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diff --git a/Thermodynamics_Demystified_by_Merle_C._Potter/Chapter4_2.ipynb b/Thermodynamics_Demystified_by_Merle_C._Potter/Chapter4_2.ipynb new file mode 100755 index 00000000..e17b48f1 --- /dev/null +++ b/Thermodynamics_Demystified_by_Merle_C._Potter/Chapter4_2.ipynb @@ -0,0 +1,719 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:abd0537a9b98c6cb49f3032d9229bd28210acff6f4a7d9f1c5434293bf15ef92"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4:The First Law of Thermodynamics"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.1:PG-62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#initialization of variables\n",
+ "K=100 # spring constant in kN/m\n",
+ "d=0.8 # dispacement of spring in m\n",
+ " # to get total work we integrate from 0 to 0.8 displacement\n",
+ "x1=0; # lower limit of integration\n",
+ "x2=0.8; # upper limit of integration\n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "# we find work\n",
+ "def integrand(x,K):\n",
+ " return K*x\n",
+ "\n",
+ "W12, err = quad(integrand, x1, x2, K) # integrating to get work\n",
+ "Q12=W12; # by first law of thermodynamics\n",
+ "print \"The Heat transfer is \",int(Q12),\" J\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Heat transfer is 32 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.2:PG-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "P= 5*746 # power of fan converted in watt\n",
+ "t=1*60*60 # time converted to seconds\n",
+ "\n",
+ "# by first law of thermodynamics Q=delU + W\n",
+ "# Q=0 hence -W=delU\n",
+ "# first we find work input\n",
+ "W=-P*t # work in J\n",
+ "delU=-W # from 1st law\n",
+ "print \"The internal energy increase is \",float(delU),\" J\"\n",
+ "# The answer is approximated in textbook\n",
+ "# our answer is precise\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The internal energy increase is 13428000.0 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.3:PG-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "P=400 # pressure in kPa\n",
+ "T1=200 # initial temperature in degree celsius\n",
+ "V1= 2 # initial volume in m^3\n",
+ "Q=3500 # heat added in kJ\n",
+ "v1=0.5342 # specific volume of steam at 200 degree celcius and 0.4 Mpa pressure from table C.3\n",
+ "u1=2647 # specific internal energy in kJ/kg @ pressure = 0.4 MPa\n",
+ "m=V1/v1 # mass in kg\n",
+ "# we have a relation Between u2 and v2 from 1st law of thermodynamics\n",
+ "v2=1.06 # specific volume at state 2 by trial and error and interpolation\n",
+ "V2=m*v2 \n",
+ "u2=((3500-400*(V2-V1))/m)+2647 # specific internal energy for v2=1.06 by trial and error\n",
+ "\n",
+ "# on interpolation from steam table at 0.4 MPa we get temperature \n",
+ "T2=644 # temperature in degree celsius\n",
+ "print \"The temperature for u2=\",round(u2),\" kJ and v2 =\",round(v2,3),\" kg/m^3 is \\n \",int(T2),\" degree celsius\"\n",
+ "# this numerical is solved by trial and error thus refer to Appendix C"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The temperature for u2= 3372.0 kJ and v2 = 1.06 kg/m^3 is \n",
+ " 644 degree celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.4:PG-67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# initialization of variables\n",
+ "P=400 # pressure in kPa\n",
+ "T1=200 # initial tmperature in degree celsius\n",
+ "V=2 # initial volume in m^3\n",
+ "Q=3500 # heat added in kJ\n",
+ "\n",
+ "#solution\n",
+ "h1=2860 # initial enthalpy @ 200*C and 400 kPa from steam table\n",
+ "v=0.5342 # specific volume from steam table C.3 \n",
+ "m=V/v;\n",
+ "h2=(Q/m)+h1; # final enthalpy in kJ/kg from energy equation\n",
+ "\n",
+ "# NOW USING THIS ENTHLAPY AND INTERPOlATING FROM STEAM TABLE\n",
+ "\n",
+ "T2=600+(92.6/224)*100\n",
+ "\n",
+ "print \"The Final temperature is \",int(T2),\" degree Celsius\"\n",
+ "# result is obtained from interpolation on steam table\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Final temperature is 641 degree Celsius\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.5:PG-71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "T1=300 # initial temperature in degree celsius\n",
+ "T2=700 # final temperature in degree celsius\n",
+ "P=150 # pressure in kPa\n",
+ "m=3 # mass of steam in kg\n",
+ "\n",
+ "# solution\n",
+ "# part (a)\n",
+ "from scipy.integrate import quad\n",
+ "\n",
+ "# now we make function to integrate\n",
+ "def integrand(T):\n",
+ " return 2.07+(T-400)/1480\n",
+ "\n",
+ "I, err = quad(integrand, T1, T2) # integrating specific heat over temperature range\n",
+ "delH=m*I #integrate('2.07+(T-400)/1480','T',T1,T2) # expressing as function of temperature and integrating\n",
+ "\n",
+ "print\" The change in Enthalpy is \",int(delH),\" kJ \\n\"\n",
+ " \n",
+ "# part(b)\n",
+ "CPavg=delH/(m*(T2-T1)) # avg value of specific heat at constant pressure\n",
+ "print \" The average value of Cp is \",round(CPavg,2),\" kJ/kg.*C\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The change in Enthalpy is 2565 kJ \n",
+ "\n",
+ " The average value of Cp is 2.14 kJ/kg.*C\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.6,PG-72"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "m=1 # mass of nitrogen in kg\n",
+ "T1=300 # initial temperature in Kelvin\n",
+ "T2=1200 # final temperature in Kelvin\n",
+ "M=28.0 # in kg/kmol\n",
+ "# part(a)\n",
+ "# the enthalpy change is found from gas table in App.E\n",
+ "delh=36777-8723 # from gas table\n",
+ "delH=delh/M \n",
+ "print \" The entalpy change from gas table is \",round(delH),\" kJ/kg \\n\"\n",
+ "\n",
+ "# part (b) \n",
+ "Cp=1.042 # from table B.2\n",
+ "delH=Cp*(T2-T1)\n",
+ "print \" The entalpy change by assuming constant specific heat is \",round(delH),\" kJ/kg\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The entalpy change from gas table is 1002.0 kJ/kg \n",
+ "\n",
+ " The entalpy change by assuming constant specific heat is 938.0 kJ/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.7:PG-76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "x=0.7 # quality of steam\n",
+ "P1=200 # initial pressure in kPa\n",
+ "P2=800 # final pressure in kPa\n",
+ "V=2 # volume in m^3\n",
+ "# The values are taken from TABLE C.2\n",
+ "vf1=0.0010 # specific volume of saturated liquid at 200 kPa\n",
+ "vg1=0.8857 # specific volume of saturated gas at 200 kPa\n",
+ "uf1=504.5 # specific internal energy of saturated liquid @ state 1\n",
+ "ug1=2529.5 # speciific internal energy of saturated gas @ state 1\n",
+ "\n",
+ "v1=vf1+x*(vg1-vf1); # specific volume of vapour\n",
+ "m=V/v1\n",
+ "\n",
+ "u1=uf1+x*(ug1-uf1) # specific internal energy of vapour @ state 1\n",
+ "v2=v1 # constant volume process\n",
+ "u2=((0.6761-0.6203)*(3661-3853)/(0.6761-0.6181))+3853 # from steam table @ 800kPa by interpolating\n",
+ "Q=m*(u2-u1) # heat transfer\n",
+ "print \"The heat transfer is \",round(Q,3),\" kJ\"\n",
+ "# The answer in the textbook is approximated"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The heat transfer is 5630.537 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.8:PG-76"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "V=0.02 # volume in m^3\n",
+ "P=400 # pressure in kPa\n",
+ "T1=50+273 # initial temperature in kelvin\n",
+ "T2=700+273 # final temperature in kelvin\n",
+ "Q=50 # heat added in kJ\n",
+ "R=287 # constant for air\n",
+ "Cp=1 # constant for specific heat of air\n",
+ "\n",
+ "# using the ideal gas equation\n",
+ "\n",
+ "m=P*1000*V/(R*T1) # mass of air in kg\n",
+ "W=Q-(m*Cp*(T2-T1)) # work done from first law\n",
+ "# result\n",
+ "print \"The Paddle work is \",round(W,2),\" kJ\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Paddle work is -6.09 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.9,PG-77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "V1=2 # initial volume in m^3\n",
+ "V2=0.2 # final volume in m^3\n",
+ "T1=20+273 # temperature in kelvin\n",
+ "P=200 # pressure in kPa\n",
+ "R=0.287 # constant for air\n",
+ "gama=1.4 # polytropic index for air\n",
+ "Cv=0.717 # specific heat at constant volume for air\n",
+ "\n",
+ "#solution\n",
+ "\n",
+ "#using the ideal gas equation\n",
+ "m=(P*V1)/(R*T1) # mass in kg\n",
+ "# process is adiabatic thus\n",
+ "T2=T1*((V1/V2)**(gama-1)) # final temperature\n",
+ "\n",
+ "W=-m*Cv*(T2-T1) # work from first law\n",
+ "print \"The Work is \",int(W),\" kJ\"\n",
+ "# solution is approximated in textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Work is -1510 kJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.10:PG-79"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# initialization of variables\n",
+ "P1=2000.0 # initial pressure in kPa\n",
+ "T1=600.0 # initial temperature in degree celsius\n",
+ "p2=600.0 # final pressure in kPa\n",
+ "T2=200.0 # final temperature in degree celsius\n",
+ "d1=0.06 # diameter of inlet pipe in metre\n",
+ "d2=0.120 # diameter of outlet pipe in metre\n",
+ "V1=20.0 # velocity at inlet in m/s\n",
+ "\n",
+ "# solution\n",
+ "# from superheat table C.3 values are noted\n",
+ "v1=0.1996 # specific volume of superheated steam @ 600*C and 2000 kPa\n",
+ "v2=0.3520 # specific volume of superheated steam @ 200*C and 2000 kPa\n",
+ "rho1=1/v1 # initial density\n",
+ "rho2=1/v2 # final density\n",
+ "A1=(math.pi*d1**2)/4 # inlet area\n",
+ "A2=(math.pi*d2**2)/4 # exit area\n",
+ "\n",
+ "V2=(rho1*A1*V1)/(rho2*A2) # from continuity equation\n",
+ "print \" The Exit velocity is \",round(V2,2),\" m/s \\n\"\n",
+ "\n",
+ "mdot=rho1*A1*V1 # mass flow rate\n",
+ "print\" The mass flow rate is \",round(mdot,3),\" kg/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The Exit velocity is 8.82 m/s \n",
+ "\n",
+ " The mass flow rate is 0.283 kg/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.11:PG-82"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "P1=8000 # initial pressure in kPa\n",
+ "T1=300 # temperature in degree celsius\n",
+ "P2=2000 # final pressure in kPa\n",
+ "\n",
+ "# solution\n",
+ "h1=2785 # specific enthalpy of steam in kJ/kg @ 8000 kPa and 300 degree celsius from steam table\n",
+ "h2=h1 # throttling process thus enthalpy is constant\n",
+ "T2=212.4 # from steam table as we know enthalpy and pressure\n",
+ "hf2=909 # specific enthalpy of saturated liquid @ 2000 kPa and 300 degree celsius\n",
+ "hg2=2799.5 # specific enthalpy of saturated gas @ 2000 kPa and 300 degree celsius\n",
+ "x2=(h2-hf2)/(hg2-hf2) # quality of steam\n",
+ "\n",
+ "vg2=0.0992 # specific volume of saturated gas @ 2000 kPa and 212.4*c\n",
+ "vf2=0.0012 # specific volume of saturated liquid @ 2000 kPa and 212.4*c\n",
+ "v2=vf2+x2*(vg2-vf2) # from properties of pure substance\n",
+ "\n",
+ "print \"The Final Temperature and Specific volume is \",round(T2,1),\"*C and \",round(v2,3),\" m^3/kg\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Final Temperature and Specific volume is 212.4 *C and 0.098 m^3/kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.12:PG-84"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# initialization of variables\n",
+ "P1=4000 # inlet pressure in kPa\n",
+ "T1=500 # inlet temperature in degree celsius\n",
+ "V1=200 # inlet steam velocity in m/s\n",
+ "d1=0.05 # inlet diameter in 'm'\n",
+ "P2=80 # exit pressure in kPa\n",
+ "d2=0.250 # exit diameter in 'm'\n",
+ "\n",
+ "# solution\n",
+ "v1=0.08643 # specific volume from steam table @ 4000 kPa and 500*C\n",
+ "v2=2.087 # specific volume from steam table @ 80 kPa and 500*C\n",
+ "rho1=1/v1 # density at inlet\n",
+ "rho2=1/v2 # density at outlet\n",
+ "A1=(math.pi*d1**2)/4 # inlet area\n",
+ "A2=(math.pi*d2**2)/4\n",
+ "mdot=rho1*A1*V1 # mass flow rate\n",
+ "mdot=round(mdot,3) # rounding to 3 significant digits\n",
+ "\n",
+ "#now using table C.3\n",
+ "h1=3445 # initial specific enthalpy @ 4000 kPa and 500 *C \n",
+ "h2=2666 # final specific enthalpy @ 80 kPa and 500 *C\n",
+ "WT=-mdot*(h2-h1) # maximum power from first law\n",
+ "print \" The power output is \",round(WT),\" kJ/s \\n \"\n",
+ "\n",
+ "V2=(A1*V1*rho1)/(A2*rho2) \n",
+ "V2=round(V2) # rounding of digits\n",
+ "delKE=mdot*((V2**2)-(V1**2))/2 # the change in kinetic energy\n",
+ "print \" The change in K.E is \",round(delKE),\" J/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The power output is 3540.0 kJ/s \n",
+ " \n",
+ " The change in K.E is -6250.0 J/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.13:PG-85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "# initialization of variables\n",
+ "Wdot=10 # pump power in hp\n",
+ "g=9.81 # acceleration due to gravity\n",
+ "rho=1000 # density of water in kg/m^3\n",
+ "d1=0.06 # inlet dimeter in 'm'\n",
+ "d2=0.10 # oulet diamter in 'm'\n",
+ "V1=10 # velocity of water at inlet in m/s\n",
+ "\n",
+ "#solution\n",
+ "A1=math.pi*(d1**2)/4 # area of inlet\n",
+ "A2=math.pi*(d2**2)/4 # area of outlet\n",
+ "V2=A1*V1/A2 # oulet velocity from continuity equation\n",
+ "\n",
+ "mdot=rho*A1*V1 # mass flow rate\n",
+ "delP=((((Wdot*746)/mdot)-((V2**2)-V1**2)/(2*g))*rho)/1000 # change in pressure in kPa\n",
+ "print \"The rise in pressure is \",round(delP),\" kPa\"\n",
+ "# The answer is approximated in textbook , our answer is precise \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rise in pressure is 268.0 kPa\n"
+ ]
+ }
+ ],
+ "prompt_number": 80
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.14:PG-85"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "# initialization of variables\n",
+ "P1=7000.0 # inlet pressure in Pa\n",
+ "T1=420.0 # inlet temperature in degree celsius\n",
+ "V1=400.0 # inlet velocity in m/s\n",
+ "d1=0.200 # inlet diameter in 'm'\n",
+ "V2=700.0 # exit velocity in m/s\n",
+ "k=1.4 # polytopic index for air\n",
+ "Cp=1000 # specific heat at constant pressure for air in j/kg.K\n",
+ "R=287 # specific gas constant for air\n",
+ "\n",
+ "#solution\n",
+ "\n",
+ "#part (a)\n",
+ "T2=(((V1**2)-V2**2)/(2*Cp))+T1 # outlet temperature in degree celsius\n",
+ "print \" The exit temperature is \",round(T2),\" *C \\n\"\n",
+ "\n",
+ "#part (b)\n",
+ "\n",
+ "rho1=P1/(R*(T1+273)) # density at entrance\n",
+ "A1=(math.pi*d1**2)/4\n",
+ "mdot=rho1*A1*V1 # \n",
+ "print \" The mass flow rate is \",round(mdot,3),\" kg/s \\n\"\n",
+ "\n",
+ "# part (c)\n",
+ "\n",
+ "rho2=rho1*(((T2+273)/(T1+273))**(1/(k-1))) # density at exit\n",
+ "# now we find the exit diameter\n",
+ "d2=math.sqrt((rho1*V1*(d1)**2)/(rho2*V2))\n",
+ "print \" The outlet diameter is \",round(d2,3),\" m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The exit temperature is 255.0 *C \n",
+ "\n",
+ " The mass flow rate is 0.442 kg/s \n",
+ "\n",
+ " The outlet diameter is 0.212 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 87
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.15:PG-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# initialization of variables\n",
+ "mdots=100 # mass flow rate of sodium in kg/s\n",
+ "Ts1=450 # inlet temperature of sodium in degree celsius\n",
+ "Ts2=350 # exit temperature of sodium in degree celsius\n",
+ "Cp=1.25 # specific heat of sodium in KJ/kg.*C\n",
+ "Tw1=20 # inlet temperature of water in degree celsius\n",
+ "Pw=5000 # inlet pressure of water in kPa \n",
+ "\n",
+ "# solution\n",
+ "hw1=88.65 # enthalpy from table C.4\n",
+ "hw2=2794 # enthalpy from table C.3\n",
+ "mdotw=(mdots*Cp*(Ts1-Ts2))/(hw2-hw1) # mass flow rate of water\n",
+ "print \" The mass flow rate of water is \",round(mdotw,2),\" kg/s \\n\"\n",
+ "Qdot=mdotw*(hw2-hw1) # heat transfer in kW using energy equation\n",
+ "# result\n",
+ "print \" The rate of heat transfer is \",round(Qdot),\" kW\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The mass flow rate of water is 4.62 kg/s \n",
+ "\n",
+ " The rate of heat transfer is 12500.0 kW\n"
+ ]
+ }
+ ],
+ "prompt_number": 90
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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