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Elan Coil [88]
2 years ago
8

An inventor claims to have developed a food freezer that at steady state requires a power input of 0.25 kW to extract energy by

heat transfer at a rate of 3000 J/s from freezer contents at 270 K. Investigate this claim for an ambient temperature of 293 K.
Engineering
1 answer:
Serga [27]2 years ago
6 0

Answer:

investor claim is acceptable

Explanation:

given data

Win = 0.25 kW

Qc = 3000 J/s = 3kW

Th = 293 K

Tc = 270 K

solution

we get here coefficient of performance of cycle is

coefficient of performance = \frac{Qc}{Win}    ..................1

put here value and we get

coefficient of performance = \frac{3}{0.25}  

coefficient of performance = 1.2

and

coefficient of performance of ideal refrigeration is

coefficient of performance = \frac{Tc}{Th-Tc}    ..................2

coefficient of performance = \frac{270}{293-270}  

coefficient of performance = 11.74

and

we can see here that coefficient of performance of ideal refrigeration  is is more than real cycle coefficient of performance

so investor claim is acceptable

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1 year ago
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

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Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

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Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

5 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
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Answer:

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3 years ago
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3 years ago
A heat pump operates on a vapor-compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the com
Rudiy27

Answer:

Hello your question has some missing information below are the missing information

The refrigerant enters the compressor as saturated vapor at 140kPa Determine The coefficient of performance of this heat pump

answer : 2.49

Explanation:

For  vapor-compression refrigeration cycle

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