Q. 324.0( 1 Vote )

# Seven rods A, B, C, D, E, F and G are joined as shown in figure. All the rods have equal cross-sectional area A and length ℓ. The thermal conductivities of the rods are K_{A} = K_{C} = K_{D}, K_{D} = 2K_{D}, K_{E} = 3K_{D}, K_{F} = 4K_{D} and K_{D} = 5K_{D}. The rod E is kept at a constant temperature T_{1} and the rod G is kept at a constant temperature T_{2}(T_{2}> T_{1}).

(a) Show that the rod F ahs a uniform temperature T = (T_{1} + 2T_{2})/3.

(b) Find the rate of heat flowing from the source which maintains the temperature T_{2}.

Answer :

Given thermal conductivity of the respective rods as follows-

K_{A} = K_{C} = K_{0}

K_{B} = K_{D} = 2K_{0}

K_{E} = 3K_{0}, K_{F} = 4K_{0}

K_{9}= 5K_{0}

Also, length of each rod is *l*

At steady state, temperature at the ends of rod F will be same.

Let T be the temperature of rod F

(a)

Rate of heat flow through rod A + rod C

= Rate of heat flow through rod B + rod D

⇒ + = +

Substituting the values in terms of k_{0} –

⇒ + = +

⇒ 2k_{0} (T_{1}-T) = 2 2 k_{0} (T–T_{2})

⇒ T =

(b) To find the rate of flow of heat from rod G, which is at Temperature T_{2}

Looking into the above diagram, we can say that it forms a balanced Wheatstone bridge.

Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.

Hence we can remove the F for simplification

From above diagram, we can see that R_{A} and R_{B} are connected in series.

⇒ R_{AB} = R_{A} + R_{B}

And R_{C} and R_{D} are connected in series

⇒ R_{CD} = R_{C} + R_{D}

Then, R_{AB} and R_{CD} are connected in parallel

Now ,

R_{A} = , R_{B} = , R_{C} = , R_{D} =

Since R_{A ,} R_{B} are connected in series

R_{AB} = and R_{CD} =

Since R_{AB} R_{CD} are in parallel

=

=

=

Now, rate of flow of heat from the source rod

q = =

=

Hence, rate of flow of heat from the source rod is given by

q =

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