{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 11 - The steam engine" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 1: pg 326" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.1\n", " (a) The bore of the cylinder is (mm) = 239.0\n", " (b) The piston stroke is (mm) = 299.0\n", " (c) The speed of the engine is (rev/min) = 301.1\n" ] } ], "source": [ "#pg 326\n", "print('Example 11.1')\n", "import math\n", "# aim : To determine the \n", "# (a) bore of the cylinder\n", "# (b) piston stroke\n", "# (c) speed of the engine\n", "\n", "# Given values\n", "P_req = 60.;# power required to develop, [kW]\n", "P = 1.25;# boiler pressure, [MN/m^2]\n", "Pb = .13;# back pressure, [MN/m^2]\n", "cut_off = .3;# [stroke]\n", "k = .82;# diagram factor\n", "n = .78;# mechanical efficiency\n", "LN = 3.;# mean piston speed, [m/s]\n", "\n", "# solution\n", "# (a)\n", "r = 1/cut_off;# expansion ratio\n", "Pm = P/r*(1+math.log(r))-Pb;# mean effective pressure, [MN/m^2]\n", "P_ind = P_req/n;# Actual indicated power developed, [kW]\n", "P_the = P_ind/k;# Theoretical indicated power developed, [kW]\n", "\n", "# using indicated_power=Pm*LN*A\n", "# Hence\n", "A = P_the/(Pm*LN)*10**-3;# piston area,[m^2]\n", "d = math.sqrt(4*A/math.pi)*10**3;# bore ,[mm]\n", "print ' (a) The bore of the cylinder is (mm) = ',round(d)\n", "\n", "# (b)\n", "# given that stroke is 1.25 times bore\n", "L = 1.25*d;# [mm]\n", "print ' (b) The piston stroke is (mm) = ',round(L)\n", "\n", "# (c)\n", "# LN=mean piston speed, where L is stroke in meter and N is 2*rev/s,(since engine is double_acting)\n", "# hence\n", "rev_per_sec = LN/(2*L*10**-3);# [rev/s]\n", "\n", "rev_per_min = rev_per_sec*60;# [rev/min]\n", "print ' (c) The speed of the engine is (rev/min) = ',round(rev_per_min,1)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 2: pg 328" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.2\n", " (a) The diameter of the cylinder is (mm) = 189.0\n", " (b) The piston stroke is (mm) = 227.2\n", " (c) The actual steam consumption/h is (kg) = 514.3\n", " The indicated thermal efficiency is (percent) = 6.8\n", "The answers are a bit different due to rounding off error in textbook\n" ] } ], "source": [ "#pg 328\n", "print('Example 11.2')\n", "import math\n", "# aim : To determine the \n", "# (a) the diameter of the cylinder\n", "# (b) piston stroke\n", "# (c) actual steam consumption and indicated thermal efficiency\n", "\n", "# Given values\n", "P = 900.;# inlet pressure, [kN/m^2]\n", "Pb = 140.;# exhaust pressure, [kN/m^2]\n", "cut_off =.4;# [stroke]\n", "k = .8;# diagram factor\n", "rs = 1.2;# stroke to bore ratio\n", "N = 4.;# engine speed, [rev/s]\n", "ip = 22.5;# power output from the engine, [kW]\n", "\n", "# solution\n", "# (a)\n", "r = 1/cut_off;# expansion ratio\n", "Pm = P/r*(1+math.log(r))-Pb;# mean effective pressure, [kN/m^2]\n", "Pm = Pm*k;# actual mean effective pressure, [kN/m^2]\n", "\n", "# using ip=Pm*L*A*N\n", "# and L=r*d; where L is stroke and d is bore\n", "d = (ip/(Pm*rs*math.pi/4.*2*N))**(1./3);# diameter of the cylinder, [m]\n", "\n", "print ' (a) The diameter of the cylinder is (mm) = ',round(d*1000)\n", "\n", "# (b)\n", "L = rs*d;# stroke, [m]\n", "print ' (b) The piston stroke is (mm) = ',round(L*1000,1)\n", "\n", "# (c)\n", "SV = math.pi/4*d**2*L;# stroke volume, [m^3]\n", "V = SV*cut_off*2*240*60;# volume of steam consumed per hour, [m^3]\n", "v = .2148;# specific volume at 900 kN/m^2, [m^3/kg]\n", "SC = V/v;# steam consumed/h, [kg]\n", "ASC = 1.5*SC;# actual steam consumption/h, [kg]\n", "print ' (c) The actual steam consumption/h is (kg) = ',round(ASC,1)\n", "\n", "m_dot = ASC/3600.;# steam consumption,[kg/s] \n", "# from steam table\n", "hg = 2772.1;# specific enthalpy of inlet steam, [kJ/kg]\n", "hfe = 458.4;# specific liquid enthalpy at exhaust pressure, [kJ/kg]\n", "\n", "ITE = ip/(m_dot*(hg-hfe));# indicated thermal efficiency\n", "print ' The indicated thermal efficiency is (percent) = ',round(ITE*100,1)\n", "\n", "print 'The answers are a bit different due to rounding off error in textbook'\n", "# End\n", " " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 3: pg 330" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.3\n", " (a) The diagram factor is = 0.85\n", " (b) The indicated thermal efficiency is (percent) = 15.0\n" ] } ], "source": [ "#pg 330\n", "print('Example 11.3');\n", "\n", "# aim : To determine\n", "# (a) the diagram factor\n", "# (b) the indicated thermal efficiency of the engine\n", "import math\n", "# given values\n", "d = 250.*10**-3;# cylinder diameter, [m]\n", "L = 375.*10**-3;# length of stroke, [m]\n", "P = 1000.;# steam pressure , [kPa]\n", "x = .96;# dryness fraction of steam\n", "Pb = 55;# exhaust pressure, [kPa]\n", "r = 6.;# expansion ratio\n", "ip = 45.;# indicated power developed, [kW]\n", "N = 3.5;# speed of engine, [rev/s]\n", "m = 460.;# steam consumption, [kg/h]\n", "\n", "# solution\n", "# (a)\n", "Pm = P/r*(1+math.log(r))-Pb;# [kN/m**3]\n", "A = math.pi*(d)**2/4;# area, [m**2]\n", "tip = Pm*L*A*N*2;# theoretical indicated power, [kW]\n", "k = ip/tip;# diagram factor\n", "print ' (a) The diagram factor is = ',round(k,2)\n", "\n", "# (b)\n", "# from steam table at 1 MN/m**2\n", "hf = 762.6;# [kJ/kg]\n", "hfg = 2013.6;# [kJ/kg]\n", "# so \n", "h1 = hf+x*hfg;# specific enthalpy of steam at 1MN/m**2, [kJ/kg]\n", "# minimum specific enthalpy in engine at 55 kPa \n", "hf = 350.6;# [kJ/kg]\n", "# maximum energy available in engine is\n", "h = h1-hf;# [kJ/kg]\n", "ITE = ip/(m*h/3600)*100;# indicated thermal efficiency\n", "print ' (b) The indicated thermal efficiency is (percent) = ',round(ITE)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 4: pg 333" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.4\n", "The steam consumption is (kg/h) = 213.0\n" ] } ], "source": [ "#pg 333\n", "print('Example 11.4');\n", "\n", "# aim : To determine\n", "# steam consumption\n", "\n", "# given values\n", "P1 = 11.;# power, [kW]\n", "m1 = 276.;# steam use/h when developing power P1,[kW]\n", "ip = 8.;# indicated power output, [kW]\n", "B = 45.;# steam used/h at no load, [kg]\n", "\n", "# solution\n", "# using graph of Fig.11.9 \n", "A = (m1-B)/P1;# slop of line, [kg/kWh]\n", "W = A*ip+B;# output, [kg/h]\n", "#results\n", "print 'The steam consumption is (kg/h) = ',W\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 5: pg 338" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.5\n", " (a) The intermediate pressure is (kN/m^2) = 455.0\n", " (b) The indicated power is (kW) = 110.5\n", " (c) The steam consumption of the engine is (kg/h) = 1016.0\n" ] } ], "source": [ "#pg 338\n", "print('Example 11.5');\n", "from scipy.optimize import brentq\n", "# aim : To determine\n", "# (a) the intermediate pressure\n", "# (b) the indicated power output\n", "# (c) the steam consumption of the engine\n", "import math\n", "# given values\n", "P1 = 1400.;# initial pressure, [kN/m**2]\n", "x = .9;# dryness fraction\n", "P5 = 35.;# exhaust pressure\n", "k = .8;# diagram factor of low-pressure cylindaer\n", "L = 350.*10**-3;# stroke of both the cylinder, [m]\n", "dhp = 200.*10**-3;# diameter of high pressure cylinder, [m]\n", "dlp = 300.*10**-3;# diameter of low-pressure cylinder, [m]\n", "N = 300.;# engine speed, [rev/min]\n", "\n", "# solution\n", "# taking reference Fig.11.13\n", "Ahp = math.pi/4*dhp**2;# area of high-pressure cylinder, [m**2]\n", "Alp = math.pi/4*dlp**2;# area of low-pressure cylinder, [m**2]\n", "# for equal initial piston loads\n", "# (P1-P7)Ahp=(P7-P5)Alp\n", "def f(P7):\n", "\treturn (P1-P7)*Ahp - (P7-P5)*Alp\n", "#deff('[x]=f(P7)','x=(P1-P7)*Ahp-(P7-P5)*Alp');\n", "P7 = brentq(f,0,1000);# intermediate pressure, [kN/m**2]\n", "print ' (a) The intermediate pressure is (kN/m^2) = ',P7\n", "\n", "# (b)\n", "V6 = Ahp*L;# volume of high-pressure cylinder, [m**3]\n", "P2 = P1;\n", "P6 = P7;\n", "# using P2*V2=P6*V6\n", "V2 = P6*V6/P2; # [m**3]\n", "V1 = Alp*L;# volume of low-pressure cylinder, [m**3]\n", "R = V1/V2;# expansion ratio\n", "Pm = P1/R*(1+math.log(R))-P5;# effective pressure of low-pressure cylinder, [kn/m**2]\n", "Pm = k*Pm;# actual effective pressure, [kN/m**2]\n", "ip = Pm*L*Alp*N*2/60.;# indicated power, [kW]\n", "print ' (b) The indicated power is (kW) = ',round(ip,1)\n", "\n", "# (c) \n", "COV = V1/ R;# cut-off volume in high-pressure cylinder, [m**3]\n", "V = COV*N*2*60;# volume of steam admitted/h\n", "# from steam table\n", "vg = .1407;# [m**3/kg]\n", "AV = x*vg;# specific volume of admission steam, [m**3/kg]\n", "m = V/AV;# steam consumption, [kg/h]\n", "print ' (c) The steam consumption of the engine is (kg/h) = ',round(m)\n", "\n", "# End \n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 6: pg 340" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.6\n", " (a) The indicated power is (kW) = 440.0\n", " (b) The diameter of high-pressure cylinder is (mm) = 383.0\n", " (c) The intermediate opressure is (kN/m^2) = 338.0\n" ] } ], "source": [ "#pg 340\n", "print('Example 11.6');\n", "\n", "# aim : To determine\n", "# (a) the indicated power output\n", "# (b) the diameter of high-pressure cylinder of the engine\n", "# (c) the intermediate pressure\n", "import math\n", "from math import sqrt, exp, log, pi\n", "# given values\n", "P = 1100.;# initial pressure, [kN/m**2]\n", "Pb = 28.;# exhaust pressure\n", "k = .82;# diagram factor of low-pressure cylindaer\n", "L = 600.*10**-3;# stroke of both the cylinder, [m]\n", "dlp = 600.*10**-3;# diameter of low-pressure cylinder, [m]\n", "N = 4.;# engine speed, [rev/s]\n", "R = 8.;# expansion ratio\n", "\n", "# solution\n", "# taking reference Fig.11.13\n", "# (a)\n", "Pm = P/R*(1+log(R))-Pb;# effective pressure of low-pressure cylinder, [kn/m**2]\n", "Pm = k*Pm;# actual effective pressure, [kN/m**2]\n", "Alp = pi/4*dlp**2;# area of low-pressure cylinder, [m**2]\n", "ip = Pm*L*Alp*N*2;# indicated power, [kW]\n", "print ' (a) The indicated power is (kW) = ',round(ip)\n", "\n", "# (b)\n", "# work done by both cylinder is same as area of diagram\n", "w = Pm*Alp*L;# [kJ]\n", "W = w/2;# work done/cylinder, [kJ]\n", "V2 = Alp*L/8;# volume, [m63]\n", "P2 = P;# [kN/m**2]\n", "# using area A1267=P2*V2*log(V6/V2)=W\n", "V6 = V2*exp(W/(P2*V2));# intermediate volume, [m**3]\n", "# using Ahp*L=%pi/4*dhp**2*L=V6\n", "dhp = sqrt(V6*4/L/pi);# diameter of high-pressure cylinder, [m]\n", "print ' (b) The diameter of high-pressure cylinder is (mm) = ',round(dhp*1000)\n", "\n", "# (c)\n", "# using P2*V2=P6*V6\n", "P6 = P2*V2/V6; # intermediate pressure, [kN/m**2]\n", "print ' (c) The intermediate opressure is (kN/m^2) = ',round(P6)\n", "\n", "# End \n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 7: pg 342" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.7\n", " (a) The engine speed is (rev/s) = 6.52\n", " (b) The diameter of the high pressure cylinder is (mm) = 179.0\n" ] } ], "source": [ "#pg 342\n", "print('Example 11.7');\n", "\n", "# aim : To determine\n", "# (a) The speed of the engine\n", "# (b) the diameter of the high pressure cylinder\n", "import math\n", "from math import sqrt,log, exp,pi\n", "# given values\n", "ip = 230.;# indicated power, [kW]\n", "P = 1400.;# admission pressure, [kN/m**2]\n", "Pb = 35.;# exhaust pressure, [kN/m**2]\n", "R = 12.5;# expansion ratio\n", "d1 = 400.*10**-3;# diameter of low pressure cylinder, [m]\n", "L = 500.*10**-3;# stroke of both the cylinder, [m]\n", "k = .78;# diagram factor\n", "rv = 2.5;# expansion ratio of high pressure cylinder\n", "\n", "# solution\n", "# (a)\n", "Pm = P/R*(1+log(R))-Pb;# mean effective pressure in low pressure cylinder, [kN/m**2]\n", "ipt = ip/k;# theoretical indicated power, [kw]\n", "# using ip=Pm*L*A*N\n", "A = pi/4*d1**2;# area , [m**2]\n", "N = ipt/(Pm*L*A*2);# speed, [rev/s]\n", "print ' (a) The engine speed is (rev/s) = ',round(N,2)\n", "\n", "# (b)\n", "Vl = A*L;# volume of low pressure cylinder, [m**3]\n", "COV = Vl/R;# cutt off volume of hp cylinder, [m**3]\n", "V = COV*rv;# total volume, [m**3]\n", "\n", "# V = %pi/4*d**2*L, so\n", "d = sqrt(4*V/pi/L);# diameter of high pressure cylinder, [m]\n", "print ' (b) The diameter of the high pressure cylinder is (mm) = ',round(d*1000)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 8: pg 344" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.8\n", " (a) The actual mean effective pressure referred to LP cylinder is (kN/m^2) = 234.1\n", " The hypothetical mean effective pressure referred to LP cylinder is (kN/m^2) = 287.0\n", " (b) The overall diagram factor is = 0.814\n", " (c) The indicated power is (kW) = 143.0\n" ] } ], "source": [ "#pg 344\n", "print('Example 11.8');\n", "\n", "# aim : To determine\n", "# (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder\n", "# (b) the overall diagram factor\n", "# (c) the indicated power \n", "import math\n", "from math import pi,sqrt,log\n", "# given values\n", "P = 1100.;# steam supply pressure, [kN/m**2]\n", "Pb = 32.;# back pressure, [kN/m**2]\n", "d1 = 300.*10**-3;# cylinder1 diameter, [m]\n", "d2 = 600.*10**-3;# cylinder2 diameter, [m]\n", "L = 400.*10**-3;# common stroke of both cylinder, [m]\n", "\n", "A1 = 12.5;# average area of indicated diagram for HP, [cm**2]\n", "A2 = 11.4;# average area of indicated diagram for LP, [cm**2]\n", "\n", "P1 = 270.;# indicator calibration, [kN/m**2/ cm]\n", "P2 = 80.;# spring calibration, [kN/m**2/ cm]\n", "N = 2.7;# engine speed, [rev/s]\n", "l = .75;# length of both diagram, [m]\n", "\n", "# solution\n", "# (a)\n", "# for HP cylinder\n", "Pmh = P1*A1/7.5;# [kN/m**2]\n", "F = Pmh*pi/4*d1**2;# force on HP, [kN]\n", "PmH = Pmh*(d1/d2)**2;# pressure referred to LP cylinder, [kN/m**2]\n", "PmL = P2*A2/7.5;# pressure for LP cylinder, [kN/m**2]\n", "PmA = PmH+PmL;# actual effective pressure referred to LP cylinder, [kN/m**2]\n", "\n", "Ah = pi/4*d1**2;# area of HP cylinder, [m**2]\n", "Vh = Ah*L;# volume of HP cylinder, [m**3]\n", "CVh = Vh/3;# cut-off volume of HP cylinder, [m**3]\n", "Al = pi/4*d2**2;# area of LP cylinder, [m**2]\n", "Vl = Al*L;# volume of LP cylinder, [m**3]\n", "\n", "R = Vl/CVh;# expansion ratio\n", "Pm = P/R*(1+log(R))-Pb;# hypothetical mean effective pressure referred to LP cylinder, [kN/m**2]\n", "\n", "print ' (a) The actual mean effective pressure referred to LP cylinder is (kN/m^2) = ',PmA\n", "print ' The hypothetical mean effective pressure referred to LP cylinder is (kN/m^2) = ',round(Pm)\n", "\n", "# (a)\n", "ko = PmA/Pm;# overall diagram factor\n", "print ' (b) The overall diagram factor is = ',round(ko,3)\n", "\n", "# (c) \n", "ip = PmA*L*Al*N*2;# indicated power, [kW]\n", "print ' (c) The indicated power is (kW) = ',round(ip)\n", "\n", "# End\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example 9: pg 345" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Example 11.9\n", " (a) The actual mean effective pressure referred to LP cylinder is (kN/m^2) = 219.6\n", " The hypothetical mean effective pressure referred to LP cylinder is (kN/m^2) = 326.0\n", " (b) The overall diagram factor is = 0.673\n", " (c) The pecentage of the total indicated power developed in HP cylinder is (percent) = 29.9\n", " The pecentage of the total indicated power developed in IP cylinder is (percent) = 30.1\n", " The pecentage of the total indicated power developed in LP cylinder is (percent) = 40.1\n" ] } ], "source": [ "#pg 345\n", "print('Example 11.9');\n", "\n", "# aim : To determine\n", "# (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder\n", "# (b) the overall diagram factor\n", "# (c) the pecentage of the total indicated power developed in each cylinder\n", "from math import pi,log,sqrt\n", "# given values\n", "P = 1400.;# steam supply pressure, [kN/m**2]\n", "Pb = 20.;# back pressure, [kN/m**2]\n", "Chp = .6;# cut-off in HP cylinder, [stroke]\n", "dh = 300.*10**-3;# HP diameter, [m]\n", "di = 500.*10**-3;# IP diameter, [m]\n", "dl = 900.*10**-3;# LP diameter, [m]\n", "\n", "Pm1 = 590.;# actual pressure of HP cylinder, [kN/m**2]\n", "Pm2 = 214.;# actual pressure of IP cylinder, [kN/m**2]\n", "Pm3 = 88.;# actual pressure of LP cylinder, [kN/m**2]\n", "\n", "# solution\n", "# (a)\n", "# for HP cylinder\n", "PmH = Pm1*(dh/dl)**2;# PmH referred to LP cylinder, [kN/m**2]\n", "# for IP cylinder\n", "PmI = Pm2*(di/dl)**2;# PmI referred to LP cylinder, [kN/m**2]\n", "PmA = PmH+PmI+Pm3;# actual mean effective pressure referred to LP cylinder, [kN/m**2]\n", "\n", "R = dl**2/(dh**2*Chp);# expansion ratio\n", "Pm = P/R*(1+log(R))-Pb;# hypothetical mean effective pressure referred to LP cylinder, [kN/m**2]\n", "\n", "print ' (a) The actual mean effective pressure referred to LP cylinder is (kN/m^2) = ',round(PmA,2)\n", "print ' The hypothetical mean effective pressure referred to LP cylinder is (kN/m^2) = ',round(Pm)\n", "\n", "# (b)\n", "ko = PmA/Pm;# overall diagram factor\n", "print ' (b) The overall diagram factor is = ',round(ko,3)\n", "\n", "# (c)\n", "HP = PmH/PmA*100;# %age of indicated power developed in HP\n", "IP = PmI/PmA*100; # %age of indicated power developed in IP\n", "LP = Pm3/PmA*100; # %age of indicated power developed in LP\n", "print ' (c) The pecentage of the total indicated power developed in HP cylinder is (percent) = ',round(HP,1)\n", "print ' The pecentage of the total indicated power developed in IP cylinder is (percent) = ',round(IP,1)\n", "print ' The pecentage of the total indicated power developed in LP cylinder is (percent) = ',round(LP,1)\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 }