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

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:  

$\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s}$

$W_{real} = \eta_t*W_s$

$W_{real} = 0.8*100 kW$

$W_{real} = 80 kW$

Part 2

Coefficient of performance COP is defined by:

$COP = \frac{Q_{out}}{W}$

$Q_{out} = W*COP$

$Q_{out} = 5 kW*2.2$

$Q_{out} = 11 kW$

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)

$COP = \frac{Q_{in}}{W}$

$COP = \frac{3 kW}{5 kW}$

$COP = 0.6$