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.

Answer 1

Answer:

$C_{day} = 2.261\,USD$

Explanation:

The heat transfer rate rejected to the refrigeration system is:

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

$\dot Q_{L} = 3.768\,kW$

The electric power needed to make refrigeration possible is:

$\dot W = \frac{\dot Q_{L}}{COP_{R}}$

$\dot W = \frac{3.768\,kW}{2.8}$

$\dot W = 1.346\,kW$

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:

$C_{day} = c\cdot \Delta E_{day}$

$C_{day} = (0.07\,\frac{USD}{kWh} )\cdot (32.304\,kWh)$

$C_{day} = 2.261\,USD$

Answer 2

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