Pump Specifications Turbine Specifications Requirements and

Pump Specifications Turbine Specifications Requirements and Assumptions 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 enviromental concerns the outlet temperature of the chill water should not exceed 20C. 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 section. From the given information of the pump and turbine determine ALL of the following. (a) Determine the condenser pressure in KPa. (b) Determine the boiler pressure in MPa. (c) Determine the pump outlet temperature in Celsius. (d) Determine the turbine inlet temperature in Celsius. (e) Determine the turbine outlet quality. (f) Determine the turbine outlet temperature in Celsius. (g) Determine the mass flow rate for the Rankine Cycle, in kg/s. (h) Determine the rate of heat into the Rankine Cycle, in MW. (i) Determine the net power produced by the Rankine Cycle, in MW (j) Determine the overall thermal efficiency of the Rankine Cycle, as a fraction (0 to 1). (k) Determine the mass flow rate of the cooling water, in kg/s. Determine the cooling water outlet temperature in Celsius. I need step by step solutions for A through L. Thanks in advance.

Solution

solution:

here for given rankine cycle operated steam turbine cycle we have to choose mass flow rate of steam on the basis of steam quality at outlet,thermal efficiency and net work done and other constaraints all forming linear system of equation in terms of enthalphy of system and we can obtain solution as follows.

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

6)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

7)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

8)in thisway we get net power output to be completely W=20 MW and output quality as x=.9