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-{
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "# Chapter 1 - General Introduction"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 1: pg 11"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 1,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.1\n",
- "The Work done is (MJ) = 0.98\n"
- ]
- }
- ],
- "source": [
- "#pg 11\n",
- "#calculate the work done\n",
- "print 'Example 1.1'\n",
- "\n",
- "# Given values\n",
- "P = 700.; #pressure,[kN/m**2]\n",
- "V1 = .28; #initial volume,[m**3]\n",
- "V2 = 1.68; #final volume,[m**3]\n",
- "\n",
- "#solution\n",
- "\n",
- "W = P*(V2-V1);# # Formula for work done at constant pressure is, [kJ]\n",
- "\n",
- "#results\n",
- "print 'The Work done is (MJ) = ',W*10**-3\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 2: pg 13"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 2,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.2\n",
- "The new volume of the gas is (m^3) = 0.0355\n"
- ]
- }
- ],
- "source": [
- "#pg 13\n",
- "#calculate the new volume\n",
- "print 'Example 1.2'\n",
- "\n",
- "#Given values\n",
- "P1 = 138.; # initial pressure,[kN/m**2]\n",
- "V1 = .112; #initial volume,[m**3]\n",
- "P2 = 690; # final pressure,[kN/m**2]\n",
- "Gama=1.4; # heat capacity ratio\n",
- "\n",
- "# solution\n",
- "\n",
- "# since gas is following, PV**1.4=constant,hence\n",
- "\n",
- "V2 =V1*(P1/P2)**(1/Gama); # final volume, [m**3] \n",
- "\n",
- "#results\n",
- "print 'The new volume of the gas is (m^3) = ',round(V2,4)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 3: pg 15"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 3,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.3\n",
- "Final Volume (m^3) = 0.077\n",
- "The Work done by gas during expansion is (kJ) = 37.2\n"
- ]
- }
- ],
- "source": [
- "#pg 15\n",
- "#calculate the work done by gas\n",
- "print 'Example 1.3'\n",
- "\n",
- "# Given values\n",
- "P1 = 2070; # initial pressure, [kN/m^2]\n",
- "V1 = .014; # initial volume, [m^3]\n",
- "P2 = 207.; # final pressure, [kN/m^2]\n",
- "n=1.35; # polytropic index\n",
- "\n",
- "# solution\n",
- "\n",
- "# since gas is following PV^n=constant\n",
- "# hence \n",
- "\n",
- "V2 = V1*(P1/P2)**(1/n); # final volume, [m^3]\n",
- "\n",
- "# calculation of workdone\n",
- "\n",
- "W=(P1*V1-P2*V2)/(1.35-1); # using work done formula for polytropic process, [kJ]\n",
- "\n",
- "#results\n",
- "print 'Final Volume (m^3) = ',round(V2,3)\n",
- "print 'The Work done by gas during expansion is (kJ) = ',round(W,1)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 4: pg 17"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 4,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.4\n",
- " The final pressure (kN/m^2) = 800.0\n",
- " Work done on the gas (kJ) = -11.64\n"
- ]
- }
- ],
- "source": [
- "#pg 17\n",
- "#calculate the final pressure and work done\n",
- "print 'Example 1.4'\n",
- "import math\n",
- "\n",
- "# Given values\n",
- "P1 = 100; # initial pressure, [kN/m^2]\n",
- "V1 = .056; # initial volume, [m^3]\n",
- "V2 = .007; # final volume, [m^3]\n",
- "\n",
- "# To know P2\n",
- "# since process is hyperbolic so, PV=constant\n",
- "# hence\n",
- "\n",
- "P2 = P1*V1/V2; # final pressure, [kN/m^2]\n",
- "\n",
- "# calculation of workdone\n",
- "W = P1*V1*math.log(V2/V1); # formula for work done in this process, [kJ]\n",
- "\n",
- "#results\n",
- "print ' The final pressure (kN/m^2) = ',P2\n",
- "print ' Work done on the gas (kJ) = ',round(W,2)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 5: pg 21"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 5,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.5\n",
- "The heat required (kJ) = 191.25\n"
- ]
- }
- ],
- "source": [
- "#pg 21\n",
- "#calculate the heat required\n",
- "print 'Example 1.5'\n",
- "\n",
- "# Given values\n",
- "m = 5.; # mass, [kg]\n",
- "t1 = 15.; # inital temperature, [C]\n",
- "t2 = 100.; # final temperature, [C]\n",
- "c = 450.; # specific heat capacity, [J/kg K]\n",
- "\n",
- "# solution\n",
- "\n",
- "# using heat transfer equation,[1]\n",
- "Q = m*c*(t2-t1); # [J]\n",
- "#results\n",
- "print 'The heat required (kJ) = ',round(Q*10**-3,2)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 6: pg 22"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 7,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.6\n",
- "Required heat transfer to accomplish the change (kJ) = 1814.4\n"
- ]
- }
- ],
- "source": [
- "#pg 22\n",
- "print 'Example 1.6'\n",
- "\n",
- "#Calculate the required heat transfer \n",
- "# Given values\n",
- "m_cop = 2.; # mass of copper vessel, [kg]\n",
- "m_wat = 6.; # mass of water, [kg]\n",
- "c_wat = 4.19; # specific heat capacity of water, [kJ/kg K]\n",
- "\n",
- "t1 = 20.; # initial temperature, [C]\n",
- "t2 = 90.; # final temperature, [C]\n",
- "\n",
- "# From the table of average specific heat capacities\n",
- "c_cop = .390; # specific heat capacity of copper,[kJ/kg k]\n",
- "\n",
- "# solution\n",
- "Q_cop = m_cop*c_cop*(t2-t1); # heat required by copper vessel, [kJ]\n",
- "\n",
- "Q_wat = m_wat*c_wat*(t2-t1); # heat required by water, [kJ]\n",
- "\n",
- "# since there is no heat loss,so total heat transfer is sum of both\n",
- "Q_total = Q_cop+Q_wat ; # [kJ]\n",
- "\n",
- "#results\n",
- "print 'Required heat transfer to accomplish the change (kJ) = ',Q_total\n",
- "\n",
- "#End"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 7: pg 22"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 8,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.7\n",
- "The final temperature is (C) = 56.9\n"
- ]
- }
- ],
- "source": [
- "#pg 22\n",
- "print('Example 1.7');\n",
- "#calculate the final temperature\n",
- "\n",
- "# Given values\n",
- "m = 10.; # mass of iron casting, [kg]\n",
- "t1 = 200.; # initial temperature, [C]\n",
- "Q = -715.5; # [kJ], since heat is lost in this process\n",
- "\n",
- "# From the table of average specific heat capacities\n",
- "c = .50; # specific heat capacity of casting iron, [kJ/kg K]\n",
- "\n",
- "# solution\n",
- "# using heat equation\n",
- "# Q = m*c*(t2-t1)\n",
- "\n",
- "t2 = t1+Q/(m*c); # [C]\n",
- "\n",
- "#results\n",
- "print 'The final temperature is (C) = ',t2\n",
- "\n",
- "# End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 8: pg 23"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 9,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.8\n",
- "The specific heat capacity of the liquid is (kJ/kg K) = 2.1\n"
- ]
- }
- ],
- "source": [
- "#pg 23\n",
- "#calculate the specific heat capacity\n",
- "print('Example 1.8');\n",
- "# Given values\n",
- "m = 4.; # mass of the liquid, [kg]\n",
- "t1 = 15.; # initial temperature, [C]\n",
- "t2 = 100.; # final temperature, [C]\n",
- "Q = 714.; # [kJ],required heat to accomplish this change\n",
- "\n",
- "# solution\n",
- "# using heat equation\n",
- "# Q=m*c*(t2-t1)\n",
- "\n",
- "# calculation of c\n",
- "c=Q/(m*(t2-t1)); # heat capacity, [kJ/kg K] \n",
- "\n",
- "#results\n",
- "print 'The specific heat capacity of the liquid is (kJ/kg K) = ',c\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 9: pg 27"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 10,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.9\n",
- "The power output of the engine is (kJ) = 48.7\n",
- "The energy rejected by the engine is (MJ/min) = 11.7\n"
- ]
- }
- ],
- "source": [
- "#pg 27\n",
- "#calculate the energy rejected by the engine\n",
- "print('Example 1.9');\n",
- "\n",
- "\n",
- "# Given values\n",
- "m_dot = 20.4; # mass flowrate of petrol, [kg/h]\n",
- "c = 43.; # calorific value of petrol, [MJ/kg]\n",
- "n = .2; # Thermal efficiency of engine\n",
- "\n",
- "# solution\n",
- "m_dot = 20.4/3600; # [kg/s]\n",
- "c = 43*10**6; # [J/kg]\n",
- "\n",
- "# power output\n",
- "P_out = n*m_dot*c; # [W]\n",
- "\n",
- " \n",
- "# power rejected\n",
- "\n",
- "P_rej = m_dot*c*(1-n); # [W]\n",
- "P_rej = P_rej*60*10**-6; # [MJ/min]\n",
- "\n",
- "#results\n",
- "print 'The power output of the engine is (kJ) = ',round(P_out*10**-3,1)\n",
- "print 'The energy rejected by the engine is (MJ/min) = ',round(P_rej,1)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 10: pg 28"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 11,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.10\n",
- "Thermal efficiency of the plant = 0.173\n"
- ]
- }
- ],
- "source": [
- "#pg 28\n",
- "print('Example 1.10');\n",
- "#calculate the thermal efficiency\n",
- "\n",
- "\n",
- "# Given values\n",
- "m_dot = 3.045; # use of coal, [tonne/h]\n",
- "c = 28; # calorific value of the coal, [MJ/kg]\n",
- "P_out = 4.1; # output of turbine, [MW]\n",
- "\n",
- "# solution\n",
- "m_dot = m_dot*10**3/3600; # [kg/s]\n",
- "\n",
- "P_in = m_dot*c; # power input by coal, [MW]\n",
- "\n",
- "n = P_out/P_in; # thermal efficiency formula\n",
- "\n",
- "#results\n",
- "print 'Thermal efficiency of the plant = ',round(n,3)\n",
- "\n",
- "#End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 11: pg 29"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 13,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.11\n",
- "The power output of the engine (kW) = 12.5\n"
- ]
- }
- ],
- "source": [
- "#pg 29\n",
- "#calculate the power output of the engine\n",
- "print('Example 1.11');\n",
- "\n",
- "\n",
- "# Given values\n",
- "v = 50.; # speed, [km/h]\n",
- "F = 900.; # Resistance to the motion of a car\n",
- "\n",
- "# solution\n",
- "v = v*10**3/3600; # [m/s]\n",
- "Power = F*v; # Power formula, [W]\n",
- "\n",
- "print 'The power output of the engine (kW) = ',Power*10**-3\n",
- " \n",
- "# End"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 12: pg 31"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 14,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.12\n",
- "The power output from the engine (kW) = 15.79\n"
- ]
- }
- ],
- "source": [
- "#pg 31\n",
- "#calculate the power output from the engine\n",
- "\n",
- "print('Example 1.12');\n",
- "\n",
- "# Given values\n",
- "V = 230.; # volatage, [volts]\n",
- "I = 60.; # current, [amps]\n",
- "n_gen = .95; # efficiency of generator\n",
- "n_eng = .92; # efficiency of engine\n",
- "\n",
- "# solution\n",
- "\n",
- "P_gen = V*I; # Power delivered by generator, [W]\n",
- "P_gen=P_gen*10**-3; # [kW]\n",
- "\n",
- "P_in_eng=P_gen/n_gen;#Power input from engine,[kW]\n",
- "\n",
- "P_out_eng=P_in_eng/n_eng;#Power output from engine,[kW]\n",
- "\n",
- "#results\n",
- "print 'The power output from the engine (kW) = ',round(P_out_eng,2)\n",
- "\n",
- "# End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 13: pg 32"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 15,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.13\n",
- "The current taken by heater (amps) = 17.4\n"
- ]
- }
- ],
- "source": [
- "#pg 32\n",
- "#calculate the current taken by heater\n",
- "print('Example 1.13');\n",
- "\n",
- "\n",
- "\n",
- "# Given values\n",
- "V = 230.; # Voltage, [volts]\n",
- "W = 4.; # Power of heater, [kW]\n",
- "\n",
- "# solution\n",
- "\n",
- "# using equation P=VI\n",
- "I = W/V; # current, [K amps]\n",
- "#results\n",
- "print 'The current taken by heater (amps) = ',round(I*10**3,1)\n",
- "\n",
- "# End\n"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "## Example 14: pg 32"
- ]
- },
- {
- "cell_type": "code",
- "execution_count": 16,
- "metadata": {
- "collapsed": false
- },
- "outputs": [
- {
- "name": "stdout",
- "output_type": "stream",
- "text": [
- "Example 1.14\n",
- "Mass of coal burnt by the power station in 1 hour (tonne) = 218.0\n"
- ]
- }
- ],
- "source": [
- "#pg 32\n",
- "#calculate the mass of coal burnt\n",
- "print('Example 1.14');\n",
- "\n",
- "# Given values\n",
- "P_out = 500.; # output of power station, [MW]\n",
- "c = 29.5; # calorific value of coal, [MJ/kg]\n",
- "r=.28; \n",
- "\n",
- "# solution\n",
- "\n",
- "# since P represents only 28 percent of energy available from coal\n",
- "P_coal = P_out/r; # [MW]\n",
- " \n",
- "m_coal = P_coal/c; # Mass of coal used, [kg/s]\n",
- "m_coal = m_coal*3600; # [kg/h]\n",
- "\n",
- "#After one hour\n",
- "m_coal = m_coal*1*10**-3; # [tonne]\n",
- "#results\n",
- "print 'Mass of coal burnt by the power station in 1 hour (tonne) = ',round(m_coal,0)\n",
- "\n",
- "# End\n"
- ]
- }
- ],
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- "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": "2.7.11"
- }
- },
- "nbformat": 4,
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generated by cgit (git 2.34.1) at 2025-06-23 06:30:53 +0000