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Crank
3 years ago
12

: The interior wall of a furnace is maintained at a temperature of 900 0C. The wall is 60 cm thick, 1 m wide, 1.5 m broad of mat

erial whose thermal conductivity is 0.4 W/m K. The temperature of the outside surface of the wall is 150 0C. Determine the heat through the wall. Also determine thermal resistance to heat flow.
Engineering
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

<em>Heat is lost at the rate of 750 J/s or W</em>

<em>The thermal resistance is 1 K/W</em>

Explanation:

interior temperature T_{2} = 900 °C

wall thickness t = 60 cm = 0.6 m

width = 1 m

breadth = 1.5 m

thermal conductivity k = 0.4 W/m-K

outside temperature T_{1} = 150 °C

heat through the wall = ?

The area of the wall A = w x b = 1 x 1.5 = 1.5 m^2

Temperature difference dt = T_{2} - T_{1} = 900 - 150 = 750 °C

note that dt is also equal to 750 K since to convert from °C to K we'll have to add 273 to both temperature, which will still cancel out when we subtract the two temperatures.

To get the heat that escapes through the wall, we use the equation

Q = Ak\frac{dt}{t}

substituting values, we have

Q = 1.5 x 0.4 x \frac{750}{0.6} = <em>750 J/s or W</em>

Thermal resistance R_{t} = \frac{dt}{Q}

R_{t} = 750/750 =<em> 1 K/W</em>

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The drag force, Fd, imposed by the surrounding air on a
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Answer:

a)  23.551 hp

b)  516.89 hp

Explanation:

<u>given:</u>

F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2

<u>required:</u>

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<u>solution:</u>

(F_{d})_{a}  =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}.............(1)

by substituting in the equation (1)

         =353.27 lbf

(F_{d})_{b}  =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}..........(2)

by substituting in the equation (2)

         = 2769.29 lbf

power is defined by

             P=F.V

     P_{a}=353.27*36.67

           =12954.411 lbf.ft/s

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           =23.551 hp

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           = 284323 lbf.ft/s

           = 284323*.001818

           = 516.89 hp

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