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Consider a typical steam Rankine power cycle. The steam generator and condenser

ID: 1862645 • Letter: C

Question

Consider a typical steam Rankine power cycle. The steam generator and condenser operate at 80 bar

and 0.10 bar respectively. The main liquid pump in the cycle has an isentropic efficiency of 70 % and is driven

by a 1500 kW electric motor. The pump operates adiabatically at steady-state steady-flow conditions with

negligible changes in kinetic and potential energy. The pump inlet and exit pipes are the same size. The

average flow velocity in the pipes is specified as 10 m/s.

The liquid may be treated as incompressible with ? = 1000 kg/m3

& C = 4.18 kJ/kg-K.

a) Find how much liquid is pumped,

m& = _______ (kg/s) and the pipe diameter, D = _______ (m).

b) Find the pump exit temperature, T4 = _____ (oC), and enthalpy, h4 = ________ (kJ/kg).

c) Find the entropy production rate, 3 ?& 4 = _________ (kJ/s-K).

Consider a typical steam Rankine power cycle. The steam generator and condenser operate at 80 bar and 0.10 bar respectively. The main liquid pump in the cycle has an isentropic efficiency of 70 % and is driven by a 1500 kW electric motor. The pump operates adiabatically at steady-state steady-flow conditions with negligible changes in kinetic and potential energy. The pump inlet and exit pipes are the same size. The average flow velocity in the pipes is specified as 10 m/s. The liquid may be treated as incompressible with rho = 1000 kg/m3 & C = 4.18 kJ/kg-K. Find how much liquid is pumped, = (kg/s) and the pipe diameter, D = (m). Find the pump exit temperature, T4 = (degree C), and enthalpy, h4 = (kJ/kg). Find the entropy production rate, 3 4 = (kJ/s-K).

Explanation / Answer

1500*1000*0.7 = (80 - 0.1)*10^5*V' V' = 0.1314 m^3/s a) mass flow rate = V' * density = 0.1314*1000 = 133.14 kg/s b) since, pump operates adiabatically at steady-state steady-flow conditions dq = du + dw 0 = M'*Cp*(T4 - T3) - 1500*1000*0.7 (T4 - T3) = 1500000*0.7/(4.18*1000*133.14) = 1.8867 T4 = 1.8867 + T3 (where T3 is ambient temperature) T4 = 26.8867 degree celcius for T3 = 25 degree celcius h4 = 4.18*(273.13 + 26.8867) = 1254.07 kJ/kg c) entropy produce = Cp*ln(T4/T3) - R*ln(P4/P3) = 4.18*ln(300.0167/298.13) - R*ln(80/0.1)