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Animated Solution for Physics - Properties of Solids and Liquids: A point source of heat of power $P$ is placed at the centre of a spherical shell of mean radius $R$ . The material of the shell has thermal conductivity $K$ . If the temperature difference between the outer and inner surface of the shell is not to exceed $T$ , the thickness of the shell should not be less than …… .

Q: What if the shell was thick?

For a thick shell with inner radius $r_1$ and outer radius $r_2$, the thermal resistance is exactly $\frac{r_2 - r_1}{4\pi K r_1 r_2}$. For a thin shell, $r_1 \approx r_2 \approx R$, which simplifies to $\frac{t}{4\pi K R^2}$.

Q: Can we use the formula for a flat slab here?

Yes! Because the shell is thin ($t \ll R$), the curvature effects are negligible. Locally, any small section of the shell behaves like a flat slab of thickness $t$ and area $A$.

Visualized Solution

\text{System Setup}

\text{Point source of power } P \text{ at the center.}
\text{Spherical shell of mean radius } R \text{ and thickness } t.
\text{Heat flows radially outward.}

\text{Thermal Resistance}

\text{For a thin spherical shell } (t \ll R), \text{ the cross-sectional area is approximately constant.}
A \approx 4\pi R^2
\text{Thermal resistance: } R_{\text{th}} = \frac{t}{KA}

\text{Heat Transfer Equation}

\text{Rate of heat transfer equals the power of the source in steady state.}
H = P
\text{Ohm's law for heat: } \Delta T = H R_{\text{th}}

\text{Substituting Values}

\Delta T = P \left( \frac{t}{K(4\pi R^2)} \right)
\Delta T = \frac{P t}{4\pi K R^2}

\text{Applying the Constraint}

\text{Given condition: The temperature difference must not exceed } T.
\Delta T \le T
\frac{P t}{4\pi K R^2} \le T

\text{Solving for Thickness}

t \le \frac{4\pi K T R^2}{P}
\text{The maximum allowed thickness is } \frac{4\pi K T R^2}{P}.

\text{Conclusion \& Typo Note}

\text{The question asks for the thickness to "not be less than", which is physically contradictory.}
\text{Correct interpretation: thickness should not exceed } \frac{4\pi K T R^2}{P}.

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The Sigma Insight: Heat Transfer

Solution Diagram

The problem asks us to find the condition on the thickness of a spherical shell such that the temperature difference across it does not exceed a given value

T

The Physical Setup Imagine a point source of heat generating power $P$ located exactly at the center of a spherical shell

The shell has a mean radius

R

and a thermal conductivity

K

. Because the source is constantly generating heat, this heat must flow radially outward through the material of the shell to reach the surroundings.

In a steady state, the rate at which heat flows through the shell (the heat current,

H

) must be exactly equal to the power generated by the source:

H = P

Thermal Resistance of a Thin Shell

To find the temperature difference across the shell, we can use the thermal equivalent of Ohm's law:

Δ T = H \cdot R_{th}

where

R_{th}

is the thermal resistance of the shell.

For a thin shell of thickness

t

and mean radius

R

, the inner and outer surface areas are approximately the same. We can treat it locally like a flat slab of area

A = 4 π R^{2}

. The thermal resistance is given by:

R_{th} = \frac{t}{K A} = \frac{t}{K ( 4 π R ^{2} )}

The Heat Transfer Equation

Substituting the thermal resistance and the heat current into our temperature difference equation, we get:

Δ T = P (\frac{t}{4 π K R ^{2}})

The problem states that this temperature difference must not exceed

T

. Therefore, we set up the inequality:

Δ T \leq T

\frac{P t}{4 π K R ^{2}} \leq T

Solving for the thickness

t

, we find:

t \leq \frac{4 π K T R ^{2}}{P}

The Typo in the Problem You might have noticed a contradiction in the problem's wording

It asks for the thickness to "not be less than" a certain value. However, our physical intuition and the math tell us that a thicker shell provides more thermal resistance, which would increase the temperature difference. To keep the temperature difference below

T

, the thickness must be restricted to a maximum value. Therefore, the correct phrasing should be that the thickness "should not exceed"

\frac{4 π K T R ^{2}}{P}

Common Traps & FAQs

Why do we use the mean radius

R

to calculate the area?

Since the shell is thin, the inner and outer radii are very close to

R

. The surface area

4 π r^{2}

doesn't change significantly across the thickness, so we can approximate it as

4 π R^{2}

What if the shell was thick?

For a thick shell with inner radius

r_{1}

and outer radius

r_{2}

, the thermal resistance is exactly

\frac{r _{2} - r _{1}}{4 π K r _{1} r _{2}}

. For a thin shell,

r_{1} \approx r_{2} \approx R

, which simplifies to

\frac{t}{4 π K R ^{2}}

Why is the heat current equal to the power of the source?

In a steady state, there is no accumulation of heat within the shell. All the energy generated by the point source per second must pass through the shell to the outside.

Why does the question's wording 'should not be less than' contradict physics?

Thermal resistance is directly proportional to thickness. A thicker shell provides more resistance, which leads to a larger temperature difference for the same heat current. To keep the temperature difference below a certain limit, the thickness must be *less* than a maximum value, not more.

Can we use the formula for a flat slab here?

Yes! Because the shell is thin (

t ≪ R

), the curvature effects are negligible. Locally, any small section of the shell behaves like a flat slab of thickness

t

and area

A

Visualized Solution

\text{System Setup}

\text{Thermal Resistance}

\text{Heat Transfer Equation}

\text{Substituting Values}

\text{Applying the Constraint}

\text{Solving for Thickness}

\text{Conclusion \& Typo Note}

The Sigma Insight: Heat Transfer

The Physical Setup Imagine a point source of heat generating power $P$ located exactly at the center of a spherical shell

Thermal Resistance of a Thin Shell

The Heat Transfer Equation

The Typo in the Problem You might have noticed a contradiction in the problem's wording

Similar Questions

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Visualized Solution

\text{System Setup}

\text{Thermal Resistance}

\text{Heat Transfer Equation}

\text{Substituting Values}

\text{Applying the Constraint}

\text{Solving for Thickness}

\text{Conclusion \& Typo Note}

The Sigma Insight: Heat Transfer

The Physical Setup Imagine a point source of heat generating power .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P located exactly at the center of a spherical shell

Thermal Resistance of a Thin Shell

The Heat Transfer Equation

The Typo in the Problem You might have noticed a contradiction in the problem's wording

Similar Questions

The Physical Setup Imagine a point source of heat generating power $P$ located exactly at the center of a spherical shell