hello, below is my question. ive tried to work it but everything i find involves

Question

hello, below is my question. ive tried to work it but everything i find involves more information and material than i have been given / learned. Will be rated highly if you help me! thank you! GIVEN: In a steam power plant, water enters the pump as saturated liquid, then leaves the turbine as saturated vapor. The pressure levels for the two devices are 5MPa and 10 kPa. FIND: The ratio of the turbine work to the pump work is to be determined. HINT: use the formula for reversible steady flow work, assuming the processes reversible and neglecting losses. thank you!

Explanation / Answer

TURBINE ;

at 10 Kpa, enthalpy = h2 = hv = 2584.9 KJ/Kg

and entropy = s2 = sg = 8.1507 KJ/Kg.K

now s2 = s1 = 8.1507 KJ.Kg.K as the process is reversible (1 - is the inlet of pump)

so at P1 = 5 MPa nd s1 = 8.1507 KJ/Kg.K,from steam table : h1 = 3999 KJ/Kg.

so turbine work = h1 - h2 = 3999 - 2584.9 = 1414.1 KJ/Kg

PUMP :

at 10 Kpa , specific volume = v1 = vf at 10 Kpa = 0.001 m^3/kg

so pump work = 0.01 x (5000 - 10) = 4.99 KJ/Kg

so ratio = 1414.1/3.99 = 354.41

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