{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 17 - Engine and plant trails" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 1: pg 589 " ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.1\n", " The Indicated power is (kW) = 26.2\n", " The Brake power is (kW) = 22.0\n", " The mechanical efficiency is (percent) = 837.0\n", "Energy can be tabulated as :-\n", "----------------------------------------------------------------------------------------------------\n", " kJ/s Percentage \n", "----------------------------------------------------------------------------------------------------\n", " Energy from fuel 88.0 100.0 \n", " Energy to brake power 22.0 25.0 \n", " Energy to coolant 20.7 23.5 \n", " Energy to exhaust 33.6 38.2 \n", " Energy to suroundings,etc. 11.8 13.4\n" ] } ], "source": [ "#pg 589\n", "print('Example 17.1');\n", "\n", "# aim : To determine\n", "# the indicated and brake output and the mechanicl efficiency\n", "# draw up an overall energy balance and as % age\n", "import math\n", "# given values\n", "h = 21;# height of indicator diagram, [mm]\n", "ic = 27;# indicator calibration, [kN/m**2 per mm]\n", "sv = 14*10**-3;# swept volume of the cylinder;,[m**3]\n", "N = 6.6;# speed of engine, [rev/s]\n", "ebl = 77;# effective brake load, [kg]\n", "ebr = .7;# effective brake radious, [m]\n", "fc = .002;# fuel consumption, [kg/s]\n", "CV = 44000;# calorific value of fuel, [kJ/kg]\n", "cwc = .15;# cooling water circulation, [kg/s]\n", "Ti = 38;# cooling water inlet temperature, [C]\n", "To = 71;# cooling water outlet temperature, [C]\n", "c = 4.18;# specific heat capacity of water, [kJ/kg]\n", "eeg = 33.6;# energy to exhaust gases, [kJ/s]\n", "g = 9.81;# gravitational acceleration, [m/s**2]\n", "\n", "# solution\n", "PM = ic*h;# mean effective pressure, [kN/m**2]\n", "LA = sv;# swept volume of the cylinder, [m**3]\n", "ip = PM*LA*N/2;# indicated power,[kW]\n", "T = ebl*g*ebr;# torque, [N*m]\n", "bp = 2*math.pi*N*T;# brake power, [W]\n", "n_mech = bp/ip;# mechanical efficiency\n", "print ' The Indicated power is (kW) = ',round(ip,2)\n", "print ' The Brake power is (kW) = ',round(bp*10**-3)\n", "print ' The mechanical efficiency is (percent) = ',round(n_mech)\n", "\n", "ef = CV*fc;# energy from fuel, [kJ/s]\n", "eb = bp*10**-3;# energy to brake power,[kJ/s]\n", "ec = cwc*c*(To-Ti);# energy to coolant,[kJ/s]\n", "es = ef-(eb+ec+eeg);# energy to surrounding,[kJ/s]\n", "\n", "print('Energy can be tabulated as :-');\n", "print('----------------------------------------------------------------------------------------------------');\n", "print(' kJ/s Percentage ')\n", "print('----------------------------------------------------------------------------------------------------');\n", "print ' Energy from fuel ',ef,' ',ef/ef*100,'\\n Energy to brake power ',round(eb),' ',round(eb/ef*100),'\\n Energy to coolant ',round(ec,1),' ',round(ec/ef*100,1),' \\n Energy to exhaust ',eeg,' ',round(eeg/ef*100,1),'\\n Energy to suroundings,etc. ',round(es,1),' ',round(es/ef*100,1)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 2: pg 591" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.2\n", " (a) The brake power is (kW) = 14.657\n", " (b) The indicated power is (kW) = 18.2\n", " (c) The mechanical efficiency is (percent) = 80.4\n", " (d) The indicated thermal efficiency is (percent) = 12.94\n", " (e) The brake steam consumption is (kg/kWh) = 13.75\n", " (f) Energy supplied/min is (kJ) = 9092.0\n", " Energy to bp/min is (kJ) = 879.0\n", " Energy to condenser cooling water/min is (kJ) = 5196.0\n", " Energy to condensate/min is (kJ) = 534.0\n", " Energy to surrounding, etc/min is (kJ) = 2483.0\n", "answer in the book is misprinted for Es\n" ] } ], "source": [ "#pg 591\n", "print('Example 17.2');\n", "import math\n", "# aim : To determine\n", "# (a) bp\n", "# (b) ip\n", "# (c) mechanical efficiency\n", "# (d) indicated thermal efficiency\n", "# (e) brake specific steam consumption\n", "# (f) draw up complete energy account for the test one-minute basis taking 0 C as datum\n", "\n", "# given values\n", "d = 200.*10**-3;# cylinder diameter, [mm]\n", "L = 250.*10**-3;# stroke, [mm]\n", "N = 5.;# speed, [rev/s]\n", "r = .75/2;# effective radious of brake wheel, [m]\n", "Ps = 800.;# stop valve pressure, [kN/m**2]\n", "x = .97;# dryness fraction of steam\n", "BL = 136.;# brake load, [kg]\n", "SL = 90.;# spring balance load, [N]\n", "PM = 232.;# mean effective pressure, [kN/m**2]\n", "Pc = 10.;# condenser pressure, [kN/m**2]\n", "m_dot = 3.36;# steam consumption, [kg/min]\n", "CC = 113.;# condenser cooling water, [kg/min]\n", "Tr = 11.;# temperature rise of condenser cooling water, [K]\n", "Tc = 38.;# condensate temperature, [C]\n", "C = 4.18;# heat capacity of water, [kJ/kg K]\n", "g = 9.81;# gravitational acceleration, [m/s**2]\n", "\n", "# solution\n", "# from steam table\n", "# at 800 kN/m**2\n", "tf1 = 170.4;# saturation temperature, [C]\n", "hf1 = 720.9;# [kJ/kg]\n", "hfg1 = 2046.5;# [kJ/kg]\n", "hg1 = 2767.5;# [kJ/kg]\n", "vg1 = .2403;# [m**3/kg]\n", "\n", "# at 10 kN/m**2\n", "tf2 = 45.8;# saturation temperature, [C]\n", "hf2 = 191.8;# [kJ/kg]\n", "hfg2 = 2392.9;# [kJ/kg]\n", "hg2 = 2584.8;# [kJ/kg]\n", "vg2 = 14.67;# [m**3/kg]\n", "\n", "# (a)\n", "T = (BL*g-SL)*r;# torque, [Nm]\n", "bp = 2*math.pi*N*T*10**-3;# brake power,[W]\n", "print ' (a) The brake power is (kW) = ',round(bp,3)\n", "\n", "# (b)\n", "A = math.pi*d**2/4;# area, [m**2]\n", "ip = PM*L*A*N*2;# double-acting so*2, [kW]\n", "print ' (b) The indicated power is (kW) = ',round(ip,1)\n", "\n", "# (c)\n", "n_mec = bp/ip;# mechanical efficiency\n", "print ' (c) The mechanical efficiency is (percent) = ',round(n_mec*100,1)\n", "\n", "# (d)\n", "h = hf1+x*hfg1;# [kJ/kg]\n", "hf = hf2;\n", "ITE = ip/((m_dot/60)*(h-hf));# indicated thermal efficiency\n", "print ' (d) The indicated thermal efficiency is (percent) = ',round(ITE*100,2)\n", "# (e)\n", "Bsc=m_dot*60/bp;# brake specific steam consumption, [kg/kWh]\n", "print ' (e) The brake steam consumption is (kg/kWh) = ',round(Bsc,2)\n", "\n", "# (f)\n", "# energy balanvce reckoned from 0 C\n", "Es = m_dot*h;# energy supplied, [kJ]\n", "Eb = bp*60;# energy to bp, [kJ]\n", "Ecc = CC*C*Tr;# energy to condensate cooling water, [kJ]\n", "Ec = m_dot*C*Tc;# energy to condensate, [kJ]\n", "Ese = Es-Eb-Ecc-Ec;# energy to surrounding,etc, [kJ]\n", "\n", "print ' (f) Energy supplied/min is (kJ) = ',round(Es)\n", "\n", "print ' Energy to bp/min is (kJ) = ',round(Eb)\n", "print ' Energy to condenser cooling water/min is (kJ) = ',round(Ecc)\n", "print ' Energy to condensate/min is (kJ) = ',round(Ec)\n", "print ' Energy to surrounding, etc/min is (kJ) = ',round(Ese)\n", "\n", "print 'answer in the book is misprinted for Es'\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3: pg 593" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.3\n", " (a) The Brake power is (kW) = 60.5\n", " (b) The brake specific fuel consumption is (kg/kWh) = 0.309\n", " (c) The indicated thermal efficiency is (percent) = 33.2\n", " (d) Energy from fuel is (kJ) = 13184.0\n", " Energy to brake power is (kJ) = 3629.0\n", " Energy to cooling water is (kJ) = 4038.0\n", " Energy to exhaust is (kJ) = 3739.0\n", " Energy to surrounding, etc is (kJ) = 1778.0\n", "The answer is a bit different due to rounding off error in textbook\n" ] } ], "source": [ "#pg 593\n", "print('Example 17.3');\n", "\n", "# aim : To determine\n", "# (a) the brake power\n", "# (b) the brake specific fuel consumption\n", "# (c) the indicated thermal efficiency\n", "# (d) the energy balance, expressing the various items\n", "import math\n", "# given values\n", "t = 30.;# duration of trial, [min]\n", "N = 1750.;# speed of engine, [rev/min]\n", "T = 330.;# brake torque, [Nm]\n", "mf = 9.35;# fuel consumption, [kg]\n", "CV = 42300.;# calorific value of fuel, [kJ/kg]\n", "cwc = 483.;# jacket cooling water circulation, [kg]\n", "Ti = 17.;# inlet temperature, [C]\n", "To = 77.;# outlet temperature, [C]\n", "ma = 182.;# air consumption, [kg]\n", "Te = 486.;# exhaust temperature, [C]\n", "Ta = 17.;# atmospheric temperature, [C]\n", "n_mec = .83;# mechanical efficiency\n", "c = 1.25;# mean specific heat capacity of exhaust gas, [kJ/kg K]\n", "C = 4.18;# specific heat capacity, [kJ/kg K]\n", "\n", "# solution\n", "# (a)\n", "bp = 2*math.pi*N*T/60*10**-3;# brake power, [kW]\n", "print ' (a) The Brake power is (kW) = ',round(bp,1)\n", "\n", "# (b)\n", "bsf = mf*2/bp;#brake specific fuel consumption, [kg/kWh]\n", "print ' (b) The brake specific fuel consumption is (kg/kWh) = ',round(bsf,3)\n", "\n", "# (c)\n", "ip = bp/n_mec;# indicated power, [kW]\n", "ITE = ip/(2*mf*CV/3600);# indicated thermal efficiency\n", "print ' (c) The indicated thermal efficiency is (percent) = ',round(ITE*100,1)\n", "\n", "# (d)\n", "# taking basis one minute \n", "ef = CV*mf/30;# energy from fuel, [kJ]\n", "eb = bp*60;# energy to brake power,[kJ]\n", "ec = cwc/30*C*(To-Ti);# energy to cooling water,[kJ]\n", "ee = (ma+mf)/30*c*(Te-Ta);# energy to exhaust, [kJ]\n", "es = ef-(eb+ec+ee);# energy to surrounding,etc,[kJ]\n", "\n", "print ' (d) Energy from fuel is (kJ) = ',round(ef)\n", "print ' Energy to brake power is (kJ) = ',round(eb)\n", "print ' Energy to cooling water is (kJ) = ',round(ec)\n", "print ' Energy to exhaust is (kJ) = ',round(ee)\n", "print ' Energy to surrounding, etc is (kJ) = ',round(es)\n", " \n", "print 'The answer is a bit different due to rounding off error in textbook'\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4: pg 594" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.4\n", " (a) The indicated power of the engine is (kW) = 69.9\n", " (b) The mechanical efficiency of the engine is (percent) = 74.4\n" ] } ], "source": [ "#pg 594\n", "print('Example 17.4');\n", "\n", "# aim : To determine\n", "# (a) the indicated power of the engine\n", "# (b) the mechanical efficiency of the engine\n", "\n", "# given values\n", "bp = 52;# brake power output, [kW]\n", "bp1 = 40.5;# brake power of cylinder cut1, [kW]\n", "bp2 = 40.2;# brake power of cylinder cut2, [kW]\n", "bp3 = 40.1;# brake power of cylinder cut3, [kW]\n", "bp4 = 40.6;# brake power of cylinder cut4, [kW]\n", "bp5 = 40.7;# brake power of cylinder cut5, [kW]\n", "bp6 = 40.0;# brake power of cylinder cut6, [kW]\n", "\n", "# sollution\n", "ip1 = bp-bp1;# indicated power of cylinder cut1, [kW]\n", "ip2 = bp-bp2;# indicated power of cylinder cut2, [kW]\n", "ip3 = bp-bp3;# indicated power of cylinder cut3, [kW]\n", "ip4 = bp-bp4;# indicated power of cylinder cut4, [kW]\n", "ip5 = bp-bp5;# indicated power of cylinder cut5, [kW]\n", "ip6 = bp-bp6;# indicated power of cylinder cut6, [kW]\n", "\n", "ip = ip1+ip2+ip3+ip4+ip5+ip6;# indicated power of engine,[kW]\n", "print ' (a) The indicated power of the engine is (kW) = ',ip\n", "\n", "# (b)\n", "n_mec = bp/ip;# mechanical efficiency\n", "print ' (b) The mechanical efficiency of the engine is (percent) = ',round(n_mec*100,1)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 5: pg 595" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.5\n", " The Brake power is (kW) = 29.3\n", " The Indicated power is (kW) = 37.3\n", " The mechanical efficiency is (percent) = 78.8\n", "Energy can be tabulated as :-\n", "----------------------------------------------------------------------------------------------------\n", " kJ/s Percentage \n", "----------------------------------------------------------------------------------------------------\n", " Energy from fuel 135.3 100.0 \n", " Energy to brake power 29.3 21.7 \n", " Energy to exhaust 35.4 26.0 \n", " Energy to coolant 44.5 32.9 \n", " Energy to suroundings,etc. 26.1 19.3\n", "there is minor variation in the result reported in the book due to rounding off error\n" ] } ], "source": [ "#pg 595\n", "print('Example 17.5');\n", "\n", "# aim : To determine\n", "# the brake power,indicated power and mechanicl efficiency\n", "# draw up an energy balance and as % age of the energy supplied\n", "\n", "# given values\n", "N = 50.;# speed, [rev/s]\n", "BL = 267.;# break load.,[N]\n", "BL1 = 178.;# break load of cylinder cut1, [N]\n", "BL2 = 187.;# break load of cylinder cut2, [N]\n", "BL3 = 182.;# break load of cylinder cut3, [N]\n", "BL4 = 182.;# break load of cylinder cut4, [N]\n", "\n", "FC = .568/130;# fuel consumption, [L/s]\n", "s = .72;# specific gravity of fuel\n", "CV = 43000;# calorific value of fuel, [kJ/kg]\n", "\n", "Te = 760;# exhaust temperature, [C]\n", "c = 1.015;# specific heat capacity of exhaust gas, [kJ/kg K]\n", "Ti = 18;# cooling water inlet temperature, [C]\n", "To = 56;# cooling water outlet temperature, [C]\n", "mw = .28;# cooling water flow rate, [kg/s]\n", "Ta = 21;# ambient tempearture, [C]\n", "C = 4.18;# specific heat capacity of cooling water, [kJ/kg K]\n", "\n", "# solution\n", "bp = BL*N/455;# brake power of engine, [kW]\n", "bp1 = BL1*N/455;# brake power of cylinder cut1, [kW]\n", "i1 = bp-bp1;# indicated power of cylinder cut1, [kW]\n", "bp2 = BL2*N/455;# brake power of cylinder cut2, [kW]\n", "i2 = bp-bp2;# indicated power of cylinder cut2, [kW]\n", "bp3 = BL3*N/455;# brake power of cylinder cut3, [kW]\n", "i3 = bp-bp3;# indicated power of cylinder cut3, [kW]\n", "bp4 = BL4*N/455;# brake power of cylinder cut4, [kW]\n", "i4 = bp-bp4;# indicated power of cylinder cut4, [kW]\n", "\n", "ip = i1+i2+i3+i4;# indicated power of engine, [kW]\n", "n_mec = bp/ip;# mechanical efficiency\n", "\n", "print ' The Brake power is (kW) = ',round(bp,1)\n", "print ' The Indicated power is (kW) = ',round(ip,1)\n", "print ' The mechanical efficiency is (percent) = ',round(n_mec*100,1)\n", "\n", "mf = FC*s;# mass of fuel/s, [kg]\n", "ef = CV*mf;# energy from fuel/s, [kJ]\n", "me = 15*mf;# mass of exhaust/s,[kg],(given in condition)\n", "ee = me*c*(Te-Ta);# energy to exhaust/s,[kJ]\n", "ec = mw*C*(To-Ti);# energy to cooling water/s,[kJ]\n", "es = ef-(ee+ec+bp);# energy to surrounding,etc/s,[kJ]\n", "\n", "print('Energy can be tabulated as :-');\n", "print('----------------------------------------------------------------------------------------------------');\n", "print(' kJ/s Percentage ')\n", "print('----------------------------------------------------------------------------------------------------');\n", "print ' Energy from fuel ',round(ef,1),' ',ef/ef*100,'\\n Energy to brake power ',round(bp,1),' ',round(bp/ef*100.,1),'\\n Energy to exhaust ',round(ee,1),' ',round(ee/ef*100),'\\n Energy to coolant ',round(ec,1),' ',round(ec/ef*100,1),'\\n Energy to suroundings,etc. ',round(es,1),' ',round(es/ef*100,1)\n", "\n", "print 'there is minor variation in the result reported in the book due to rounding off error'\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 6: pg 596" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 17.6\n", " (a) The brake power is (MW) = 23.719\n", " (b) The fuel consumption is (tonne/h) = 4.74\n", " (c) The brake thermal efficiency is (percent) = 42.0\n" ] } ], "source": [ "#pg 596\n", "print('Example 17.6');\n", "\n", "# aim : To determine \n", "# (a) the break power of engine\n", "# (b) the fuel consumption of the engine\n", "# (c) the brake thermal efficiency of the engine\n", "import math\n", "# given values\n", "d = 850*10**-3;# bore , [m]\n", "L = 2200*10**-3;# stroke, [m]\n", "PMb = 15;# BMEP of cylinder, [bar]\n", "N = 95./60;# speed of engine, [rev/s]\n", "sfc = .2;# specific fuel oil consumption, [kg/kWh]\n", "CV = 43000;# calorific value of the fuel oil, [kJ/kg]\n", "\n", "# solution\n", "# (a)\n", "A = math.pi*d**2/4;# area, [m**2]\n", "bp = PMb*L*A*N*8/10;# brake power,[MW]\n", "print ' (a) The brake power is (MW) = ',round(bp,3)\n", "\n", "# (b)\n", "FC = bp*sfc;# fuel consumption, [kg/h]\n", "print ' (b) The fuel consumption is (tonne/h) = ',round(FC,2)\n", "\n", "# (c)\n", "mf = FC/3600;# fuel used, [kg/s]\n", "n_the = bp/(mf*CV);# brake thermal efficiency\n", "print ' (c) The brake thermal efficiency is (percent) = ',round(n_the*100)\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 }