Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities k1 and k2. The equivalent conductivity of the combination is

A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities k1 and k2. The equivalent conductivity of the combination is

Grade:11

2 Answers

Bhagyesh
11 Points
4 years ago
H = H1 + H2 for slabs in parallel. ∆T.KA/L = ∆T.K1.A/2L + ∆T.K2.A/2LHere for k1 and k2 we take A/2. That`s imp. So simplify and get K = K1 + K2 / 2
Rishi Sharma
askIITians Faculty 646 Points
one year ago
Dear Student,
Please find below the solution to your problem.

H = H1 + H2 for slabs in parallel.
∆T.KA/L = ∆T.K1.A/2L + ∆T.K2.A/2L
Here for k1 and k2 we take A/2.
So simplify and get K = (K1 + K2) / 2

Thanks and Regards

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free