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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 _____.

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xero099

Answer:

[tex]C_{day} = 2.261\,USD[/tex]

Explanation:

The heat transfer rate rejected to the refrigeration system is:

[tex]\dot Q_{L} = (0.25\,\frac{kg}{s} )\cdot (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (30\,^{\textdegree}C-15\,^{\textdegree}C)[/tex]

[tex]\dot Q_{L} = 3.768\,kW[/tex]

The electric power needed to make refrigeration possible is:

[tex]\dot W = \frac{\dot Q_{L}}{COP_{R}}[/tex]

[tex]\dot W = \frac{3.768\,kW}{2.8}[/tex]

[tex]\dot W = 1.346\,kW[/tex]

The daily energy consumption is:

[tex]\Delta E_{day} = (1.346\,kW)\cdot (86400\,s)[/tex]

[tex]\Delta E_{day} = 116294.4\,kJ[/tex]

[tex]\Delta E_{day} = 32.304\,kWh[/tex]

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

[tex]C_{day} = c\cdot \Delta E_{day}[/tex]

[tex]C_{day} = (0.07\,\frac{USD}{kWh} )\cdot (32.304\,kWh)[/tex]

[tex]C_{day} = 2.261\,USD[/tex]

ogbe2k3

Answer:

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

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