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Question
Seven rods A, B, C, D, E, F and G are joined as shown in the figure. All the rods have equal cross-sectional area A and length l. The thermal conductivities of the rods are KA = KC = K0, KB = KD = 2K0, KE = 3K0, KF = 4K0 and KG = 5K0. 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 has a uniform temperature T = (T1 + 2T2)/3. (b) Find the rate of heat flowing from the source which maintains the temperature T2.
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Solution
Given:
KA = KC = K0
KB = KD = 2K0
KE = 3K0, KF = 4K0
K9= 5K0
Here, K denotes the thermal conductivity of the respective rods.
In steady state, temperature at the ends of rod F will be same.
Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D
`(K_A·A·(T-T_1))/ ( l ) + (K_c·A(T-T_1))/(l) + (K_B·A( T_2 - T ))/l`
2k0 ( T - T1 ) = 2 × 2 K0 ( T2 - T)
Temp of rod F = T = `(T_1 + 2T_2)/3`
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T2 is given by
Rate of flow of heat, q = `(DeltaT)/("Thermal resistance")`
First, we will find the effective thermal resistance of the circuit.
From the diagram, we can see 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, for simplification, we can remove this branch.
From the diagram, we find that RA and RB are connected in series.
∴ RAB = RA + RB
RC and RD are also connected in series.
∴ RCD = RC + RD
Then, RAB and RCD are in parallel connection.
`R_A = l/(K_0A)`
`R_B = l/(2K_0A)`
`R_C = l /( K_0A)`
`R_D = l/(2K_0A)`
`R_{AB} = (3l)/(2k_0A)`
`R_{CD} = (3l)/(2K_0A)`
`R_"eff" = ((3l)/(2K_0A)xx(3l)/(2K_0A))/ (((3l)) /( 2K_0A) + (3l)/(2K_0A))`
`=(3l)/(4K_0A)`
`⇒ q = (DeltaT)/("R_eff")`
`= (T_1 - T_2)/((3l)/ (4K_0A))`
`⇒ (4K_0A(T_1 - T_2))/(3L)`
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