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

Win = 64 W ... Option C

Step-by-step 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

Answer 2

The correct option is;

(c) 64W

Step-by-step explanation:

Here we have the Coefficient Of Performance, COP given by

`COP = (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 = (197.39 )/(W) = 3.1` which gives

`W = (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.