7.61 Steam enters a turbine operating at steady state at 6 MPa, 500°C with a mas

Question

7.61 Steam enters a turbine operating at steady state at 6 MPa, 500°C with a mass flow rate of 400 kg/s. Saturated vapor exits at 8 kPa. Heat transfer from the turbine to its surroundings takes place at a rate of 8 MW at an average surface temperature of 180°C. The effects of motion and gravity are negligible.

(a) For a control volume enclosing the turbine, determine the power developed and the rate of exergy destruction, each in MW.

ans. --> W_cv = 330.1 MW, E_d = 167.1

(b) If the turbine is located in a facility where the ambient temperature is 27°C determine the rate of exergy destruction for an enlarged control volume that includes the turbine and its immediate surroundings so the heat transfer takes place at the ambient temperature. Explain why the exergy destruction values of parts (a) and (b) differ.

ans. --> 169.8 MW

Please show all work.

Explanation / Answer

solution:

1)here by steady flow energy equtaion for turbine

W'=m'(h1-h2)

h1=3422.2 kj/kg

h2=2577.1 kj/kg from table

work done=W'=338.04 MW

but heat rejected=Wrejected=Wexergy=8 MW

hence useful work is

W'useful=W'max-Wrejected=338.04-8=330.04 MW

2)here by second law of thermodynamics entropy generation is given by

S'=Qrejected/Tavg+m'(S2-S1)

s1=6.882

S2=8.23

we get that for Tavg=180+273=453 k

S'=556.86 kj/kg k

where exergy destruction is given by

Ed=Tavg*S'=453*.55686=252.257 MW

for Tavg=300 K

S1'=565.86 kj/kg k

exergy geberation as

Ed1=S1'*Tavg=.56586*300=169.76 MW

as entropy production rate is almost same for both cases but exergy which work potential and directly proportional to temperature at which work rejection to be done and hence useful work would be more fore second case as exergy is reduced.

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