Question

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to 6°C at an average rate of 10 kg/h. If the COP of this refrigerator is 3.1, the required power input to this refrigerator is
(a) 197W
(b) 612W
(c) 64W
(d) 109W
(e) 403W

Answer 1

Answer:

The correct option is;

(c) 64W

Explanation:

Here we have the Coefficient Of Performance, COP given by

$COP = \frac{Q_{cold}}{W} = 3.1$

The heat change from 23° to 6°C for a mass of 10 kg/h which is equivalent to 10/(60×60) kg/s or 2.78 g/s we have

$Q_{cold}$ = m·c·ΔT = 2.78 × 4.18 × (23 - 6) = 197.39 J

Therefore, plugging in the value for  $Q_{cold}$ in the COP equation we get;

$COP = \frac{197.39 }{W} = 3.1$ which gives

$W = \frac{197.39 }{3.1} = 63.674 \ J$

Since we were working with mass flow rate then the power input is the same as the work done per second and the power input to the refrigerator = 63.674 J/s ≈ 64 W.

The power input to the refrigerator is approximately 64 W.

Answer 2

Answer:

Win = 64 W ... Option C

Explanation:

Given:-

- The water is cooled in the refrigerator with delta temperature, ΔT=(23 - 6 )

- The flow rate of the refrigerated water is flow ( m ) = 10 kg/h

- The COP of the refrigerator is  = 3.1:

Find:-

the required power input to this refrigerator is

Solution:-

- The COP - The coefficient of performance of a refrigerator is a quantity that defines the efficiency of the system. The COP is given as:

                                         

                                COP = QL / Win

Where,

             QL : The rate of heat loss

             Win : The input power required

- The rate of heat loss can be determined from first law of thermodynamics.

                               Qin - Wout = flow (m)*c*ΔT

Where,

               Qin = - QL ... Heat lost.

               c : The heat capacity of water = 4,200 J / kg°C

- There is no work being done on the system so, Wout = 0

                              -QL = flow (m)*c*ΔT

                              -QL = ( 10 / 3600 )*4200*( 6 - 23 )

                               QL = 198.33 W

- The required power input ( Qin ) would be:

                              Win = QL / COP

                               Win = 198.33 / 3.1

                                      = 63.97 W ≈ 64 W