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3 years ago

Can a real refrigerator have higher COP than the COP of the Carnot refrigerator?

Engineering

2 answers:

astra-53 [7]3 years ago

5 0

Answer:

Explanation:

As we know that Carnot refrigerator is an ideal refrigerator in which all processes are reversible and COP of refrigerator depends only on temperature limit of evaporator and condenser.

On the other hand a real refrigerator consist irreversibility that is why the COP of real refrigerator is less that Carnot refrigerator.

So we can say that COP of Carnot refrigerator is always more that COP of real refrigerator.

Flauer [41]3 years ago

4 0

Answer: NO

Explanation:

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Wet steam at 15 bar is throttled adiabatically in a steady-flow process to 2 bar. The resulting stream has a temperature of 130°

cricket20 [7]

Answer:

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

Explanation:

The adiabatic throttling process is modelled after the First Law of Thermodynamics:

$m\cdot (h_{in} - h_{out}) = 0$

$h_{in} = h_{out}$

Properties of water at inlet and outlet are obtained from steam tables:

State 1 - Inlet (Liquid-Vapor Mixture)

$P = 1500\,kPa$

$T = 198.29\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 6.3068\,\frac{kJ}{kg\cdot K}$

$x = 0.967$

State 2 - Outlet (Superheated Vapor)

$P = 200\,kPa$

$T = 130\,^{\textdegree}C$

$h = 2726.9\,\frac{kJ}{kg}$

$s = 7.1776\,\frac{kJ}{kg\cdot K}$

The change of entropy of the steam is derived of the Second Law of Thermodynamics:

$\Delta s = 7.1776\,\frac{kJ}{kg\cdot K} - 6.3068\, \frac{kJ}{kg\cdot K}$

$\Delta s = 0.8708\,\frac{kJ}{kg\cdot K}$

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3 years ago

Select the correct answer.<br> Which equation gives you the amount of work performed?

Fantom [35]

Answer:

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4 years ago

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Anvisha [2.4K]

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4 years ago

Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa

DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

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coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l ) = ∝ * ΔT

( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )

∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

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sergey [27]

Answer:

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