path: root/Thermodynamics_An_Engineering_Approach/Chapter4.ipynb

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  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 4: Energy Analysis of Closed Systems"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-2 ,Page No.169"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#Given values\n",
      "m=10;#mass in lbm\n",
      "Po=60;#steam oressure in psia\n",
      "T1=320;#intial temp in F\n",
      "T2=400;#final temp in F\n",
      "\n",
      "#from Table A\u20136E\n",
      "v1=7.4863;#at 60 psia and 320 F\n",
      "v2=8.3548;#at 60 psia and 400 F\n",
      "\n",
      "#calculations\n",
      "#W = P dV which on integrating gives W = m * P * (V2 - V1)\n",
      "W=m*Po*(v2-v1)/5.404;#coverting into Btu from psia-ft^3\n",
      "print'work done by the steam during this process %f Btu'%round(W,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "work done by the steam during this process 96.400000 Btu\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-3 ,Page No.170"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import log\n",
      "#given data\n",
      "P1=100;#pressure in kPa\n",
      "V1=0.4;#intial vol in m^3\n",
      "V2=0.1;#final vol in m^3\n",
      "\n",
      "#calculations\n",
      "#for isothermal W = P1*V1* ln(V2/V1)\n",
      "W=P1*V1*log(V2/V1);\n",
      "print'the work done during this process %f kJ'%round(W,1)"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the work done during this process -55.500000 kJ\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-4 ,Page No.171"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "V1=0.05;#Volumne of gas in m^3\n",
      "P1=200;#Pressure in kPa\n",
      "k=150;#Spring constant in kN/m\n",
      "A=0.25;#Cross-sectional area in m^2\n",
      "\n",
      "#calculations\n",
      "\n",
      "#Part - a\n",
      "V2=2*V1;\n",
      "x2=(V2-V1)/A;#printlacement of spring\n",
      "F=k*x2;#compression force\n",
      "P2=P1+F/A;#additional pressure is equivalent the compression of spring\n",
      "print'the final pressure inside the cylinder %i kPa'%P2;\n",
      "\n",
      "#Part - b\n",
      "#work done is equivalent to the area of the P-V curve of Fig 4-10\n",
      "W=(P1+P2)/2*(V2-V1);#area of trapezoid = 1/2 * sum of parallel sides * dist. b/w them\n",
      "print'the total work done by the gas %i kJ'%W;\n",
      "\n",
      "#Part - c\n",
      "x1=0;#intial compression of spring\n",
      "Wsp=0.5*k*(x2**2-x1**2);\n",
      "print'the fraction of this work done against the spring to compress it %i kJ'%Wsp\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final pressure inside the cylinder 320 kPa\n",
        "the total work done by the gas 13 kJ\n",
        "the fraction of this work done against the spring to compress it 3 kJ\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-5 ,Page No.174"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given values\n",
      "m=0.025;#mass of saturated water vapor in kg\n",
      "V=120;#rated voltage of heater in V\n",
      "I=0.2;#rated current in A\n",
      "t=300;#total time taken in sec\n",
      "P1=300;#constant pressure in kPa\n",
      "Qout=3.7;#heat lost in kJ\n",
      "\n",
      "#from Table A\u20135\n",
      "#at P1 the conditon is sat. vap\n",
      "h1=2724.9;\n",
      "\n",
      "#Calculations\n",
      "\n",
      "#Part - a\n",
      "#therotical proving\n",
      "\n",
      "#Part - b\n",
      "We=V*I*t/1000;#electrical work in kJ\n",
      "#from eqn 4 -18 i.e derived in earler part\n",
      "#it states it Ein - Eout = Esystem\n",
      "# it applies as Win - Qout = H = m (h2 - h1)\n",
      "h2=(We-Qout)/m+h1;\n",
      "##from Table A\u20135\n",
      "#at h2 we get\n",
      "P2=300;\n",
      "T=200;\n",
      "print'the final temperature of the steam %i C'%T\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final temperature of the steam 200 C\n"
       ]
      }
     ],
     "prompt_number": 8
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-6 ,Page No.176"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "m=5.0;#mass of water in kg\n",
      "P1=200;#pressure on one side in kPa\n",
      "T=25;#temperature in C\n",
      "\n",
      "#from Table A\u20134\n",
      "#the liq. is in compressed state at 200 kPa and 25 C\n",
      "vf=0.001;\n",
      "vg=43.340;\n",
      "uf=104.83;\n",
      "ufg=2304.3;\n",
      "v1=vf;\n",
      "u1=uf;\n",
      "\n",
      "#calculations\n",
      "\n",
      "#Part - a\n",
      "V1=m*v1;\n",
      "Vtank=2*V1;\n",
      "print'the volume of the tank %f m^3'%round(Vtank,2);\n",
      "\n",
      "#Part - b\n",
      "V2=Vtank;\n",
      "v2=V2/m;\n",
      "#from Table A\u20134 \n",
      "# at T=25 vf=0.101003 m^3/kg and vg=43.340 m^3/kg\n",
      "# vf<v2<vg therefore it is saturated liquid\u2013vapor mixture\n",
      "P2=3.1698;\n",
      "print'the final pressure %f kPa'%round(P2,4);\n",
      "\n",
      "#Part - c\n",
      "#Ein - Eout = Esystem\n",
      "#Qin= dU = m(u2 - u1)\n",
      "x2=(v2-vf)/(vg-vf);\n",
      "u2=uf+x2*ufg;\n",
      "Qin=m*(u2-u1);\n",
      "print'the heat transfer for this process %f kJ'%Qin\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the volume of the tank 0.010000 m^3\n",
        "the final pressure 3.169800 kPa\n",
        "the heat transfer for this process 0.265846 kJ\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-7 ,Page No.183"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "import math\n",
      "from scipy.integrate import quad \n",
      "from pylab import *\n",
      "\n",
      "#given data\n",
      "T1=300;#intial temp of air in K\n",
      "P=200;#pressure in kPa\n",
      "T2=600;#final temp in K\n",
      "M=28.97;#molecular weight in kg/kmol\n",
      "Ru=8.314;\n",
      "\n",
      "#Part - a\n",
      "#from Table A\u201317\n",
      "u1=214.07;\n",
      "u2=434.78;\n",
      "du=u2-u1;#change in internal energy\n",
      "print'change in internal energy from data from the air table %f kJ/kg'%round(du,2);\n",
      "\n",
      "#Part - b\n",
      "#from Table A\u20132c\n",
      "a=28.11;\n",
      "b=0.1967*10**-2;\n",
      "c=0.4802*10**-5;\n",
      "d=-1.966*10**-9;\n",
      "#  by equation Cp(T)=a+bT+cT^2+dT^3\n",
      "def intgrnd1(T): \n",
      " return ((a-Ru)+b*T+c*T**2+d*T**3)\n",
      "dU, err = quad(intgrnd1, T1, T2) \n",
      "du=dU/M;\n",
      "print'change in internal energy the functional form of the specific heat %f kJ/kg'%round(du,1); \n",
      "\n",
      "#Part - c\n",
      "#from Table A\u20132b\n",
      "Cavg=0.733;\n",
      "du=Cavg*(T2-T1);\n",
      "print'change in internal energy the functional form the average specific heat value %i kJ/kg'%round(du,0);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "change in internal energy from data from the air table 220.710000 kJ/kg\n",
        "change in internal energy the functional form of the specific heat 222.500000 kJ/kg\n",
        "change in internal energy the functional form the average specific heat value 220 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-8 ,Page No.184"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "m=1.5;#mass in lbm\n",
      "T1=80;#temperature in F\n",
      "P1=50;#pressure in psia\n",
      "W=0.02;#power rating in hp\n",
      "t=30.0/60;#converting into hrs from min\n",
      "\n",
      "#from Table A\u20132Ea\n",
      "Cv=0.753;\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part a\n",
      "Wsh=W*t*2545;#in Btu\n",
      "#Ein - Eout = Esystem\n",
      "#Wsh = dU = m (u2 - u1) = m * Cv * (T2 - T1)\n",
      "T2= Wsh/(m*Cv)+T1;\n",
      "print'the final temperature %f F'%round(T2,1);\n",
      "\n",
      "#part b\n",
      "#using ideal gas eqn\n",
      "# P1 * V1 / T1 = P2 * T2 /V2\n",
      "P2= 50 * (T2 +460)/ (T1+460);\n",
      "# temp should in R therefore + 460\n",
      "print'the final pressure %f psia'%round(P2,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final temperature 102.500000 F\n",
        "the final pressure 52.100000 psia\n"
       ]
      }
     ],
     "prompt_number": 21
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-9 ,Page No.185"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "V1=0.5;#volumne of nitrogen gas in m^3\n",
      "P=400;#pressure in kPa\n",
      "T1=27;#temp in C\n",
      "I=2;#rated current in A\n",
      "t=5*60;#converting into s from min\n",
      "V=120;#rated voltage in V\n",
      "Qout=2800/1000;#in kJ\n",
      "R=0.297;\n",
      "\n",
      "#from Table A\u20132a\n",
      "Cp=1.039;\n",
      " \n",
      "#calculations\n",
      "P1=P;\n",
      "We=V*I*t/1000;#in kJ\n",
      "m=P1*V1/(R*(T1+273));\n",
      "#Ein - Eout = Esystem\n",
      "# We,in - Qout = dH =  m (h2 - h1) = m * Cp * (T2 - T1)\n",
      "T2=(We-Qout)/(m*Cp)+T1;\n",
      "print'the final temperature of nitrogen in %i C'%T2\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final temperature of nitrogen in 57 C\n"
       ]
      }
     ],
     "prompt_number": 25
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-10 ,Page No.187"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "P1=150;#initially air pressure in kPa\n",
      "P2=350;#final pressure in kPa\n",
      "T1=27+273;#temperatuere in K\n",
      "V1=400.0/1000;#volumne in m^3\n",
      "R=0.287;\n",
      "\n",
      "#from Table A\u201317\n",
      "u1=214.07;\n",
      "u2=1113.52;\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part a\n",
      "V2=2*V1;\n",
      "#using ideal gas eqn\n",
      "# P1 * V1 / T1 = P2 * T2 /V2\n",
      "T2=P2*V2*T1/(P1*V1);\n",
      "print'the final temperature %i K'%T2;\n",
      "\n",
      "#part b\n",
      "# Work done is Pdv\n",
      "W=P2*(V2-V1);\n",
      "print'the work done by the air %i kJ'%W;\n",
      "\n",
      "#part c\n",
      "#Ein - Eout = Esystem\n",
      "#Qin - Wout = dU = m(u2 - u1)\n",
      "m= P1* V1 /(T1 * R);\n",
      "Q= m*(u2 - u1)+ W;\n",
      "print'the total heat transferred to the air %i kJ'%round(Q)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the final temperature 1400 K\n",
        "the work done by the air 140 kJ\n",
        "the total heat transferred to the air 767 kJ\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-11 ,Page No.190"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "T=100;#twmperature of liquid water in C\n",
      "P=15;#pressure in MPa\n",
      "\n",
      "#from Table A\u20137\n",
      "#at P=15 mPa and T = 100 C\n",
      "hg=430.39;\n",
      "hf=419.17\n",
      "vf=0.001;\n",
      "Psat=101.42;#in kPa\n",
      "\n",
      "#calculations\n",
      "\n",
      "#part a\n",
      "h=hg;\n",
      "print'enthalpy of liquid water by using compressed liquid tables %f kJ/kg'%round(h,2);\n",
      "\n",
      "#part b\n",
      "#Approximating the compressed liquid as a saturated liquid at 100\u00b0C\n",
      "h=hf;\n",
      "print'enthalpy of liquid water by approximating it as a saturated liquid %f kJ/kg'%round(h,2);\n",
      "\n",
      "#part c\n",
      "h=hf + vf*(P*1000 - Psat );\n",
      "print'enthalpy of liquid water by using the correction given by Eq. 4\u201338 %f kJ/kg'%round(h,2);\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "enthalpy of liquid water by using compressed liquid tables 430.390000 kJ/kg\n",
        "enthalpy of liquid water by approximating it as a saturated liquid 419.170000 kJ/kg\n",
        "enthalpy of liquid water by using the correction given by Eq. 4\u201338 434.070000 kJ/kg\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-12 ,Page No.191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#suffix i for iron\n",
      "#suffix w for water\n",
      "#given data\n",
      "mi=50;#mass in Kg\n",
      "T1i=80;#temperature in C \n",
      "Vw=0.5;#volumne in m^3\n",
      "T1w=25;#temperature in C \n",
      "v=0.001;#specific volume of liquid water at or about room temperature\n",
      "\n",
      "#from Table A\u20133\n",
      "ci=0.45;\n",
      "cw=4.18;\n",
      "\n",
      "#calculations\n",
      "mw=Vw/v;\n",
      "#Ein - Eout = Esystem\n",
      "# du = 0 i.e (mcdT)iron + (mcdT)water = 0\n",
      "# mi * ci * (T - T1i) + mw *cw * (T-T1w)\n",
      "#on rearranging above equn\n",
      "T= (mi*ci*T1i + mw*cw*T1w)/(mi*ci+mw*cw);\n",
      "print'the temperature when thermal equilibrium is reached %f C'%round(T,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the temperature when thermal equilibrium is reached 25.600000 C\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-13 ,Page No.191"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "from math import sqrt\n",
      "\n",
      "#given data\n",
      "maf=0.15;#mass of affected tissue in kg\n",
      "caf=3.8;#specific heat of the tissue in kJ/Kg C\n",
      "dTaf=1.8;#suffix af for affected tissue\n",
      "mh=1.2;#suffix h for hand ; mass of hand in kg\n",
      "\n",
      "#calculations\n",
      "#Ein - Eout = Esystem\n",
      "#dUaffected tissue - KEhand = 0\n",
      "#from above equation we can deduce that\n",
      "Vhand= sqrt(2*maf*caf*dTaf*1000/mh);#for conversion factor mutiplying by 1000 to get m^2/s^2\n",
      "print'the velocity of the hand just before impact %f m/s'%round(Vhand,1);"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "the velocity of the hand just before impact 41.400000 m/s\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-14 ,Page No.199"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "m=90;#mass of man in kg\n",
      "\n",
      "#from Tables 4\u20131 and 4\u20132\n",
      "Ehb=275;#hamburger\n",
      "Ef=250;#fries\n",
      "Ec=87;#cola\n",
      "\n",
      "#calculation\n",
      "\n",
      "#part a\n",
      "Ein=2*Ehb+Ef+Ec;\n",
      "#The rate of energy output for a 68-kg man watching TV is to be 72 Calories/h\n",
      "Eout=m*72.0/68;\n",
      "t=Ein/Eout;\n",
      "print'by watching TV %f hours'%round(t,1);\n",
      "\n",
      "#part b\n",
      "#The rate of energy output for a 68-kg man watching TV is to be 860 Calories/h\n",
      "Eout=m*860.0/68;\n",
      "t=Ein/Eout*60#converting in min\n",
      "t=ceil(t);\n",
      "print'by fast swimming %f mins'%t;\n",
      "\n",
      "#for last question\n",
      "print('answers be for a 45-kg man energy takes twice as long in each case');\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "by watching TV 9.300000 hours\n",
        "by fast swimming 47.000000 mins\n",
        "answers be for a 45-kg man energy takes twice as long in each case\n"
       ]
      }
     ],
     "prompt_number": 37
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4-15 ,Page No.199"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#given data\n",
      "E=75;#in Cal/day\n",
      "\n",
      "#calculation\n",
      "Ereduced=E*365.0;\n",
      "#The metabolizable energy content of 1 kg of body fat is 33,100 kJ\n",
      "Ec=33100.0;\n",
      "mfat=Ereduced/Ec*4.1868;\n",
      "print'weight this person will lose in one year %f kg'%round(mfat,2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "weight this person will lose in one year 3.460000 kg\n"
       ]
      }
     ],
     "prompt_number": 41
    }
   ],
   "metadata": {}
  }
 ]
}