First slide

Heat transfer

Question

A wall has two layers A and B, each made of different material, “A” has thickness 10 cm, while “B” has thickness 20cm, their coefficients of conductivities are in the ratio of 3:1, a constant temperature difference of $35^{\circ} c$ exists across the wall, the difference of temperature across the layer “A” is

Moderate

A

$28^{\circ} c$
B

$14^{\circ} c$
C

$5^{\circ} c$
D

${3.5}^{\circ} c$

Solution

$r a t e o f t r a n s f e r o f h e a t i s \frac{Q}{t} = \frac{{KA(θ}_{1} {-θ}_{2})}{l} here k is coefficient of thermal conductivity A i s a r e a o f c r o s s s e c t i o n θ_{1} {-θ}_{2} is temperature difference l i s l e n g t h o f w i d t h a t s t e a d y s t a t e, h e a t t r a n s f e r i s c o n s \tan t t e m p e r a t u r e a c r o s s l a y e r A i s θ_{1} -θ t e m p e r a t u r e a c r o s s l a y e r B i s θ {-θ}_{2} \frac{Q}{t} = \frac{K_{1} {A(θ}_{1} -θ)}{l_{1}} = \frac{K_{2} {A(θ-θ}_{2})}{l_{2}} 3 (θ_{1} -θ)= \frac{1}{2} (θ - θ_{2})---(1) (θ_{1} {-θ}_{2} {)=35}^{0} C⇒ θ_{2} = θ_{1} - 35 ---(2) o n s u b s t i t u t i n g t h e v a l u e o f θ_{2} f r o m e q n (2) i n e q n (1) \Rightarrow 3 θ_{1} -3θ= \frac{1}{2} θ - \frac{1}{2} (θ_{1} - 35) \Rightarrow 6 θ_{1} -6θ= θ - θ_{1} + 35 \Rightarrow 7 (θ_{1} - θ) = 35 θ_{1} - θ = 5^{0} C = t e m p e r a t u r e d i f f e r e n c e a c r o s s l a y e r A$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App