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 KA = KC = KD, KD = 2KD, KE = 3KD, KF = 4KD and KD = 5KD. The rod E is kept at a constant temperature T1 and the rod G is kept at a constant temperature T2(T2> T1).

(a) Show that the rod F ahs a uniform temperature T = (T1 + 2T2)/3.

(b) Find the rate of heat flowing from the source which maintains the temperature T2.

Answer :

Given thermal conductivity of the respective rods as follows-

KA = KC = K0
KB = KD = 2K0
KE = 3K0, KF = 4K0
K9= 5K0

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


Rate of heat flow through rod A + rod C

= Rate of heat flow through rod B + rod D

+ = +

Substituting the values in terms of k0

+ = +

2k0 (T1-T) = 2 2 k0 (T–T2)

T =

(b) To find the rate of flow of heat from rod G, which is at Temperature T2

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 RA and RB are connected in series.


And RC and RD are connected in series


Then, RAB and RCD are connected in parallel

Now ,

RA = , RB = , RC = , RD =

Since RA , RB are connected in series

RAB = and RCD =

Since RAB RCD 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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