Q.444. Thermal Conduction.

Question 444.
Figure (a) shows 2 metal bars X and Y which have the same cross-sectional and length joined in series. The thermal conduction of Y is three times of X.
a. If the rods are well lagged, find the temperature at the common junction.
b. The rods are then joined in parallel as shown in figure(b). Compare the rate of heat flow in figure(b) with that in figure(a) if the rods are well-lagged in both cases.

Answer 444.
a) Let T = temperature at the common junction.
Heat flow rate through both the blocks will be the same
=> Q/t = kA(T - 0)/L = 3kA(100 - T)/L
=> T = 300 - 3T
=> T = 75° C.

b)
Part a) is a case of series connection and that of b) is of parallel connection.
For part a),
Equivalent thermal resistance = L/kA + L/3kA = 4L/3kA
For part b),
equivalent thermal resistance = 1 / [1/(L/kA) + 1/(L/3kA)] = 1 / (4kA/L) = L/(4kA)
=> For a),
(Q/t)_a = 100 / (4L/3kA) = 75kA/L
and for b)
(Q/t)_b = 100 / (L/4kA) = 400kA/L
=> ratio of heat flow rate of b) to a)
[400kA/L) / (75kA/L)
= 16/3.

Link to YA!