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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Chapter 11 : Diesel engine and Gas Turbine Power Plants"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.1 Pg: 766"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Net increase in brake power is 28.67 kW\n"
]
}
],
"source": [
"from __future__ import division\n",
"#Input data\n",
"C=3.5#Capacity in litres\n",
"P=13.1#Indicated power in kW/m**3\n",
"N=3600#Speed in rpm\n",
"ve=82#Volumetric efficiency in percent\n",
"p1=1.013#Pressure in bar\n",
"T1=25+273#Temperature in K\n",
"rp=1.75#Pressure ratio\n",
"ie=70#Isentropic efficiency in percent\n",
"me=80#Mechanical efficiency in percent\n",
"g=1.4#Ratio of specific heats\n",
"R=0.287#Gas constant in kJ/kg.K\n",
"Cp=1.005#Specific heat in kJ/kg.K\n",
"\n",
"#Calculations\n",
"EC=(C/1000)#Engine capacity in m**3\n",
"Vs=(N/2)*EC#Swept volume in m**3\n",
"Vui=(ve/100)*Vs#Unsupercharged induced volume in m**3/min\n",
"dp=(rp*p1)#Blower delivery pressure in bar\n",
"T2sT1=(rp)**((g-1)/g)#Ratio of temperatures\n",
"T2s=(T2sT1*T1)#Temperature in K\n",
"dT21=(T2s-T1)/(ie/100)#Difference in temperature in K\n",
"T2=dT21+T1#Temperature in K\n",
"EV=(Vs*dp*T1)/(p1*T2)#Equivalent volume in m**3/min\n",
"iiv=EV-Vui#Increase in induced volume in m**3/min\n",
"iip=(P*iiv)#Increase in indicated power in kW\n",
"iipi=((dp-p1)*100*Vs)/60#Increase in induced power due to increase in induction pressure in kW\n",
"tiip=iip+iipi#Total increase in indicated power in kW\n",
"tibp=tiip*(me/100)#Total increase in brake power in kW\n",
"ma=(dp*100*Vs)/(60*R*T2)#Mass of air in kg/s\n",
"WI=(ma*Cp*(T2-T1))#Work input to heater in kW\n",
"Pb=(WI/(me/100))#Power required in kW\n",
"NI=tibp-Pb#Net increase in brake power in kW\n",
"\n",
"#Output\n",
"print \"Net increase in brake power is %3.2f kW\"%(NI)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.2 Pg: 768"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" (a) the temperature of air leaving the compressor is 129.76 degree C \n",
" (b) the temperature of gases leaving the turbine is 482.26 degree C \n",
" (c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is 15.08 percent\n"
]
}
],
"source": [
"#Input data\n",
"p1=0.97#Pressure in bar\n",
"t1=30+273#Temperature in K\n",
"p2=2.1#Pressure in bar\n",
"af=18#Air fuel ratio\n",
"t3=580+273#Temperature in K\n",
"p3=1.9#Pressure in bar\n",
"p4=1.06#Pressure in bar\n",
"iec=0.75#Isentropic efficiency of compressor\n",
"iet=0.85#Isentropic efficiency of turbine\n",
"cpa=1.01#Specific heat for air in kJ/kg.K\n",
"ga=1.4#Ratio of specific heats\n",
"cpex=1.15#Specific heat in kJ/kg.K\n",
"gex=1.33#Ratio of specific heats\n",
"\n",
"#Calculations\n",
"t2s=t1*(p2/p1)**((ga-1)/ga)#Tempeature in K\n",
"t21=(t2s-t1)/iec#Temperature in K\n",
"t2=t21+t1#Temperature in K\n",
"T2=t2-273#Temperature in degree C\n",
"t3t4s=(p3/p4)**((gex-1)/gex)#Ratio of temperatures\n",
"t4s=(t3/t3t4s)#Temperature in K\n",
"t4=t3-((t3-t4s)*iet)#Temperature in K\n",
"T4=t4-273#Temperature in degree C\n",
"mp=(((cpex*(1+(1/af))*(t3-t4))-(cpa*(t2-t1)))/(cpex*(1+(1/af))*(t3-t4)))*100#Percentage of mechanical power loss\n",
"\n",
"#Output\n",
"print \" (a) the temperature of air leaving the compressor is %3.2f degree C \\n (b) the temperature of gases leaving the turbine is %3.2f degree C \\n (c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is %3.2f percent\"%(T2,T4,mp)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.3 Pg: 770"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" ENERGY BALANCE SHEET \n",
" (in kW) (in percent)\n",
" 1. Brake power 53.12 37.69 \n",
" 2. Heat carried away by exhaust gases 29.09 20.64 \n",
" 3. Heat lost to jacket cooling water 45.10 32.00 \n",
" 4. Heat loss unaccounted 13.64 9.67 \n",
" Total 140.94 100.00\n"
]
}
],
"source": [
"from __future__ import division\n",
"#Input data\n",
"a=215#Current in A\n",
"v=210#Voltage in V\n",
"e=85#Efficiency in percent\n",
"q=11.8#Quantity of fuel supplied in kg/h\n",
"cv=43#Calorific value in MJ/kg\n",
"af=18#Air fuel ratio\n",
"w=560#Water in litres/h\n",
"tw=38#Temeparature in degree C\n",
"te=97#Temeparature in degree C\n",
"cp=1.04#Specific heat in kJ/kg.K\n",
"ta=30#Temeparature in degree C\n",
"l=32#Percentage lost \n",
"sw=4.187#Specific heat in kJ/kg.K\n",
"\n",
"#Calculations\n",
"P=(a*v)/1000#Power in kW\n",
"BP=(P/(e/100))#Brake power in kW\n",
"E=(q/3600)*cv*1000#Energy supplied in kW\n",
"mg=(q/3600)*(1+af)#Rate of gases in kg/s\n",
"he=(mg*cp*(te-ta))+((w/3600)*sw*tw)#Heat carried away by exhaust gases in kW\n",
"hj=(l/100)*E#Heat lost to jacket cooling water in kW\n",
"pBP=(BP/E)*100#Percentage\n",
"pE=(E/E)*100#Percentage\n",
"phe=(he/E)*100#Percenatge\n",
"phj=(hj/E)*100#Percenatge \n",
"\n",
"#Output\n",
"print \" ENERGY BALANCE SHEET \\n (in kW) (in percent)\\n 1. Brake power %3.2f %3.2f \\n 2. Heat carried away by exhaust gases %3.2f %3.2f \\n 3. Heat lost to jacket cooling water %3.2f %3.2f \\n 4. Heat loss unaccounted %3.2f %3.2f \\n Total %3.2f %3.2f\"%(BP,pBP,he,phe,hj,phj,(E-(BP+he+hj)),(((E-(BP+he+hj))/E)*100),E,(pBP+phe+phj+(((E-(BP+he+hj))/E)*100)))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.4 Pg: 771"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" Indicated power is 18.07 kW \n",
" Brake power is 12.811 kW \n",
"\n",
" ENERGY BALANCE SHEET \n",
" (in kJ) (in percent)\n",
" 1. Energy equivalent in ip 21685 31.59 \n",
" 2. Energy carried away by cooling water 16748 24.40 \n",
" 3. Energy carried away by exhaust gases 18887 27.52 \n",
" 4. Unaccounted for energy loss 11320 16.49 \n",
" Total 68640 100.00\n"
]
}
],
"source": [
"from __future__ import division\n",
"#Input data\n",
"t=20#Trial time in minutes\n",
"NL=680#Net brake load in N\n",
"mep=3#Mean effective pressure in bar\n",
"N=360#Speed in rpm\n",
"Fc=1.56#Fuel consumption in kg\n",
"cw=160#Cooling water in kg\n",
"Tw=32#Temperature of water at inlet in degree C\n",
"Wo=57#Water outlet temperature in degree C\n",
"a=30#Air in kg\n",
"Ta=27#Room temperature in degree C\n",
"Te=310#Exhaust gas temperature in degree C\n",
"d=210#Bore in mm\n",
"l=290#Stroke in mm\n",
"bd=1#Brake diameter in m\n",
"cv=44#Calorific value in MJ/kg\n",
"st=1.3#Steam formed in kg per kg fuel in the exhaust\n",
"cp=2.093#Specific heat of steam in exhaust in kJ/kg.K\n",
"cpx=1.01#Specific heat of dry exhaust gases in kJ/kg.K\n",
"cpw=4.187#Specific heat of water in kJ/kg.K\n",
"\n",
"#Calculations\n",
"ip=(mep*100*(l/1000)*(3.14/4)*(d/1000)**2*N)/60#Indicated Power in kW\n",
"bp=((2*3.14*N*(NL*(1/2)))/60)/1000#Brake power in kW\n",
"nm=(bp/ip)*100#Mechanical efficiency in percent\n",
"qs=(Fc*cv*10**3)#Heat supplied in kJ\n",
"qip=(ip*t*60)#Heat equivalent of ip in kJ\n",
"qcw=(cw*cpw*(Wo-Tw))#Heat carried away by cooling water in kJ\n",
"tm=(Fc*a)#Toatl mass of exhaust gas in kg\n",
"ms=(st*Fc)#Mass of steam formed in kg\n",
"mde=(tm-ms)#Mass of dry exhaust gas in kg\n",
"Ed=(mde*cpx*(Te-Ta))#Energy carried away by dry exhaust gases in kJ\n",
"Es=(ms*((cpw*(100-Ta))+2257.9+(cp*(Te-100))))#Energy carried away by steam in kJ\n",
"TE=(Ed+Es)#Total energy carried away by exhaust gases in kJ\n",
"ue=(qs-(qip+qcw+TE))#Unaccounted energy in kJ\n",
"pqip=(qip/qs)*100#Percentage\n",
"pqcw=(qcw/qs)*100#Percentage\n",
"pTE=(TE/qs)*100#Percentage\n",
"pue=(ue/qs)*100#Percentage\n",
"\n",
"#Output\n",
"print \" Indicated power is %3.2f kW \\n Brake power is %3.3f kW \\n\\n ENERGY BALANCE SHEET \\n (in kJ) (in percent)\\n 1. Energy equivalent in ip %3.0f %3.2f \\n 2. Energy carried away by cooling water %3.0f %3.2f \\n 3. Energy carried away by exhaust gases %3.0f %3.2f \\n 4. Unaccounted for energy loss %3.0f %3.2f \\n Total %3.0f %3.2f\"%(ip,bp,qip,pqip,qcw,pqcw,TE,pTE,ue,pue,qs,(pqip+pqcw+pTE+pue))"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.5 Pg: 796"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" (a)The flow velocity is 393 m/s \n",
" (b) The blade angles at the tip are : \n",
" Fixed blades (root) are 57.28 degrees and 23.88 degrees \n",
" Moving blades (root) are 38.78 degrees and 50.10 degrees \n",
" Fixed blades (tip) are 47.38 degrees and 17.17 degrees \n",
" Moving blades (tip) are 0.45 degrees and 54.23 degrees \n",
" (c) The degree of reaction at : \n",
" the tip is 64 percent \n",
" the root is 26 percent\n"
]
}
],
"source": [
"from math import tan,sin,atan,degrees,pi,cos,sqrt\n",
"#Input data\n",
"Vbm=360#Blade velocity in m/s\n",
"b1=20#Blade angle at inlet in degrees\n",
"a2=b1#Angle in degrees\n",
"b2=52#Blade angle at exit in degrees\n",
"a1=b2#Angle in degrees\n",
"R=50#Degree of reaction in percent\n",
"dm=0.45#Mean diameter of the blade in m\n",
"bh=0.08#Mean blade height in m\n",
"\n",
"#Calculations\n",
"Vf=(Vbm/(tan(pi/180*b2)-tan(pi/180*b1)))#Velocity in m/s\n",
"rt=(dm/2)+(bh/2)#Mean radius in m\n",
"Vbt=(Vbm*(rt/(dm/2)))#Velocity in m/s\n",
"Vw1m=Vf*tan(pi/180*a1)#Velocity in m/s\n",
"Vw1t=(Vw1m*((dm/2)/rt))#Velocity in m/s\n",
"dVw1=(Vf*(tan(pi/180*b1)+tan(pi/180*b2))*Vbm)/Vbt#Velocity in m/s\n",
"rr=(dm/2)-(bh/2)#Radius in m\n",
"Vbr=(Vbm*(rr/(dm/2)))#Velocity in m/s\n",
"Vw1r=(Vw1m*((dm/2)/rr))#Velocity in m/s\n",
"Vr2=Vf/cos(pi/180*b2)#Velocity in m/s\n",
"dVwr=((Vw1m+((Vr2*sin(pi/180*b2))-Vbm))*Vbm)/Vbr#Velocity in m/s\n",
"a1r=degrees(atan(Vw1r/Vf))#Angle in degrees\n",
"a2r=degrees(atan((dVwr-Vw1r)/Vf))#Angle in degrees\n",
"b1r=degrees(atan((Vw1r-Vbr)/Vf))#Angle in degrees\n",
"b2r=degrees(atan((Vbr+(Vf*tan(pi/180*a2r)))/Vf))#Angle in degrees\n",
"a1t=degrees(atan(Vw1t/Vf))#Angle in degrees\n",
"a2t=degrees(atan((dVw1-Vw1t)/Vf))#Angle in degrees\n",
"b1t=degrees(atan((Vw1t-Vbt)/Vf))#Angle in degrees\n",
"b2t=degrees(atan((Vbt+(Vf*tan(pi/180*a2t)))/Vf))#Angle in degrees\n",
"Rt=((Vf*(tan(pi/180*b2t)-tan(pi/180*b1t)))/(2*Vbt))*100#Degree of reaction at the tip in percent\n",
"Rr=((Vf*(tan(pi/180*b2r)-tan(pi/180*b1r)))/(2*Vbr))*100#Degree of reaction at the root in percent\n",
"\n",
"#Output\n",
"print \" (a)The flow velocity is %3.0f m/s \\n (b) The blade angles at the tip are : \\n Fixed blades (root) are %3.2f degrees and %3.2f degrees \\n Moving blades (root) are %3.2f degrees and %3.2f degrees \\n Fixed blades (tip) are %3.2f degrees and %3.2f degrees \\n Moving blades (tip) are %3.2f degrees and %3.2f degrees \\n (c) The degree of reaction at : \\n the tip is %3.0f percent \\n the root is %3.0f percent\"%(Vf,a1r,a2r,b1r,b2r,a1t,a2t,b1t,b2t,Rt,Rr)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.6 Pg: 799"
]
},
{
"cell_type": "code",
"execution_count": 20,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Impeller tip diameter is 549 mm\n"
]
}
],
"source": [
"from math import tan,sin,atan,degrees,pi,cos,sqrt\n",
"#Input data\n",
"N=16000#Speed in rpm\n",
"T1=17+273#Temperature in K\n",
"rp=4#Pressure ratio\n",
"In=82#Isentropic efficiency in percent\n",
"s=0.85#Slip factor\n",
"a=20#Angle in degrees\n",
"d=200#Diameter in mm\n",
"V=120#Velocity in m/s\n",
"cp=1.005#Specific heat in kJ/kg.K\n",
"g=1.4#Ratio of specific heats\n",
"\n",
"#Calculations\n",
"T2sT1=(rp)**((g-1)/g)#Temperature ratio\n",
"T2s=T1*T2sT1#Temeprature in K\n",
"dTs=(T2s-T1)#Temperature difference in K\n",
"dT=dTs/(In/100)#Temperature difference in K\n",
"Wc=(cp*dT)#Power input in kJ/kg\n",
"Vb1=(3.14*(d/1000)*N)/60#Velocity in m/s\n",
"Vw1=(V*sin(pi/180*a))#Pre-whirl velocity in m/s\n",
"Vb2=sqrt(((Wc*1000)+(Vb1*Vw1))/s)#Velocity in m/s\n",
"d2=((Vb2*60)/(3.14*N))*1000#Tip diameter in mm\n",
"\n",
"#Output\n",
"print \"Impeller tip diameter is %3.0f mm\"%(d2)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Ex: 11.7 Pg: 801"
]
},
{
"cell_type": "code",
"execution_count": 22,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Workdone factor of the compressor is 0.86\n"
]
}
],
"source": [
"from math import tan,sin,atan,degrees,pi,cos,sqrt\n",
"#Input data\n",
"T1=25+273#Temperature in K\n",
"rp=6#Pressure ratio\n",
"Vb=220#Mean velocity in m/s\n",
"b1=45#Angle in degrees\n",
"a2=b1#Angle in degrees\n",
"b2=15#Angle in degrees\n",
"a1=b2#Angle in degrees\n",
"R=50#Degree of reaction in percent\n",
"n=10#Number of stages\n",
"In=83#Isentropic efficiency in percent\n",
"cp=1.005#Specific heat in kJ/kg.K\n",
"g=1.4#Ratio of specific heats\n",
"\n",
"#Calculations\n",
"V1=(Vb/(sin(pi/180*b2)+(cos(pi/180*a1)*tan(pi/180*a2))))#Velocity in m/s\n",
"V2=(V1*cos(pi/180*b2))/cos(pi/180*b1)#Velocity in m/s\n",
"dVw=(V2*sin(pi/180*a2))-(V1*sin(pi/180*a1))#Velocity in m/s.Textbook answer is wrong. Correct answer is 127 m/s\n",
"T2sT1=rp**((g-1)/g)#Temperature ratio\n",
"T2s=(T2sT1*T1)#Temperature in K\n",
"dTs=(T2s-T1)#Temperature difference in K\n",
"dT=(dTs/(In/100))#Temperature difference in K\n",
"W=(cp*dT)#Workdone in kJ/kg\n",
"w=(W*10**3)/(Vb*dVw*n)#Work done factor\n",
"\n",
"#Output\n",
"print \"Workdone factor of the compressor is %3.2f\"%(w)"
]
},
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"source": [
"## Ex: 11.8 Pg: 802"
]
},
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{
"name": "stdout",
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"text": [
" (a) the air fuel ratio is 75.55 \n",
" (b) the cycle efficiency is 41.8 percent \n",
" (c) the power supplied by the plant is 41 MW \n",
" (d) the specific fuel consumption of the plant is 0.244 kg/kW.h and the fuel consumption per hour is 10007.26 kg\n"
]
}
],
"source": [
"from math import sqrt,pi\n",
"#Input data\n",
"p1=1#Pressure in bar\n",
"T1=20+273#Temperature in K\n",
"Tm=900+273#Maximum temperature in K\n",
"rp=6#Pressure ratio\n",
"e=0.7#Effectiveness of regenerator\n",
"ma=210#Rate of air flow in kg/s\n",
"CV=40800#Calorific value in kJ/kg\n",
"ic=0.82#Isentropic efficiencies of both the compressors\n",
"it=0.92#Isentropic efficiencies of both the turbine\n",
"cn=0.95#Combustion efficiency \n",
"mn=0.96#Mechanical efficiency\n",
"gn=0.95#Generator efficiency\n",
"cp=1.005#Specific heat of air in kJ/kg.K\n",
"cpg=1.08#Specific heat of gas in kJ/kg.K\n",
"g1=1.4#Ratio of specific heats for air\n",
"g=1.33#Ratio of specific heats for gas\n",
"\n",
"#Calculations\n",
"pi=sqrt(p1*rp)#Intermediate pressure in bar\n",
"T2sT1=(pi/p1)**((g1-1)/g1)#Temperature ratio\n",
"T2s=(T2sT1*T1)#temperature in K\n",
"T2=((T2s-T1)/ic)+T1#Temperature in K\n",
"T4s=(T1*(rp/pi)**((g1-1)/g1))#Temperature in K\n",
"T4=((T4s-T1)/ic)+T1#Temperature in K\n",
"T7s=(Tm/(rp/p1)**((g-1)/g))#Temperature in K\n",
"T7=Tm-(it*(Tm-T7s))#Temperature in K\n",
"T5=(e*(T7-T4))+T4#Temperature in K\n",
"mf=1/((cp*(Tm-T5))/((CV*cn)-(cp*(Tm-T5))))#Air fuel ratio\n",
"Wgt=((1+(1/mf))*cpg*(Tm-T7))#Workdone by turbine in kJ/kg of air\n",
"Wc=(cp*((T2-T1)+(T4-T1)))#Workdone by compressor in kJ/kg of air\n",
"Wnet=(Wgt-Wc)#Net workdone in kJ/kg of air\n",
"Q=(CV*cn)/mf#Heat supplied in kJ/kg of air\n",
"ncy=(Wnet/Q)*100#Cycle efficiency in percent\n",
"PO=(Wnet*ma*mn*gn)/10**3#Power output in MW\n",
"Fc=(ma*3600*(1/mf))#Fuel consumption per hour in kg\n",
"SFC=(Fc/(PO*10**3))#Specific fuel consumption in kg/kW.h\n",
"\n",
"#Output\n",
"print \" (a) the air fuel ratio is %3.2f \\n (b) the cycle efficiency is %3.1f percent \\n (c) the power supplied by the plant is %3.0f MW \\n (d) the specific fuel consumption of the plant is %3.3f kg/kW.h and the fuel consumption per hour is %3.2f kg\"%(mf,ncy,PO,SFC,Fc)\n"
]
}
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|