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julia-pushkina [17]
2 years ago
13

Explain how the ideal vapour compression refrigerator cycle is an improvement upon the Carnot cycle. Use any two of the problems

to explain.
Engineering
1 answer:
Grace [21]2 years ago
8 0
  • answer explanation:
  • <u>Basically, the Carnot cycle is turning a heat differential into mechanical energy, then using mechanical energy to recreate the heat differential. A heat pump only turns mechanical energy into a heat differential. It exchanges the gas being compressed on an ongoing basis and does not extract mechanical energy from it, this continuous loss of mechanical energy input is what allows continuous generation of a heat differential.</u>

  • <u>Also, the Carnot cycle is an idealized theoretical model, not a practically achievable engineering goal. It illustrates the essential mechanism by which heat differentials can be turned into mechanical energy and vice versa, it has to ignore the fact of inefficiency in conversion between mechanical energy and heat differential for the sake of illustrating a set of ideal processes. A real refrigeration system has to deal with the question of whether the system is pumping away more heat than it generates itself through inefficiency.</u>
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On highways, the far left lane is usually the<u> fastest</u> moving traffic.

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he desk has a weight of 80 lb and a centerof gravity at G. Determine the initial acceleration of a desk when the man applies eno
-BARSIC- [3]

Answer:

N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb

Explanation:

Na and Nb are the vertical reactions on each of the two legs at A and at B

For the horizontal forces:

Fcos(30)-0.5N_a-0.5N_b=0\\0.5N_a+0.5N_b= Fcos(30)\\N_a+N_b= 2Fcos(30)

For the vertical forces:

N_a+N_b-Fsin(30)-75=0\\N_a+N_b=Fsin(30)+80

Therefore equating both equations:

2Fcos(30)=Fsin(30)+80\\F(2cos(30)-sin(30))=80\\F=\frac{80}{2cos(30)-sin(30)} =64.93N

After the desk star to slide:

sum of all vertical force = ma , therefore:

N_a+N_b-64.93sin(30)-80=0\\N_a+N_b=64.93sin(30)+80

sum of all horizontal force = ma

64.93cos(30)-0.2N_a-0.2N_b=\frac{80lb}{32.2ft/s^2}a\\ 0.2(N_a+N_b)=64.93cos(30)-\frac{80lb}{32.2ft/s^2}a\\N_a+N_b=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}a}{0.2}=324.65-12.42a

equating both equations:

324.65-12.42a=64.93sin(30)+80\\12.42a=324.65-64.93sin(30)-80\\12.42a=212.185\\a=17.08ft/s^2

From the moment equation:

4N_b-80(2)-64.93(3)=\frac{-80}{32.2} (17.08)(2)\\N_b=67.48lb

N_a=\frac{64.93cos(30)-\frac{80lb}{32.2ft/s^2}(17.08)}{0.2}-67.48 = 45.04lb

For each leg: N_a=\frac{45.04}{2}=22.52lb,N_b=\frac{67.48}{2} =33.74lb

7 0
3 years ago
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