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


(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 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.


RAB = RA + RB


And RC and RD are connected in series



RCD = RC + RD


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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