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Question

Find the equivalent thermal conductivity of the given rod.


A

2K3

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B

3K

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C

3K2

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D

4K3

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Solution

The correct option is D

4K3


Step 1: Given Data

Let, A=area, T1 is the temperature of the sink, T2 is the temperature of the source, T is the temperature of the junction, L be the thickness of the junction

Let the equivalent coefficient of thermal conductivity be K.

The given thermal conductivities are K and 2K each having length L.

Let H1 be the amount of heat taken from the source and H2 be the amount of heat given to the sink that is transferred in time t.

Step 2: Express the Thermal Rates of the Rod

Thermal conductivity is given as,

â–³Qâ–³t=-Kâ–³Tâ–³x

where â–³Q is the change in heat and â–³x is the change in length.

For the first conductor, the thermal rate is given by

H1t=KAT-T1L 1

For the second conductor, the thermal rate is given by

H2t=2KAT2-TL 2

Step 3: Evaluate Temperature of the Junction

The two slabs are connected in series so the heat rate flowing through them will be the same

So, H1t=H2t

⇒KAT-T1L=2KAT2-TL

⇒ T-T1=2T2-T

⇒ 3T=2T2+T1

⇒ T=2T2+T13

Step 4: Find the Equivalent Thermal Conductivity

Adding equations 1 and 2 we get,

H1+H2t=KAT-T1+2KAT2-TL

=KAL132T2+T1-T1+2T2-132T2+T1

=KAL2T2+T1-3T1+6T2-4T2-2T13

=43K.ALT2-T1

Hence, the equivalent thermal conductivity is 4K3.

∴Option d is the correct answer.


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