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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Appendix B"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 3"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Mass of vapour: 2.77 Kg\n"
+ ]
+ }
+ ],
+ "source": [
+ "# -*- coding: utf8 -*-\n",
+ "from __future__ import division\n",
+ "#Example: 13.3\n",
+ "'''Consider 100 m 3 of an air–water vapor mixture at 0.1 MPa, 35◦C, and 70% relative\n",
+ "humidity. Calculate the humidity ratio, dew point, mass of air, and mass of vapor.'''\n",
+ "\n",
+ "#Variable Declaration: \n",
+ "r = 0.70\t\t\t#relative humidity\n",
+ "Pg = 5.628\t\t\t#saturation pressure in kPa\n",
+ "P = 100\t\t\t\t#net pressure kPa \n",
+ "V = 100\t\t\t\t#volume in m**3\n",
+ "Ra = 0.287\t\t\t#gas constant for water vapour\n",
+ "T = 308.2\t\t\t#Temperature in K\n",
+ "\n",
+ "#Calculations:\n",
+ "Pv = r*Pg\t\t\t#vapour pressure in kPa\n",
+ "Pa = P-Pv\t\t\t#Partial pressure of air\n",
+ "w = 0.622*Pv/Pa\t\t#humidity ratio formula\n",
+ "ma = Pa*V/(Ra*T)\t#mass in kg\n",
+ "mv = w*ma\t\t\t#mass of vapour\n",
+ "\n",
+ "#Results:\n",
+ "print 'Mass of vapour: ', round(mv,2),'Kg'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 4"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Mass of vapour condense 2.175 Kg\n"
+ ]
+ }
+ ],
+ "source": [
+ "# -*- coding: utf8 -*-\n",
+ "from __future__ import division\n",
+ "#Example: 13.4\n",
+ "'''Calculate the amount of water vapor condensed if the mixture of Example 13.3 is cooled\n",
+ "to 5◦C in a constant-pressure process.'''\n",
+ "\n",
+ "#Variable Declaration: \n",
+ "w1 = 0.0255\t\t#w1 = w, humidity ratio at initial temperature\n",
+ "ma = 108.6\t\t#mass of air in kg\n",
+ "P = 100\t\t\t#kPa net pressure\n",
+ "Pg2 = 0.8721\n",
+ "\n",
+ "#Calculations:\n",
+ "Pv2 = Pg2\n",
+ "w2 = 0.622*Pv2/(P-Pg2)\n",
+ "mc = ma*(w1-w2)\n",
+ "\n",
+ "#Results:\n",
+ "print 'Mass of vapour condense',round(mc,3) ,'Kg'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Heat transferred per unit mass: -41.65 kJ/kg of dry air\n"
+ ]
+ }
+ ],
+ "source": [
+ "# -*- coding: utf8 -*-\n",
+ "from __future__ import division\n",
+ "#Example: 13.5\n",
+ "''' An air-conditioning unit is shown in Fig. 13.5, with pressure, temperature, and relative\n",
+ "humidity data. Calculate the heat transfer per kilogram of dry air, assuming that changes\n",
+ "in kinetic energy are negligible.'''\n",
+ "\n",
+ "#Variable Declaration: \n",
+ "r1 = 0.80\t\t\t\t#realtive humidity at state 1\n",
+ "Pg1 = 4.246\t\t\t\t#saturation pressure of vapour in kPa\n",
+ "P1 = 105\t\t\t\t#net pressure at state 1 in kPa\n",
+ "P2 = 100\t\t\t\t#net pressure at state 2 in kPa\n",
+ "r2 = 0.95\t\t\t\t#relative humidity at state 2\n",
+ "Pg2 = 1.7051\t\t\t#saturation pressure of vapour in kPa\n",
+ "T1 = 30\t\t\t\t\t#C\n",
+ "T2 = 15\t\t\t\t\t#C\n",
+ "Cp = 1.004\t\t\t\t#specific heat of water vapour in kJ/kg\n",
+ "hv2 = 2528.9\t\t\t#enthalpy of vapourisation of vapour in kJ/kg\n",
+ "hv1 = 2556.3\t\t\t#enthalpy of vapourisation of vapour in kJ/kg\n",
+ "hl2 = 62.99\t\t\t\n",
+ "\n",
+ "#Calculations:\n",
+ "Pv1 = r1*Pg1\t\t\t#partial pressure of vapour in kPa\n",
+ "w1 = 0.622*Pv1/(P1-Pv1)\t#humidity ratio at state 1\n",
+ "Pv2 = r2*Pg2\t\t\t#partial pressure of vapour in kPa\n",
+ "w2 = 0.622*Pv2/(P2-Pv2)\t#humidity ratio at state 2\n",
+ "q = Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2)\t\t#kJ/kg\n",
+ "\n",
+ "#Results:\n",
+ "print 'Heat transferred per unit mass: ',round(q,2),' kJ/kg of dry air'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Heat transferred -339.1 kJ\n"
+ ]
+ }
+ ],
+ "source": [
+ "# -*- coding: utf8 -*-\n",
+ "from __future__ import division\n",
+ "#Example: 13.6\n",
+ "''' A tank has a volume of 0.5 m 3 and contains nitrogen and water vapor. The temperature\n",
+ "of the mixture is 50◦C, and the total pressure is 2 MPa. The partial pressure of the water\n",
+ "vapor is 5 kPa. Calculate the heat transfer when the contents of the tank are cooled to\n",
+ "10◦C.'''\n",
+ "\n",
+ "#Variable Declaration: \n",
+ "Pn2 = 1995\t\t\t\t#Pressure of nitrogen in kPa\n",
+ "V = 0.5\t\t\t\t\t#Volume in m**3\n",
+ "Rn2 = 0.2968\t\t\t#Gas constant for nitrogen in kJ/kg.K\n",
+ "Rv = 0.4615\t\t\t\t#gas constant for vapour\n",
+ "T1 = 323.2\t\t\t\t#Temperature in K\n",
+ "T2 = 283.2\t\t\t\t#Temperature in K\n",
+ "Pv1 = 5\t\t\t\t\t#Pressure of water vapour in kPa at state 1\n",
+ "Pv2 = 1.2276\t\t\t#Pressure of water vapour in kPa at state 2\n",
+ "uv1 = 2443.1\t\t\t#specific internal energy of vapour in kJ/kg at state 1\n",
+ "uv2 = 2389.2\t\t\t#specific internal energy of vapour in kJ/kg at state 2\n",
+ "ul2 = 42.0\t\t\t\t#specific internal energy of liquid water in kJ/kg\n",
+ "Cv = 0.745\t\t\t\t#specific heat at constant volume in kJ/kg.K\n",
+ "\n",
+ "#Calculations:\n",
+ "mn2 = Pn2*V/(Rn2*T1)\t#mass of nitrogen\n",
+ "mv1 = Pv1*V/(Rv*T1)\t\t#mass of vapour in kg\n",
+ "mv2 = Pv2*V/(Rv*T2)\t\t#mass of vapour in kg\n",
+ "ml2 = mv1-mv2\t\t\t#mass of liquid condensed n kg\n",
+ "Q = mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1\n",
+ "\n",
+ "#Results:\n",
+ "print 'Heat transferred',round(Q,1),'kJ'"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Example 7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Relative humidity: 0.0107\n",
+ "Humidity ratio: 0.399\n"
+ ]
+ }
+ ],
+ "source": [
+ "# -*- coding: utf8 -*-\n",
+ "from __future__ import division\n",
+ "#Example: 13.7\n",
+ "'''The pressure of the mixture entering and leaving the adiabatic saturator is 0.1 MPa, the\n",
+ "entering temperature is 30◦C, and the temperature leaving is 20◦C, which is the adiabatic\n",
+ "saturation temperature. Calculate the humidity ratio and relative humidity of the air–water\n",
+ "vapor mixture entering.'''\n",
+ "\n",
+ "#Variable Declaration: \n",
+ "P = 100\t\t\t\t#net pressure n kPa \n",
+ "Pg2 = 2.339\t\t\t#saturation pressure of vapour in kPa\n",
+ "Cpa = 1.004\t\t\t#specific heat n kJ/kg/K\n",
+ "T2 = 20\t\t\t\t#final temp in C\n",
+ "T1 = 30\t\t\t\t#initial temp in C\n",
+ "Hfg2 = 2454.1\t\t#specific heat difference at state 2 in kJ/kg\n",
+ "hv1 = 2556.3\t\t#enthalpy of water vapour at state 1 in kJ/kg\n",
+ "hl2 = 83.96\t\t\t#enthalpy of liquid water in kJ/kg\n",
+ "Pg1 = 4.246\t\t\t#saturation pressure at state 1 in kPa\n",
+ "\n",
+ "#Calculations:\n",
+ "Pv2 = Pg2\t\t\t#partial pressure of vapour\n",
+ "w2 = 0.622*Pv2/(P-Pg2)\n",
+ "w1 = (Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2)\n",
+ "Pv1 = 100*w1/(0.622+w1)\n",
+ "r = Pv1/Pg1\t\t\t#humidity ratio\n",
+ "\n",
+ "#Results:\n",
+ "print 'Relative humidity: ',round(w1,4)\n",
+ "print 'Humidity ratio: ',round(r,3)"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 2",
+ "language": "python",
+ "name": "python2"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 2
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython2",
+ "version": "2.7.6"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}