A household refrigerator is maintained at temperature of $2^oc$. Every time the door is opened, warm material is placed inside introducing an average of 420KJ, but making only a small change in the temperature of the refrigerator. The door is opened 20 times a day and the refrigerator operates at 15% of the Ideal COP. The cost of the work is 4 rupees per kWh.The atmosphere is at $30^oC/$ - Mumbai university > MECH > SEM 3 > THERMO

Marks: 10M

Year: Dec 2014

Given:

Q = 420 kJ

N = 20

Th = 303K; Tl = 275K

Cost = 4rs/kWh

Cop = 0.15 COPmax

Solution:

COPmax = Tl/(Th-Tl) = 275/(303 – 275) = 275/28

COP = 0.15 x 275/28 = 1.473

Also, COP = Q / W

W = 420/1.473 = 285.13kJ

So, each time the door is opened, 285.13 kJ of work is to be done by fridge and that much electrical energy is consumed. The fridge is opened 20 times a day or 600 times a month.

So, monthly consumption = 285.13 x 600 = 171079.4 kJ

Cost = 4 rs/ kWh = 4/3600 rs/kJ

Bill = cost x consumption

= 4/3600 x 171079.4 = 190 rupees