{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 9: Propulsion"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Work per kilogram of steam for this isentropic process:  377.5 KJ/Kg-K\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.1\n",
    "'''Steam enters a steam turbine at a pressure of 1 MPa, a temperature of 300\n",
    "◦C, and avelocity of 50 m/s. The steam leaves the turbine at a pressure of 150 kPa and a velocity of\n",
    "200 m/s. Determine the work per kilogram of steam flowing through the turbine, assuming\n",
    "the process to be reversible and adiabatic.\n",
    "Control volume: Turbine.\n",
    "Sketch: Fig. 9.2.\n",
    "Inlet state: Fixed (Fig. 9.2).\n",
    "Exit state: P e , V e known.\n",
    "Process: Steady state, reversible and adiabatic.\n",
    "Model: Steam tables.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "hi = 3051.2 \t\t#initial specific heat of enthalpy of steam in kJ/kg\n",
    "si = 7.1228 \t\t#initial specific entropy of steam in kJ/kg-K\n",
    "Pe = 0.15 \t\t\t#final pressure in MPa\n",
    "se = si \t\t\t#specific entropy in final state in kJ/kg-K\n",
    "sf = 1.4335 \t\t#in kJ/kg-K\n",
    "sfg = 5.7897 \t\t#in kJ/kg-K\n",
    "vi = 50 \t\t\t#velocity with which steam enters turbine in m/s\n",
    "ve = 200 \t\t\t#velocity with which steam leaves the turbine in m/s\n",
    "hf = 467.1 \t\t\t#in kJ/kg\n",
    "hfg = 2226.5 \t\t#in kJ/kg\n",
    "\n",
    "#Calculations:\n",
    "xe = (se-sf)/sfg \t#quality of steam in final state\n",
    "he = hf+xe*hfg \t\t#final specific heat of enthalpy of steam in kJ/kg\n",
    "w = hi+vi**2/(2*1000)-he-ve**2/(2*1000) \t\t#work of steam for isentropic process in kJ/kg\n",
    "\n",
    "#Results\n",
    "print \"Work per kilogram of steam for this isentropic process: \",round(w,1),\"KJ/Kg-K\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Exit velocity of the steam from the nozzle: 737.0 m/s\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.2\n",
    "''' Consider the reversible adiabatic flow of steam through a nozzle. Steam enters the nozzle\n",
    "at 1 MPa and 300◦C, with a velocity of 30 m/s. The pressure of the steam at the nozzle\n",
    "exit is 0.3 MPa. Determine the exit velocity of the steam from the nozzle, assuming a\n",
    "reversible, adiabatic, steady-state process.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "hi = 3051.2 \t\t#initial specific heat of enthalpy in kJ/kg\n",
    "si = 7.1228 \t\t#initial specific entropy in kJ/kg-K\n",
    "Pe = 0.3 \t\t\t#final pressure in MPa\n",
    "he = 2780.2 \t\t#final specific heat of enthalpy in kJ/kg-K\n",
    "Te = 159.1 \t\t\t#final temperature in celsius\n",
    "vi = 30 \t\t\t#velocity with which steam enters the nozzle in m/s\n",
    "\n",
    "#Calculations:\n",
    "se = si \t\t\t#final specific entropy \n",
    "ve = ((2*(hi-he)+(vi**2/1000))*1000)**0.5 \t\t#final velocity of steam with which it exits in m/s\n",
    "\n",
    "#Results\n",
    "print \"Exit velocity of the steam from the nozzle:\",round(ve),'m/s'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Specific work required:  -271.0 KJ/Kg\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.4\n",
    "''' An air compressor in a gas station (see Fig. 9.4) takes in a flow of ambient air at 100 kPa,\n",
    "290 K and compresses it to 1000 kPa in a reversible adiabatic process. We want to know\n",
    "the specific work required and the exit air temperature.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "Cp = 1.004 \t\t#specific heat of air at constant pressure in kJ/kg-K\n",
    "Ti = 290 \t\t#initial temperature in kelvins\n",
    "Pi = 100 \t\t#initial pressure in kPa\n",
    "Pe = 1000 \t\t#final pressure in kPa\n",
    "k = 1.4 \n",
    "\n",
    "#Calculations:\n",
    "Te = Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins\n",
    "we = Cp*(Ti-Te) \t\t#required specific work in kJ/kg\n",
    "\n",
    "#Results:\n",
    "print \"Specific work required: \",round(we),'KJ/Kg'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.5"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Entropy generated in this process:  0.072 KW/K\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.5\n",
    "''' A de-superheater works by injecting liquid water into a flow of superheated steam. With\n",
    "2 kg/s at 300 kPa, 200◦C, steam flowing in, what mass flow rate of liquid water at 20◦C\n",
    "should be added to generate saturated vapor at 300 kPa? We also want to know the rate of\n",
    "entropy generation in the process.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "h1 = 2865.54 \t\t#specific heat of enthalpy at state 1 in kJ/kg\n",
    "h2 = 83.94 \t\t\t#specific heat of enthalpy  at state 2 in kJ/kg\n",
    "h3 = 2725.3 \t\t#specific heat of enthalpy at state 3 in kJ?kg\n",
    "s1 = 7.3115 \t\t#specific entropy at state 1 in kJ/kg-K\n",
    "s2 = 0.2966 \t\t#specific entropy at state 2 in kJ/kg-K\n",
    "s3 = 6.9918 \t\t#specific entropy at state 3in kJ/kg-K\n",
    "m1 = 2 \t\t\t\t#mass flow rate at state 1 in kg/s\n",
    "\n",
    "#Calculations:\n",
    "m2 = m1*(h1-h3)/(h3-h2)\t#mass flow rate at state 2 in kg/s\n",
    "m3 = m1+m2 \t\t\t\t#mass flow rate at state 3 in kg/s\n",
    "Sgen = m3*s3-m1*s1-m2*s2#entropy generation in the process\n",
    "\n",
    "#Results:\n",
    "print \"Entropy generated in this process: \",round(Sgen,3),'KW/K'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.6"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Total amount of work required to fill the tank:  -31.9 KJ\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "from math import log\n",
    "#Example: 9.6\n",
    "'''Assume an air tank has 40 L of 100 kPa air at ambient temperature 17◦C. The adiabatic\n",
    "and reversible compressor is started so that it charges the tank up to a pressure of 1000\n",
    "kPa and then it shuts off. We want to know how hot the air in the tank gets and the total\n",
    "amount of work required to fill the tank.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "T1 = 17+273 \t\t#initial temperature of tank in Kelvins\n",
    "sT1 = 6.83521 \t\t#specific entropy in kJ/kg-K\n",
    "R = 0.287 \t\t\t#gas constant in kJ/kg-K\n",
    "P1 = 100 \t\t\t#initial pressure in kPa\n",
    "P2 = 1000 \t\t\t#final pressure in kPa\n",
    "T2 = 555.7 \t\t\t#from interplotation \n",
    "V1 = 0.04 \t\t\t#volume of tank in m**3\n",
    "V2 = V1 \t\t\t#final volume is equal to initial volume\n",
    "u1 = 207.19 \t\t#initial specific heat of enthalpy in kJ/kg\n",
    "u2 = 401.49 \t\t#final specific heat of enthalpy in kJ/kg\n",
    "hin = 290.43 \t\t#in kJ/kg\n",
    "\n",
    "#Calculations:\n",
    "sT2 = sT1+R*log(P2/P1)\t#specific entropy at temperature T2 in kJ/kg-K\n",
    "m1 = P1*V1/(R*T1) \t#initial mass of air in tank in kg\n",
    "m2 = P2*V2/(R*T2) \t#final mass of air in tank in kg\n",
    "Min = m2-m1 \t\t#in kg\n",
    "W12 = Min*hin+m1*u1-m2*u2 \t\t#work required to fill the tank in kJ\n",
    "\n",
    "#Results:\n",
    "print \"Total amount of work required to fill the tank: \",round(W12,1),'KJ'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.7"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Work required to pump water isentropically:  4.92 KJ/Kg\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.7\n",
    "''' Calculate the work per kilogram to pump water isentropically from 100 kPa, 30◦C to 5 MPa.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "P1 = 100 \t\t\t#initial pressure in kPa\n",
    "P2 = 5000 \t\t\t#final pressure in kPa\n",
    "v = 0.001004 \t\t#specific volume in m**3/kg\n",
    "\n",
    "#Calculations:\n",
    "w = v*(P2-P1) \t\t#work required to pump water isentropically\n",
    "\n",
    "#Results:\n",
    "print \"Work required to pump water isentropically: \",round(w,2),'KJ/Kg'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.8"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Ideal nozzle can generate upto Ve =  20.0 m/s in the exit flow.\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.8\n",
    "'''Consider a nozzle used to spray liquid water. If the line pressure is 300 kPa and the water\n",
    "temperature is 20◦C, how high a velocity can an ideal nozzle generate in the exit flow?'''\n",
    "\n",
    "#Variable Declaration: \n",
    "vf = 0.001002 \t#in m**3/kg\n",
    "Pi = 300 \t\t#Line pressure in kPa\n",
    "Po = 100 \t\t#in kPa\n",
    "\n",
    "#Calculations:\n",
    "v = vf\n",
    "Ve = (2*v*(Pi-Po)*1000)**0.5 \t\t#velocity in the exit flow\n",
    "\n",
    "#Results:\n",
    "print \"Ideal nozzle can generate upto Ve = \",round(Ve),'m/s in the exit flow.'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.9"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Rate of entropy generation for this process:  0.00166 KW/K\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.9\n",
    "'''Saturated vapor R-410a enters the uninsulated compressor of a home central air-\n",
    "conditioning system at 5◦C. The flow rate of refrigerant through the compressor is 0.08\n",
    "kg/s, and the electrical power input is 3 kW. The exit state is 65◦C, 3000 kPa. Any heat transfer from the compressor is with the ambient at 30◦C. Determine the rate of entropy\n",
    "generation for this process.\n",
    "Control volume: Compressor out to ambient T 0 .\n",
    "Inlet state: T i , x i known; state fixed.\n",
    "Exit state: P e , T e known; state fixed.\n",
    "Process: Steady-state, single fluid flow.\n",
    "Model: R-410a tables, B.4.'''\n",
    "#Variable Declaration: \n",
    "hi = 280.6 \t\t\t\t#in kJ/kg\n",
    "he = 307.8 \t\t\t\t#in kJ/kg\n",
    "si = 1.0272 \t\t\t#in kJ/kg\n",
    "se = 1.0140 \t\t\t#in kJ/kg\n",
    "m = 0.08 \t\t\t\t#flow rate of refrigerant in kg/s\n",
    "P = 3 \t\t\t\t\t#electrical power input in kW\n",
    "To = 30 \t\t\t\t#in °C\n",
    "\n",
    "#Calculations:\n",
    "Qcv = m*(he-hi)-P \t\t#in kW\n",
    "Sgen = m*(se-si)-Qcv/(To+273.2)\t#Rate of entropy generation \n",
    "\n",
    "#Results:\n",
    "print \"Rate of entropy generation for this process: \",round(Sgen,5),'KW/K'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.10"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Efficiency of the turbine:  80.9 %\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.10\n",
    "''' A steam turbine receives steam at a pressure of 1 MPa and a temperature of 300\n",
    "◦C. The steam leaves the turbine at a pressure of 15 kPa. The work output of the turbine is measured\n",
    "and is found to be 600 kJ/kg of steam flowing through the turbine. Determine the efficiency\n",
    "of the turbine.\n",
    "Control volume: Turbine.\n",
    "Inlet state: P i , T i known; state fixed.\n",
    "Exit state: P e known.\n",
    "Process: Steady state.\n",
    "Model: Steam tables.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "hi = 3051.2 \t\t#initial specific heat of enthalpy in kJ/kg\n",
    "si = 7.1228 \t\t#initial specific entropy in kJ/kg-K\n",
    "sf = 0.7548 \t\t#in kJ/kg-K\n",
    "sfg = 7.2536 \t\t#in kJ/kg-K\n",
    "hf = 225.9 \t\t#in kJ/kg\n",
    "hfg = 2373.1 \t\t#in kJ/kg\n",
    "wa = 600 \t\t#actual work output of turbine in kJ/kg\n",
    "\n",
    "#Calculations:\n",
    "ses = si \t\t#final specific entropy is same as the initial\n",
    "xes = (si-sf)/sfg \t\t#quality of steam when it leaves the turbine\n",
    "hes = hf+xes*hfg \t\t#final specific heat of enthalpy in kJ/kg\n",
    "ws = hi-hes \t\t#work output of turbine calculated ideally  in kJ/kg\n",
    "nturbine = wa/ws \t\t#efiiciency of turbine \n",
    "\n",
    "#Results:\n",
    "print \"Efficiency of the turbine: \",round(nturbine*100,1),'%'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.11"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Turbine inlet pressure:  2995.0 Kpa\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "from math import e\n",
    "#Example: 9.11\n",
    "''' Air enters a gas turbine at 1600 K and exits at 100 kPa, 830 K.The turbine efficiency is\n",
    "estimated to be 85%. What is the turbine inlet pressure?\n",
    "Control volume: Turbine.\n",
    "Inlet state: T i known.\n",
    "Exit state: P e , T e known; state fixed.\n",
    "Process: Steady state.\n",
    "Model: Air tables, Table A.7.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "hi = 1757.3 \t\t#initial specific heat of enthalpy of air in  kJ/kg\n",
    "si = 8.6905 \t\t#initial specifc entropy of airin kJ/kg-K\n",
    "he = 855.3 \t\t\t#final specific heat of enthalpy of air in kJ/kg\n",
    "n = 0.85 \t\t\t#efficiency of turbine \n",
    "Tes = 683.7 \t\t#final temperature in kelvins from air tables\n",
    "ses = 7.7148 \t\t#in kJ/kg-K\n",
    "R = 0.287 \t\t\t#gas constant in kJ/kg-K\n",
    "\n",
    "#Calculations:\n",
    "w = hi-he \t\t\t#actual work done by turbine in kJ/kg\n",
    "ws = w/n \t\t\t#ideal work done by turbine in kJ/kg\n",
    "hes = hi-ws \t\t#from first law of isentropic process\n",
    "Pi = 100/e**((si-ses)/-R) #turbine inlet pressure in kPa\n",
    "\n",
    "#Results:\n",
    "print \"Turbine inlet pressure: \",round(Pi),'Kpa'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 9.12"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Required work input:  -53.0 KJ/Kg\n",
      "Exit temperature:  352.6 K\n"
     ]
    }
   ],
   "source": [
    "# -*- coding: utf8 -*-\n",
    "from __future__ import division\n",
    "#Example: 9.12\n",
    "'''Air enters an automotive supercharger at 100 kPa, 300 K and is compressed to 150 kPa.\n",
    "The efficiency is 70%. What is the required work input per kilogram of air? What is the\n",
    "exit temperature?\n",
    "Control volume: Supercharger (compressor).\n",
    "Inlet state: P i , T i known; state fixed.\n",
    "Exit state: P e known.\n",
    "Process: Steady state.\n",
    "Model: Ideal gas, 300 K specific heat, Table A.5.'''\n",
    "\n",
    "#Variable Declaration: \n",
    "Pe = 150 \t\t#final pressure of air in kPa\n",
    "Pi = 100 \t\t#initial presure of air in kPa\n",
    "k = 1.4\n",
    "Ti = 300 \t\t#initial temperature of air in kelvis\n",
    "n = 0.7 \t\t#efficiency of automotive supercharger \n",
    "\n",
    "#Calculations:\n",
    "Tes = Ti*(Pe/Pi)**((k-1)/k) \t\t#from second law\n",
    "ws = 1.004*(Ti-Tes) \t\t#from first law of isentropic process\n",
    "w = ws/n \t\t#real work input in kJ/kg\n",
    "Te = Ti-w/1.004 \t\t#temperature at supercharger exit in K\n",
    "\n",
    "#Results:\n",
    "print \"Required work input: \",round(w),'KJ/Kg'\n",
    "print \"Exit temperature: \",round(Te,1),'K'"
   ]
  }
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