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Gelneren [198K]
3 years ago
7

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm-diameter tubes through which refrigerant passes at

a temperature of −18°C. Air is cooled as it fows over the tubes, maintaining a surface convection coeffcient of 100 W/m2 ⋅ K, and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of −3°C, what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator’s defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range 0 ≤δ ≤ 4 mm. Frost may be assumed to have a thermal conductivity of 0.4 W/m ⋅ K. (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which T[infinity] = 20°C and natural convection maintains a convection coeffcient of 2 W/m2 ⋅ K, how long will it take for the frost to melt? The frost may be assumed to have a mass density of 700 kg/m3 and a latent heat of fusion of 334 kJ/kg
Engineering
1 answer:
docker41 [41]3 years ago
6 0
1.01 W/M that’s the answer
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Answer:

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3 0
2 years ago
A concentrated load P is applied to the upper end of a 1.47-m-long pipe. The outside diameter of the pipe is D = 112 mm and the
myrzilka [38]

Answer:

Pmax = 38251.73 N

Explanation:

Given info

L = 1.47 m

D = 112 mm ⇒ R = D/2 = 112/2 mm = 56 mm

d = 101 mm  ⇒ r = D/2 = 101/2 mm = 50.5 mm

a) We can apply the following equation in order to get Q (First Moment of Area):

Q = 2*(A₁*y₁-A₂*y₂)

where

A₁ = π*R² = π*(56 mm)² = 3136 π mm²  

y₁ = 4*R/(3*π) = 4*56/(3*π) mm = 224/(3*π) mm

A₂ = π*r² = π*(50.5 mm)² = 2550.25 π mm²

y₂ = 4*r/(3*π) = 4*50.5/(3*π) mm = 202/(3*π) mm

then

Q = 2*(3136 π mm²*224/(3*π) mm-2550.25 π mm²*202/(3*π) mm)

⇒ Q = 62437.833 mm³

b) If  τallow = 83 MPa = 83 N/mm²

P = ?

We can use the equation

τ = V*Q / (t*I)   ⇒  V = τ*t*I / Q

where

t = D - d = 112 mm - 101 mm = 11 mm

I = (π/64)*(D⁴-d⁴) = (π/64)*((112 mm)⁴-(101 mm)⁴) = 2615942.11 mm⁴

Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

⇒  Q = (4/3)*((56 mm)³-(50.5 mm)³) = 62437.833 mm³

then we have

V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

⇒ V = 2942.255 N

Finally Pmax = V = 38251.73 N

6 0
3 years ago
A(n) ____________________ measures the resistance to current flow in a circuit.
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Answer: OHMMETER & MEGOHMMETER:

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6 0
2 years ago
A small lake with volume of 160,000 m^3 receives agricultural drainage waters that contain 150 mg / L total dissolved solids (TD
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Answer:

Explanation:

Given that : -

The desirable limit is 500 mg / l , but

allowable upto 2000 mg / l.

The take volume is V = 160.000 m3

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Code of tpr frpm drawn = 150 x 60, 000 x 1000

Ci = 9000 kg / gr

Cl = 100 x 160,000 x 1000

Cl = 16, 000 kg

Since allowable limit = 2000 mg / l

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Tr=<u>304,000</u>

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=33.77

5 0
3 years ago
Read 2 more answers
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otez555 [7]

Answer:

Define Variables and Use List methods to do the following

Explanation:

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3 0
2 years ago
Read 2 more answers
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