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Rina8888 [55]
3 years ago
12

A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and

the surrounding air is at 75°F. Determine (a) the reversible power, (b) the irreversibility, and (c) the second-law efficiency of this freezer.
Engineering
1 answer:
Igoryamba3 years ago
8 0

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

TL = low temperature of freezer = 20 °F

TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

W actual, power input = 0.7 hp

conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL /  COP r  where 75 Btu/min = 1.76856651 hp  where W actual = 0.70 hp

a) W rev =  1.76856651 hp  /  8.73  = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual =  0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

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Read 2 more answers
In the circuit given below, R1 = 17 kΩ, R2 = 74 kΩ, and R3 = 5 MΩ. Calculate the gain 1formula58.mml when the switch is in posit
Elenna [48]

Answer:a

a) Vo/Vi = - 3.4

b) Vo/Vi = - 14.8

c) Vo/Vi = - 1000

Explanation:

a)

R1 = 17kΩ

for ideal op-amp

Va≈Vb=0 so Va=0

(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0

sin we know Va≈Vb=0

so

-Vi/5kΩ + -Vo/17kΩ = 0

Vo/Vi = - 17k/5k

Vo/Vi = -3.4

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b)

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for ideal op-amp

Va≈Vb=0 so Va=0

so

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Vo/Vi = - 74kΩ/5kΩ

Vo/Vi = - 14.8

║Vo/Vi ║ = 14.8  ( negative sign phase inversion)

c)

Also for ideal op-amp

Va≈Vb=0 so Va=0

Now for position 3 we apply nodal analysis we got at position 1

(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0           ( 5MΩ = 5000kΩ )

so

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Vo/Vi = - 5000kΩ/5kΩ

Vo/Vi = - 1000

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