{ "metadata": { "name": "", "signature": "sha256:6c08b023c55a26622d295a66797c942f394a0d06e61a13446845d2989d5e21c3" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 20:Internal Combustion Engines" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.2:pg-852" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "# Four cylinder engine\n", "BP = 30.0 # Power developed by engine in kW\n", "N = 2500.0 # Speed in rpm\n", "P_m = 800.0 # Mean effective pressure for each cylinder in kN/m**2\n", "n_m = 0.8 # Mechanical efficiency\n", "r = 1.5 # Stroke to bore ratio\n", "n_b = 0.28 # Brake thermal efficiency\n", "c_v = 44.0 # Heating value of petrol in MJ/kg\n", "print \"\\n Example 20.2\\n\"\n", "IP = BP/n_m\n", "d = ((IP*1000*60)/(P_m*1000*r*(math.pi/4)*N*4))**(1.0/3.0)\n", "L = r*d\n", "m_f = BP/(c_v*1000*n_b)\n", "bsfc = m_f*3600/BP\n", "print \"\\n Diameter of cylinder = \",d*10**2 ,\" cm\\n Stroke of each cylinder = \",L*100 ,\" cm\\n Brake specific fuel consumption = \",bsfc ,\" kg/kWh\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.2\n", "\n", "\n", " Diameter of cylinder = 6.20350490899 cm\n", " Stroke of each cylinder = 9.30525736349 cm\n", " Brake specific fuel consumption = 0.292207792208 kg/kWh\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.1:pg-851" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "d = 6.5# Diametre in cm\n", "L = 9.5 # Stroke in cm\n", "T = 64.0 # Torque in Nm\n", "N = 3000.0 # Speed in rpm\n", "V_c = 63.0 # Clearance volume in cm**3\n", "r = 0.5 # Brake efficiency ratio\n", "c_v = 42.0 # Calorific value of gasoline in MJ/kg\n", "print \"\\n Example 20.1\\n\"\n", "V_s = (math.pi/4)*(d**2)*(L)\n", "r_k = (V_s+V_c)/V_c\n", "n_as = 1- (1.0/(r_k**(0.4)))\n", "n_b = r*n_as\n", "BP = (2*math.pi*T*N)/60000\n", "m_f = (BP*3600)/(n_b*c_v*1000)# in kg/h\n", "BMEP = BP*60*2/((math.pi/4)*4*(d**2)*L*N*10**(-6))\n", "print \"\\n Fuel consumption of the engine = \",m_f ,\" Kg/h\\n BMEP=\",BMEP ,\" kN/m**2\"\n", "#The answers vary due to round off error\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.1\n", "\n", "\n", " Fuel consumption of the engine = 6.73508593048 Kg/h\n", " BMEP= 637.807536593 kN/m**2\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.3:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "F = 680.0 # Net brake load in N\n", "N = 360.0 # \n", "d = 10.0# Bore in cm\n", "L = 15.0 # Stroke in cm\n", "T = 58.0 # Torque in Nm\n", "v = 300.0 # Speed in m/min\n", "n_m = 0.8 # Mechanical efficiency\n", "n_th = 0.4 # Indicated thermal efficiency\n", "c_v = 44.0 # Calorific value of gasoline in MJ/kg\n", "print \"\\n Example 20.3\\n\"\n", "N = v/(2*L*(10**(-2)))\n", "BP = (2*math.pi*T*N)/60000\n", "IP = BP/n_m\n", "p_m = (IP*60)/(L*(math.pi/4)*(d**2)*N*10**(-6))\n", "m_f = (IP*3600)/(n_th*c_v*1000)\n", "bsfc = m_f/BP\n", "print \"\\n Indicated power = \",IP ,\" kW\\n Indicate mean effective pressure = \",p_m ,\" kN/m**2\\n Fuel consumption per kWh on brake power output = \",bsfc ,\" Kg/kWh\"\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.3\n", "\n", "\n", " Indicated power = 7.59218224618 kW\n", " Indicate mean effective pressure = 386.666666667 kN/m**2\n", " Fuel consumption per kWh on brake power output = 0.255681818182 Kg/kWh\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.4:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "T = 20.0 # Time in minute\n", "F = 680.0 # Net brake load in N\n", "N = 360.0 # Speed in rpm\n", "mep = 3.0 # Mean effective pressure in bar\n", "f = 1.56 # Fuel consumption in kg\n", "m_w = 160.0 # Cooling water in kg\n", "t = 57.0 # Water inlet temperature in degree centigrade\n", "r = 30.0 # Air used per kg of fuel\n", "t_r = 27.0 # Room temperature in degree centigrade\n", "t_e = 310.0 # Exhaust gas temperature in degree centigrade\n", "d = 210.0 # Bore in mm\n", "L = 290.0 # Stroke in mm\n", "D = 1.0 # Brake diameter in m\n", "cv = 44.0 # Calorific value in MJ/kg\n", "m_s = 1.3 # Steam formed per kg fuel in the exhaust in kg\n", "s = 2.093 # Specific heat of steam in the exhaust in kJ/kgK\n", "s_d = 1.01 # Specific heat of dry exhaust gases in kJ/kgK\n", "print \"\\n Example 20.4\\n\"\n", "i_p = mep*100*L*(10**-3)*(math.pi/4)*((d*(10**-3))**2)*N/60\n", "b_p = (2*math.pi*(F*(D/2))*N)/60000\n", "n_m = b_p / i_p\n", "h = f*cv*1000\n", "i_pe = i_p*T*60\n", "e_w = m_w * 4.187*(t-32)\n", "m_t = f*r + f\n", "m_s_ = m_s*f\n", "m_d = m_t - m_s_\n", "e_d = m_d * s_d * (t_e-t_r)\n", "e_s = m_s_*(4.187*(100-t_r) + 2257.9 +s*(t_e-100))\n", "e_t = e_s + e_d\n", "e_Un = h - (i_pe + e_w + e_t)\n", "print \"\\n Indicated power = \",i_p ,\" kW\\n Brake power = \",b_p ,\" kW\"\n", "print \"\\n Energy release by combustion of fuel is \",h ,\" kJ \\n 1. Energy equivalent of ip is \",i_pe ,\" kJ (\",(i_pe/h)*100 ,\" percent)\\n 2. Energy carried away by cooling water is \",e_w ,\" kJ (\",(e_w/h)*100 ,\" percent),\\n 3. Energy carried away by exhaust gases is \",e_t ,\" kJ (\",(e_t/h)*100 ,\" percent),\\n 4. Unaccounted energy loss (by difference) is \",e_Un ,\" kJ (\",(e_Un/h)*100 ,\" percent)\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.4\n", "\n", "\n", " Indicated power = 18.080022801 kW\n", " Brake power = 12.8176980266 kW\n", "\n", " Energy release by combustion of fuel is 68640.0 kJ \n", " 1. Energy equivalent of ip is 21696.0273613 kJ ( 31.6084314704 percent)\n", " 2. Energy carried away by cooling water is 16748.0 kJ ( 24.3997668998 percent),\n", " 3. Energy carried away by exhaust gases is 19333.323828 kJ ( 28.1662643182 percent),\n", " 4. Unaccounted energy loss (by difference) is 10862.6488107 kJ ( 15.8255373117 percent)\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.5:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "F = 610.0 # Net brake load in N\n", "N = 350.0 # Speed in rpm\n", "d = 20.0 # Bore in cm\n", "L = 30.0 # Stroke in cm\n", "imep = 275.0 # Mean effective pressure in kN/m**2\n", "D = 1.0 # Brake diameter in m\n", "m_o = 4.25 # Oil consumption in kg/h\n", "cv = 44.0 # Calorific value in MJ/kg\n", "print \"\\n Example 20.5\\n\"\n", "i_p = imep*1000*L*(10**-2)*(math.pi/4)*((d*(10**-2))**2)*N/60000\n", "b_p = (2*math.pi*(F*(D/2))*N)/60000\n", "n_m = b_p / i_p\n", "n_th = i_p *3600/(m_o*cv*1000)\n", "n_br = n_th*n_m\n", "print \"\\n Indicated power = \",i_p ,\" kW\\n Brake power = \",b_p ,\" kW\\n Mechanical efficiency = \",n_m*100 ,\" percent,\\n Indicated thermal efficiency = \",n_th*100 ,\" percent,\\n Brake thermal efficiency = \",n_br*100 ,\" percent\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.5\n", "\n", "\n", " Indicated power = 15.1189146454 kW\n", " Brake power = 11.178833859 kW\n", " Mechanical efficiency = 73.9393939394 percent,\n", " Indicated thermal efficiency = 29.1059319377 percent,\n", " Brake thermal efficiency = 21.5207496751 percent\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.6:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "no = 6.0 # No of cylinders\n", "Vs = 1.75 # Stroke volume in litres\n", "P = 26.25 # Power developed in kW\n", "N = 506.0 # Speed in rpm\n", "mep = 600.0 # Mean effectine pressure in kN/m**2\n", "print \"\\n Example 20.6\\n\"\n", "n = P*60000/(no*mep*1000*Vs*(10**-3))\n", "n_e = N/2\n", "n_m = n_e - n\n", "print \"\\nAvg no of misfire = \",n_m\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.6\n", "\n", "\n", "Avg no of misfire = 3.0\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.7:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "Bp = 110 # Brake power in kW\n", "n_m = 0.8 # Mechanical efficiency of the engine\n", "m_f = 50 # Fuel required for engine in kg/h\n", "r_f = 5 # Reduced engine friction in kW\n", "print \"\\n Example 20.7\\n\"\n", "Ip = Bp/n_m\n", "Fp = Ip-Bp\n", "Fp_n = Fp-r_f\n", "Ip_new = Bp + Fp_n\n", "m_f_new = Ip_new * m_f/ Ip\n", "s_f = m_f- m_f_new\n", "print \"\\nSaving in fuel = \",s_f ,\" kg/h\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.7\n", "\n", "\n", "Saving in fuel = 1.81818181818 kg/h\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.8:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "Bp = 14.7 # Brake power when all cylinder operating in kW\n", "Bp1 = 10.14 # Brake power with cylinder no. 1 cut out in kW\n", "Bp2 = 10.3 # Brake power with cylinder no. 2 cut out in kW\n", "Bp3 = 10.36 # Brake power with cylinder no. 3 cut out in kW\n", "Bp4 = 10.21 # Brake power with cylinder no. 4 cut out in kW\n", "m_f = 5.5 # Fuel consumption in kg/h\n", "cv = 42 # Calorific value MJ/kg\n", "d = 8 # Diameter of cylinder in cm\n", "L = 10 # Stroke of cylinder in cm\n", "Vc = 0.1 # Clearance volume in litre\n", "print \"\\n Example 20.8\\n\"\n", "Ip1 = Bp-Bp1\n", "Ip2 = Bp-Bp2\n", "Ip3 = Bp-Bp3\n", "Ip4 = Bp-Bp4\n", "Ip = Ip1+Ip2+Ip3+Ip4\n", "n_m = Bp/Ip\n", "Vs = (math.pi/4)*((d*(10**-2))**2)*(L*(10**-2))\n", "r_k = (Vs+(Vc*(10**-3)))/(Vc*(10**-3))\n", "n_ase = 1- (1/(r_k**(1.4-1)))\n", "n_th = Ip*3600/(m_f*cv*1000)\n", "R_e = n_th/n_ase\n", "print \"\\n Mechanical efficiency = \", ,\" percent,\\n Relative efficiency on indicated power basis = \", ,\" percent\",n_m*100,R_e*100)\n", "#The value of answer is different because of round off error\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [], "prompt_number": 0 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.9:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "Bp = 28.35 # Brake power in kW\n", "N = 1500.0 # Speed in rpm\n", "x = 20.0 # Rich percent of mixture\n", "t = 15.5 # Temperature in degree centrigrde\n", "p = 760 # Pressure in mm of mercury\n", "f = 0.7 # Fraction of volume of air in th cylinder relative to swept volume\n", "R = 14.8 # Theoratical Air fuel ratio\n", "d = 82.0 # Diameter of cylinder in mm\n", "L = 130.0 # Stroke of cylinder in mm\n", "cv = 44.0 # Heating value of petrol in MJ/kg\n", "n_m = 0.9 # Mechanical efficiency of the engine\n", "print \"\\n Example 20.9\\n\"\n", "Ip = Bp/n_m\n", "p_ = 101.325 # In kN/m**2 as p = 760 mm mercury\n", "v_a = f*(math.pi/4)*((d*(10**-3))**2)*(L*(10**-3))*(N/2)*4\n", "m = p_*(v_a)/(0.287*(t+273))\n", "m_f = (m/R)*(1+x/100)\n", "n_th = Ip*3600/(m_f*cv*1000*60)\n", "bmep = Bp*60/((math.pi/4)*((d*(10**-3))**2)*(L*10**-3)*(N/2)*4)\n", "print \"\\n Indicated thermal efficiency = \",n_th*100 ,\" percent,\\n Brake mean effective preassure = \",bmep ,\" kN/m**2\"\n", "#The value of answer is different because of round off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.9\n", "\n", "\n", " Indicated thermal efficiency = 30.0275891939 percent,\n", " Brake mean effective preassure = 825.889834193 kN/m**2\n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.10:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "d = 25.0 # Throat diameter in mm\n", "D = 1.2 # Main jet diameter in mm\n", "c_d = 0.85 # Cofficient of discharge for the venturi \n", "C_d = 0.65 # Cofficient of discharge for fuel jet\n", "h = 6.0 # Height of the throat from gasoline surface in mm\n", "p_1 = 1.0 # Ambient pressure in bar\n", "T = 300.0 # Ambient temperature in K\n", "Ro_f = 760.0 # Density in kg/m**3\n", "print \"\\n Example 20.10\\n\"\n", "delta_p = h*(10**-3)*Ro_f*9.81\n", "p_2 = p_1-delta_p*(10**-5)\n", "Ro_air = p_1*(10**5)/(287*T)\n", "v = (2*delta_p/Ro_air)**(1.0/2.0)\n", "print \"\\n Minimum velocity of air required to start the flow = \",v ,\" m/s\"\n", "#The value of answer is different because of round off error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.10\n", "\n", "\n", " Minimum velocity of air required to start the flow = 8.77674536488 m/s\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.11:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "Bp = 40.0 # Brake power when all cylinder operating in kW\n", "N = 2000.0 # Speed in rpm\n", "Bp1 = 32.2 # Brake power with cylinder no. 1 cut out in kW\n", "Bp2 = 32.0 # Brake power with cylinder no. 2 cut out in kW\n", "Bp3 = 32.5 # Brake power with cylinder no. 3 cut out in kW\n", "Bp4 = 32.4 # Brake power with cylinder no. 4 cut out in kW\n", "Bp5 = 32.1 # Brake power with cylinder no. 5 cut out in kW\n", "Bp6 = 32.3 # Brake power with cylinder no. 6 cut out in kW\n", "d = 100.0 # Diameter of cylinder in mm\n", "L = 125.0 # Stroke of cylinder in mm\n", "Vc = 0.000123 # Clearance volume in m**3\n", "m_f = 9.0 # Fuel consumption in kg/h\n", "cv = 40.0 # Heating value in MJ/kg\n", "print \"\\n Example 20.11\\n\"\n", "Ip1 = Bp-Bp1\n", "Ip2 = Bp-Bp2\n", "Ip3 = Bp-Bp3\n", "Ip4 = Bp-Bp4\n", "Ip5 = Bp-Bp5\n", "Ip6 = Bp-Bp6\n", "Ip = Ip1+Ip2+Ip3+Ip4+Ip5+Ip6\n", "n_m = Bp/Ip\n", "bmep = Bp*2*60/(L*(10**-3)*((d*(10**-3))**2)*(math.pi/4)*N)\n", "Vs = (math.pi/4)*((d*(10**-3))**2)*(L*(10**-3))\n", "r_k = (Vs+Vc)/Vc\n", "n_ase = 1- (1/(r_k**(1.4-1)))\n", "n_th = Ip*3600/(m_f*cv*1000)\n", "R_e = n_th/n_ase\n", "print \"\\n Mechanical efficiency = \",n_m*100 ,\" percent,\\n Brake mean effective pressure = \",bmep*(10**-2) ,\" bar\\n Air standard ratio = \",n_ase*100 ,\" percent,\\n Brake thermal efficiency is \",n_th*100 ,\" percent,\\n Relative efficiency = \",R_e*100 ,\" percent\"\n", "#The value of answer for air standard efficiency is different because of round off error\n", "# Answer given in the book for bmep is 3.055 bar which is wrong.\n", "# Answer given in the book for brake thermal efficiency is 40 percent which is wrong.\n", "# Answer given in the book for relative efficiency is 68.6 percent which is wrong.\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.11\n", "\n", "\n", " Mechanical efficiency = 86.0215053763 percent,\n", " Brake mean effective pressure = 24.4461992589 bar\n", " Air standard ratio = 58.4417930454 percent,\n", " Brake thermal efficiency is 46.5 percent,\n", " Relative efficiency = 79.5663472609 percent\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.12:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "p1 = 0.95 # Pressure in bar\n", "t1 = 25 # Temperature in degree centigrade\n", "p2 = 2 # Delivery pressure in bar\n", "r = 18 # Air fuel ratio\n", "t3 = 600 # Temperature of gasses leaving the engine in degree centigrade\n", "p3 = 1.8 # Pressure of gasses leaving the engine in bar\n", "p4 = 1.04 # Pressure at the inlet of turbine in bar\n", "n_c = 0.75 # Efficiency of compresor\n", "n_t = 0.85 # Efficiency of turbine\n", "Cp = 1.005 # Heat capacity of air in kJ/kgK\n", "Cp_ = 1.15 # Heat capacity of gasses in kJ/kgK\n", "gama = 1.4 # Adiabatic index for air\n", "print \"\\n Example 20.12\\n\"\n", "T2_s = (t1+273)*(p2/p1)**((gama-1)/gama)\n", "T2 = (t1+273)+((T2_s-(t1+273))/n_c)\n", "Wc = Cp*(T2-(t1+273))\n", "T4_s = (t3+273)*((p4/p3)**((gama-1)/gama))\n", "T4 = (t3+273)-((t3+273)-T4_s)*n_t\n", "Wt = (1+(1/r))*Cp_*((t3+273)-T4)\n", "n = (Wt-Wc)/Wt\n", "print \"\\n Power lost as a percentage of the power produced by the turbine = \",n*100 ,\" percent\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.12\n", "\n", "\n", " Power lost as a percentage of the power produced by the turbine = 23.5485226573 percent\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.13:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "Bp = 250.0 # Power developed by the engine in kW\n", "n = 6.0 # No of cylinders \n", "N = 2000.0 # Speed in rpm\n", "bsfc = 0.2 # Specific fuel consumption in kg/kWh\n", "P = 35.0 # Pressure at the begining of the injection in bar\n", "p_max = 55.0 # Maximum cylinder pressure in bar\n", "p = 180.0 # Expected pressure for injection in bar\n", "P_max = 520.0 # Maximum pressure at the injection in bar\n", "c_d = 0.78 # Cofficient of discharge\n", "s = 0.85 # Specific gravity of fuel oil\n", "p_atm = 1.0 # Atmospheric pressure in bar\n", "theta = 18.0 # Crank angle in degree\n", "print \"\\n Example 20.13\\n\"\n", "Bp_cy = Bp/n\n", "m_f = Bp_cy*bsfc/60 # in kg/min\n", "f_c = m_f*(2/N)\n", "T = theta/(360*(N/60))\n", "delta_p = p-P\n", "delta_p_ = P_max-p_max\n", "avg_delta_p = (delta_p+delta_p_)/2\n", "v = c_d*sqrt((2*(avg_delta_p)*(10**5))/(s*1000))\n", "V = m_f*(10**-3)/(s*1000)\n", "A = V/(v*T)\n", "print \"\\n Total orifice area per injector = \",A*10**6 ,\" mm**2\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.13\n", "\n", "\n", " Total orifice area per injector = 0.521323450963 mm**2\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.14:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "n=1.3 # Polytropic index\n", "p1 = 140.0 # Pressure at point one in kN/m**2\n", "p2 = 360.0 # Pressure at point two in kN/m**2\n", "r_e = 0.4 # Relative efficiency\n", "cv = 18840 # Calorific value in kJ/m**2\n", "print \"\\n Example 20.14\\n\"\n", "r = (((p2/p1)**(1/n))-1)/((0.75-0.25*((p2/p1)**(1.0/n))))\n", "r_k = r+1\n", "n_ase = 1.0-(1.0/((r_k)**(0.4)))\n", "n_th = r_e*n_ase\n", "V_f = n_th*cv/3600\n", "print \"\\n Thermal efficiency = \",n_th*100 ,\" percemt,\\n Gas consumption per kWh on indicated power basis = \",V_f ,\" m**3/kWh\"\n", "#The value of answer is different because of round off error\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.14\n", "\n", "\n", " Thermal efficiency = 19.8935818353 percemt,\n", " Gas consumption per kWh on indicated power basis = 1.04109744938 m**3/kWh\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.15:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "# Given that\n", "d = 180.0 # Bore in mm\n", "L = 200.0 # Stroke in mm\n", "Bp = 245.0 # Brake power in kW\n", "N = 1500.0 # Speed in rpm\n", "mep = 8.0 # Mean effective pressure in bar\n", "m_f = 70.0 # Fuel consumption in kg/h\n", "cv = 42.0 # Heating value of fuel in MJ/kg\n", "m_h = 0.12 # Fraction of hydrogen content by mass\n", "m_a = 26.0 # Air consumption in kg/min\n", "m_w = 82.0 # Mass of cooling water in kg/min\n", "delta_t = 44 # Cooling water temperature rise in degree centigrade\n", "m_o = 50.0 # Cooling oil circulated through the engine in kg/min\n", "delta_T = 24 # Cooling oil temperature rise in degree centigrade\n", "s_o = 2.1 # Specific heat of cooling oil in kJ/kgK\n", "t = 30.0 # Room temperature in degree centigrade\n", "t_e = 400.0 # Exhaust gas temperature on degree centigrade\n", "c_p_de = 1.045 # Heat capacity of dry exhaust gas in kJ/kgK\n", "p = 0.035 # Partial pressure of steam in exhaust gas in bar\n", "print \"\\n Example 20.15\\n\"\n", "h = m_f*cv*1000/3600\n", "Ip = mep*(10**5)*L*(10**-3)*(math.pi/4)*((d*(10**-3))**2)*N*6/(2*60000)\n", "n_m = Bp/Ip\n", "h_w = (m_w/60)*(4.187*delta_t)\n", "h_o = (m_o/60)*(s_o*delta_T)\n", "m_e = m_f/60 + m_a\n", "m_v = m_h*9*(m_f/60)\n", "m_de = (m_e-m_v)/60\n", "H = 3060 # From the steam table the enthalpy of steam at the exhaust contion(0.035 bar) in kJ/kg\n", "h_s = (m_v/60)*H\n", "h_de = (m_de)*(c_p_de)*(t_e-t)\n", "h_su = h - (Bp+h_w+h_s+h_o+h_de)\n", "print \"\\n Mechanical efficiency = \",n_m*100 ,\" percemt\"\n", "print \"\\n Energy Balance\"\n", "print \"\\n Input Output\"\n", "print \"\\n Heat supplied by fuel \",h ,\" kW -\"\n", "print \"\\n Useful work(BP) - \",Bp ,\" kW\"\n", "print \"\\n Heat carried by cooling water - \",h_w ,\" kW\"\n", "print \"\\n Heat carried by steam - \",h_s ,\" kW\"\n", "print \"\\n Heat carried by cooling oil - \",h_o ,\" kW\"\n", "print \"\\n Heat carried by dry exhaust gas - \",h_de ,\" kW\"\n", "print \"\\n Heat transferred to surroundings - \",h_su ,\" kW\"\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.15\n", "\n", "\n", " Mechanical efficiency = 80.2324301595 percemt\n", "\n", " Energy Balance\n", "\n", " Input Output\n", "\n", " Heat supplied by fuel 816.666666667 kW -\n", "\n", " Useful work(BP) - 245.0 kW\n", "\n", " Heat carried by cooling water - 251.778266667 kW\n", "\n", " Heat carried by steam - 64.26 kW\n", "\n", " Heat carried by cooling oil - 42.0 kW\n", "\n", " Heat carried by dry exhaust gas - 166.946877778 kW\n", "\n", " Heat transferred to surroundings - 46.6815222222 kW\n" ] } ], "prompt_number": 22 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex20.16:pg-853" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given that\n", "N = 3000 # Speed in rpm\n", "T = 66.5 # Torque in Nm\n", "d = 60 # Bore in mm\n", "L = 100 # Stroke in mm\n", "Vc = 60 # Clearance volume in cc\n", "r_e = 0.5 # Relative efficiency\n", "cv = 42 # Calorific value in MJ/kg\n", "print \"\\n Example 20.16\\n\"\n", "Vs = (math.pi/4)*((60*(10**-3))**2)*(L*(10**-3))\n", "r_k = (Vs+(Vc*(10**-6)))/(Vc*(10**-6))\n", "n_ase = 1-(1/(r_k**(0.4)))\n", "n_br = n_ase*r_e\n", "Bp = (2*(math.pi)*T*N)/(60000)\n", "m_f = Bp*3600/(cv*1000*n_br)\n", "bmep = Bp*60000/(Vs*(N/2))\n", "print \"\\n Fuel consumption = \",m_f ,\" kg/h,\\n Brake mean effective pressure = \",bmep*(10**-5) ,\" bar\"\n", "#The answer given in the book for bmep has calculation error\n", "# The answer has round off error for fuel consumption" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "\n", " Example 20.16\n", "\n", "\n", " Fuel consumption = 7.13500385939 kg/h,\n", " Brake mean effective pressure = 29.5555555556 bar\n" ] } ], "prompt_number": 23 } ], "metadata": {} } ] }