Steam at a temp. of 500 degrees celsius and pressure of 20mpa enters a rankine cycle and expands adiabatically to a condenser pressure of 100kpa. the adiabatic pump consumes 35 kj/kg of energy to bring the pressure of the liquid saturated water (x = 0) that is exiting the condenser up to the pressure of the boiler. the thermal efficiency of the power plant is 0.40. 1. draw a t-s diagram that shows 1 complete cycle of this power plant. 2. draw the power plant with its compartments and label them. 3. how much heat (in kj/kg) is put into the cycle during the heat absorption process? d) what is the isentropic efficiency of the turbine? (ignore changes in potential or kinetic energy). e) what is the isentropic efficiency of the pump?

The answers t o question a and b are attached. The heat into the cycle  is 2788.69 kJ/kg. The isentropic efficiency of the turbine and pump is 0.94 and 0.59 respectively.

Explanation:

c) We determine the enthalpies at various points in the cycle:

State 1: P1 =100 kPa,  saturated liquid x=0 ∴ From steam tables

h1 = 417.51 kJ/kg, v1 = 0.001043 m^3/kg

State 2: P2 = 20 MPa , s2= s1 (isentropic compression)

We know that the work of the pump is 35 kJ/kg therefore we can do an energy balance arround the pump:

h2 = h1 + w,pump

h2 = 417.51 + 35 = 452.51 kJ/kg

State 3: P3 = 20 MPa, T3 = 500°C ∴ From steam tables for superheated vapor h3 = 3241.2 kJ/kg, s3 = 6.1446 kJ/kg.K

State 4: P4 = 100 kPa (saturated liquid-vapor mixture)

We have to determined the quality (x4) of the saturated liquid-vapor mixture and we know that s3=s4= 6.1446 kJ/kg.K

∴ x4 = (s4-sf)/sfg  from steam tables sf = 1.3028 kJ/kg.K, sfg = 6.0562 kJ/kg./k

x4 = 0.799 ∴ From the saturated water tables h4 = hf + x4×hfg = 419.17 + 0.799×2256.4 = 2222.03 kJ/kg

We can do an energy balance around the boiler:

q,in = h3-h2 =  3241.2 -  452.51 = 2788.69 kJ/kg

d) the isentropic efficiency of the turbine:

The isentropic efficiency of the turbine is the actual work of the turbine divided by the isentropic work of the turbine.

n, turbine = w,actual/w,isentropic

we know that he thermal efficiency is n=0.4 = w,net/q,in

∴ w, net = 0.4×2788.69 = 1115.476 kJ/kg

we know that w, net is the total work of the pump and turbine:

w, net = w, pump - w, turbine

w, isentropic turbine = 35 - 1115.476 kJ/kg = - 1080.48 kJ/kg = 1080.48 kJ/kg of work out

This is the actual work of the turbine:

The work of the turbine can be done using the energy balance around the turbine:

h3 = w, turbine + h4

w= h3 - h4 = 3241.2 - 2222.03= 1019.17 kJ/kg

∴ the isentropic efficiency = n, isentropic = 1019.17/1080.48 = 0.94

e) The isentropic work of the pump can be determined from the energy balance around the pump:

h1 + w, pump = h2

w, pump =  452.51 kJ/kg - 417.51 kJ/kg = 35 kJ/kg

w, isentropic pump = v1×(P1-P2) = 0.001043× (20000-100) = 20.7557 kJ/kg

Therefore the isentropic efficiency of the pump is n = 20.7557/35 = 0.59