Water is the working fluid in an ideal Rankine cycle.
Saturated vapor enters the turbine at 16MPa, and the
condenserpressure is 8kPa. The mass flow rate of steam entering
theturbine is 120 kg/s. Determine
a) the net power developed, in kW
b) the rate of heat transfer to the steam passing through
theboiler, in kW
c) the thermal efficiency
d) the mass flow rate of condenser of cooling water, inkg/s,
if the cooling water undergoes a temperature increase of 18degree
Celcius with negligible pressure change in passing throughthe
condenser.

A rankine cycle is a generation cycle using water as a working fluid, when heat enters the boiler the water undergoes a series of changes in state and energy until generating power through the turbine.

This cycle is composed of four main components, the boiler, the pump, the turbine and the condenser as shown in the attached image

To solve any problem regarding the rankine cycle, enthalpies in all states must be calculated using the thermodynamic tables and taking into account the following.

• The pressure of state 1 and 4 are equal

• The pressure of state 2 and 3 are equal

• State 1 is superheated steam

• State 2 is in saturation state

• State 3 is saturated liquid at the lowest pressure

• State 4 is equal to state 3 because the work of the pump is negligible.

Once all enthalpies are found, the following equations are used using the first law of thermodynamics

Wout = m (h1-h2)

Qin = m (h1-h4)

Win = m (h4-h3)

Qout = m (h2-h1)

The efficiency is calculated as the power obtained on the heat that enters

Efficiency = Wout / Qin

Efficiency = (h1-h2) / (h1-h4)

first we calculate the enthalpies in all states

h1=245.2kJ/Kg

h2=227.9KJ/Kg

h3=h4=173.9KJ/Kg

a)we use the ecuation for Wout

Wout = m (h1-h2)

Wout = 120 (245.2-227.9) =2076kW

b) we use the ecuation for Qin

Qin = m (h1-h4)

Qin = 120 (245.2-173.9) =8556Kw

C) we use the ecuation for efficiency

Efficiency = Wout / Qin

Efficiency =2076/8556=0.2426=24.26%

D)f

For this part, the first thing we should do is calculate Qout later using the first law of thermodynamics in the condenser

Qout = m (h2-h1)

Qout = 120 (227.9-173.9) =6480kW

The heat of the condenser is equal to that of water.

Qout=m Cp ΔT

m=Qout/(CpΔT)

Cp=4.182KJ/kgK

m=6480/(4.182x18)=86kg/S