Answer:
Explanation:
From the given information:
Water freezing temp. ![T_L = 32 ^0 \ F](https://tex.z-dn.net/?f=T_L%20%3D%2032%20%5E0%20%5C%20F)
Heat rejected temp ![T_H = 72 ^0 \ F](https://tex.z-dn.net/?f=T_H%20%3D%2072%20%5E0%20%5C%20F)
Recall that:
The coefficient of performance is:
![COP_{ref} = \dfrac{T_L}{T_H - T_L} \\ \\ = \dfrac{32+460}{(72 +460) -(32+460)} \\ \\ =\dfrac{492}{532 -492} \\ \\ = \dfrac{492}{40} \\ \\ COP_{ref} = 12.3](https://tex.z-dn.net/?f=COP_%7Bref%7D%20%3D%20%5Cdfrac%7BT_L%7D%7BT_H%20-%20T_L%7D%20%5C%5C%20%5C%5C%20%20%3D%20%5Cdfrac%7B32%2B460%7D%7B%2872%20%2B460%29%20-%2832%2B460%29%7D%20%5C%5C%20%5C%5C%20%3D%5Cdfrac%7B492%7D%7B532%20-492%7D%20%5C%5C%20%5C%5C%20%3D%20%5Cdfrac%7B492%7D%7B40%7D%20%5C%5C%20%5C%5C%20COP_%7Bref%7D%20%3D%2012.3)
Again:
The efficiency given by COP can be represented by:
![COP = \dfrac{Q_L}{W} \\ \\ W = \dfrac{Q_L}{COP} \\ \\ W = \dfrac{2000 \ lbm \times 144 \ Btu/lbm}{12.3} \\ \\ W = 23414.63 \ Btu](https://tex.z-dn.net/?f=COP%20%3D%20%5Cdfrac%7BQ_L%7D%7BW%7D%20%5C%5C%20%5C%5C%20W%20%3D%20%5Cdfrac%7BQ_L%7D%7BCOP%7D%20%5C%5C%20%5C%5C%20W%20%3D%20%5Cdfrac%7B2000%20%5C%20lbm%20%5Ctimes%20144%20%5C%20Btu%2Flbm%7D%7B12.3%7D%20%5C%5C%20%5C%5C%20W%20%3D%2023414.63%20%5C%20Btu)
Finally; the power input in an hour can be determined by using the formula:
![Power= \dfrac{W}{t} \\ \\ Power = \dfrac{23414.63 \ Btu}{ 1 \ hr} \\ \\ Power = \dfrac{23414.634 \times 1055.056 \ J}{1 \times 3600} \\ \\ Power = 6.86 \ kW](https://tex.z-dn.net/?f=Power%3D%20%5Cdfrac%7BW%7D%7Bt%7D%20%5C%5C%20%5C%5C%20%20Power%20%3D%20%5Cdfrac%7B23414.63%20%5C%20Btu%7D%7B%201%20%5C%20hr%7D%20%5C%5C%20%5C%5C%20%20Power%20%3D%20%5Cdfrac%7B23414.634%20%5Ctimes%201055.056%20%5C%20J%7D%7B1%20%5Ctimes%203600%7D%20%5C%5C%20%5C%5C%20%20Power%20%20%3D%206.86%20%5C%20kW)
In hp; since 1 kW = 1.34102 hp
6.86kW will be = (6.86 × 1.34102) hp
= 9.199 hp
Answer:
The module is why it’s goin to work
Explanation:
Answer: General purpose branch circuit
Explanation:
General purpose branch circuit are the type of circuits that are used mainly to supply light to two or more receptacle outlets for small appliances. This circuits are about 120v can be used either in residential, commercial and industrial buildings.
Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.
Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.
Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:
- <u><em>Decrease the length of the pipes</em></u>: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
- <u><em>Reduce the surface roughness of the pipes</em></u>: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
- <u><em>Increase the pipe diameter: </em></u>By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.
You can learn more about friction losses at
brainly.com/question/13348561
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