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Liono4ka [1.6K]
3 years ago
5

An irreversible heat pump and a Carnot heat pump operate between the same two thermal reservoirs. Which heat pump has higher COP

?
Engineering
1 answer:
Shkiper50 [21]3 years ago
6 0

Answer:

Carnot heat pump

Explanation:

Carnot heat pump is an ideal heat pump in which all processes are reversible and that consume minimum amount of work to and produces maximum amount of heating effect compare to all real engine.And that is why COP of Carnot heat pump is more as compare to real heat pump.

All real heat pump are not perfectly reversible heat pump So this is also called irreversible heat pump .Due to irreversibility the COP of  irreversible heat pump is always  less than the COP of  Carnot heat pump.

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What colour is best for radiative heat transfer? a. Black b. Brown c. Blue d. White
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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. A
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An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone. A police car observed the automobile. At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile at a constant acceleration of 1.96 m/s2 . The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s2

Find the total time required for the police car  to over take the automobile.

Answer:

15.02 sec

Explanation:

The total time required for the police car to overtake the automobile is related to the distance covered by both  cars which is equal from instant point of abreast.

So; we can say :

D_{pursuit} =D_{police}

By using the second equation of motion to find the distance S;

S= ut + \dfrac{1}{2}at^2

D_{pursuit} = (15.65 *12 )+(15.65 (t)+ (\dfrac{1}{2}*(-3.05)t^2)

D_{pursuit} = (187.8)+(15.65 \ t)-0.5*(3.05)t^2)

D_{pursuit} = (187.8+15.65 \ t-1.525 t^2)

D_{police} = ut _P + \dfrac{1}{2}at_p^2

where ;

u  = 0

D_{police} =  \dfrac{1}{2}at_p^2

D_{police} =  \dfrac{1}{2}*(1.96)*(t+12)^2

D_{police} = 0.98*(t+12)^2

D_{police} = 0.98*(t^2 + 144 + 24t)

D_{police} = 0.98t^2 + 141.12 + 23.52t

Recall that:

D_{pursuit} =D_{police}

(187.8+15.65 \ t-1.525 t^2)=  0.98t^2 + 141.12 + 23.52t

(187.8 - 141.12)  + (15.65 \ t  -  23.52t)  -( 1.525 t^2    - 0.98t^2)  =   0

= 46.68 - 7.85 t -2.505 t² = 0

Solving by using quadratic equation;

t = -6.16 OR  t = 3.02

Since we can only take consideration of the value with a  positive integer only; then t = 3.02 secs

From the question; The motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit;

Therefore ; the total time  required for the police car  to over take the automobile = 12 s + 3.02 s

Total time  required for the police car  to over take the automobile = 15.02 sec

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