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Mandarinka [93]
3 years ago
8

A furnace wall consisting of 0.25 m of fire clay brick, 0.20 m of kaolin, and a 0.10‐m outer layer of masonry brick is exposed t

o furnace gas at 1370 K with air at 300 K adjacent to the outside wall. The inside and outside convective heat‐transfer coefficients are 115 and 23 W/m2 ⋅ K, respectively. Determine the heat loss per square foot of wall and the temperature of the outside wall surface under these conditions.

Engineering
1 answer:
dalvyx [7]3 years ago
4 0

Answer:

q=1910.71 W/m²

T=383.07 K

Explanation:

Given that

L₁=0.25 m  ,K₁=1.13 W/(m.K)

L₂=0.2 m   ,  K₂=1.45 W/(m.K)

L₃= 0.1 m   , K₃=0.66 W/(m.K)

h₁=115 W/(m².K)

h₂=23 W/(m².K)

The total thermal resistance

R=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}+\dfrac{1}{h_1A}+\dfrac{1}{h_2A}\ K/W

Now by putting the values

Put A= 1 m²      ( To find heat transfer per unit area)

R=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}+\dfrac{1}{h_1A}+\dfrac{1}{h_2A}\ K/W

R=\dfrac{0.25}{1.13}+\dfrac{0.2}{1.45}+\dfrac{0.1}{0.66}+\dfrac{1}{115}+\dfrac{1}{23}\ K/W

R= 0.56 K/W

We know that

q= ΔT/R

Here  ΔT = 1370 - 300 K =1070 K

Now by putting the values

q= 1070/0.56

q=1910.71 W/m²

Lets T is the outside surface temperature

q = h₂ ( T- 300)

Now by putting the values

1910.71 = 23 ( T- 300)

T=383.07 K

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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
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Answer:

a. Solid length Ls = 2.6 in

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Factor of safety FOS = 2.04

Explanation:

Given details

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d = 0.2 in,

D = 2 in,

n = 12 coils,

Lo = 5 in

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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.

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