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Varvara68 [4.7K]
3 years ago
13

Oil at 150 C flows slowly through a long, thin-walled pipe of 30-mm inner diameter. The pipe is suspended in a room for which th

e air temperature is 20 C and the convection coefficient at the outer tube surface is 11 W/m2 K. Estimate the heat loss per unit length of tube.
Physics
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

<em>1.01 W/m</em>

Explanation:

diameter of the pipe d = 30 mm = 0.03 m

radius of the pipe r = d/2 = 0.015 m

external air temperature Ta = 20 °C

temperature of pipe wall Tw = 150 °C

convection coefficient at outer tube surface h = 11 W/m^2-K

From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.

area of the pipe per unit length A = \pi r ^{2} = 7.069*10^{-4} m^2/m

convectional heat loss Q = Ah(Tw - Ta)

Q = 7.069 x 10^-4 x 11 x (150 - 20)

Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>

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Answer:

Explanation:

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Slow down the descent speed.

Parachutes are an obvious method for slowing the decent speed, as long as the design includes a way to keep the parachute open.

Cushion the egg so that something other than the egg itself absorbs the impact of landing.

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2 years ago
Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of
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T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

Explanation:

The formula we are going to use is:

\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)

Where:

ε is the emissivity

σ is the Stefan constant

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Calculating The area:

A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2

σ= 5.67*10^{-8}\ W/m^2.K^4

ε =0.95

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T_s=328 K

\dot Q_{rad=100 W

100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K

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T_{surr}=296.289-273\\T_{surr}=23.289^oC

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