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svetoff [14.1K]
3 years ago
8

A heated plate is built into the wall of a large room. The room is maintained at 300 K. The exposed surface of the plate is main

tained at 400 K by a heater attached to the back of the plate. Assume the sides of the plate and back of the heater are perfectly insulated. The exposed surface of the plate (area 0.1 m^2 ) has an emissivity of 0.3 and a convection heat transfer coefficient of 6 W/m^2K with the room and air temperature of 300 K. Show resistance circuit for this problem and draw a picture to show relevant heat modes on system to receive full points. Calculate the following:
a. Find the net radiant heat flux from the plate surface, q1"=____
b. The required heater power, P=______
c. The radiosity of the plate surface, J1=
Engineering
1 answer:
myrzilka [38]3 years ago
7 0

Answer:

B

Explanation:

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Getting the bottom of your feet burned when walking on hot sand is due to a form of energy transmission known as conduction.

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In Science, there are three (3) main types of energy transmission and these include the following:

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In this scenario, we can infer and logically conclude that burning the bottom of your feet when walking on hot sand is primarily due to a form of energy transmission known as conduction because it involves the transfer of thermal energy (heat) due to the movement of particles.

Read more on heat conduction here: brainly.com/question/12072129

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Because assembly language is so close in nature to machine language, it is referred to as a ____________.
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Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

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3 years ago
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