Question

In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. What is the shaft work generated per kilogram of steam if the efficiency of the turbine is 0.56

Answer 1

The shaft work generated per kilogram is
`-1.3 (MJ)/(kg)`

Step-by-step explanation:

Given:

Temperature
`T = 1273.15` K

Initial Pressure
`P_(1) = 1.8` MPa

Final pressure
`P_(2) = 0.1` MPa

From the table superheated,

`h_(i) = 4635`
`(K J)/(Kg)` and
`h_(f) = 2706.54`
`(K J)/(Kg)`

Work done by shaft is,

`W = h_(f) - h_(i)`

`W = 2706.54 - 4635`

`W = -1928.46 (kJ)/(kg)`

But here efficiency is 0.56,

So work generated per kg is,

Work =
`0.56 *(- 1928.46)`

Work =
`-1.3`
`(MJ)/(kg)`

Therefore, the shaft work generated per kilogram is
`-1.3 (MJ)/(kg)`