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Zepler [3.9K]
3 years ago
6

4. An aluminum alloy fin of 12 mm thick, 10 mm width and 50 mm long protrudes from a wall, which is maintained at 120C. The amb

ient air temperature is 22C. The heat transfer coefficient and conductivity of the fin material are 140 W/m2K and 55 W/mk respectively. Determine a. Temperature at the end of the fin b. Temperature at the middle of the fin. c. Calculate the heat dissipation energy of the fin
Engineering
1 answer:
Minchanka [31]3 years ago
7 0

Answer:

a) 84.034°C

b) 92.56°C

c) ≈ 88 watts

Explanation:

Thickness of aluminum alloy fin = 12 mm

width = 10 mm

length = 50 mm

Ambient air temperature = 22°C

Temperature of aluminum alloy is maintained at 120°C

<u>a) Determine temperature at end of fin</u>

m = √ hp/Ka

   = √( 140*2 ) / ( 12 * 10^-3 * 55 )

   = √ 280 / 0.66 = 20.60

Attached below is the remaining answers

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4 0
3 years ago
An axial compressive load of 708 kN is applied to a cylindrical component, 81 mm in diameter and 418 mm long, made of aluminium.
dalvyx [7]

Answer:

The compressive stress of aplying a force of 708 kN in a 81 mm diamter cylindrical component is 0.137 kN/mm^2 or 137465051 Pa (= 137.5 MPa)

Explanation:

The compressive stress in a cylindrical  component can be calculated aby dividing the compressive force F to the cross sectional area A:

fc= F/A

If the stress is wanted in Pascals (Pa), F and A must be in Newtons and square meters respectively.

For acylindrical component the cross sectional area A is:

A=πR^

If the diameter of the component is 81 mm, the radius is the half:

R=81mm /2 = 40.5 mm

Then A result:

A= 3.14 * (40.5 mm)^2  = 5150.4 mm^2

In square meters:

A= 3.14 * (0.0405 m)^2  = 0.005150 m^2

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