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-{
- "cells": [
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# Chapter 7 - Thermodynamics to flow processes"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.1 Page: 256"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 2,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.1 - Page: 256\n",
-      "\n",
-      "\n",
-      "Flow rate is 0.1237 m/s\n",
-      "\n",
-      "Velocity of water at the outlet is 3.938 m/s\n"
-     ]
-    }
-   ],
-   "source": [
-    "from __future__ import division\n",
-    "from math import pi\n",
-    "print \"Example: 7.1 - Page: 256\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "# *****Data******#\n",
-    "d1 = 0.15## [inlet dia, m]\n",
-    "d2 = 0.20## [outlet dia, m]\n",
-    "U1 = 7## [inlet velocity, m/s]\n",
-    "#****************#\n",
-    "\n",
-    "# From Fig. 7.2 (Pg 256)\n",
-    "# At the inlet:\n",
-    "A1 = (pi/4)*d1**2## [square m]\n",
-    "# At the outlet:\n",
-    "A2 = (pi/4)*d2**2## [square m]\n",
-    "Q = A1*U1## [cubic m/s]\n",
-    "print \"Flow rate is %.4f m/s\\n\"%(Q)#\n",
-    "# Using Continuity Eqn.\n",
-    "# density1*U1*A1 = Density2*U2*A2\n",
-    "# For water: Density1 = Density2. Therefore:\n",
-    "U2 = A1*U1/A2#\n",
-    "print \"Velocity of water at the outlet is %.3f m/s\"%(U2)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.2 Page: 257"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 3,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.2 - Page: 257\n",
-      "\n",
-      "\n",
-      "Discharge through the 10 cm pipe is 0.0942 cubic m/sec\n",
-      "\n",
-      "Average velocity in the 15 cm pipe is 6.38 m/s\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.2 - Page: 257\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "d1 = 0.2## [m]\n",
-    "d2 = 0.15## [m]\n",
-    "d3 = 0.1## [m]\n",
-    "U1 = 3## [m/s]\n",
-    "U2 = 2.5## [m/s]\n",
-    "#**************#\n",
-    "\n",
-    "# From Fig. 7.3 (Pg: 257)\n",
-    "# For pipe I:\n",
-    "A1 = (pi/4)*d1**2## [square m]\n",
-    "Q1 = A1*U1## [cubic m/s]\n",
-    "# For pipe II:\n",
-    "A2 = (pi/4)*d2**2## [square m]\n",
-    "Q2 = A2*U2## [cubic m/s]\n",
-    "# For pipe III:\n",
-    "A3 = (pi/4)*d3**2## [square m]\n",
-    "# From continuity Eqn.:\n",
-    "Q3 = Q1 - Q2## [cubic m/s]\n",
-    "U3 = Q3/A3## [m/s]\n",
-    "print \"Discharge through the 10 cm pipe is %.4f cubic m/sec\\n\"%(Q1)#\n",
-    "print \"Average velocity in the 15 cm pipe is %.2f m/s\"%(U3)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.3 Page: 262"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 4,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.3 - Page: 262\n",
-      "\n",
-      "\n",
-      "Pressure at section 2 is 5.20 bar\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.3 - Page: 262\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "d1 = 0.3## [m]\n",
-    "d2 = 0715##[m]\n",
-    "Q = 40/1000## [cubic m/s]\n",
-    "Z1 = 8## [m]\n",
-    "Z2 = 6## [m]\n",
-    "P1 = 5*10**5## [Pa]\n",
-    "density = 1000## [kg/cubic m]\n",
-    "g = 9.81## [m/square s]\n",
-    "#*************#\n",
-    "\n",
-    "# From Fig. 7.3 (Pg: 262)\n",
-    "A1 = (pi/4)*d1**2## [square m]\n",
-    "A2 = (pi/4)*d2**2## [square m]\n",
-    "U1 = Q/A1## [m/s]\n",
-    "U2 = Q/A2## [m/s]\n",
-    "# Applying Bernoulli's equations at sections 1 & 2:\n",
-    "P2 =  ((U1**2/(2*g) + Z1 + P1/(density*g)) - (U2**2/(2*g) + Z2))*(density*g)## [Pa]\n",
-    "print \"Pressure at section 2 is %.2f bar\"%(P2/10**5)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.4 Page: 268"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 5,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.4 - Page: 268\n",
-      "\n",
-      "\n",
-      "Power Requirement of the compressor is 3.79 kW\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.4 - Page: 268\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "P1 = 100## [kPa]\n",
-    "T1 = 320## [K]\n",
-    "P2 = 600## [kPa]\n",
-    "T2 = 430## [K]\n",
-    "m_dot = 0.03## [kg/s]\n",
-    "Qout = 15## [kJ/kg]\n",
-    "#*************#\n",
-    "\n",
-    "# The energy balance around the compressor:\n",
-    "# dE_System/dt = Ein - Eout\n",
-    "# Since it is a steady state process:\n",
-    "# dE_Sysytem/dt = 0\n",
-    "# Ein = Eout\n",
-    "# Win + m_dot*H1 = Qout + m_dot*H2\n",
-    "# Since, Qout = Qout/m\n",
-    "# Win = m_dot*(Qout + (H2 - H1))\n",
-    "# From enthalpy chart of air:\n",
-    "H1 = 320.20## [Enthalpy of air at 320 K, kJ/kg]\n",
-    "H2 = 431.43## [Enthalpy of air at 430 K, kJ/kg]\n",
-    "Win = m_dot*(Qout + (H2 - H1))## [kW]\n",
-    "print \"Power Requirement of the compressor is %.2f kW\"%(Win)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.5 Page: 269"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 1,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.5 - Page: 269\n",
-      "\n",
-      "\n",
-      "Work done by reversible adiabatic compression when gama = 1.4 is -96.53 J/g\n",
-      "\n",
-      "Work done by isothermal compression is -76.03 J/g\n",
-      "\n",
-      "Work done in single stage compression is -19.49 kW\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import log\n",
-    "print \"Example: 7.5 - Page: 269\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "P1 = 100## [kPa]\n",
-    "T1 = 250## [K]\n",
-    "Q = 0.1## [cubic m/s]\n",
-    "P2 = 500## [kPa]\n",
-    "M = 44## [g/mol]\n",
-    "R = 8.314## [J/mol K]\n",
-    "#****************#\n",
-    "\n",
-    "# Solution (a)\n",
-    "# Work done by reversible adiabatic compression, gama = 1.4#\n",
-    "gama = 1.4#\n",
-    "T2 = T1*(P2/P1)**((gama - 1)/gama)## [K]\n",
-    "Wad = (gama*R/(gama - 1))*(T1 - T2)## [J/mol]\n",
-    "Wad = Wad/M## [J/g]\n",
-    "print \"Work done by reversible adiabatic compression when gama = 1.4 is %.2f J/g\\n\"%(Wad)#\n",
-    "\n",
-    "# Solution (b)\n",
-    "# Work done by isothermal compression:\n",
-    "Wiso = - (R*T1)*log(P2/P1)## [J/mol]\n",
-    "Wiso = Wiso/M## [J/g]\n",
-    "print \"Work done by isothermal compression is %.2f J/g\\n\"%(Wiso)#\n",
-    "\n",
-    "# Solution (c)\n",
-    "# Work done in single stage compression, gama = 1.3:\n",
-    "gama = 1.3#\n",
-    "V1 = Q## [cubic m]\n",
-    "Wsingle_stage = (gama*P1*V1/(gama - 1))*(1-(P2/P1)**((gama - 1)/gama))## [kW]\n",
-    "print \"Work done in single stage compression is %.2f kW\"%(Wsingle_stage)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.7 Page: 274"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 7,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.7 - Page: 274\n",
-      "\n",
-      "\n",
-      "The mass flow rate of water is 448.70 kg/min\n",
-      "\n",
-      "The rate of heat transfer is 18777.00 kJ/min\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.7 - Page: 274\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "T_steam1 = 50## [OC]\n",
-    "T_steam2 = 30## [OC]\n",
-    "msteam_dot = 10## [kg/min]\n",
-    "T_water1 = 15## [OC]\n",
-    "T_water2 = 25## [OC]\n",
-    "#***************#\n",
-    "\n",
-    "# Solution (a)\n",
-    "# From the Stem Table:\n",
-    "H1 = 2645.9## [kJ/kg, At 50 OC]\n",
-    "H2 = 768.2## [kJ/kg, At 30 OC]\n",
-    "H3 = 62.982## [kJ/kg, At 15 OC]\n",
-    "H4 = 104.83## [kJ/kg, At 25 OC]\n",
-    "# The mass & Energy balance of the above flow gives:\n",
-    "mwater_dot = msteam_dot*(H1 - H2)/(H4 - H3)## [kg/min]\n",
-    "print \"The mass flow rate of water is %.2f kg/min\\n\"%(mwater_dot)#\n",
-    "\n",
-    "# Solution (b)\n",
-    "Qdot = mwater_dot*(H4 - H3)## [kJ/min]\n",
-    "print \"The rate of heat transfer is %.2f kJ/min\"%(Qdot)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.8 Page: 279"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 9,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.8 - Page: 279\n",
-      "\n",
-      "\n",
-      "Outlrt velocity is 603.3 m/s\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "from math import sqrt\n",
-    "print \"Example: 7.8 - Page: 279\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "P1 = 500## [kPa]\n",
-    "T1 = 623## [K]\n",
-    "mdot = 12## [kg/s]\n",
-    "P2 = 500## [kPa]\n",
-    "T2 = 523## [K]\n",
-    "Qdot = -120## [kW]\n",
-    "H1 = 3168## [kJ/kg]\n",
-    "H2 = 2976## [kJ/kg]\n",
-    "#************#\n",
-    "\n",
-    "Q = Qdot/mdot## [kJ/kg]\n",
-    "# By energy balance:\n",
-    "# (deltaU**2/2) + g*deltaZ + deltaH = Q - Ws\n",
-    "# Considering negligible change in P.E., deltaZ = 0 & Ws = 0.\n",
-    "# (U2**2 - U1**2)/2 + deltaH = Q\n",
-    "deltaH = H2 - H1## [kJ/kg]\n",
-    "# Neglecting inlet velocity.\n",
-    "U2 = sqrt(2*(Q - deltaH)*1000)## [m/s]\n",
-    "print \"Outlrt velocity is %.1f m/s\\n\"%(U2)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.9 Page: 279"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 10,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.9 - Page: 279\n",
-      "\n",
-      "\n",
-      "Critical Ratio is 0.53\n",
-      "\n",
-      "The discharge velocity is 1174.46 m/s\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.9 - Page: 279\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "Pin = 1000## [kPa]\n",
-    "Tin = 600## [K]\n",
-    "Uin = 50## [m/s]\n",
-    "gama = 1.4#\n",
-    "M = 17## [g/mol]\n",
-    "R = 8314## [kJ/mol K]\n",
-    "MachNumber = 2#\n",
-    "#***************#\n",
-    "\n",
-    "# Solution (i)\n",
-    "# Using Eqn. (7.36):\n",
-    "Critical_Ratio = (2/(gama + 1))**(gama/(gama - 1))#\n",
-    "print \"Critical Ratio is %.2f\\n\"%(Critical_Ratio)#\n",
-    "\n",
-    "# Solution (ii)\n",
-    "PV_in = R*Tin/M## [square m]\n",
-    "Uthroat = sqrt(Uin**2 + (2*gama*PV_in/(gama - 1))*(1-(Critical_Ratio)**((gama - 1)/gama)))## [m/s]\n",
-    "Uout = MachNumber*Uthroat## [m/s]\n",
-    "print \"The discharge velocity is %.2f m/s\"%(Uout)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.10 Page: 280"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 11,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.10 - Page: 280\n",
-      "\n",
-      "\n",
-      "(a) Final Velocity is 901.11 m/s\n",
-      "\n",
-      "(b) Final Velocity is 1553.06 m/s\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.10 - Page: 280\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "P1 = 800## [kPa]\n",
-    "T1 = 773## [K]\n",
-    "H1 = 3480## [kJ/kg]\n",
-    "P2 = 100## [kPa]\n",
-    "T2 = 573## [K]\n",
-    "H2 = 3074## [kJ/kg]\n",
-    "#***************#\n",
-    "\n",
-    "# Solution (a)\n",
-    "# Velocity of the fluid exiting the nozzle:\n",
-    "# U2 = sqrt(U1**2 + 2*(H1 - H2))\n",
-    "# Neglecting initial velocity:\n",
-    "U2 = sqrt(2*(H1 - H2)*1000)## [m/s]\n",
-    "print \"(a) Final Velocity is %.2f m/s\\n\"%(U2)#\n",
-    "\n",
-    "# Solution (b)\n",
-    "U1 = 40## [m/s]\n",
-    "U2 = sqrt((U1**2 + 2*(H1 - H2))*1000)## [m/s]\n",
-    "print \"(b) Final Velocity is %.2f m/s\\n\"%(U2)#"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Example: 7.11 Page: 281"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 12,
-   "metadata": {
-    "collapsed": false
-   },
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "Example: 7.11 - Page: 281\n",
-      "\n",
-      "\n",
-      "Mass flow rate of the steam is 17.50 kg/s\n",
-      "\n",
-      "The temperature of the steam leaving the outlet is 393.38 K\n",
-      "\n",
-      "Area of diffuser outlet is 0.28 square m\n",
-      "\n"
-     ]
-    }
-   ],
-   "source": [
-    "print \"Example: 7.11 - Page: 281\\n\\n\"\n",
-    "\n",
-    "# Solution\n",
-    "\n",
-    "#*****Data******#\n",
-    "P1 = 100## [kPa]\n",
-    "T1 = 200## [OC]\n",
-    "U1 = 190## [m/s]\n",
-    "A1 = 2000/10**4## [square m]\n",
-    "U2 = 70## [m/s]\n",
-    "P2 = 200## [kPa]\n",
-    "Qdot = 100## [kW]\n",
-    "V1 = 2.172## [cubic m/kg]\n",
-    "H1 = 2875.3## [kJ/kg]\n",
-    "#***************#\n",
-    "\n",
-    "# Solution (a)\n",
-    "mdot = U1*A1/V1## [kg/s]\n",
-    "print \"Mass flow rate of the steam is %.2f kg/s\\n\"%(mdot)#\n",
-    "\n",
-    "# Solution (b)\n",
-    "# Amount of heat transferred to the surrounding per unit steam:\n",
-    "Q = Qdot/mdot## [kJ/kg]\n",
-    "# The Enthalpy at the diffuser outlet can be estimated as:\n",
-    "H2 = Q + H1 + (U1**2 - U2**2)/2## [kJ/kg]\n",
-    "# From the steam table:\n",
-    "T2 = 393.38## [K]\n",
-    "V2 = 1.123## [cubic m/kg]\n",
-    "print \"The temperature of the steam leaving the outlet is %.2f K\\n\"%(T2)#\n",
-    "\n",
-    "# Solution (c)\n",
-    "A2 = V2*mdot/U2## [square m]\n",
-    "print \"Area of diffuser outlet is %.2f square m\\n\"%(A2)#"
-   ]
-  }
- ],
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-  "kernelspec": {
-   "display_name": "Python 2",
-   "language": "python",
-   "name": "python2"
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-   "codemirror_mode": {
-    "name": "ipython",
-    "version": 2
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-   "nbconvert_exporter": "python",
-   "pygments_lexer": "ipython2",
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