Steady State Conduction Questions and Answers

Answer: B

The critical radius of insulation for a pipe is \(k/h_{o}\).

For insulation of outer radius r on a pipe, the total resistance per unit length is conduction through the insulation plus convection from its outer surface:

\[ R' = \frac{\ln(r/r_{i})}{2\pi k} + \frac{1}{2\pi r h_{o}} \]

Setting \(dR'/dr = 0\) gives \(\frac{1}{kr} = \frac{1}{h_{o}r^{2}}\), so \(r_{cr} = k/h_{o}\). Below this radius, adding insulation actually increases heat loss, because the larger surface gains more convection than the extra conduction resistance costs.

\(2k/h_{o}\) is the critical radius for a sphere. \(h_{o}/k\) and \(2h_{o}/k\) invert the ratio and do not even have the units of length.

Answer: A

The ratio is about 2.58: the insulation more than doubles the heat loss.

Bare wire, radius 1 mm, loses heat by convection only:

\[ \begin{aligned} q_{0} &= h(2\pi r_{1})\Delta T \\ &= 25(2)(3.14)(0.001)(100) = 15.7\ \text{W/m} \end{aligned} \]

Insulated wire, outer radius 3 mm, has conduction and convection resistances in series (per metre):

\[ \begin{aligned} R &= \frac{\ln(3/1)}{2\pi k} + \frac{1}{h\,2\pi r_{2}} \\ &= \frac{1.0986}{3.14} + \frac{1}{0.471} = 0.350 + 2.123 \\ &= 2.473\ \text{K.m/W} \\ q &= \frac{100}{2.473} = 40.4\ \text{W/m} \end{aligned} \]

\[ \frac{q}{q_{0}} = \frac{40.4}{15.7} = 2.58 \]

The reason is the critical radius of insulation, \(k/h = 0.5/25 = 20\) mm. Below it, adding insulation increases the outer surface faster than it adds resistance, so the loss rises. 0.39 is the inverse ratio, which would be right only if insulation always reduced heat loss.

Answer: C

The temperature stays \(T_{A}\).

The total heat generated in the rod, per unit cross-section, is

\[ \int_{0}^{L}\cos\frac{2\pi x}{L}\,dx = 0 \]

The end \(x = 0\) and the sides are insulated, so in steady state the heat leaving at \(x = L\) must equal this net generation, zero. The convective flux \(h(T(L) - T_{\infty})\) is therefore zero, so \(T(L) = T_{\infty}\) whatever the value of \(h\).

The temperature inside follows from \(kT'' = -\dot{q}\) with \(T'(0) = 0\) and \(T(L) = T_{\infty}\), none of which involves \(h\); at mid-length it is \(T_{\infty} - L^{2}/(2\pi^{2}k)\). Doubling \(h\) changes nothing, so the temperature stays \(T_{A}\).

The other options assume a net heat flow through the convective end, which is zero here.

Answer: B

The temperature distribution is plot (B): rising steeply near x = 0 and flattening towards x = L.

With no heat sources, the steady heat flux through the slab is the same at every x:

\[ \begin{aligned} q &= -k\frac{dT}{dx} = \text{constant} \\ \frac{dT}{dx} &= \frac{\text{const}}{A + Bx} \end{aligned} \]

Since T rises from 0 to \(T_{0}\), the slope is positive, and because k = A + Bx grows with x, the slope gets smaller as x increases. The profile is therefore steep where the conductivity is low and gentle where it is high: a curve that is concave down.

The straight line (A) is the constant-conductivity case. Plot C has the slope increasing with x, which would need k to decrease. Plot D has a temperature that falls in the middle, which would need heat to flow both ways at once, impossible without sources.

Answer: B

Non-isotropic conductivity is shown by wood.

Wood is fibrous: heat flows more easily along the grain than across it, typically about twice as fast, so its thermal conductivity depends on direction. A material like this is called anisotropic.

Lead, copper and brass are metals with a crystalline structure that, in bulk, conducts heat equally in all directions, so they are isotropic for heat conduction.

Answer: D

The insulation thickness for maximum heat dissipation is 15 mm.

For a cylinder, heat loss is greatest when the outer radius of the insulation equals the critical radius:

\[ \begin{aligned} r_{c} &= \frac{k}{h} = \frac{0.5}{20} = 0.025\ \text{m} = 25\ \text{mm} \\ t &= r_{c} - r = 25 - 10 = 15\ \text{mm} \end{aligned} \]

The wire's radius is 10 mm, half its 20 mm diameter. 25 mm is the critical radius itself, not the thickness, and 30 mm and 20 mm come from using the diameter in place of the radius.

Answer: C

The heat loss is \(128\ \text{W/m}^{2}\).

Neglecting the inside air resistance, the heat passes through the wall and the outside air film in series. Per square metre:

\[ \begin{aligned} R &= \frac{0.18}{0.9} + \frac{1}{20} = 0.20 + 0.05 = 0.25 \\ q &= \frac{27 - (-5)}{0.25} = \frac{32}{0.25} = 128 \end{aligned} \]

Using the wall resistance alone gives \(32/0.2 = 160\), which forgets the outside film. 88 and 110 do not follow from these resistances.

Answer: B

The heat flux is \(2.466\ \text{MW/m}^{2}\).

For steady one-dimensional conduction through a plane wall, Fourier's law gives the heat transfer per unit area:

\[ \begin{aligned} q &= \frac{k (T_{1} - T_{2})}{L} \\ &= \frac{370 \times (350 - 50)}{0.045} \\ &= \frac{111{,}000}{0.045} \\ &\approx 2.466 \times 10^{6}\ \text{W/m}^{2} \end{aligned} \]

That is 2.466 MW/m\(^{2}\). The thickness must be in metres: using 45 instead of 0.045 makes the answer a thousand times too small, 0.0025 MW/m\(^{2}\). Copper's high conductivity is why the flux is so large for a 300°C difference.