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lawyer [7]
3 years ago
5

A treated aluminum plate is irradiated with G-1000 Wm2 while at a temperature of 400 K The plate is treated with a surface coati

ng that provides an absorptivity of 0.14 and an emissivity of 0.76. Find:
(a) The radiosity, J, of the plate surface
(b) The net radiation heat flux at the top surface
Engineering
1 answer:
klemol [59]3 years ago
7 0

Answer:

(a) The radiosity, J, of the plate surface is 1343.2 W/m²

(b) The net radiation heat flux at the top surface is 1863.2 W/m²

Explanation:

J = εσT⁴ + (1 - ε)G

where;

J is the  radiosity of the plate surface

ε is the emissivity  

σ is the stefan-Boltzmann constant

T is the temperature of the surface

G is the irradiance of the surface

J = 0.76 x 5.67 x 10⁻⁸ x (400)⁴ + (1 - 0.76) x 1000

J = 1343.2 W/m²

Part (b) net radiation heat flux at the top surface

this is equivalent to sum of reflected and emitted heat.

Rnet = G - (Rσ + Rt) + Rε

Rnet = G + Rε - (Rσ + Rt)

Rnet = G + εσT⁴ -  (1 - ε)G

Rnet = 1000 + 1103.2 - 240

Rnet = 1863.2 W/m²

Therefore, the net radiation heat flux at the top surface is 1863.2 W/m²

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private int x;

private int y;

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public Rectangle(int x, int y, int w, int h)

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public int getX()

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public int getY()

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return y;

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return width;

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public int getHeight()

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return height;

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public void changeSize(int w,int h)

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width=w;

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======================

//main.java

class Main {

public static void main(String[] args) {

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//print info of rect using toString method

System.out.println(rect.toString());

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}

==========================================================================================

//Output

Coordinate is (5,12) and dimension is : 4x8 Area is 32

Coordinate is (5,12) and dimension is : 3x10 Area is 30

========================================================================================

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