Question

A cylindrical tube of diameter 3 m and length 20 m is lined with 3 cm of insulating material of conductivity 10^–4cal cm^–1 s^–10 C^–1. It is maintained at 20ŗC, although the outer temperature is – 30ŗC. What is the rate of heating required to maintain the inner temperature –

• A. 1298 J/s
• B. 12980 J/s
• C. 12098 J/s
• D. 3100 J/s

Answer 1

Answer: (B)

12980 J/s

Answer 2

R = 1.5 m, length = 20 m.

Consider a cylinder of radius = R and thickness = dR.

Rate of flow of heat

Q = KA$\frac{dt}{dR}$

dt = temp difference in thickness dR.

A = 2pRl, Q = K(2pR)$\frac{\mathcal{l}dt}{dR}$

The rate of flow of heat kept constant to maintain temperature inside.

or $\frac{dR}{R}$=$\frac{k2\pi\mathcal{l}}{Q}$dt

$\int_{R}^{R'}\frac{dR}{R}$= $\frac { \mathrm { k } 2 \pi \ell } { \mathrm { Q } }$ $\int_{t_{1}}^{t_{2}}{dt}$

$\left\lbrack \log R \right\rbrack_{R}^{R'}$=$\frac{k2\pi\mathcal{l}}{Q}$ (t₂ – t₁)

Q =$\frac{k2\pi\mathcal{l}(20 - ( - 30))}{\log\frac{1.53}{1.50}}$Q = 3100 cal/sec.

Q = 12980 J/s.