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Turbin

- The document describes a steam turbine with a rated capacity of 56,400 kW operating with steam conditions of 8,600 kPa and 500°C at the inlet and 10 kPa at the outlet. - Given a turbine efficiency of 0.75, the state of steam (enthalpy and quality) at the outlet is calculated to be 2,436.62 kJ/kg and 0.939, respectively. - The mass flow rate of steam through the turbine is calculated to be 59.05 kg/s.

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Nurul Ramadhani
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283 views3 pages

Turbin

- The document describes a steam turbine with a rated capacity of 56,400 kW operating with steam conditions of 8,600 kPa and 500°C at the inlet and 10 kPa at the outlet. - Given a turbine efficiency of 0.75, the state of steam (enthalpy and quality) at the outlet is calculated to be 2,436.62 kJ/kg and 0.939, respectively. - The mass flow rate of steam through the turbine is calculated to be 59.05 kg/s.

Uploaded by

Nurul Ramadhani
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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TURBIN/EXPANDER

turbin Ws

∆Ep + ∆Ek +∆H = Q – Ws

-W = ∆H

η = ∆H/∆Hs = -Ws/ws (isentropic)

Example 7.6

A Steam turbin with rated capacity of 56.400 kW (56.400 kJ s—1) operates with steam at inlet
conditions of 8.600 kPa and 500°C, and discharges into a condenser at a pressure of 10 kPa.
Assuming a turbine efficiency of 0,75, determine the state of the steam at discharge and the mass
rate of flow of the steam.

Answer

Diketahui :

P = 8.600 kPa

T = 500°C

η = 0,75

H1 = 3391,6 kJ/kg

S1 = 6,6858 kJ/kg

Penyelesaian :

S1 = S2’
Pada P = 10 kPa

S2L = 0,6514

S2v= 8,1481

S2’= S2L + x2’ (S2v - S2L)

6,6858 = 0,6514 + x2’ (8,1481 – 0,6514)

6,6858 - 0,6514 = 8,1481 x2’ – 0,6514 x2’

6,034 = 7,4967 x2’

x2’ = 0,8049

Pada P = 10 kPa

H2L = 192,5

H2V = 2585,1

H2’ = H2L + x2’ (H2V - H2L)

H2’ = 192,5 + 0,8049 (2585,1 – 192,5)

H2’ = 2118,30 kJ/kg

 (∆H)s = H2’ – H1

= 2118,30 – 3391,6

= -1273,3 kJ/kg

 ∆H = η x (∆H)s

= 0,75 x (-1273,3)

= -954,975 kJ/kg

 ∆H = H2 – H1

H2 = -954,975 +3391,6

= 2436,62 kJ/kg
 H2 = H2L + x (H2V - H2L)
2436,62 = 192,5 + x (2585,1 – 192,5)
2436,62 = 192,5 + 2392,6 x
2436,62 -192,5 = 2392,6 x
2244,12 = 2392,6 x
x = 0,939

 -Ws = ∆H
Ws = -56400 kJ/s
-Ws = ∆H
-56400 kJ/s = m (H2 – H1)
-56500 kJ/s = m (2436,62 – 3391,6) Kj/kg
m = 59,05 kg/s

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