First slide

Heat transfer

Question

Two rods are taken having same length and area and of the same material are joined side by side, heat is allowed to flow through them for 12 minutes. If now the rods are joined in parallel, then how much time will it take to flow the same amount of heat?

Moderate

A

24 minutes
B

3 minutes
C

12 minutes
D

6 minutes

Solution

in Series time of flow of heat is
$t_{1} =$ 12 minutes
$\begin{array}{rcl} a m o u n t o f h e a t t h a t f l o w s i n t w o r o d s w h e n i n s e r i e s i s \\ Q_{1} = \frac{K_{s e r i e s} {A(θ}_{1} {-θ}_{2})}{2l} t_{1} ----(1) \\ h e r e K_{s e r i e s} i s c o e f f i c i e n t o f t h e r m a l c o n d u c t i v i t y w h e n r o d s a r e i n s e r i e s \\ A i s a r e a o f c r o s s s e c t i o n \\ θ_{1}, θ_{2} i s t e m p e r a t u r e s a t t h e i r e a c h e n d \\ l i s l e n g t h o f r o d \\ t_{1} i s t i m e o f f l o w o f h e a t \\ I n s e r i e s, e f f e c t i v e c o e f f i c e i n t o f t h e r m a l c o n d u c t i v i t y i s K_{s e r i e s}, \\ \frac{l + l}{K_{s e r i e s}} & = & \frac{l}{K} + \frac{l}{K} \\ K_{s e r i e s} = K \\ w h e n r o d s a r e i n p a r a l l e l, e f f e c t i v e c o e f f i c e i n t o f t h e r m a l c o n d u c t i v i t y i s \\ K_{p a r a l l e l} = \frac{K_{1} + K_{2}}{2} \\ K_{p a r a l l e l} = \frac{K + K}{2} \\ K_{p a r a l l e l} = K \\ w h e n r o d s a r e t a k e n i n p a r a l l e l, a m o u n t o f h e a t t h a t f l o w s i n t i m e t_{2} i s \\ Q_{2} & = & \frac{K_{p a r a l l e l} {2A(θ}_{1} {-θ}_{2})}{l} t_{2} ----(2) \\ t o f i n d t_{2} equate eqn(1) and eqn(2) \\ Q_{1} & = & Q_{2} \\ \frac{K_{s e r i e s} {A(θ}_{1} {-θ}_{2})}{2l} t_{1} = \frac{K_{p a r a l l e l} {2A(θ}_{1} {-θ}_{2})}{l} t_{2} \\ h e r e K_{s e r i e s} {=K}_{p a r a l l e l} \\ o n s o l v i n g w e g e t \\ t_{2} =3 minutes \end{array}$ ---(1)

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