For insulating a spherical container (negligible thickness) of di

| For insulating a spherical container (negligible thickness) of diameter 0.5 m, two layers of thickness 50 mm each and thermal conductivities k₁ = 0.06 W/m K and k₂ = 0.1 W/m K are readily available. The liquid to be stored is at a temperature of 400 K and the outside surface temperature is always at 300 K. Then

A. Heat lost by the liquid if K2 is inside and K1 is outside is 86.2 W

B. The interface temperature between two insulation layers for K2 inside and K1 outside is 327 K

C. For better insulation, lesser thermal conductivity material (K1) should be inside.

D. For better insulation, higher thermal conductivity material (K2) should be inside.

Please scroll down to see the correct answer and solution guide.

Right Answer is:

SOLUTION

Concept:

Heat conduction through concentric spheres.

For the above setup,

The heat conducted is given by

\(Q = \frac{{{\rm{\Delta }}{T_{overall}}}}{{{\rm{\Sigma }}\;{R_{th}}}}\)

where ΣR_th = R_th1 + R_th2

At the interface,

\(Q = \frac{{{T_1} - {T_i}}}{{{R_{th1}}}} \Rightarrow {T_i} = {T_1} - Q.{R_{th1}}\)

For spherical layers, thermal resistance is given by

\({R_{th}} = \frac{{{r_2} - {r_1}}}{{4\pi k{r_1}{r_2}}}\)

Schematic:

Calculation:

Given:

r₁ = 0.25 m; r₂ = 0.25 + 0.05 = 0.30 m; r₃ = 0.3 + 0.05 = 0.35 m;

T₁ = 400 K; T₂ = 300 K; k₁ = 0.06 W/m K; k₂ = 0.1 W/m K;

If k₂ is inside and k₁ is outside,

\({R_{th1}} = \frac{{{r_2} - {r_1}}}{{4\pi {k_2}{r_1}{r_2}}} = \frac{{0.3 - 0.25}}{{4\pi \times 0.1 \times 0.3 \times 0.25}} = 0.53\;K/W\)

\({R_{th2}} = \frac{{{r_3} - {r_2}}}{{4\pi {k_1}{r_3}{r_2}}} = \frac{{0.35 - 0.30}}{{4\pi \times 0.06 \times 0.35 \times 0.30}} = 0.63\;K/W\)

Total resistance = 1.16 K/W;

Now heat conducted is

\(Q = \frac{{400 - 300}}{{1.16}} = 86.2\;W\) (Option 1)

Interface temperature is

T_i = 400 – 86.2 × 0.53 = 354.31 K (Option 2 is wrong)

For better insulation, the lesser thermal conductivity should be inside as it offers more resistance. As thermal resistance R_th is inversely proportional to thermal conductivity (k).

If k₁ is inside and k₂ is ouside,

Total resistance = R_th1 + R_th2

\({R_{tot}} = \frac{{{r_2} - {r_1}}}{{4\pi {k_1}{r_1}{r_2}}} + \frac{{{r_3} - {r_2}}}{{4\pi {k_2}{r_1}{r_2}}}\)

\(\Rightarrow {R_{tot}} = \frac{{0.3 - 0.25}}{{4\pi \times 0.06 \times 0.3 \times 0.25}} + \frac{{0.35 - 0.30}}{{4\pi \times 0.1 \times 0.35 \times 0.30}} = 1.26\;K/W\)

∴ (R_tot)_{2nd case}> (R_tot)_{1st case}(Option 3)