A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of 70°C is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of 1 mm and thermal conductivity of 0.05 W/m·K, while the solid plate is 13 mm thick and has a thermal conductivity of 1.2 W/m·K. Inside the heated chamber, the convection heat transfer coefficient is 70 W/m2·K. If the bottom surface of the solid plate is maintained at 52°C.

Required:
Determine the temperature inside the heated chamber and the surface temperature of the transparent film. Assume thermal contact resistance is negligible.

Answers

Answer 1

Answer:

1.) 103.23 degree centigrade

2.) 126.96 degree centigrade

Explanation:

Given that a transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of 70°C is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of 1 mm and thermal conductivity of 0.05 W/m·K, while the solid plate is 13 mm thick and has a thermal conductivity of 1.2 W/m·K. Inside the heated chamber, the convection heat transfer coefficient is 70 W/m2·K. If the bottom surface of the solid plate is maintained at 52°C.

To determine the temperature inside the heated Chamber, let us first calculate the heat transfer rate per unit area through the plate by using the formula

Heat transfer rate R = k( Tb - T2)/L

Where

k = 1.2 W/mA.K

Tb = 70 degree

T2 = 52 degree

L = 13mm = 13/1000 = 0.013m

substitute all the parameters into the formula above.

R = 1.2 x ( 70 - 52 )/ 0.013

R = 1.2 x (18/0.013)

R = 1661. 5 W/m^2

the surface temperature of the transparent film will be

Ts = Tb + (RLf/Kf)

Ts = 70 + (1661.5 x 0.001)/0.05

Ts = 70 + (1.6615)/0.05

Ts = 70 + 33.23

Ts = 103.23 degree centigrade

the temperature inside the heated Chamber will be calculated by using the formula

Ti = Ts + (R/h)

Ti = 103.23 + (1661.5/70)

Ti = 103.23 + 23.74

Ti = 126.97 degree centigrade


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Answers

Answer:

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Explanation:

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Required:
Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperatureis maintained at Tt < 85°C for the unfinned and finned packages.

Answers

Answer:

For unfined package = 1.30 × 10^-3 W.

Dor fined package = 8.64 × 10^-3 W.

Explanation:

STEP ONE: The first thing to do is to determine the cross sectional area and input the value into the convectional resistance formula/equation in order to determine the heat rate for the unfined package.

Thus, Area = (dimension of the electronic device) ^2. = (10 × 10^-6)^2 m.

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Required:
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Answers

Answer:

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Explanation:

STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.

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