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poizon [28]
3 years ago
15

Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a

t 0.08 MPa. If the refrigerant undergoes a throttling process, what is the quality of the refrigerant exiting the expansion valve?

Engineering
1 answer:
AveGali [126]3 years ago
5 0

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2)  + x(2) * h(fg2)

so

exit quality  = \frac{h1 - h(f2)}{h(fg2)}

exit quality  = \frac{332.17- 9.04}{1382.73)}

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

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A steel wire is suspended vertically from its upper end. The wire is 400 ft long and has a diameter of 3/16 in. The unit weight
jek_recluse [69]

Answer:

a) the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire is 624.83 lb

Explanation:

Given the data in the question;

Length of wire L = 400 ft = ( 400 × 12 )in = 4800 in

Diameter d = 3/16 in

Unit weight w = 490 pcf

First we determine the area of the wire;

A = π/4 × d²

we substitute

A = π/4 × (3/16)²

A = 0.0276 in²

Next we get the Volume

V = Area × Length of wire

we substitute

V = 0.0276 × 4800

V = 132.48 in³

Weight of the steel wire will be;

W = Unit weight × Volume

we substitute

W = 490 × ( 132.48 / 12³ )

W = 490 × 0.076666

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σ_w = W / A

we substitute

σ_w = 37.57 / 0.0276

= 1361.23 psi

Therefore, the maximum tensile stress due to the weight of the wire is 1361.23 psi

b) the maximum load P that could be supported at the lower end of the wire. Allowable tensile stress is 24,000 psi

Maximum load P that the wire can safely support its lower end will be;

P = ( σ_{all - σ_w )A

we substitute

P = ( 24000 - 1361.23  )0.0276

P = 22638.77 × 0.0276

P = 624.83 lb

Therefore, the maximum load P that could be supported at the lower end of the wire is 624.83 lb

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2 years ago
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