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   "source": [
    "# Chapter 12 Non steady flow friction and availibility"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.1 Pg:482"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Work done in case 1 = 572 Btu\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "p1=100 #psia\n",
    "p2=14.7 #psia\n",
    "k=1.4\n",
    "T1=700 #R\n",
    "R=10.73/29\n",
    "V=50\n",
    "cv=0.171\n",
    "cp=0.24\n",
    "R2=1.986/29\n",
    "#calculations\n",
    "T2=T1/ (p1/p2)**((k-1)/k)\n",
    "m1=p1*V/(R*T1)\n",
    "m2=p2*V/(R*T2)\n",
    "Wrev= cv*(m1*T1 - m2*T2) - (m1-m2)*(T2)*cp\n",
    "#results\n",
    "print \"Work done in case 1 = %d Btu\"%(Wrev)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.2 Pg:482"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The friction of the process per pound of air = 18.6 Btu/lbm\n",
      "\n",
      " Loss of available energy = -16.20 Btu/lbm\n"
     ]
    }
   ],
   "source": [
    "from __future__ import division\n",
    "from scipy import log\n",
    "#Initialization of variables\n",
    "p1=100 #psia\n",
    "p2=10 #psia\n",
    "n=1.3\n",
    "T1=800 #R\n",
    "cv=0.172\n",
    "R=1.986/29\n",
    "T0=537 #R\n",
    "cp=0.24\n",
    "#calculations\n",
    "T2=T1*(p2/p1)**((n-1)/n)\n",
    "dwir=cv*(T1-T2)\n",
    "dwr=R*(T2-T1)/(1-n)\n",
    "dq=dwr-dwir\n",
    "dI=-T0*(cp*log(T2/T1) - R*log(p2/p1))\n",
    "#results\n",
    "print \"The friction of the process per pound of air = %.1f Btu/lbm\"%(dq)\n",
    "print \"\\n Loss of available energy = %.2f Btu/lbm\"%(dI)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.3 Pg:483"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Friction = 72.9 ft-lbf/lbm\n",
      "\n",
      " Available energy loss in case a = -72.9 ft-lbf/lbm\n",
      "\n",
      " Available energy loss in case b = -145.9 ft-lbf/lbm\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "ms=10 #lbm\n",
    "den=62.3 #lbm/ft**3\n",
    "A1=0.0218 #ft**2\n",
    "A2=0.00545 #ft**2\n",
    "p2=50 #psia\n",
    "p1=100 #psia\n",
    "gc=32.2 #ft/s**2\n",
    "dz=30 #ft\n",
    "T0=537 #R\n",
    "T1=620 #R\n",
    "T2=420 #R\n",
    "#calculations\n",
    "V1=ms/(A1*den)\n",
    "V2=ms/(A2*den)\n",
    "df=-144/den*(p2-p1) - (V2**2 -V1**2)/(2*gc) - dz\n",
    "dI=-T0/T1 *df\n",
    "dI2= -T0/T2 *df\n",
    "#results\n",
    "print \"Friction = %.1f ft-lbf/lbm\"%(df)\n",
    "print \"\\n Available energy loss in case a = %.1f ft-lbf/lbm\"%(dI)\n",
    "print \"\\n Available energy loss in case b = %.1f ft-lbf/lbm\"%(dI2)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.4 Pg:484"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "From fig 12.4,\n",
      "Pressure drop = 0.00 lbf/ft**2 100 ft\n",
      "The answer in the textbook is wrong. Please use a calculator to verify it.\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "r=2.5 #in\n",
    "mf=160 #cfm\n",
    "rho=1/14\n",
    "mu=0.0000121\n",
    "v=14 #ft**3/lbm\n",
    "g=32.2 #ft/s**2\n",
    "z=100 #ft\n",
    "#calculations\n",
    "A=3.14*(r/12)**2\n",
    "V=mf/A /60\n",
    "Re=(2*r/12)*V*rho/mu\n",
    "print \"From fig 12.4,\"\n",
    "f=0.0225/4\n",
    "dp=4*f*(rho)*(V/v)**2 /(2*g*(2*r/12)) *z\n",
    "#dp=2.32\n",
    "#results\n",
    "print \"Pressure drop = %.2f lbf/ft**2 100 ft\"%(dp)\n",
    "print \"The answer in the textbook is wrong. Please use a calculator to verify it.\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.5 Pg:485"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Mass rate of air flow = 161 cfm\n",
      "The answer is a bit different due to rounding off error in textbook\n"
     ]
    }
   ],
   "source": [
    "from math import pi,sqrt\n",
    "#Initialization of variables\n",
    "D=0.0724 #ft\n",
    "gc=32.2 #ft/s**2\n",
    "rho=1.0/14\n",
    "L=100 #ft\n",
    "mu2=1.46*10**(-10)\n",
    "dp=2.32\n",
    "dia=5.0 #in\n",
    "rho2=48500.0\n",
    "vol=14.0 #ft**3/lbm\n",
    "#calculations\n",
    "ref=D**3 *2*dp*gc*rho/(mu2*L)\n",
    "mf=rho2*pi/4 *(dia/12) *sqrt(mu2)\n",
    "mfr=mf*vol*60\n",
    "#results\n",
    "print \"Mass rate of air flow = %d cfm\"%(mfr)\n",
    "print \"The answer is a bit different due to rounding off error in textbook\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.6 Pg:486"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Loss of available energy = -725 Btu/lbm mixture \n",
      "\n",
      " Effectiveness of combustion = 0.409 \n"
     ]
    }
   ],
   "source": [
    "from math import log\n",
    "#Initialization of variables\n",
    "cp=0.25\n",
    "T=3460 #R\n",
    "T0=520 #R\n",
    "dG=1228 #Btu/lbm\n",
    "#calculations\n",
    "hf=cp*(T-T0)-T0*cp*log(T/T0)\n",
    "dC=hf-dG\n",
    "Ec=hf/dG\n",
    "#results\n",
    "print \"Loss of available energy = %d Btu/lbm mixture \"%(dC)\n",
    "print \"\\n Effectiveness of combustion = %.3f \"%(Ec)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.7 Pg:487"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Loss of available energy = -219.3 Btu/lbm mixture \n",
      "\n",
      " Efficiency of cycle = 0.563 \n",
      "\n",
      " Effectiveness of overall cycle = 0.23\n",
      "The answer is a bit different due to rounding off error in textbook\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "cp1=0.25\n",
    "T=3460 #R\n",
    "T0=946.2 #R\n",
    "T00=520 #R\n",
    "dG=1228 #Btu/lbm\n",
    "cp=0.45\n",
    "#calculations\n",
    "dqa=cp1*(T-T0)\n",
    "w=cp*dqa\n",
    "hf=cp1*(T-T00)-T00*cp1*log(T/T00)\n",
    "heat=w-hf\n",
    "eff=w/hf\n",
    "epower=w/dG\n",
    "#results\n",
    "print \"Loss of available energy = %.1f Btu/lbm mixture \"%(heat)\n",
    "print \"\\n Efficiency of cycle = %.3f \"%(eff)\n",
    "print \"\\n Effectiveness of overall cycle = %.2f\"%(epower)\n",
    "print \"The answer is a bit different due to rounding off error in textbook\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Ex:12.8 Pg:487"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "All the values are obtained from Mollier chart,\n",
      "Engine efficiency = 74.3 percent\n",
      "\n",
      " Effectiveness = 80.8 percent\n",
      "\n",
      " Loss of available energy  = -32.6 Btu/lbm\n"
     ]
    }
   ],
   "source": [
    "#Initialization of variables\n",
    "p1=400 #psia\n",
    "t1=600 #F\n",
    "h1=1306.9 #Btu/lbm\n",
    "b1=480.9 #Btu/lbm\n",
    "p2=50 #psia\n",
    "h2=1122 #Btu/lbm\n",
    "h3=1169.5 #Btu/lbm\n",
    "b3=310.9 #Btu/lbm\n",
    "#calculations\n",
    "print \"All the values are obtained from Mollier chart,\"\n",
    "dw13=h1-h3\n",
    "dw12=h1-h2\n",
    "dasf=b3-b1\n",
    "etae=dw13/dw12\n",
    "eta=abs(dw13/dasf)\n",
    "dq=dw13+dasf\n",
    "#results\n",
    "print \"Engine efficiency = %.1f percent\"%(etae*100)\n",
    "print \"\\n Effectiveness = %.1f percent\"%(eta*100)\n",
    "print \"\\n Loss of available energy  = %.1f Btu/lbm\"%(dq)"
   ]
  }
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