Question

During an experiment conducted in a room at 25°C, a laboratory assistant measures that a refrigerator that draws 2 kW of power has removed 30,000 kJ of heat from the refrigerated space, which is maintained at -30°C. The running time of the refrigerator during the experiment was 20 min. Determine if these measurements are reasonable. g

Answer 1

Answer:

Not reasonable.

Explanation:

To solve this problem it is necessary to take into account the concepts related to the performance of a reversible refrigerator. The coefficient of performance is basically defined as the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. For a reversible refrigerator the coefficient is given by

$COP_{R,rev} = \frac{1}{\frac{T_1}{T_2}-1}$

Where,

$T_1 =$High temperature

$T_2 =$Low Temperature

With our values previous given we can find it:

$T_2 = -30\°C = (-30+273)$

$T_2 = 243K$

$T_1 = 25\°C = (25+273)$

$T_1 = 298K$

With these values we can now calculate the coefficient of performance:

$COP_{R,rev} = \frac{1}{\frac{298}{243}-1}$

$COP_{R,rev} = 4.42$

At the same time we can calculate the work consumption of the refrigerator, this is

$W = \dot{W}\Delta t$

Where,

$\dot{W} =$ Required power input

t = time to remove heat from a cool to water medium

$W = 2kJ/s * 20 min$

$W = 2kJ/s * 1200s$

$W = 2400kJ$

In this way we can calculate the coefficient of the refrigerator directly:

$COP_R = \frac{Q_L}{W}$

Where,

Q = Amoun of heat rejected

$COP_R = \frac{30000}{2400}$

$COP_R = 12.5$

Comparing the values of both coefficients we have that the experiments are NOT reasonable, because the coefficient of a refrigerator is high compared to  coefficient of reversible refrigerator.