Question

A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 25 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump. Use steam tables.

Answer 1

Answer:

power produced by turbine = 79656.15 KW

pump work =227.57 KW

Explanation:

here At T1 = 700 C and P1 = 4 Mpa, from steam tabel : h1 =3906.41 KJ/Kg and s1 = 7.62 KJ/Kg.K

now s1 = s2 = 7.62 KJ/Kg.K as 1-2 is isentropic

so at P2 = 20 Kpa and s2 = 7.62 KJ/Kg.K , fom steam table : h2 = 2513.33 KJ/Kg

power produced by turbine = m.( h1 - h2) = 57.18 x ( 3906.41 - 2513.33) = 79656.15 KW

at P3 = 20 Kpa, 3 = vf = 0.001 m^3/kg

so pump work = v3.dP = 57.18 x 0.001 x ( 4000 - 20) =227.57 KW

Answer 2

Answer:

PT = 70 000kw

Pp = 202.4kw

Explanation:

The full details of the explanation is attached.