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(30 points) We are designing a geothermal flash steam power plant that pumps water from underground geothermal reservoirs at
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Answer #1

T ​​​​​​1 =  180°C

P 1 = 5 MPa

Turbine output = 0.1 MPa

A) feed of flash tank = 10 Kg/s

Fraction of it forms saturated steam and rest leaves of it water

From perry handbook steam tables

Properties of water at 180°C and 5 Mpa

H = 765.219 KJ/kg

s = 2.1340 KJ/kg K

U = 758.599 KJ/kg

Output of turbine is 0.1 MPa

Properties of saturated steam at 0.1 MPa

T - 99.6°C

H = 2674.9 KJ/kg

s= 7.3588 KJ/kg K

U = 2505.5 KJ/kg

Enthaply of water at feed of flash drum

= 10(765.219) = 7652.19 KJ/s

Enthaply of saturated steam = 2674.9 KJ/kg

mass of steam produced

7652.19 = m (2674.9)

m = 2.860 Kg/s

Flowrate of stream leaving flash tank = 2.860 Kg/s

B ) output from the turbine is saturated vapor

For reversible process

∆sT = 0

∆ssystem = swater - ssteam = 2.1340-7.3588 = -5. 224

∆ssurrounding =( H/T) water - (H/T) steam

swater = (765.219/(180+273)) = 1.689 KJ/kg K

ssteam = (2674.9)/(273+99.6) ) = 7.17 KJ/kg K

∆Ssurrounding = 7.17-1.689+5.48 KJ/kg K

∆ST = -5. 22 + 5.48 = 0.22

∆ST>0 , hence the process strictly speaking is irreversible but it is close to reversibility

C) The process is well insulated hence adiabatic

at adiabatic conditions Q= 0

∆U = W

∆U = m​​​​​​wu1 - m su2

mw = 10 kg/S

ms = 2.860 Kg/s

u1 = 758.99 KJ/kg

u2 = 2505.5 KJ/kg

∆U = 10(758.99) -(2.860) (2505.5) = 424.17 KJ

∆U = W = 424.17 KJ

please upvote if helpful

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