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snow_lady [41]
3 years ago
6

A composite wall consists of 20 mm thick steel plate backed by insulation brick (k = 0.39 W/mK) of 50 cm thickness and overlaid

by mineral wool of 20 cm thickness (k = 0.05 W/mK) and 70 cm layer of brick of (k = 0.39 W/mK). The inside is exposed to convection at 650°C with h = 65 W/ m2K. The outside is exposed to air at 35°C with a convection coefficient of 15 W/m2K. Determine the heat loss per unit area, interface temperatures and temperature gradients in each materials.

Engineering
1 answer:
jeyben [28]3 years ago
6 0

Answer:

Heat loss=85.9W/m^2

ΔT1(Steel)=0.04C

ΔT2(Brick1)=110.13C

ΔT3(Mwood)=343.6C

ΔT1(Brick2)=154.18C

Explanation:

raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.

To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.

I attached the procedure

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Answer:

  • nP  ≈ 4.9
  • nL =  1.50

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

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Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

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nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

nL = \frac{SpAt - Fi}{CP} = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

3 0
3 years ago
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