A composite block is made of slabs A, B, C, D and E of different thermal conductivities (given in terms of a constant K and sizes (given in terms of length, L)) as shown in the figure. All slabs are of same width. Heat ‘Q’ flows only from left to right through the blocks. Then in steady state [2011]

Question

Smart Achievers · Accepted Answer

(1, 3, 4)

According to question, heat Q flows only from left to right through the blocks. Hence heat flow through slab A and E are the same.

$We know that thermal resistance R = \frac{ℓ}{K A}$

$Let the width of slabs be W . Then$

$R_{A} = \frac{L}{2 K (4 L) W} = \frac{1}{8 K W}, R_{B} = \frac{4 L}{3 K (L W)} = \frac{4}{3 K W}$

$R_{C} = \frac{4 L}{4 K (2 L W)} = \frac{1}{2 K W}, R_{D} = \frac{4 L}{5 K (L W)} = \frac{4}{5 K W}$

$R_{E} = \frac{L}{6 K (4 L W)} = \frac{1}{24 K W}$

$Now, Δ T = Q R$

Since the resistance to heat flow is least for slab E, the temperature difference across E is smallest.

Also

$Q_{C} = \frac{Δ T_{C}}{R_{C}} = \frac{Δ T_{C}}{1 / 2 K W} = 2 K W (Δ T_{C})$

$Q_{B} = \frac{Δ T_{B}}{R_{B}} = \frac{Δ T_{C}}{4 / 3 K W} = \frac{3 K W (Δ T_{C})}{4}$ $[∵ Δ T_{B} = Δ T_{C}]$

$Q_{D} = \frac{Δ T_{D}}{R_{D}} = \frac{Δ T_{C}}{4 / 5 K W} = \frac{5 K W (Δ T_{C})}{4}$ $[∵ Δ T_{D} = Δ T_{C}]$

$Q_{B} + Q_{D} = \frac{3 K W (Δ T_{C})}{4} + \frac{5 K W (Δ T_{C})}{4}$

$= \frac{8 K W (Δ T_{C})}{4} = 2 K W (Δ T_{C}) = Q_{C}$

$i.e., Q_{C} = Q_{B} + Q_{D}$

Solution

1 heat flow through A and E slabs are same.

2 heat flow through slab E is maximum.

3 temperature difference across slab E is smallest.

4 heat flow through C = heat flow through B + heat flow through D.

Want Us to Email you About Special Offers & Updates?