A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to the outside air. At steady‐state, the AC cycle requires 0.434kW and has a coefficient of performance (COP) of 6.22. Determine the rate at which the energy is removed from the room air, in kW. If electricity is valued at $0.10/kw-hr, determine the cost of operating the unit for 24hrs.

Answers

Answer 1

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by  = 6.22

Therefore he heat removed is given by , [tex]$Q_H = 6.22 \times 0.434$[/tex]

                                                                     [tex]$Q_H = 2.7 \ kW $[/tex]

Now if the electricity is valued at  0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

                                                                                            = 6.48

Therefore the operating cost = $ 6.48 for 24 hours.


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