Question

A steam turbine has isentropic efficiency of 0.8. Isentropically, it is supposed to deliver work of 100 kW. What is the actual work delivered by the turbine? A heat pump has a COP of 2.2. It takes 5 kW electric power. What is the heat delivery rate to the room being heated in kW? Heat pump is used for winter heating of a room. A refrigerator takes 5 kW electric power. It extracts 3 kW of heat from the space being cooled a. What is the heat delivery rate to the surroundings in kW? b. What is the COP of the refrigerator?

Answer 1

80 kW; 11 kW; 8 kW; 0.6

Step-by-step explanation:

Part 1

Isentropic turbine efficiency:

`\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = (W_(real))/(W_s)`

`W_(real) = \eta_t*W_s`

`W_(real) = 0.8*100 kW`

`W_(real) = 80 kW`

Part 2

Coefficient of performance COP is defined by:

`COP = (Q_(out))/(W)`

`Q_(out) = W*COP`

`Q_(out) = 5 kW*2.2`

`Q_(out) = 11 kW`

Part 3

(a)

Energy balance for a refrigeration cycle gives:

`Q_(in) + W = Q_(out)`

`3 kW + 5 kW = Q_(out)`

`8 kW = Q_(out)`

(b)

`COP = (Q_(in))/(W)`

`COP = (3 kW)/(5 kW)`

`COP = 0.6`