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Mashcka [7]
3 years ago
5

An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

aintain its temperature constant at 20. if the temperature of the outside is 35 the power required to operate this air conditioning system is
(a) 0.58 kW
(b) 3.20 kW
(c) 1.56 kW
(d) 2.26 kW
(e) 1.64 kW
Engineering
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

COP = \frac{T_{L}}{T_{H}-T_{L}}

COP = \frac{293.15\,K}{308.15\,K-293.15\,K}

COP = 19.543

Lastly, the power required to operate the air conditioning system is:

\dot W = \frac{\dot Q_{L}}{COP}

\dot W = \frac{32\,kW}{19.543}

\dot W = 1.637\,kW

Hence, the answer is E.

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Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
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Answer:

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Explanation:

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Answer:

I think

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The recommended time weighted average air concentration for occupational exposure to water soluble hexavalent chromium (Cr VI) i
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Answer:

CDI= 1.393 × 10^-3 mg/kg.d

Explanation:

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8 0
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In the context of mechanical systems, what does the term efficiency mean? OA the factor by which a machine multiplies a force B.
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