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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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Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

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Answer:

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Answer:

no it has to be removed

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