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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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The soil borrow material to be used to construct a highway embankment has a mass unit weight of 107.0 lb/cf and a water content
MrRissso [65]

Answer:

Option D

Explanation:

Given information

Bulk unit weight of 107.0 lb/cf

Water content of 7.3%,=0.073

Specific gravity of the soil solids is 2.62

Specifications

Dry unit weight is 113 lb/cf  

Water content is 6%.

Volume of embankment is 440,000-cy

Borrow material

Dry_{unit,weight}=\frac {bulk_{unit,weight}}{1+water_{content}}=\frac {107}{1+0.073}= 99.72041 lb/cf  

Embankment

Considering that the volume of embankment is inversely proportional to the dry unit weight

\frac {V_{embankment}}{V_{borrow}}=\frac {Dry_{borrow}}{Dry_{embankment}}

Therefore, V_{borrow}=V_{embankment} *\frac {Dry_{embarkement}}{Dry_{borrow}}

V_{borrow}=440,000-cy*\frac {113 lb/cf }{99.72041 lb/cf }= 498594-cy

Therefore, volume of borrow material is 498594-cy

(b)

The weight of water in embankment is found by multiplying the moisture content and dry unit weight.

Assuming that all the specifications are achieved, weight of water in embankment=0.06*113=6.78 lb/cf

Since 1 yd^{3}= 27 ft^{3}

The embankment requires water of  6.78*27*440000= 80546400 lb

Borrow materials’ water will also be 0.073*99.72041=7.27959 lb/cf

Borrow material requires water of 7.27959*27*498594=97998120 lb

Extra water between borrow material and embankment=97998120 lb-80546400 lb=17451720 lb

Unit_{weight}=\frac {17451720}{498594}=35.00186 lb

1 gallon is approximately 8.35 yd^{3} hence

\frac {35.00186 lb/yd^{3}}{8.35}=4.19184 gallons/yd^{3}

That's approximately 4.2 gallons

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4.2 A vapor compression refrigeration machine uses 30kW of electric power to produce 50 tons of cooling. What is
stellarik [79]

Answer:

5.833

Explanation:

Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.

COP=\frac {RE}{P} where RE is refrigeration effect and P is power input

Here, the power input is given as 30 kW

We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW

Now the COP=\frac {175}{30}=5.833

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