IIT JEE , AIEEE Forum - Physics , Thermal Physics

ramya_kapoor

a cylinder has inner radius r1 and outer radius r2....temperature of iiner surface t1 and outer surface temp. t2...

length of cylinder is l....derive expression for the heat current ie..

Q/

T............THX

akku

Hi ramya,
FOR RADIAL FLOW OF HEAT

H=Q/t= -kAdt/dx
Let there b a cylindrical element @ distance r frm axis of cylinder
H.dr= -k (2

rl)dT
Hdr/r= -k2

ldT
now integrate both sides from r1 to r2 and t1to t2
H=k2

l(t1-t2)/ln(r2/r1) where ln=natural log= log base e

FOR AXIAL FLOW OF HEAT
H=K

(r2^2-r1^2)(t1-t2)/l

PLS RATE !

rphy

agree with akku..

joyfrancis

Well I have a different answer

Q/t = (KA

T) / l

note that in the above equation A is the lateral surface area of the cylinder and l is the length travelled is nothing but (r2-r1) tsince the heat flow is radially outwards.

so this way

Q/t = 2LKpi(t2-t1)(r2-r1)/(r2-r1)

= 2LKpi(t2-t1)(r2-r1)

ramya_kapoor

thx akkuuuuuuuuu

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