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diff --git a/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter1.ipynb b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter1.ipynb new file mode 100644 index 00000000..53149a6f --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter1.ipynb @@ -0,0 +1,703 @@ +{ + "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" + ] + } + ], + "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": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |