Q. 16

Three rods of lengths 20 cm each and area of cross-section 1 cm2 are joined to form a triangle ABC. The conductivities of the rods are KAB = 50 J s–1 m–1 °C–1, KBC = 200 J s–1 m–1 °C–1 and KAC = 400 J s–1 m–1 °C–1. The junctions A, B and C are maintained at 40°C, 80°C and 80°C respectively. Find the rate of heat flowing through the rods AB, AC and BC.

Answer :


Given:
Length of all the rods: x= AB=BC=AC = 20 cm = 0.2 m
Area of cross section of these rods: A = 1 cm2 = 0.0001 m2.
Thermal conductivity of rod AB : KAB = 50 J s–1 m–1 °C–1
Thermal conductivity of rod BC : KBC = 200 J s–1 m–1 °C–1
Thermal conductivity of rod AC : KAC = 400 J s–1 m–1 °C–1
Temperature at A : T1 = 40 °C
Temperature at B : T2 = 80 °C
Temperature at C : T3 = 80 °C
Formula used:
Rate of amount of heat flowing or heat current is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness or length of the material.
(1)
Rate of heat flowing in the rod AB is



Hence, rate of heat flowing through the rod Ab is 1 J/s or 1 W
(2)
Rate of heat flowing in the rod BC is



Hence, the rate of heat flowing through the rod BC is 0 as both the ends of the rods are maintained at same temperature.
(3)
Rate of heat flowing in the rod BC is



Hence, the rate of heat flowing through the rod AC is 8 J/s.


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