# Chapter 11 : Diesel engine and Gas Turbine Power Plants¶

## Ex: 11.1 Pg: 766¶

In [15]:
from __future__ import division
#Input data
C=3.5#Capacity in litres
P=13.1#Indicated power in kW/m**3
N=3600#Speed in rpm
ve=82#Volumetric efficiency in percent
p1=1.013#Pressure in bar
T1=25+273#Temperature in K
rp=1.75#Pressure ratio
ie=70#Isentropic efficiency in percent
me=80#Mechanical efficiency in percent
g=1.4#Ratio of specific heats
R=0.287#Gas constant in kJ/kg.K
Cp=1.005#Specific heat in kJ/kg.K

#Calculations
EC=(C/1000)#Engine capacity in m**3
Vs=(N/2)*EC#Swept volume in m**3
Vui=(ve/100)*Vs#Unsupercharged induced volume in m**3/min
dp=(rp*p1)#Blower delivery pressure in bar
T2sT1=(rp)**((g-1)/g)#Ratio of temperatures
T2s=(T2sT1*T1)#Temperature in K
dT21=(T2s-T1)/(ie/100)#Difference in temperature in K
T2=dT21+T1#Temperature in K
EV=(Vs*dp*T1)/(p1*T2)#Equivalent volume in m**3/min
iiv=EV-Vui#Increase in induced volume in m**3/min
iip=(P*iiv)#Increase in indicated power in kW
iipi=((dp-p1)*100*Vs)/60#Increase in induced power due to increase in induction pressure in kW
tiip=iip+iipi#Total increase in indicated power in kW
tibp=tiip*(me/100)#Total increase in brake power in kW
ma=(dp*100*Vs)/(60*R*T2)#Mass of air in kg/s
WI=(ma*Cp*(T2-T1))#Work input to heater in kW
Pb=(WI/(me/100))#Power required in kW
NI=tibp-Pb#Net increase in brake power in kW

#Output
print "Net increase in brake power is %3.2f kW"%(NI)

Net increase in brake power is 28.67 kW


## Ex: 11.2 Pg: 768¶

In [16]:
#Input data
p1=0.97#Pressure in bar
t1=30+273#Temperature in K
p2=2.1#Pressure in bar
af=18#Air fuel ratio
t3=580+273#Temperature in K
p3=1.9#Pressure in bar
p4=1.06#Pressure in bar
iec=0.75#Isentropic efficiency of compressor
iet=0.85#Isentropic efficiency of turbine
cpa=1.01#Specific heat for air in kJ/kg.K
ga=1.4#Ratio of specific heats
cpex=1.15#Specific heat in kJ/kg.K
gex=1.33#Ratio of specific heats

#Calculations
t2s=t1*(p2/p1)**((ga-1)/ga)#Tempeature in K
t21=(t2s-t1)/iec#Temperature in K
t2=t21+t1#Temperature in K
T2=t2-273#Temperature in degree C
t3t4s=(p3/p4)**((gex-1)/gex)#Ratio of temperatures
t4s=(t3/t3t4s)#Temperature in K
t4=t3-((t3-t4s)*iet)#Temperature in K
T4=t4-273#Temperature in degree C
mp=(((cpex*(1+(1/af))*(t3-t4))-(cpa*(t2-t1)))/(cpex*(1+(1/af))*(t3-t4)))*100#Percentage of mechanical power loss

#Output
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)

 (a) the temperature of air leaving the compressor is 129.76 degree C
(b) the temperature of gases leaving the turbine is 482.26 degree C
(c) the mechanical power loss in the turbocharger as a percentage of the power generated in the turbine is 15.08 percent


## Ex: 11.3 Pg: 770¶

In [17]:
from __future__ import division
#Input data
a=215#Current in A
v=210#Voltage in V
e=85#Efficiency in percent
q=11.8#Quantity of fuel supplied in kg/h
cv=43#Calorific value in MJ/kg
af=18#Air fuel ratio
w=560#Water in litres/h
tw=38#Temeparature in degree C
te=97#Temeparature in degree C
cp=1.04#Specific heat in kJ/kg.K
ta=30#Temeparature in degree C
l=32#Percentage lost
sw=4.187#Specific heat in kJ/kg.K

#Calculations
P=(a*v)/1000#Power in kW
BP=(P/(e/100))#Brake power in kW
E=(q/3600)*cv*1000#Energy supplied in kW
mg=(q/3600)*(1+af)#Rate of gases in kg/s
he=(mg*cp*(te-ta))+((w/3600)*sw*tw)#Heat carried away by exhaust gases in kW
hj=(l/100)*E#Heat lost to jacket cooling water in kW
pBP=(BP/E)*100#Percentage
pE=(E/E)*100#Percentage
phe=(he/E)*100#Percenatge
phj=(hj/E)*100#Percenatge

#Output
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)))

                           ENERGY BALANCE SHEET
(in kW)            (in percent)
1. Brake power                               53.12              37.69
2. Heat carried away by exhaust gases        29.09              20.64
3. Heat lost to jacket cooling water         45.10              32.00
4. Heat loss unaccounted                     13.64              9.67
Total                            140.94             100.00


## Ex: 11.4 Pg: 771¶

In [18]:
from __future__ import division
#Input data
t=20#Trial time in minutes
mep=3#Mean effective pressure in bar
N=360#Speed in rpm
Fc=1.56#Fuel consumption in kg
cw=160#Cooling water in kg
Tw=32#Temperature of water at inlet in degree C
Wo=57#Water outlet temperature in degree C
a=30#Air in kg
Ta=27#Room temperature in degree C
Te=310#Exhaust gas temperature in degree C
d=210#Bore in mm
l=290#Stroke in mm
bd=1#Brake diameter in m
cv=44#Calorific value in MJ/kg
st=1.3#Steam formed in kg per kg fuel in the exhaust
cp=2.093#Specific heat of steam in exhaust in kJ/kg.K
cpx=1.01#Specific heat of dry exhaust gases in kJ/kg.K
cpw=4.187#Specific heat of water in kJ/kg.K

#Calculations
ip=(mep*100*(l/1000)*(3.14/4)*(d/1000)**2*N)/60#Indicated Power in kW
bp=((2*3.14*N*(NL*(1/2)))/60)/1000#Brake power in kW
nm=(bp/ip)*100#Mechanical efficiency in percent
qs=(Fc*cv*10**3)#Heat supplied in kJ
qip=(ip*t*60)#Heat equivalent of ip in kJ
qcw=(cw*cpw*(Wo-Tw))#Heat carried away by cooling water in kJ
tm=(Fc*a)#Toatl mass of exhaust gas in kg
ms=(st*Fc)#Mass of steam formed in kg
mde=(tm-ms)#Mass of dry exhaust gas in kg
Ed=(mde*cpx*(Te-Ta))#Energy carried away by dry exhaust gases in kJ
Es=(ms*((cpw*(100-Ta))+2257.9+(cp*(Te-100))))#Energy carried away by steam in kJ
TE=(Ed+Es)#Total energy carried away by exhaust gases in kJ
ue=(qs-(qip+qcw+TE))#Unaccounted energy in kJ
pqip=(qip/qs)*100#Percentage
pqcw=(qcw/qs)*100#Percentage
pTE=(TE/qs)*100#Percentage
pue=(ue/qs)*100#Percentage

#Output
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))

 Indicated power is 18.07 kW
Brake power is 12.811 kW

ENERGY BALANCE SHEET
(in kJ)            (in percent)
1. Energy equivalent in ip                     21685                 31.59
2. Energy carried away by cooling water        16748                 24.40
3. Energy carried away by exhaust gases        18887                 27.52
4. Unaccounted for energy loss                 11320                 16.49
Total                              68640                 100.00


## Ex: 11.5 Pg: 796¶

In [19]:
from math import tan,sin,atan,degrees,pi,cos,sqrt
#Input data
b1=20#Blade angle at inlet in degrees
a2=b1#Angle in degrees
b2=52#Blade angle at exit in degrees
a1=b2#Angle in degrees
R=50#Degree of reaction in percent
dm=0.45#Mean diameter of the blade in m

#Calculations
Vf=(Vbm/(tan(pi/180*b2)-tan(pi/180*b1)))#Velocity in m/s
Vbt=(Vbm*(rt/(dm/2)))#Velocity in m/s
Vw1m=Vf*tan(pi/180*a1)#Velocity in m/s
Vw1t=(Vw1m*((dm/2)/rt))#Velocity in m/s
dVw1=(Vf*(tan(pi/180*b1)+tan(pi/180*b2))*Vbm)/Vbt#Velocity in m/s
Vbr=(Vbm*(rr/(dm/2)))#Velocity in m/s
Vw1r=(Vw1m*((dm/2)/rr))#Velocity in m/s
Vr2=Vf/cos(pi/180*b2)#Velocity in m/s
dVwr=((Vw1m+((Vr2*sin(pi/180*b2))-Vbm))*Vbm)/Vbr#Velocity in m/s
a1r=degrees(atan(Vw1r/Vf))#Angle in degrees
a2r=degrees(atan((dVwr-Vw1r)/Vf))#Angle in degrees
b1r=degrees(atan((Vw1r-Vbr)/Vf))#Angle in degrees
b2r=degrees(atan((Vbr+(Vf*tan(pi/180*a2r)))/Vf))#Angle in degrees
a1t=degrees(atan(Vw1t/Vf))#Angle in degrees
a2t=degrees(atan((dVw1-Vw1t)/Vf))#Angle in degrees
b1t=degrees(atan((Vw1t-Vbt)/Vf))#Angle in degrees
b2t=degrees(atan((Vbt+(Vf*tan(pi/180*a2t)))/Vf))#Angle in degrees
Rt=((Vf*(tan(pi/180*b2t)-tan(pi/180*b1t)))/(2*Vbt))*100#Degree of reaction at the tip in percent
Rr=((Vf*(tan(pi/180*b2r)-tan(pi/180*b1r)))/(2*Vbr))*100#Degree of reaction at the root in percent

#Output
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)

 (a)The flow velocity is 393 m/s
(b) The blade angles at the tip are :
Fixed blades (root) are 57.28 degrees and 23.88 degrees
Moving blades (root) are 38.78 degrees and 50.10 degrees
Fixed blades (tip) are 47.38 degrees and 17.17 degrees
Moving blades (tip) are 0.45 degrees and 54.23 degrees
(c) The degree of reaction at :
the tip is  64 percent
the root is  26 percent


## Ex: 11.6 Pg: 799¶

In [20]:
from math import tan,sin,atan,degrees,pi,cos,sqrt
#Input data
N=16000#Speed in rpm
T1=17+273#Temperature in K
rp=4#Pressure ratio
In=82#Isentropic efficiency in percent
s=0.85#Slip factor
a=20#Angle in degrees
d=200#Diameter in mm
V=120#Velocity in m/s
cp=1.005#Specific heat in kJ/kg.K
g=1.4#Ratio of specific heats

#Calculations
T2sT1=(rp)**((g-1)/g)#Temperature ratio
T2s=T1*T2sT1#Temeprature in K
dTs=(T2s-T1)#Temperature difference in K
dT=dTs/(In/100)#Temperature difference in K
Wc=(cp*dT)#Power input in kJ/kg
Vb1=(3.14*(d/1000)*N)/60#Velocity in m/s
Vw1=(V*sin(pi/180*a))#Pre-whirl velocity in m/s
Vb2=sqrt(((Wc*1000)+(Vb1*Vw1))/s)#Velocity in m/s
d2=((Vb2*60)/(3.14*N))*1000#Tip diameter in mm

#Output
print "Impeller tip diameter is %3.0f mm"%(d2)

Impeller tip diameter is 549 mm


## Ex: 11.7 Pg: 801¶

In [22]:
from math import tan,sin,atan,degrees,pi,cos,sqrt
#Input data
T1=25+273#Temperature in K
rp=6#Pressure ratio
Vb=220#Mean velocity in m/s
b1=45#Angle in degrees
a2=b1#Angle in degrees
b2=15#Angle in degrees
a1=b2#Angle in degrees
R=50#Degree of reaction in percent
n=10#Number of stages
In=83#Isentropic efficiency in percent
cp=1.005#Specific heat in kJ/kg.K
g=1.4#Ratio of specific heats

#Calculations
V1=(Vb/(sin(pi/180*b2)+(cos(pi/180*a1)*tan(pi/180*a2))))#Velocity in m/s
V2=(V1*cos(pi/180*b2))/cos(pi/180*b1)#Velocity in m/s
T2sT1=rp**((g-1)/g)#Temperature ratio
T2s=(T2sT1*T1)#Temperature in K
dTs=(T2s-T1)#Temperature difference in K
dT=(dTs/(In/100))#Temperature difference in K
W=(cp*dT)#Workdone in kJ/kg
w=(W*10**3)/(Vb*dVw*n)#Work done factor

#Output
print "Workdone factor of the compressor is %3.2f"%(w)

Workdone factor of the compressor is 0.86


## Ex: 11.8 Pg: 802¶

In [2]:
from math import sqrt,pi
#Input data
p1=1#Pressure in bar
T1=20+273#Temperature in K
Tm=900+273#Maximum temperature in K
rp=6#Pressure ratio
e=0.7#Effectiveness of regenerator
ma=210#Rate of air flow in kg/s
CV=40800#Calorific value in kJ/kg
ic=0.82#Isentropic efficiencies of both the compressors
it=0.92#Isentropic efficiencies of both the turbine
cn=0.95#Combustion efficiency
mn=0.96#Mechanical efficiency
gn=0.95#Generator efficiency
cp=1.005#Specific heat of air in kJ/kg.K
cpg=1.08#Specific heat of gas in kJ/kg.K
g1=1.4#Ratio of specific heats for air
g=1.33#Ratio of specific heats for gas

#Calculations
pi=sqrt(p1*rp)#Intermediate pressure in bar
T2sT1=(pi/p1)**((g1-1)/g1)#Temperature ratio
T2s=(T2sT1*T1)#temperature in K
T2=((T2s-T1)/ic)+T1#Temperature in K
T4s=(T1*(rp/pi)**((g1-1)/g1))#Temperature in K
T4=((T4s-T1)/ic)+T1#Temperature in K
T7s=(Tm/(rp/p1)**((g-1)/g))#Temperature in K
T7=Tm-(it*(Tm-T7s))#Temperature in K
T5=(e*(T7-T4))+T4#Temperature in K
mf=1/((cp*(Tm-T5))/((CV*cn)-(cp*(Tm-T5))))#Air fuel ratio
Wgt=((1+(1/mf))*cpg*(Tm-T7))#Workdone by turbine in kJ/kg of air
Wc=(cp*((T2-T1)+(T4-T1)))#Workdone by compressor in kJ/kg of air
Wnet=(Wgt-Wc)#Net workdone in kJ/kg of air
Q=(CV*cn)/mf#Heat supplied in kJ/kg of air
ncy=(Wnet/Q)*100#Cycle efficiency in percent
PO=(Wnet*ma*mn*gn)/10**3#Power output in MW
Fc=(ma*3600*(1/mf))#Fuel consumption per hour in kg
SFC=(Fc/(PO*10**3))#Specific fuel consumption in kg/kW.h

#Output
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)

 (a) the air fuel ratio is 75.55
(b) the cycle efficiency is 41.8 percent
(c) the power supplied by the plant is  41 MW
(d) the specific fuel consumption of the plant is 0.244 kg/kW.h and the fuel consumption per hour is 10007.26 kg