Question

Superheated steam enters a turbine at 3 MPa, 550oC, and exits at 0.01 MPa.
a) If the process is reversible adiabatic (isentropic), find the final
temperature (T2s), the final enthalpy (h2s) of the steam, and the turbine
work (Wt,s).
b) What is Sgen for the above process?
c) If the isentropic efficiency is 90%, find the actual final temperature (T2a)
and calculate Sgen?
d) Plot process (a) and (c) on a Ts diagram.

Answer 1

P₁ = 3 MPa

T₁ = 550°C

P₂ = 0.01 MPa

A) the process is reversible adiabatic (isentropic)

Then s₁=s₂

From steam tables in Perry handbook

At P₁ = 3 MPa and T₁= 550°C

H₁ = 3569.75 KJ/Kg

S₁ = 7.3767 KJ/kg K

Since

S₁ = S₂

S₂ = 7.3767 KJ/Kg K

P₂= 0.01 MPa

From steam tables

T_2s= 45.806°C = 318.806 K

H_L = 191.81 KJ/Kg

H_G = 2583.9 KJ/Kg

S_g = 8.1488 KJ/Kg K

S_L = 0.64920 KJ/kg K

The state of steam is between saturated vapor and saturated liquid

To find quality of steam

xS_g+(1-x) S_L = S_2s

x(8.1488) +(1-x) (0.64920) = 7.3767

x = 0.8970

H_2s= xH_g+ (1-x) H_L

H_2s= 0.8970(2583.9) +(1-0.8970) (191.81)

H_2s = 2337.514 KJ/Kg

Work done

W_s = H₁-H_2s

W_s = 3569.75-(2337.514) = +1232.236 KJ/Kg

B) S_gen = S₂-S₁ = 0

C) isentropic efficiency = 90%

Efficiency = (H₁-H₂) /(H₁-H_2s)

0.90 = (H₁-H₂) /(3569.75-2337.514)

H₂ = 2460.737 KJ/Kg

P = 0.01 MPa

H ₂= 2460.737 KJ/Kg

xH_g + (1-x) (H_L) = H₂

x(2583.9) +(1-x) (191.81) = 2460.737

x = 0.9485

Actual quality = 0.9485

From steam table we can see that steam at this state is in between saturated liquid and saturated vapor hence temperature does not change during this phase change

T₂ =45.806°C

S₂ = xS_g+ (1-x) (S_L) = 0.9485(8.1488) +(1-0.9485) (0.64920)

= 7.7625 KJ/Kg K

S₁ = 7.3767 KJ/Kg K

S_gen = S₂-S₁

S_gen = 7.7625-7.3767 = 0.38587 KJ/Kg K

d)

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