpa;\n", "Cwater=mdotw*cpw;\n", "#C=Cmin/Cmax\n", "Cmin=min(Cwater,Cair);\n", "Cmax=max(Cwater,Cair);\n", "C=Cmin/Cmax;\n", "#NTU is number of transfer units\n", "#NTU=(U*A)/Cmin\n", "print\"NTU is defined as (U*A)/Cmin \"\n", "NTU=(U*A)/(Cmin*1000)\n", "print\"NTU=\",NTU\n", "#To determine the effectiveness of heat exchanger we have to find out the suitable expression \n", "#For this type of heat exchanger The effectiveness(eff)is determined by (1-e**((NTU**.22*(e**-(C*NTU**0.78)-1)/C)\n", "print\"The effectiveness of heat exchanger is\"\n", "eff=(1-math.e**((NTU**0.22))*(math.e**(-C*NTU**0.78)-1)/C)\n", "print\"eff=\",eff\n", "#Hence The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci)in W.\n", "print\"The total heat transfer rate(Q)=eff*Cmin*(Thi-Tci) in W\"\n", "Q=eff*Cmin*1000*(Thi-Tci)\n", "print\"Q=\",Q\n", "#The exit temprature(Tho) of air is given by Thi-(Q/(mdota*cpa))\n", "print\"The exit temprature of air in °C \"\n", "Tho=Thi-(Q/(mdota*1000*cpa))#NOTE:-The answer slightly varies from the answer in book(i.e Tho=26°C) because the value of Q taken in book is approximated to 1*10**6W.\n", "print\"Tho=\",Tho" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex10.8:pg-437" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\n", "(a)Considering a parallel flow arrangement \n", "Tho= 50\n", "The minimum flow rate required for the oil in kg/s\n", "mdoth= 2.22340425532\n", "(b)Theoretical question\n", "If LMTD--->0,Then for a finite value of heat transfer rate U*A--->infinity.For a given finite length this implies value of U which is not possible.\n", "(c)Let us consider a counter flow arrangement\n", "The minimum flow rate required for the oil in kg/s\n", "Ch= 2.78666666667\n", "Cc= 8.36\n", "Effectiveness of heat exchanger is \n", "eff= 1.0\n" ] } ], "source": [ "import math\n", " \n", "print\"Introduction to heat transfer by S.K.Som, Chapter 10, Example 8\"\n", "#A double pipe heat exchanger of length(L)=0.30m is to be used to heat water(specific heat,cc=4.18kJ/(kg*K)) and mass flow rate(mdotw=2kg/s)\n", "L=0.30;\n", "cc=4.18;\n", "mdotw=2;\n", "#The water enters at temprature(Tci)=25°C and leaves at temprature(Tco)=50°C\n", "#The flow rate of oil is mdoth\n", "Tci=25;\n", "Tco=50; \n", "#The oil used as hot fluid has(specific heat,ch=1.88kJ/(kg*K)) and has inlet temprature(Thi)=100°C \n", "ch=1.88;\n", "Thi=100;\n", "print\"(a)Considering a parallel flow arrangement \"\n", "#For minimum value of mdoth\n", "#The theoretical minimum value of outlet temprature of hot fluid(Tho) under this situation is equal to Tco\n", "Tho=Tco;\n", "print\"Tho=\",Tho\n", "#The mass flow rate of oil is given by energy balance as mdoth=(mdotw*cpw*(Tco-Tci))/(cph*(Thi-Tho))\n", "print\"The minimum flow rate required for the oil in kg/s\"\n", "mdoth=(mdotw*cc*(Tco-Tci))/(ch*(Thi-Tho))\n", "print\"mdoth=\",mdoth\n", "print\"(b)Theoretical question\"\n", "print\"If LMTD--->0,Then for a finite value of heat transfer rate U*A--->infinity.For a given finite length this implies value of U which is not possible.\"\n", "print\"(c)Let us consider a counter flow arrangement\"\n", "#In this case value of Tho=Tci.\n", "Tho=Tci;\n", "#The mass flow rate of oil is given by energy balance as mdoth=(mdotw*cpw*(Tco-Tci))/(cph*(Thi-Tho))\n", "print\"The minimum flow rate required for the oil in kg/s\"\n", "mdoth=(mdotw*cc*(Tco-Tci))/(ch*(Thi-Tci))\n", "#Now Heat capacities are Ch=mdoth*ch and Cc=mdotw*cc\n", "Ch=mdoth*ch; \n", "Cc=mdotw*cc;\n", "print\"Ch=\",Ch\n", "print\"Cc=\",Cc\n", "Cmin=min(Ch,Cc);#minimum heat capacity in Ch and Cc \n", "#Effectiveness of heat exchanger is eff.\n", "#Tho=Tci for this kind of arrangement\n", "Tho=Tci;\n", "print\"Effectiveness of heat exchanger is \"\n", "eff=(mdoth*ch*(Thi-Tho))/(mdoth*ch*(Thi-Tci))\n", "print\"eff=\",eff" ] } ], "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 }