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olchik [2.2K]
3 years ago
5

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin

g its waste heat to cooling water that enters the condenser at 188C at a rate of 0.25 kg/s and leaves at 268C. The refrigerant enters the condenser at 1.2 MPa and 658C and leaves at 428C. The inlet state of the compressor is 60 kPa and 2348C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator, and (d) the theoretical maximum refrigeration load for the same power input to the compressor.

Engineering
1 answer:
lord [1]3 years ago
5 0

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

              b) 5.398 kw

              c) 2.14

              d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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A 10MHz clock frequency is applied to a cascaded counter consisting of a modulus-5 counter, a modulus-8 counter and two modulus-
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If a clock frequency is applied to a cascaded counter, The lowest output frequency available will be

  • The lowest output frequency will be = 2.5kHz

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<h3>Lowest Frequency</h3>

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2 years ago
A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of
Naya [18.7K]

Answer:

a. Solid length Ls = 2.6 in

b. Force necessary for deflection Fs = 67.2Ibf

Factor of safety FOS = 2.04

Explanation:

Given details

Oil-tempered wire,

d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

(a) Find the solid length

Ls = d (n + 1)

= 0.2(12 + 1) = 2.6 in Ans

(b) Find the force necessary to deflect the spring to its solid length.

N = n - 2 = 12 - 2 = 10 coils

Take G = 11.2 Mpsi

K = (d^4*G)/(8D^3N)

K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in

Fs = k*Ys = k (Lo - Ls )

= 28(5 - 2.6) = 67.2 lbf Ans.

c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

For C = D/d = 2/0.2 = 10

Kb = (4C + 2)/(4C - 3)

= (4*10 + 2)/(4*10 - 3) = 1.135

Tau ts = Kb {(8FD)/(Πd^3)}

= 1.135 {(8*67.2*2)/(Π*2^3)}

= 48.56 * 10^6 psi

Let m = 0.187,

A = 147 kpsi.inm^3

Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

Ssy = 0.50 Sut

= 0.50(198.6) = 99.3 kpsi

FOS = Ssy/ts

= 99.3/48.56 = 2.04 Ans.

7 0
3 years ago
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lys-0071 [83]

Answer:

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Aluminum is used in a telescope, rather than the silver, which is the most reflective material, because silver requires special conditions to be able to work, and silver tarnishes when exposed to air containing sulfur, and silver is susceptible to corrode.

3 0
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