Calculate the amount of work input a refrigerator needs to m

Calculate the amount of work input a refrigerator needs to make ice cubes out of a tray of 0.25 kg of liquid water at 10 C. Assume the refrigerators has a coefficient of performance equal to 3.5. The motor-compressor is 750 W. Assume this only cooling load on the refrigerators. How long does it take to make these ice cubes?

Solution

Solution:

C.V. Water in tray. We neglect tray mass.

Energy Eq.:    m(u2 u1) = 1Q2 1W2

Process : P = constant = Po

1W2 = P dV = Po m(v2 v1)

1Q2 = m(u2 u1) + 1W2 = m(h2 h1)

Tbl. B.1.1 : h1 = 41.99 kJ/kg,

Tbl. B.1.5 : h2 = - 333.6 kJ/kg

1Q2 = 0.25(-333.4 – 41.99 ) = - 93.848 kJ

Consider now refrigerator

= QL/W

W = QL/ = - 1Q2/ = 93.848/3.5 = 26.81 kJ

For the motor to transfer that amount of energy the time is found as

W = W . dt = W. t

t = W/W. = (26.81 × 1000)/750 = 35.75 s

Comment: We neglected a baseload of the refrigerator so not all the 750 W are available to make ice, also our coefficient of performance is very optimistic and finally the heat transfer is a transient process. All this means that it will take much more time to make ice-cubes.