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Thermodynamics - Rankine Cycle - Saturated steam
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Thermodynamics - Rankine Cycle - Saturated steam
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Jose Alfonso F. FabeBSME 2 – Thermodynamics 2
Level 4: Rankine Cycle
a.) Ideal Rankine Cycle with boiler producing saturated steam.
Water is the working fluid in an ideal Rankine Cycle. Saturated vapor enters the turbine at 5
𝑘𝑔
MPa, and the condenser pressure is 50 kPa. The mass flow rate of the cycle is 3 . Find the
𝑠
steam rate, turbine work, pump work, total heat added to the cycle, total heat rejected by the
condenser, power output of the cycle, net heat, and thermal efficiency.
Solution:
, At State 1:
𝑝1 = 5 𝑀𝑃𝑎
Since, it is stated in the problem that saturated vapor enters the turbine, then, it is in saturated
steam state.
From table 2.
𝑘𝐽
ℎ1 = ℎ𝑔 @5𝑀𝑃𝑎 = 2,794.3
𝑘𝑔
𝑘𝐽
𝑠1 = 𝑠𝑔 @5𝑀𝑃𝑎 = 5.9734
𝑘𝑔 ∙ 𝐾
At State 2:
𝑝2 = 50𝐾𝑃𝑎 → 0.05𝑀𝑃𝑎
𝑘𝐽
𝑠2 = 𝑠1 = 5.9734
𝑘𝑔 ∙ 𝐾
From table 2 @ 0.05 MPa
𝑘𝐽
𝑠𝑓 = 1.0910
𝑘𝑔 ∙ 𝐾
𝑘𝐽
𝑠𝑔 = 7.5939
𝑘𝑔 ∙ 𝐾
𝑠𝑓 < 𝑠2 < 𝑠𝑔 , then, it is in wet steam state.
Solve for the quality, 𝑥2;
𝑠2 = 𝑠𝑓 + 𝑥2 𝑠𝑓𝑔
𝑠2 − 𝑠𝑓
𝑥2 =
𝑠𝑓𝑔
𝑘𝐽 𝑘𝐽
5.9734 − 1.0910
𝑘𝑔 ∙ 𝐾 𝑘𝑔 ∙ 𝐾
𝑥2 =
𝑘𝐽 𝑘𝐽
7.5939 − 1.0910
𝑘𝑔 ∙ 𝐾 𝑘𝑔 ∙ 𝐾
𝑥2 = 0.7508 𝑜𝑟 75.08%%
Solving for the enthalpy in state 2, ℎ2;
𝑘𝐽 𝑘𝐽
ℎ2 = ℎ𝑓 + 𝑥2 ℎ𝑓𝑔 ; ℎ𝑓 = 340.49 & ℎ𝑓𝑔 = 2,305.4
𝑘𝑔 𝑘𝑔
𝑘𝐽 𝑘𝐽
ℎ2 = 340.49 + 0.7508(2,305.4 )
𝑘𝑔 𝑘𝑔
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