Question

11. A Carnot refrigeration cycle absorbs heat at 270 K and rejects it
at 300 K.
(a) Calculate the coefficient of performance of this refrigeration
cycle.
(b) If the cycle is absorbing 1130 kJ/min at 270 K, how many
kJ of work is required per second?
(c) If the Carnot heat pump operates between the same
temperatures as the above refrigeration cycle, what is the
coefficient of performance?
(d) How many kJ/min will the heat pump deliver at 300 K if it
absorbs 1130 kJ/min at 270 K

Answer 1

Answer:

2.33KJ/sec.

Given that,

TemperatureT

1

=270K

TemperatureT

2

=300K

Absorbed heat Q=1260KJ/min=21KJ/s

We know that,

Coefficient of performance is

C.O.P =

T

2

−T

1

T

2

C.O.P=

300−270

270

C.O.P=9

Using formula of work done

C.O.P=

W

Q

W=

9

21

W=2.33KJ/s

The work required is2.33KJ/s.

Hence, This is the required solution.

Answer 2

The coefficient of performance of the engine is 9.

The work done is 2.09 KJ/sec

The coefficient of performance of carnot engine will be 0.

The work done is 125.5KJ/sec.

Given:

Input temperature $T_{1}$ $=270K$

Output temperature $T_{2}=300K$

To find:

1. Coefficient of performance of the refrigeration cycle.
2. Work required by the cycle is it is absorbing $1130KJ/min$ at $270K$.
3. Coefficient of performance of the Carnot engine if it works between the same temperature.
4. How many $KJ/min$ will the heat pump deliver at $300K$ if it absorbs $1130KJ/min$ at $270K.$

Solution:

Step 1

We know, the coefficient of performance of any engine is given by η⁻¹ where ,η is the efficiency of the Carnot engine.

Hence,

Coefficient of performance $(COP)=\frac{T_{1} }{T_{1} -T_{2} }$

Substituting the given values in the equation, we get

$COP=\frac{270}{270-300}$

         $=9$

Hence, the coefficient of performance of the refrigeration cycle is 9.  

Step 2

We also know, a relation between work done and coefficient of performance that is

$COP=\frac{Q}{W}$  $;$ where W is the work done and Q is the heat energy

We have $Q=1130KJ/min$    $;$ or

$Q=\frac{1130}{60}KJ/sec$

Substituting the given values, we get

$9=\frac{113}{6(W)}$  

$W=\frac{113}{54}=2.09KJ/sec$

Hence, the work required is $2.09KJ/sec.$

Explanation 3

We know, the coefficient of performance of a refrigeration engine is

$COP=\frac{T_{1}- T_{2} }{T_{1} }$

If, the engine works between the same temperature or $T_{1}= T_{2}$

$COP=\frac{T_{1}- T_{1} }{T_{1} } =0$

Hence, coefficient of performance in this case will be 0.

Explanation 4

Given that

$T_{1}=270K$   $;$  $T_{2}=300K$   $;$  $Q=1130KJ/min$

Hence, we know

$COP=\frac{Q}{W}$

Substituting the given values, we get

$9=\frac{1130}{W}$

$W=125.5KJ/min$

Hence, the work delivered by the engine is $125.5KJ/min.$