Question

Air is flowing steadily through a cooling section where it is cooled from 30 °C to 15 °C at a rate of 0.25 kg/s.

If the average COP of the air-conditioner is 2.8 and the unit cost of electricity is $0.07 per kWh, the cost of electricity consumed by this air-conditioner per day (a period of 24 hours) is _____.

Answer 1

The cost of electricity consumed by the air-conditioner per day is Cday = 2.261 USD

Step-by-step explanation:

The rate of the heat transfer to the refrigeration system is:

QL = (0.25 kg/s) * (1.005 kJ/kg.0 C) * (300 C – 150 C)

QL = 3.768 kW

The power (electric) needed to make refrigeration possible is:

Ẇ = QL /COPR

Ẇ = 3.768kW/2.8

Ẇ = 1.368kW

The daily energy consumption is

∆Eday = (1.368kW) * (86400 s)

∆Eday = 116294.4 kJ

∆Eday = 32.304 kWh

Then,

The cost of electricity consumed by the air-conditioner per day is:

Cday = c . ∆Eday OR c * ∆Eday

Cday = (0.07 USD/kWh) * (32.304kWh)

Therefore,

Cday = 2.261 USD

Answer 2

`C_(day) = 2.261\,USD`

Step-by-step explanation:

The heat transfer rate rejected to the refrigeration system is:

`\dot Q_(L) = (0.25\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot ^(\textdegree)C) )\cdot (30\,^(\textdegree)C-15\,^(\textdegree)C)`

`\dot Q_(L) = 3.768\,kW`

The electric power needed to make refrigeration possible is:

`\dot W = (\dot Q_(L))/(COP_(R))`

`\dot W = (3.768\,kW)/(2.8)`

`\dot W = 1.346\,kW`

The daily energy consumption is:

`\Delta E_(day) = (1.346\,kW)\cdot (86400\,s)`

`\Delta E_(day) = 116294.4\,kJ`

`\Delta E_(day) = 32.304\,kWh`

The cost of electricity consumed by the air-conditioner per day is:

`C_(day) = c\cdot \Delta E_(day)`

`C_(day) = (0.07\,(USD)/(kWh) )\cdot (32.304\,kWh)`

`C_(day) = 2.261\,USD`