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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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Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

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here, the time-mean speed, vt is given below,

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and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

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2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

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Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

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Software development life cycle (SDLC) can be defined as a strategic process or methodology that defines the key steps or stages for creating and implementing high quality software applications.

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