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Anna [14]
4 years ago
9

A Carnot refrigeration cycle is executed in a closed system in the saturated liquid-vapor region using 3.2 kg of R-134a. The max

imum absolute temperature in the cycle is 1.1 times the minimum absolute temperature, and the net work into the cycle is 27 kJ. a. (6) Draw this cycle on p-v coordinates, labeling the isentropic compression as process 1-2. Indicate where Qin and Qout are located. b7) Calculate what Quote is, in kJ
Engineering
1 answer:
Harrizon [31]4 years ago
3 0

Answer:

Explanation:

<u>Given Data</u>

Mass of refrigerant,  m  = 3.2kg

Work input to the cycle, Wnet = 27kj

The expression for coefficient of performance

C O P  r = Q L /˙ W n e t ............(A)

The expression for maximum performance of refrigerator,

C OP r = T L /T H  − T L  . . . . . . . . . ( B )

Here,  T L

is cold temperature reservoir,  T H  is hot temperature reservoir.

Since,  T H = 2 T L . . . . . ( C )

Comparing equation (A) and (B),

Q L /˙ W n e t  = T L /T H − T L

Substituting all the respective value in above equation,

˙

Q L /27  = T L /1.1T L − T L

Q L = 270kj

The expression for heat rejected,

Q H = W n e t  +  Q L

Q H=27 + 270

QH = 340kj

<u></u>

<u></u>

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Explain combined normal and shear stresses with sketch. Write the general expression for (a) Normal and shear stresses on inclin
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a) Normal stress :

бn =[ ( бx + бy ) / 2  + ( бx - бy ) / 2  ] cos2∅ + Txysin2∅

shear stress

Tn = ( - бx - бy ) / 2  sin2∅ + Txy cos2∅

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б1 = ( бx + бy ) / 2  - \sqrt{}( ( бx - бy ) / 2 )^2 + T^2xy

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Combined normal stress and shear stress  sketches attached below

The terms in the sketch are :

бx = tensile stress in x direction

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бn = Normal stress acting on the inclined plane EF

Tn = shear stress acting on the inclined plane EF

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shear stress

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Read 2 more answers
A cubical picnic chest of length 0.5 m, constructed of sheet styrofoam of thickness 0.025 m, contains ice at 0\[Degree]C. The th
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Answer:

Rate of heat transfer is 0.56592 kg/hour

Explanation:

Q = kA(T2 - T1)/t

Q is rate of heat transfer in Watts or Joules per second

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A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2

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Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s

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