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ID: 2991555 • Letter: P
Question
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Water is the working fluid in an ideal rankine cycle. SUPERHEATED VAPOR enters the turbine at 8 MPa, 480 oC. The consenser pressure is 8 kPa. The net power output of the cycle is 100MW. Determine for the cycle:
a) rate of heat transfer to the working flui passing through the steam generator, in kW.
b) the thermal efficiency
c) the mass flow rate of condenser cooling water in kg/h, if the cooling water enters the condenser at 15 oC and exits at 35oC with negligible pressure change.
Please help me out, and please tell me where values come from. There is no table for 8MPa and 480oC, so I don't know how to find h for it!!
Explanation / Answer
From superheated steam tables, (http://eng-calculations.com/Superheated%20Steam.aspx/Superheated%20Steam.aspx)
Corresponding to 8 MPa, 480 deg C, specific enthalpy h1 = 3349.5 kJ/kg, specific entropy s1 = 6.66106 kJ/kg-K
From saturated steam tables, (http://enpub.fulton.asu.edu/ece340/pdf/steam_tables.PDF)
Corresponding to 8 kPa, sf = 0.593 kJ/kg-K and sg = 8.229 kJ/kg-K, hf = 173.8 kJ/kg, hg = 2577 kJ/kg
Isentropic expansion in turbine implies, s2 = s1.
Taking quality of steam after expansion as x, 0.593+x(8.229-0.593) = 6.66106
This gives, x = 0.795 or 79.5%
So, h2 = 173.8+0.795*(2577-173.8) = 2083.5 kJ/kg
We have, h3 = hf@8 kPa = 173.8 kJ/kg
Net power output of the cycle = m[(h1-h2)-(h4-h3)]
Assuming h4h3,
we get 100*106 = m [3349.5-2083.5]
This gives, m = 79 kg/s
(a) Rate of heat transfer to the fluid = m(h1-h4) = 79*(3349.5-173.8) = 250880 kW
(b) Efficiency = Net work/Heat input = 100*106/(250880*103) = 0.3986 = 39.86 %
(c) Heat transfer in condensor = m(h2-h3) = 79*(2083.5-173.8) = 150866.3 kW
So, mcondCp(T) = 150866.3
Taking Cp for water = 4.187 kJ/kg-K
mcond*4.187*(35-15) = 150866.3
This gives, mcond = 1801.6 kg/s