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diff --git a/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter11_1.ipynb b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter11_1.ipynb new file mode 100644 index 00000000..6707db42 --- /dev/null +++ b/Basic_Engineering_Thermodynamics_by_Rayner_Joel/Chapter11_1.ipynb @@ -0,0 +1,678 @@ +{ + "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 +} |