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+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Chapter 10:Intoduction to Internal Combustion engines"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.1;pg no: 387"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.1, Page:387 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 1\n",
+ "from stroke to bore ratio i.e L/D=1.2 and cylinder diameter=bore,i.e D=12 cm\n",
+ "stroke(L)=1.2*D in m\n",
+ "Area of indicator diagram(A)=30*10^-4 m^2\n",
+ "length of indicator diagram(l)=(1/2)*L in m\n",
+ "mean effective pressure(mep)=A*k/l in N/m^2\n",
+ "cross-section area of piston(Ap)=%pi*D^2/4 in m^2\n",
+ "for one cylinder indicated power(IP)=mep*Ap*L*N/(2*60) in W\n",
+ "for four cylinder total indicated power(IP)=4*IP in W 90477.87\n",
+ "frictional power loss(FP)=0.10*IP in W\n",
+ "brake power available(BP)=indicated power-frictional power=IP-FP in W 81430.08\n",
+ "therefore,mechanical efficiency of engine(n)=brake power/indicated power=BP/IP= 0.9\n",
+ "in percentage 90.0\n",
+ "so indicated power=90477.8 W\n",
+ "and mechanical efficiency=90%\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of indicated power,,mechanical efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 9\n",
+ "import math\n",
+ "print\"Example 10.1, Page:387 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 1\")\n",
+ "k=20.*10**6;#spring constant in N/m^2\n",
+ "N=2000.;#engine rpm\n",
+ "print(\"from stroke to bore ratio i.e L/D=1.2 and cylinder diameter=bore,i.e D=12 cm\")\n",
+ "D=12.*10**-2;#cylinder diameter in m\n",
+ "print(\"stroke(L)=1.2*D in m\")\n",
+ "L=1.2*D\n",
+ "print(\"Area of indicator diagram(A)=30*10^-4 m^2\")\n",
+ "A=30.*10**-4;\n",
+ "print(\"length of indicator diagram(l)=(1/2)*L in m\")\n",
+ "l=(1./2.)*L\n",
+ "print(\"mean effective pressure(mep)=A*k/l in N/m^2\")\n",
+ "mep=A*k/l\n",
+ "print(\"cross-section area of piston(Ap)=%pi*D^2/4 in m^2\")\n",
+ "Ap=math.pi*D**2./4.\n",
+ "print(\"for one cylinder indicated power(IP)=mep*Ap*L*N/(2*60) in W\")\n",
+ "IP=mep*Ap*L*N/(2.*60.)\n",
+ "IP=4.*IP\n",
+ "print(\"for four cylinder total indicated power(IP)=4*IP in W\"),round(IP,2)\n",
+ "print(\"frictional power loss(FP)=0.10*IP in W\")\n",
+ "FP=0.10*IP\n",
+ "BP=IP-FP\n",
+ "print(\"brake power available(BP)=indicated power-frictional power=IP-FP in W\"),round(BP,2)\n",
+ "n=BP/IP\n",
+ "print(\"therefore,mechanical efficiency of engine(n)=brake power/indicated power=BP/IP=\"),round(BP/IP,2)\n",
+ "print(\"in percentage\"),round(n*100,2)\n",
+ "print(\"so indicated power=90477.8 W\")\n",
+ "print(\"and mechanical efficiency=90%\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.2;pg no: 388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 2,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.2, Page:388 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 2\n",
+ "reciprocating pump is work absorbing machine having its mechanism similar to the piston-cylinder arrangement in IC engine.\n",
+ "mean effective pressure(mep)=A*k/l in pa\n",
+ "indicator power(IP)=Ap*L*mep*N*1*2/60 in W\n",
+ "(it is double acting so let us assume total power to be double of that in single acting) 88357.29\n",
+ "so power required to drive=88.36 KW\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of power required to drive\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.2, Page:388 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 2\")\n",
+ "A=40*10**-4;#area of indicator diagram in m^2\n",
+ "l=8*10**-2;#length of indicator diagram in m\n",
+ "D=15*10**-2;#bore of cylinder in m\n",
+ "L=20*10**-2;#stroke in m\n",
+ "k=1.5*10**8;#spring constant in pa/m\n",
+ "N=100;#pump motor rpm\n",
+ "print(\"reciprocating pump is work absorbing machine having its mechanism similar to the piston-cylinder arrangement in IC engine.\")\n",
+ "print(\"mean effective pressure(mep)=A*k/l in pa\")\n",
+ "mep=A*k/l \n",
+ "print(\"indicator power(IP)=Ap*L*mep*N*1*2/60 in W\")\n",
+ "Ap=math.pi*D**2/4\n",
+ "IP=Ap*L*mep*N*2/60\n",
+ "print(\"(it is double acting so let us assume total power to be double of that in single acting)\"),round(IP,2)\n",
+ "print(\"so power required to drive=88.36 KW\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.3;pg no: 388"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 3,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.3, Page:388 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 3\n",
+ "indicated power(IP)=brake power/mechanical efficiency in KW= 42.22\n",
+ "frictional power loss(FP)=IP-BP in KW 4.22\n",
+ "brake power at quater load(BPq)=0.25*BP in KW\n",
+ "mechanical efficiency(n1)=BPq/IP 0.69\n",
+ "in percentage 69.23\n",
+ "so indicated power=42.22 KW\n",
+ "frictional power loss=4.22 KW\n",
+ "mechanical efficiency=69.24%\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of indicated power,frictional power loss,mechanical efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "print\"Example 10.3, Page:388 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 3\")\n",
+ "n=0.9;#mechanical efficiency of engine\n",
+ "BP=38;#brake power in KW\n",
+ "IP=BP/n\n",
+ "print(\"indicated power(IP)=brake power/mechanical efficiency in KW=\"),round(IP,2)\n",
+ "FP=IP-BP\n",
+ "print(\"frictional power loss(FP)=IP-BP in KW\"),round(FP,2)\n",
+ "print(\"brake power at quater load(BPq)=0.25*BP in KW\")\n",
+ "BPq=0.25*BP\n",
+ "IP=BPq+FP;\n",
+ "n1=BPq/IP\n",
+ "print(\"mechanical efficiency(n1)=BPq/IP\"),round(n1,2)\n",
+ "print(\"in percentage\"),round(n1*100,2)\n",
+ "print(\"so indicated power=42.22 KW\")\n",
+ "print(\"frictional power loss=4.22 KW\")\n",
+ "print(\"mechanical efficiency=69.24%\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.4;pg no: 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 4,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.4, Page:389 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 4\n",
+ "brake power of engine(BP) in MW= 3120.98\n",
+ "so brake power is 3.121 MW\n",
+ "The fuel consumption in kg/hr(m)=m*BP in kg/hr\n",
+ "In order to find out brake thermal efficiency the heat input from fuel per second is required.\n",
+ "heat from fuel(Q)in KJ/s\n",
+ "Q=m*C/3600\n",
+ "energy to brake power=3120.97 KW\n",
+ "brake thermal efficiency(n)= 0.33\n",
+ "in percentage 33.49\n",
+ "so fuel consumption=780.24 kg/hr,brake thermal efficiency=33.49%\n",
+ "NOTE=>In book,it is given that brake power in MW is 31.21 but in actual it is 3120.97 KW or 3.121 MW which is corrected above.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake power,fuel consumption\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.4, Page:389 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 4\")\n",
+ "m=0.25;#specific fuel consumption in kg/KWh\n",
+ "Pb_mep=1.5*1000;#brake mean effective pressure of each cylinder in Kpa\n",
+ "N=100;#engine rpm\n",
+ "D=85*10**-2;#bore of cylinder in m\n",
+ "L=220*10**-2;#stroke in m\n",
+ "C=43*10**3;#calorific value of diesel in KJ/kg\n",
+ "A=math.pi*D**2/4;\n",
+ "BP=Pb_mep*L*A*N/60\n",
+ "print(\"brake power of engine(BP) in MW=\"),round(BP,2)\n",
+ "print(\"so brake power is 3.121 MW\")\n",
+ "print(\"The fuel consumption in kg/hr(m)=m*BP in kg/hr\")\n",
+ "m=m*BP\n",
+ "print(\"In order to find out brake thermal efficiency the heat input from fuel per second is required.\")\n",
+ "print(\"heat from fuel(Q)in KJ/s\")\n",
+ "print(\"Q=m*C/3600\")\n",
+ "Q=m*C/3600\n",
+ "print(\"energy to brake power=3120.97 KW\")\n",
+ "n=BP/Q\n",
+ "print(\"brake thermal efficiency(n)=\"),round(n,2)\n",
+ "print(\"in percentage\"),round(n*100,2)\n",
+ "print(\"so fuel consumption=780.24 kg/hr,brake thermal efficiency=33.49%\")\n",
+ "print(\"NOTE=>In book,it is given that brake power in MW is 31.21 but in actual it is 3120.97 KW or 3.121 MW which is corrected above.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.5;pg no: 389"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 5,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.5, Page:389 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 5\n",
+ "brake thermal efficiency(n)=3600/(m*C) 0.33\n",
+ "in percentage 33.49\n",
+ "brake power(BP)in KW\n",
+ "BP= 226.19\n",
+ "brake specific fuel consumption,m=mf/BP\n",
+ "so mf=m*BP in kg/hr\n",
+ "air consumption(ma)from given air fuel ratio=k*mf in kg/hr\n",
+ "ma in kg/min\n",
+ "using perfect gas equation,\n",
+ "P*Va=ma*R*T\n",
+ "sa Va=ma*R*T/P in m^3/min\n",
+ "swept volume(Vs)=%pi*D^2*L/4 in m^3\n",
+ "volumetric efficiency,n_vol=Va/(Vs*(N/2)*no. of cylinder) 1.87\n",
+ "in percentage 186.55\n",
+ "NOTE=>In this question,while calculating swept volume in book,values of D=0.30 m and L=0.4 m is taken which is wrong.Hence above solution is solve taking right values given in book which is D=0.20 m and L=0.3 m,so the volumetric efficiency vary accordingly.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake thermal efficiency,brake power,volumetric efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.5, Page:389 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 5\")\n",
+ "Pb_mep=6*10**5;#brake mean effective pressure in pa\n",
+ "N=600;#engine rpm\n",
+ "m=0.25;#specific fuel consumption in kg/KWh\n",
+ "D=20*10**-2;#bore of cylinder in m\n",
+ "L=30*10**-2;#stroke in m\n",
+ "k=26;#air to fuel ratio\n",
+ "C=43*10**3;#calorific value in KJ/kg\n",
+ "R=0.287;#gas constant in KJ/kg K\n",
+ "T=(27+273);#ambient temperature in K\n",
+ "P=1*10**2;#ambient pressure in Kpa\n",
+ "n=3600/(m*C)\n",
+ "print(\"brake thermal efficiency(n)=3600/(m*C)\"),round(n,2)\n",
+ "print(\"in percentage\"),round(n*100,2)\n",
+ "print(\"brake power(BP)in KW\")\n",
+ "A=math.pi*D**2/4;\n",
+ "BP=4*Pb_mep*L*A*N/60000\n",
+ "print(\"BP=\"),round(BP,2)\n",
+ "print(\"brake specific fuel consumption,m=mf/BP\")\n",
+ "print(\"so mf=m*BP in kg/hr\")\n",
+ "mf=m*BP\n",
+ "print(\"air consumption(ma)from given air fuel ratio=k*mf in kg/hr\")\n",
+ "ma=k*mf\n",
+ "print(\"ma in kg/min\")\n",
+ "ma=ma/60\n",
+ "print(\"using perfect gas equation,\")\n",
+ "print(\"P*Va=ma*R*T\")\n",
+ "print(\"sa Va=ma*R*T/P in m^3/min\")\n",
+ "Va=ma*R*T/P\n",
+ "print(\"swept volume(Vs)=%pi*D^2*L/4 in m^3\")\n",
+ "Vs=math.pi*D**2*L/4\n",
+ "n_vol=Va/(Vs*(N/2)*4)\n",
+ "print(\"volumetric efficiency,n_vol=Va/(Vs*(N/2)*no. of cylinder)\"),round(n_vol,2)\n",
+ "print(\"in percentage\"),round(n_vol*100,2)\n",
+ "print(\"NOTE=>In this question,while calculating swept volume in book,values of D=0.30 m and L=0.4 m is taken which is wrong.Hence above solution is solve taking right values given in book which is D=0.20 m and L=0.3 m,so the volumetric efficiency vary accordingly.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.6;pg no: 390"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 6,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.6, Page:390 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 6\n",
+ "let the bore diameter be (D) m\n",
+ "piston speed(V)=2*L*N\n",
+ "so L=V/(2*N) in m\n",
+ "volumetric efficiency,n_vol=air sucked/(swept volume * no. of cylinder)\n",
+ "so air sucked/D^2=n_vol*(%pi*L/4)*N*2 in m^3/min\n",
+ "so air sucked =274.78*D^2 m^3/min\n",
+ "air requirement(ma),kg/min=A/F ratio*fuel consumption per min\n",
+ "so ma=r*m in kg/min\n",
+ "using perfect gas equation,P*Va=ma*R*T\n",
+ "so Va=ma*R*T/P in m^3/min\n",
+ "ideally,air sucked=Va\n",
+ "so 274.78*D^2=0.906\n",
+ "D=sqrt(0.906/274.78) in m\n",
+ "indicated power(IP)=P_imep*L*A*N*no.of cylinders in KW\n",
+ "brake power=indicated power*mechanical efficiency\n",
+ "BP=IP*n_mech in KW 10.35\n",
+ "so brake power=10.34 KW\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake power\n",
+ "#intiation of all variables\n",
+ "# Chapter 9\n",
+ "import math\n",
+ "print\"Example 10.6, Page:390 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 6\")\n",
+ "N=3000;#engine rpm\n",
+ "m=5;#fuel consumption in litre/hr\n",
+ "r=19;#air-fuel ratio\n",
+ "sg=0.7;#specific gravity of fuel\n",
+ "V=500;#piston speed in m/min\n",
+ "P_imep=6*10**5;#indicated mean effective pressure in pa\n",
+ "P=1.013*10**5;#ambient pressure in pa\n",
+ "T=(15+273);#ambient temperature in K\n",
+ "n_vol=0.7;#volumetric efficiency \n",
+ "n_mech=0.8;#mechanical efficiency\n",
+ "R=0.287;#gas constant for gas in KJ/kg K\n",
+ "print(\"let the bore diameter be (D) m\")\n",
+ "print(\"piston speed(V)=2*L*N\")\n",
+ "print(\"so L=V/(2*N) in m\")\n",
+ "L=V/(2*N)\n",
+ "L=0.0833;#approx.\n",
+ "print(\"volumetric efficiency,n_vol=air sucked/(swept volume * no. of cylinder)\")\n",
+ "print(\"so air sucked/D^2=n_vol*(%pi*L/4)*N*2 in m^3/min\")\n",
+ "n_vol*(math.pi*L/4)*N*2\n",
+ "print(\"so air sucked =274.78*D^2 m^3/min\")\n",
+ "print(\"air requirement(ma),kg/min=A/F ratio*fuel consumption per min\")\n",
+ "print(\"so ma=r*m in kg/min\")\n",
+ "ma=r*m*sg/60\n",
+ "print(\"using perfect gas equation,P*Va=ma*R*T\")\n",
+ "print(\"so Va=ma*R*T/P in m^3/min\")\n",
+ "Va=ma*R*T*1000/P \n",
+ "print(\"ideally,air sucked=Va\")\n",
+ "print(\"so 274.78*D^2=0.906\")\n",
+ "print(\"D=sqrt(0.906/274.78) in m\")\n",
+ "D=math.sqrt(0.906/274.78) \n",
+ "print(\"indicated power(IP)=P_imep*L*A*N*no.of cylinders in KW\")\n",
+ "IP=P_imep*L*(math.pi*D**2/4)*(N/60)*2/1000\n",
+ "print(\"brake power=indicated power*mechanical efficiency\")\n",
+ "BP=IP*n_mech \n",
+ "print(\"BP=IP*n_mech in KW\"),round(BP,2)\n",
+ "print(\"so brake power=10.34 KW\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.7;pg no: 391"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 7,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.7, Page:391 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 7\n",
+ "After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine.\n",
+ "friction power(FP)=5 KW\n",
+ "brake power(BP) in KW= 30.82\n",
+ "indicated power(IP) in KW= 35.82\n",
+ "mechanical efficiency(n_mech)= 0.86\n",
+ "in percentage 86.04\n",
+ "brake specific fuel consumption(bsfc)=specific fuel consumption/brake power in kg/KW hr= 0.29\n",
+ "brake thermal efficiency(n_bte)= 0.29\n",
+ "in percentage 28.67\n",
+ "also,mechanical efficiency(n_mech)=brake thermal efficiency/indicated thermal efficiency\n",
+ "indicated thermal efficiency(n_ite)= 0.33\n",
+ "in percentage 33.32\n",
+ "indicated power(IP)=P_imep*L*A*N\n",
+ "so P_imep in Kpa= 76.01\n",
+ "Also,mechanical efficiency=P_bmep/P_imep\n",
+ "so P_bmep in Kpa= 65.4\n",
+ "brake power=30.82 KW\n",
+ "indicated power=35.82 KW\n",
+ "mechanical efficiency=86.04%\n",
+ "brake thermal efficiency=28.67%\n",
+ "indicated thermal efficiency=33.32%\n",
+ "brake mean effective pressure=65.39 Kpa\n",
+ "indicated mean effective pressure=76.01 Kpa\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of power and efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.7, Page:391 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 7\")\n",
+ "M=20;#load on dynamometer in kg\n",
+ "r=50*10**-2;#radius in m\n",
+ "N=3000;#speed of rotation in rpm\n",
+ "D=20*10**-2;#bore in m\n",
+ "L=30*10**-2;#stroke in m\n",
+ "m=0.15;#fuel supplying rate in kg/min\n",
+ "C=43;#calorific value of fuel in MJ/kg\n",
+ "FP=5;#friction power in KW\n",
+ "g=9.81;#acceleration due to gravity in m/s^2\n",
+ "print(\"After switching off fuel supply the capacity of motor required to run engine will be the friction power required at this speed of engine.\")\n",
+ "print(\"friction power(FP)=5 KW\")\n",
+ "BP=2*math.pi*N*(M*g*r)*10**-3/60\n",
+ "print(\"brake power(BP) in KW=\"),round(BP,2)\n",
+ "IP=BP+FP\n",
+ "print(\"indicated power(IP) in KW=\"),round(IP,2)\n",
+ "n_mech=BP/IP\n",
+ "print(\"mechanical efficiency(n_mech)=\"),round(n_mech,2)\n",
+ "print(\"in percentage\"),round(n_mech*100,2)\n",
+ "bsfc=m*60/BP\n",
+ "print(\"brake specific fuel consumption(bsfc)=specific fuel consumption/brake power in kg/KW hr=\"),round(bsfc,2)\n",
+ "n_bte=3600/(bsfc*C*1000)\n",
+ "print(\"brake thermal efficiency(n_bte)=\"),round(n_bte,2)\n",
+ "print(\"in percentage\"),round(n_bte*100,2)\n",
+ "print(\"also,mechanical efficiency(n_mech)=brake thermal efficiency/indicated thermal efficiency\")\n",
+ "n_ite=n_bte/n_mech\n",
+ "print(\"indicated thermal efficiency(n_ite)=\"),round(n_ite,2)\n",
+ "print(\"in percentage\"),round(n_ite*100,2)\n",
+ "print(\"indicated power(IP)=P_imep*L*A*N\")\n",
+ "P_imep=IP/(L*(math.pi*D**2/4)*N/60)\n",
+ "print(\"so P_imep in Kpa=\"),round(P_imep,2)\n",
+ "print(\"Also,mechanical efficiency=P_bmep/P_imep\")\n",
+ "n_mech=0.8604;#mechanical efficiency\n",
+ "P_bmep=P_imep*n_mech\n",
+ "print(\"so P_bmep in Kpa=\"),round(P_bmep,2)\n",
+ "print(\"brake power=30.82 KW\")\n",
+ "print(\"indicated power=35.82 KW\")\n",
+ "print(\"mechanical efficiency=86.04%\")\n",
+ "print(\"brake thermal efficiency=28.67%\")\n",
+ "print(\"indicated thermal efficiency=33.32%\")\n",
+ "print(\"brake mean effective pressure=65.39 Kpa\")\n",
+ "print(\"indicated mean effective pressure=76.01 Kpa\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.8;pg no: 392"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 8,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.8, Page:392 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 8\n",
+ "indicated power(IP) in KW= 282.74\n",
+ "mechanical efficiency(n_mech)=brake power/indicated power\n",
+ "so n_mech= 0.88\n",
+ "in percentage 88.42\n",
+ "brake specific fuel consumption(bsfc)=m*60/BP in kg/KW hr\n",
+ "brake thermal efficiency(n_bte)= 0.35\n",
+ "in percentage 34.88\n",
+ "swept volume(Vs)=%pi*D^2*L/4 in m^3\n",
+ "mass of air corresponding to above swept volume,using perfect gas equation\n",
+ "P*Vs=ma*R*T\n",
+ "so ma=(P*Vs)/(R*T) in kg\n",
+ "volumetric effeciency(n_vol)=mass of air taken per minute/mass corresponding to swept volume per minute\n",
+ "so mass of air taken per minute in kg/min \n",
+ "mass corresponding to swept volume per minute in kg/min\n",
+ "so volumetric efficiency 0.8333\n",
+ "in percentage 83.3333\n",
+ "so indicated power =282.74 KW,mechanical efficiency=88.42%\n",
+ "brake thermal efficiency=34.88%,volumetric efficiency=83.33%\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake thermal efficiency,volumetric effeciency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.8, Page:392 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 8\")\n",
+ "N=300.;#engine rpm\n",
+ "BP=250.;#brake power in KW\n",
+ "D=30.*10**-2;#bore in m\n",
+ "L=25.*10**-2;#stroke in m\n",
+ "m=1.;#fuel consumption in kg/min\n",
+ "r=10.;#airfuel ratio \n",
+ "P_imep=0.8;#indicated mean effective pressure in pa\n",
+ "C=43.*10**3;#calorific value of fuel in KJ/kg\n",
+ "P=1.013*10**5;#ambient pressure in K\n",
+ "R=0.287;#gas constant in KJ/kg K\n",
+ "T=(27.+273.);#ambient temperature in K\n",
+ "IP=P_imep*L*(math.pi*D**2/4)*N*4*10**3/60\n",
+ "print(\"indicated power(IP) in KW=\"),round(IP,2)\n",
+ "print(\"mechanical efficiency(n_mech)=brake power/indicated power\")\n",
+ "n_mech=BP/IP\n",
+ "print(\"so n_mech=\"),round(n_mech,2)\n",
+ "print(\"in percentage \"),round(n_mech*100,2)\n",
+ "print(\"brake specific fuel consumption(bsfc)=m*60/BP in kg/KW hr\")\n",
+ "bsfc=m*60./BP\n",
+ "n_bte=3600./(bsfc*C)\n",
+ "print(\"brake thermal efficiency(n_bte)=\"),round(n_bte,2)\n",
+ "print(\"in percentage\"),round(n_bte*100,2)\n",
+ "print(\"swept volume(Vs)=%pi*D^2*L/4 in m^3\")\n",
+ "Vs=math.pi*D**2*L/4\n",
+ "print(\"mass of air corresponding to above swept volume,using perfect gas equation\")\n",
+ "print(\"P*Vs=ma*R*T\")\n",
+ "print(\"so ma=(P*Vs)/(R*T) in kg\")\n",
+ "ma=(P*Vs)/(R*T*1000) \n",
+ "ma=0.02;#approx.\n",
+ "print(\"volumetric effeciency(n_vol)=mass of air taken per minute/mass corresponding to swept volume per minute\")\n",
+ "print(\"so mass of air taken per minute in kg/min \")\n",
+ "1*10\n",
+ "print(\"mass corresponding to swept volume per minute in kg/min\")\n",
+ "ma*4*N/2\n",
+ "print(\"so volumetric efficiency \"),round(10./12.,4)\n",
+ "print(\"in percentage\"),round((10./12.)*100.,4)\n",
+ "print(\"so indicated power =282.74 KW,mechanical efficiency=88.42%\")\n",
+ "print(\"brake thermal efficiency=34.88%,volumetric efficiency=83.33%\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.9;pg no: 393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 9,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.9, Page:393 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 9\n",
+ "indicated mean effective pressure(P_imeb)=h*k in Kpa\n",
+ "indicated power(IP)=P_imeb*L*A*N/2 in KW 9.375\n",
+ "brake power(BP)=2*%pi*N*T in KW 4.62\n",
+ "mechanical efficiency(n_mech)= 0.49\n",
+ "in percentage 49.31\n",
+ "so indicated power=9.375 KW\n",
+ "brake power=4.62 KW\n",
+ "mechanical efficiency=49.28%\n",
+ "energy liberated from fuel(Ef)=C*m/60 in KJ/s\n",
+ "energy available as brake power(BP)=4.62 KW\n",
+ "energy to coolant(Ec)=(mw/M)*Cw*(T2-T1) in KW\n",
+ "energy carried by exhaust gases(Eg)=30 KJ/s\n",
+ "unaccounted energy loss in KW= 34.75\n",
+ "NOTE=>overall energy balance sheet is attached as jpg file with this code.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of indicated power,brake power,mechanical efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.9, Page:393 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 9\")\n",
+ "h=10.;#height of indicator diagram in mm\n",
+ "k=25.;#indicator constant in KN/m^2 per mm\n",
+ "N=300.;#engine rpm\n",
+ "Vs=1.5*10**-2;#swept volume in m^3\n",
+ "M=60.;#effective brake load upon dynamometer in kg\n",
+ "r=50.*10**-2;#effective brake drum radius in m\n",
+ "m=0.12;#fuel consumption in kg/min\n",
+ "C=42.*10**3;#calorific value in KJ/kg\n",
+ "mw=6.;#circulating water rate in kg/min\n",
+ "T1=35.;#cooling water entering temperature in degree celcius\n",
+ "T2=70.;#cooling water leaving temperature in degree celcius\n",
+ "Eg=30.;#exhaust gases leaving energy in KJ/s\n",
+ "Cw=4.18;#specific heat of water in KJ/kg K\n",
+ "g=9.81;#accelaration due to gravity in m/s^2\n",
+ "print(\"indicated mean effective pressure(P_imeb)=h*k in Kpa\")\n",
+ "P_imeb=h*k\n",
+ "IP=P_imeb*Vs*N/(2*60)\n",
+ "print(\"indicated power(IP)=P_imeb*L*A*N/2 in KW\") ,round(IP,3)\n",
+ "BP=2*math.pi*N*(M*g*r*10**-3)/(2*60)\n",
+ "print(\"brake power(BP)=2*%pi*N*T in KW\"),round(BP,2)\n",
+ "n_mech=BP/IP\n",
+ "print(\"mechanical efficiency(n_mech)=\"),round(n_mech,2)\n",
+ "print(\"in percentage\"),round(n_mech*100,2)\n",
+ "print(\"so indicated power=9.375 KW\")\n",
+ "print(\"brake power=4.62 KW\")\n",
+ "print(\"mechanical efficiency=49.28%\")\n",
+ "print(\"energy liberated from fuel(Ef)=C*m/60 in KJ/s\")\n",
+ "Ef=C*m/60\n",
+ "print(\"energy available as brake power(BP)=4.62 KW\")\n",
+ "print(\"energy to coolant(Ec)=(mw/M)*Cw*(T2-T1) in KW\")\n",
+ "Ec=(mw/M)*Cw*(T2-T1)\n",
+ "print(\"energy carried by exhaust gases(Eg)=30 KJ/s\")\n",
+ "Ef-BP-Ec-Eg\n",
+ "print(\"unaccounted energy loss in KW=\"),round(Ef-BP-Ec-Eg,2)\n",
+ "print(\"NOTE=>overall energy balance sheet is attached as jpg file with this code.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.10;pg no: 394"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 10,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.10, Page:394 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 10\n",
+ "brake power(BP)=2*%pi*N*T in KW 47.12\n",
+ "so brake power=47.124 KW\n",
+ "brake specific fuel consumption(bsfc) in kg/KW hr= 0.34\n",
+ "indicated power(IP) in Kw= 52.36\n",
+ "indicated thermal efficiency(n_ite)= 0.28\n",
+ "in percentage 28.05\n",
+ "so indicated thermal efficiency=28.05%\n",
+ "heat available from fuel(Qf)=(m/mw)*C in KJ/min 11200.0\n",
+ "energy consumed as brake power(BP) in KJ/min= 2827.43\n",
+ "energy carried by cooling water(Qw) in KJ/min= 1442.1\n",
+ "energy carried by exhaust gases(Qg) in KJ/min= 4786.83\n",
+ "unaccounted energy loss in KJ/min 2143.63\n",
+ "NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake power,brake specific fuel consumption,indicated thermal efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.10, Page:394 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 10\")\n",
+ "m=4.;#mass of fuel consumed in kg\n",
+ "N=1500.;#engine rpm\n",
+ "mw=15.;#water circulation rate in kg/min\n",
+ "T1=27.;#cooling water inlet temperature in degree celcius\n",
+ "T2=50.;#cooling water outlet temperature in degree celcius\n",
+ "ma=150.;#mass of air consumed in kg\n",
+ "T_exhaust=400.;#exhaust temperature in degree celcius\n",
+ "T_atm=27.;#atmospheric temperature in degree celcius\n",
+ "Cg=1.25;#mean specific heat of exhaust gases in KJ/kg K\n",
+ "n_mech=0.9;#mechanical efficiency\n",
+ "T=300.*10**-3;#brake torque in N\n",
+ "C=42.*10**3;#calorific value in KJ/kg\n",
+ "Cw=4.18;#specific heat of water in KJ/kg K\n",
+ "BP=2.*math.pi*N*T/60\n",
+ "print(\"brake power(BP)=2*%pi*N*T in KW\"),round(BP,2)\n",
+ "print(\"so brake power=47.124 KW\")\n",
+ "bsfc=m*60/(mw*BP)\n",
+ "print(\"brake specific fuel consumption(bsfc) in kg/KW hr=\"),round(bsfc,2)\n",
+ "IP=BP/n_mech\n",
+ "print(\"indicated power(IP) in Kw=\"),round(IP,2)\n",
+ "n_ite=IP*mw*60/(m*C)\n",
+ "print(\"indicated thermal efficiency(n_ite)=\"),round(n_ite,2)\n",
+ "print(\"in percentage\"),round(n_ite*100,2)\n",
+ "print(\"so indicated thermal efficiency=28.05%\")\n",
+ "Qf=(m/mw)*C\n",
+ "print(\"heat available from fuel(Qf)=(m/mw)*C in KJ/min\"),round(Qf,2)\n",
+ "BP=BP*60 \n",
+ "print(\"energy consumed as brake power(BP) in KJ/min=\"),round(BP,2)\n",
+ "Qw=mw*Cw*(T2-T1)\n",
+ "print(\"energy carried by cooling water(Qw) in KJ/min=\"),round(Qw,2)\n",
+ "Qg=(ma+m)*Cg*(T_exhaust-T_atm)/mw\n",
+ "print(\"energy carried by exhaust gases(Qg) in KJ/min=\"),round(Qg,2)\n",
+ "Qf-(BP+Qw+Qg)\n",
+ "print(\"unaccounted energy loss in KJ/min\"),round(Qf-(BP+Qw+Qg),2)\n",
+ "print(\"NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code.\")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.11;pg no: 395"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 11,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.11, Page:395 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 11\n",
+ "indicated power of 1st cylinder=BP-BP1 in KW\n",
+ "indicated power of 2nd cylinder=BP-BP2 in KW\n",
+ "indicated power of 3rd cylinder=BP-BP3 in KW\n",
+ "indicated power of 4th cylinder=BP-BP4 in KW\n",
+ "indicated power of 5th cylinder=BP-BP5 in KW\n",
+ "indicated power of 6th cylinder=BP-BP6 in KW\n",
+ " total indicated power(IP)in KW= 61.9\n",
+ "mechanical efficiency(n_mech)= 0.81\n",
+ "in percentage 80.78\n",
+ "so indicated power=61.9 KW\n",
+ "mechanical efficiency=80.77%\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of indicated power and mechanical efficiency\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.11, Page:395 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 11\")\n",
+ "BP=50.;#brake power output at full load in KW\n",
+ "BP1=40.1;#brake power output of 1st cylinder in KW\n",
+ "BP2=39.5;#brake power output of 2nd cylinder in KW\n",
+ "BP3=39.1;#brake power output of 3rd cylinder in KW\n",
+ "BP4=39.6;#brake power output of 4th cylinder in KW\n",
+ "BP5=39.8;#brake power output of 5th cylinder in KW\n",
+ "BP6=40.;#brake power output of 6th cylinder in KW\n",
+ "print(\"indicated power of 1st cylinder=BP-BP1 in KW\")\n",
+ "BP-BP1\n",
+ "print(\"indicated power of 2nd cylinder=BP-BP2 in KW\")\n",
+ "BP-BP2\n",
+ "print(\"indicated power of 3rd cylinder=BP-BP3 in KW\")\n",
+ "BP-BP3\n",
+ "print(\"indicated power of 4th cylinder=BP-BP4 in KW\")\n",
+ "BP-BP4\n",
+ "print(\"indicated power of 5th cylinder=BP-BP5 in KW\")\n",
+ "BP-BP5\n",
+ "print(\"indicated power of 6th cylinder=BP-BP6 in KW\")\n",
+ "BP-BP6\n",
+ "IP=9.9+10.5+10.9+10.4+10.2+10\n",
+ "print(\" total indicated power(IP)in KW=\"),round(IP,2)\n",
+ "n_mech=BP/IP\n",
+ "print(\"mechanical efficiency(n_mech)=\"),round(n_mech,2)\n",
+ "print(\"in percentage\"),round(n_mech*100,2)\n",
+ "print(\"so indicated power=61.9 KW\")\n",
+ "print(\"mechanical efficiency=80.77%\")"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.12;pg no: 396"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 12,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.12, Page:396 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 12\n",
+ "brake power output of engine(BP) in KW= 19.63\n",
+ "brake power when cylinder 1 is cut(BP1) in KW= 13.74\n",
+ "so indicated power of first cylinder(IP1) in KW= 5.89\n",
+ "brake power when cylinder 2 is cut(BP2) in KW= 14.14\n",
+ "so indicated power of second cylinder(IP2) in KW= 5.5\n",
+ "brake power when cylinder 3 is cut(BP3) in KW= 14.29\n",
+ "so indicated power of third cylinder(IP3) in KW= 5.34\n",
+ "brake power when cylinder 4 is cut(BP4) in KW= 13.35\n",
+ "so indicated power of fourth cylinder(IP4) in KW= 6.28\n",
+ "now total indicated power(IP) in KW 23.01\n",
+ "engine mechanical efficiency(n_mech)= 0.85\n",
+ "in percentage 85.32\n",
+ "so BP=19.63 KW,IP=23 KW,n_mech=83.35%\n",
+ "heat liberated by fuel(Qf)=m*C in KJ/min 8127.0\n",
+ "heat carried by exhaust gases(Qg) in KJ/min= 1436.02\n",
+ "heat carried by cooling water(Qw) in KJ/min= 1730.52\n",
+ "energy to brake power(BP) in KJ/min= 1177.8\n",
+ "unaccounted losses in KJ/min 3782.66\n",
+ "NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code. \n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of brake power,indicated power,heat balance sheet\n",
+ "#intiation of all variables\n",
+ "# Chapter 9\n",
+ "import math\n",
+ "print\"Example 10.12, Page:396 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 12\")\n",
+ "N=1500.;#engine rpm at full load\n",
+ "F=250.;#brake load at full load in N\n",
+ "F1=175.;#brake reading 1 in N\n",
+ "F2=180.;#brake reading 2 in N\n",
+ "F3=182.;#brake reading 3 in N\n",
+ "F4=170.;#brake reading 4 in N\n",
+ "r=50.*10**-2;#brake drum radius in m\n",
+ "m=0.189;#fuel consumption rate in kg/min\n",
+ "C=43.*10**3;#fuel calorific value in KJ/kg\n",
+ "k=12.;#air to fuel ratio\n",
+ "T_exhaust=600.;#exhaust gas temperature in degree celcius\n",
+ "mw=18.;#cooling water flow rate in kg/min\n",
+ "T1=27.;#cooling water entering temperature in degree celcius\n",
+ "T2=50.;#cooling water leaving temperature in degree celcius\n",
+ "T_atm=27.;#atmospheric air temperature\n",
+ "Cg=1.02;#specific heat of exhaust gas in KJ/kg K\n",
+ "Cw=4.18;#specific heat of water in KJ/kg K\n",
+ "BP=2.*math.pi*N*F*r*10**-3/60.\n",
+ "print(\"brake power output of engine(BP) in KW=\"),round(BP,2)\n",
+ "BP1=2.*math.pi*N*F1*r*10**-3/60.\n",
+ "print(\"brake power when cylinder 1 is cut(BP1) in KW=\"),round(BP1,2)\n",
+ "IP1=BP-BP1\n",
+ "print(\"so indicated power of first cylinder(IP1) in KW=\"),round(IP1,2)\n",
+ "BP2=2.*math.pi*N*F2*r*10**-3/60.\n",
+ "print(\"brake power when cylinder 2 is cut(BP2) in KW=\"),round(BP2,2)\n",
+ "IP2=BP-BP2\n",
+ "print(\"so indicated power of second cylinder(IP2) in KW=\"),round(IP2,2)\n",
+ "BP3=2.*math.pi*N*F3*r*10**-3/60.\n",
+ "print(\"brake power when cylinder 3 is cut(BP3) in KW=\"),round(BP3,2)\n",
+ "IP3=BP-BP3\n",
+ "print(\"so indicated power of third cylinder(IP3) in KW=\"),round(IP3,2)\n",
+ "BP4=2.*math.pi*N*F4*r*10**-3/60.\n",
+ "print(\"brake power when cylinder 4 is cut(BP4) in KW=\"),round(BP4,2)\n",
+ "IP4=BP-BP4\n",
+ "print(\"so indicated power of fourth cylinder(IP4) in KW=\"),round(IP4,2)\n",
+ "IP=IP1+IP2+IP3+IP4\n",
+ "print(\"now total indicated power(IP) in KW\"),round(IP,2)\n",
+ "n_mech=BP/IP\n",
+ "print(\"engine mechanical efficiency(n_mech)=\"),round(n_mech,2)\n",
+ "print(\"in percentage\"),round(n_mech*100,2)\n",
+ "print(\"so BP=19.63 KW,IP=23 KW,n_mech=83.35%\")\n",
+ "Qf=m*C\n",
+ "print(\"heat liberated by fuel(Qf)=m*C in KJ/min\"),round(Qf,2)\n",
+ "Qg=(k+1)*m*Cg*(T_exhaust-T_atm)\n",
+ "print(\"heat carried by exhaust gases(Qg) in KJ/min=\"),round(Qg,2)\n",
+ "Qw=mw*Cw*(T2-T1)\n",
+ "print(\"heat carried by cooling water(Qw) in KJ/min=\"),round(Qw,2)\n",
+ "BP=19.63*60\n",
+ "print(\"energy to brake power(BP) in KJ/min=\"),round(19.63*60,2)\n",
+ "Qf-(Qg+Qw+BP)\n",
+ "print(\"unaccounted losses in KJ/min\"),round(Qf-(Qg+Qw+BP),2)\n",
+ "print(\"NOTE=>Heat balance sheet on per minute basis is attached as jpg file with this code. \")\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##example 10.13;pg no: 397"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 13,
+ "metadata": {
+ "collapsed": false
+ },
+ "outputs": [
+ {
+ "name": "stdout",
+ "output_type": "stream",
+ "text": [
+ "Example 10.13, Page:397 \n",
+ " \n",
+ "\n",
+ "Engineering Thermodynamics by Onkar Singh Chapter 10 Example 13\n",
+ "brake power(BP)=2*%pi*N*T in KW\n",
+ "indicated power(IP)=(mep*L*A*N)/60000 in KW\n",
+ "A> heat added(Q)=m*C/3600 in KJ/s 52.36\n",
+ "or Q in KJ/min\n",
+ "thermal efficiency(n_th)= 0.27\n",
+ "in percentage 26.85\n",
+ "B> heat equivalent of brake power(BP) in KJ/min= 659.76\n",
+ "C> heat loss to cooling water(Qw)=mw*Cpw*(T2-T1) in KJ/min\n",
+ "heat carried by exhaust gases=heat carried by steam in exhaust gases+heat carried by fuel gases(dry gases) in exhaust gases\n",
+ "mass of exhaust gases(mg)=mass of air+mass of fuel in kg/min\n",
+ "mg=(ma+m)/60\n",
+ "mass of steam in exhaust gases in kg/min\n",
+ "mass of dry exhaust gases in kg/min\n",
+ "D> heat carried by steam in exhaust in KJ/min 299.86\n",
+ "E> heat carried by fuel gases(dry gases) in exhaust gases(Qg) in KJ/min 782.64\n",
+ "F> unaccounted loss=A-B-C-D-E in KJ/min 537.4\n",
+ "NOTE># on per minute basis is attached as jpg file with this code.\n"
+ ]
+ }
+ ],
+ "source": [
+ "#cal of thermal efficiency and heat balance sheet\n",
+ "#intiation of all variables\n",
+ "# Chapter 10\n",
+ "import math\n",
+ "print\"Example 10.13, Page:397 \\n \\n\"\n",
+ "print(\"Engineering Thermodynamics by Onkar Singh Chapter 10 Example 13\")\n",
+ "D=20.*10**-2;#cylinder diameter in m\n",
+ "L=28.*10**-2;#stroke in m\n",
+ "m=4.22;#mass of fuel used in kg\n",
+ "C=44670.;#calorific value of fuel in KJ/kg\n",
+ "N=21000./60.;#engine rpm\n",
+ "mep=2.74*10**5;#mean effective pressure in pa\n",
+ "F=600.;#net brake load applied to a drum of 100 cm diameter in N\n",
+ "r=50.*10**-2;#brake drum radius in m\n",
+ "mw=495.;#total mass of cooling water in kg\n",
+ "T1=13.;#cooling water inlet temperature in degree celcius\n",
+ "T2=38.;#cooling water outlet temperature in degree celcius\n",
+ "ma=135.;#mass of air used in kg\n",
+ "T_air=20.;#temperature of air in test room in degree celcius\n",
+ "T_exhaust=370.;#temperature of exhaust gases in degree celcius\n",
+ "Cp_gases=1.005;#specific heat of gases in KJ/kg K\n",
+ "Cp_steam=2.093;#specific heat of steam at atmospheric pressure in KJ/kg K\n",
+ "Cpw=4.18;#specific heat of water in KJ/kg K\n",
+ "print(\"brake power(BP)=2*%pi*N*T in KW\")\n",
+ "BP=2*math.pi*N*F*r/60000\n",
+ "print(\"indicated power(IP)=(mep*L*A*N)/60000 in KW\")\n",
+ "IP=(mep*L*(math.pi*D**2/4)*N)/60000\n",
+ "Q=m*C/3600\n",
+ "print(\"A> heat added(Q)=m*C/3600 in KJ/s\"),round(Q,2)\n",
+ "print(\"or Q in KJ/min\")\n",
+ "Q=Q*60\n",
+ "Q=52.36;#heat added in KJ/s\n",
+ "n_th=IP/Q\n",
+ "print(\"thermal efficiency(n_th)= \"),round(n_th,2)\n",
+ "print(\"in percentage\"),round(n_th*100,2)\n",
+ "BP=BP*60\n",
+ "print(\"B> heat equivalent of brake power(BP) in KJ/min= \"),round(10.996*60,2)\n",
+ "print(\"C> heat loss to cooling water(Qw)=mw*Cpw*(T2-T1) in KJ/min\")\n",
+ "Qw=mw*Cpw*(T2-T1)/60\n",
+ "print(\"heat carried by exhaust gases=heat carried by steam in exhaust gases+heat carried by fuel gases(dry gases) in exhaust gases\")\n",
+ "print(\"mass of exhaust gases(mg)=mass of air+mass of fuel in kg/min\")\n",
+ "print(\"mg=(ma+m)/60\")\n",
+ "mg=(ma+m)/60\n",
+ "print(\"mass of steam in exhaust gases in kg/min\")\n",
+ "9*(0.15*m/60)\n",
+ "print(\"mass of dry exhaust gases in kg/min\")\n",
+ "mg-0.095\n",
+ "0.095*(Cpw*(100-T_air)+2256.9+Cp_steam*(T_exhaust-100))\n",
+ "print(\"D> heat carried by steam in exhaust in KJ/min\"),round(0.095*(Cpw*(100-T_air)+2256.9+Cp_steam*(T_exhaust-100)),2)\n",
+ "Qg=2.225*Cp_gases*(T_exhaust-T_air)\n",
+ "print(\"E> heat carried by fuel gases(dry gases) in exhaust gases(Qg) in KJ/min\"),round(Qg,2)\n",
+ "print(\"F> unaccounted loss=A-B-C-D-E in KJ/min\"),round(3141.79-659.76-862.13-299.86-782.64,2)\n",
+ "print(\"NOTE># on per minute basis is attached as jpg file with this code.\")\n"
+ ]
+ }
+ ],
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