Choose a suitable pump and steam turbine from those provided at the end of this
ID: 1841745 • Letter: C
Question
Choose a suitable pump and steam turbine from those provided at the end of this document to ensure at least 20 MW of power is produced for the cycle. The steam turbine exit should have at least 90% steam quality. The source for the chill water is a nearby river at 10C. Due to environmental concerns the outlet temperature of the chill water should not exceed 20 C. You must show two possible cycles with a thermal efficiency of at least 35%. Assume that the isentropic efficiency of the pump is 85% and that of the turbine is 90%. Assume that the inlet of the pump is a saturated liquid. Assume that both the boiler and condenser are isobaric devices and will be compatible with any pump and turbine selection. The chosen cycles must be possible from the given pump and turbine specifications From the given tables and assumptions calculate ALL of the following. Condenser Pressure (KPa) Boiler Pressure (MPa) Pump Outlet Temperature (Celsius) Turbine Inlet Temperature (CelsiusExplanation / Answer
solution:
1)here for selected pump we have pressure limit as
Pc=.01237 bar
Pb=130 bar
hence pump work=(Pb-Pc/10)*(1/np)
so we get that
Wp=15.292 kj/kg
2)here at inlet temperarature is
T1=10 Cor 283 k
h1=42 kj/kg
hence h2'-h1=Wp
we get that
h2'=57.29 kj/kg
3)where condition at exit steam condition at exit
x=.9
h4=h1+x(hfg)
hfg=2477.87
so we get h4
h4=2272.02 kj/kg
4)here tutbine work require is
we select mass flow rate by trial and error to be m=18.5064 kg/s
Wt=(20000/.9)+m*Wp=22505.22kw
hence for thermal efficiency n=.35
heat supplied
n=Ws/Qs
Ws=Wt-m*Wp=22222.22 kw
Qs=63492.06 kw
hence heat rejected=Qr=Qs-Ws=41269.84 kw
5)hence available equtaionare
m*(h2'-h1)=282.99 kw
m(h3-h2')=63492.06 kw
m(h3-h4)=22505.22 kw
m(h4-h1)=41269.84 kw
first two equation and last two equation are giving equal value hence balanced condition of system
from second equation
m(h3-h2')=63492.06
h3=3430.81+57.292=3488.10 kj/kg
from third equation as
m(h3-h4)=22505.22
h3=3488.09 lj/kg
5)here for
h2'=57.292 kj/kg and Pb=130 bar T2=10.75 c
for h3=3488.09 kj/lg ,Pb=130 bar,T3=830 k
6)here mass flow rate of cold water is given by
Qr=mcw*Cpw(To-Ti)
To=20c and Ti=10 C
41269.84=mcw*4.187*10
mcw=985.66 kg/s
and such high flow rate will beachieve by increasing tube of condenser heat exchanger
7)in thisway we get net power output to be completely W=20 MW and output quality as x=.9