The objective of this is to design a steam power cycle and d

The objective of this is to design a steam power cycle and develop a set of operating parameters that meet the requirements listed below. a) 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. b) The steam turbine exit should have at least 90% steam quality c) The source for the chill water is a nearby river at 10°C. Due to environmental concerns the outlet tem- perature of the chill water should not exceed 20°C. d) You must show two possible cycles with a thermal efficiency of at least 35%. Each cycle MUST use a different combination of pump and turbine. For example, the two cycles may use the same pump but different turbines, or they may use two different pumps and two different turbines. e) 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 g) Assume that both the boiler and condenser are isobaric devices and will be compatible with any pump and turbine selection. h) The chosen cycles must be possible from the given pump and turbine specifications. Results: List the information for the two cycles which produce at least 20 MW of power and have a thermal efficiency of at least 35%. The required information is provided below in a sample table. Cycle Cycle 2 Pump Model Turbine Model Condenser Pressure kPa) Boiler Pressure CMPa) Pump Outlet Temperature CC) Turbine Inlet Temperature CoC Turbine Outlet Quality Turbine Outlet Temperature CoC) Mass Flow Rate in the Rankine Cycle dkgs) Rate of Heat into the Rankine cycle OMwo Net Power Produced by the Rankine Cycle CMW) Overall Thermal Efficiency Mass Flow Rate of the Cooling Water (kg/s) Outlet Temperature of the Cooling water CoC) Pump Specifications MHD to 120 D20 18o 6300 MHA so to 120 14. 160 3600 MMK 40 to 250 75 3 230 4500

Solution

solution:

1)here pump entry temperature is 283 k hence enthalphy

h1=hf1=42 kj/kg

2)thermal efficiency is given by

n=W/Qs

W=20000 kw

n=.35

Qs=heat supply=57142.85 kw

Qs=W+Qr

Qr=37142.85 kw

where heat rejected=heat taken by cooling water

To=293 k

Ti=283 k

Qr=mcw*Cpw*(To-Ti)

Cpw=4.187

mcw=887.09 kg/s

3)here we select pump to be MMK with flow rate upto 75 m3/hr

hence Pb=130 bar

Pc=.01227 bar

hence pump work=h2\'-h1=(Pb-Pc)/10*np=15.292 kj/kg

hence h2\'=57.292 kj/kg and T2\'=10.75 c or 283.75 k

here h4 for x=.9

h4=h1f+.9hfg1=42+.9*2477.8=2272.02

Qr=m(h1-h4)

m=16.65 kg/s

Q=59.94 m3/hr hence pump selected is correct as MMK

here for pump mass flow rate selected for x=.9 steam quality is 16.65 kg/s and hence we continue with same mass flow rate

4)here we select turbine DYR with Wt=40200 kw

Wta=40200/.9=44600 kw

here for heat supplied for same flow arte is

Qs=m(h3-h2)=57142.85 kw

h3=3489.29 kj/kg

where for h3=3489.29 and Pb=130 bar

T3=830 k

here turbine work is

Wt=Wn+Wp=20000+16.65*15.292=20254.61 kw

in this way for rankine cycle efficiency of 35% we get that

T1=T4=283 k

T2\'=283.75 k

T3=830 k

Pb=130 bar

Pc=.01227 bar

m=16.65 kg/s

x=.9

mcw=887.09 kg/s

Ti=283 k and To=293 k

9)in above design power produced due to efficiency is not accounted and hence we have to increase mass flow rate to increase turbi work ,so repeated calculation areas follows

10)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

11)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

12)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

13)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

14)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

15)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

16)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

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