Question

Consider a long solid cylinder of radius R = 4cm and thermal

conductivity k = 25 W/m ℃. Heat is generated in the cylinder uniformly

at the rate of = 35 w/cm³ . The side surface of the cylinder is

maintained at a constant temperature of = 80 ℃ . The variation of

temperature in the cylinder is given by (R = outer radius)

() =

R²



[ 1 – (





)² ] + Ts

Based on this relation the value of heat flux on the side surface to the

cylinder at r = R is?​

Answer 1

Answer:

Step-by-step explanation:

Let R₁ = Initial Radius = R  

Let R₂ = Final Radius = 2R  

The two conductors are in parallel. Therefore, equivalent thermal resistance R is :-

1/R = 1/R₁ + 1/R₂ ---------- (1)

R= KA/l by Q = KA (T₁ - T₂)/ l

R₁= K₁A₁/l₁

R₂= K₂A₂/l₂

Substituting above in (1) we get,

KA/l = K₁A₁/l₁ + K₂A₂/l₂ ---------- (2)

Now, l₁ = l₂ = l

A₁ = ΠR²

A₂ = Π (2R)² - ΠR₂² = 3ΠR₂

A = Π (2R)² = 4ΠR²

Substituting in (2) we get,

K4ΠR²/ l = K₁ΠR²/l + K₂3ΠR₂/l

∴ K = (K₁ + 3K₂) /4

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Explanation: