Question

An actual Rankine Engine turbo-generator receives steam at a pressure of 10MPa and etropy of 6.931 j/kgK, the steam then expands to 500kPa with a temperature of 250°C and leaves the condenser as a saturated liquid at 150°C.

a. find the engine thermal efficiency of the actual engine

b. The actual steam rate for the engine

c. if the mechanical efficiency is 80% and the generator efficiency is 90%. What is the combined work in kW if mass steam flow rate is 5000kg/hr?

Answer 1

Given Data

According to the data, steam is received at the engine at 10 MPa (which is 100 bar) at an entropy of 6.931 J/kg-K.

The steam expands to a pressure of 500 kPa (which is 5 bar) at a temperature of 250 degree celcius.

The steam condenses in the condenser as a saturated liquid at 150 degree celcius.

Thus, according to the diagram,

Solution

From steam tables at saturated water and steam (pressure-based) at 0.5 MPa,

From the Rankine T-S diagram, we know that,

From steam tables at saturated water and steam (pressure-based) at 0.5 MPa,

We need to find out the fraction of water vapour in the liquid,

This shows that the vapour is saturated steam at the end of expansion.

This shows that the steam is superheated at state 3.

From steam tables at superheated steam at 10 MPa,

We see that the value of 6.93 is the average of values of entropy provided at 600 and 620 degree celcius.

Thus,

Similarly.

a) Thermal efficiency is given by,

b) Net work done is given by,

Mass flow rate of steam has not been provided explicitly in the problem. But provided only in part c. We will assume that the mass flow rate given in part c is applicable here as well.

Net work done in kW is given by,

Steam rate is given by,

c)

Combined work is given by,

P.S. Standard steam tables are available online to verify.