Two metal rods of equal length-one aluminum, the other stainless steel-are conne

Question

Two metal rods of equal length-one aluminum, the other stainless steel-are connected in parallel with a temperature of 22.0%u2218C at one end and 122%u2218C at the other end. Both rods have a circular cross section with a diameter of 6.00.

A)
Determine the length the rods must have if the combined rate of heat flow through them is to be 32.5J per second.

B) If the length of the rods is doubled, by what factor does the rate of heat flow change?

Please show work. Just spent an hour on this & can't figure it out.

c

Explanation / Answer

via fouriers law...

deltaQ / deltat = -k A deltaT / deltax

deltaQ / deltat = heat / time
k = conductivity of the specific material
A = cross sectional area
deltaT / deltax = change in temperature of the two ends / length of the rod

******
now.. if you had a single rod.. that would be easy right? just look up "k", calculate A and deltaT.. you were given deltaQ/deltat = 32.5J/s... so piece of cake right?

but the problem is you have two rods in parallel.. and like resistance, the total heat transfer goes like this...1 / Qt = 1 / Q1 + 1 / Q2.... ok?

so that 1 / (32.5 J/s) = [deltax / (kAl x A x deltaT) ] + [ deltax / ( kSS x A x deltaT) ]

and...
deltaT = 100.0C
KAl = 237 J/(s m K).... see the reference
KSS = 45 J/(s m K).....see the reference.. and fyi.. both #'s are highly variable.
A = pi x D^2 / 4 = 2.82x10^-3 m^2

1 / (32.5 J/s) = [deltax / (237 J/(s m K) x (2.82x10^-3 m^2) x 100C) ] + [ deltax / ( 45 J/(s m K) x (2.82x10^-3 m^2) x 100C) ]

1 / (32.5 J/s) = [deltax / (66.83 J m / s)] + [ deltax / ( 12.69 J m / s) ]

(848.07 J m / s) / (32.5 J m / s) = 12.69deltax + 66.83deltax

x = 0.328 m =32.8 cm

***********
if you double that length...

deltaQ / deltat = [65.6m / (66.83 J m / s)] + [ 65.6m / ( 12.69 J m / s) ] = 6.15 J/s

so Qfinal / Qinitial = 6.15 J/s / 32.5 J/s = 0.189

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