Question

When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner, which cools the entire house to 20°C in 38 min. If the COP of the air-conditioning system is 2.8, determine the power drawn by the air conditioner. Assume the entire mass within the house is equivalent to 800 kg of air for which cv = 0.72 kJ/kg·°C and cp = 1.0 kJ/kg·°C

Answer 1

W_in = 1.353 KW

Step-by-step explanation:

Given:

- Initial temperature of the house T_1 = 35°C

- Final temperature of the house T_2 = 20°C

- Time taken to cool the house dt = 38 min = 38×60 = 2280 s

- mass of air in the house m = 800 kg

- Specific heat at constant volume c_v = 0.72 kJ/kgK

- Specific heat at constant pressure c_p = 1.0 kJ/kgK

Find:

- Determine the power drawn by the air conditioner.

Solution:

- We will first compute the rate of heat removal from the room, we will use c_v due to a constant volume process, as follows:

Q_l = m*c_v*dT/dt

- In put values given:

Q_l = 800*0.72*(35-20) / 2280

Q_l = 3.7894 KW

- The relationship between the heat rejection and the COP of an air conditioner is given as:

COP = Q_l / W_in

W_in = Q_l / COP

W_in = 3.7894 / 2.8

W_in = 1.353 KW

- Hence the amount of power required for this process is 1.353 KW

Answer 2

1353.38 Watt

Step-by-step explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

`Q=(q)/(\Delta t)\\\Rightarrow Q=(8640000)/(2280)\\\Rightarrow Q=3789.47\ W`

`COP=(Q)/(W)\\\Rightarrow W=(Q)/(COP)\\\Rightarrow W=(3789.47)/(2.8)\\\Rightarrow W=1353.38\ W`

∴ Power drawn by the air conditioner is 1353.38 Watt