A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical

Q: Why are the thermal resistances in parallel and not in series?

Heat flows from the hot end to the cold end. Both the inner cylinder and the outer shell connect the exact same hot and cold reservoirs. Since they share the same temperature difference $\Delta T$, they are in parallel. If heat flowed radially from the center to the outside, they would be in series.

Q: Why is the area of the outer shell $3\pi R^2$ and not $4\pi R^2$?

The outer shell is a hollow cylinder. Its cross-section is a ring (annulus) with an inner radius $R$ and an outer radius $2R$. The area of this ring is the area of the large circle ($\pi (2R)^2 = 4\pi R^2$) minus the area of the empty hole in the middle ($\pi R^2$), which gives $3\pi R^2$.

Q: What is the length $l$ of the cylinder? It's not given in the problem!

You don't need to know the exact length. Because both the inner cylinder and the outer shell have the same length $l$, it appears in every term of the parallel resistance equation and beautifully cancels out in the final step.

Q: Can we just take the average of the two conductivities, like $(K_1 + K_2)/2$?

No, because the two materials do not have the same cross-sectional area. The effective conductivity is a *weighted* average based on the area. Since the outer shell has a larger area, its conductivity $K_2$ has a greater influence on the overall effective conductivity.

Q: What does "effective thermal conductivity" actually mean?

Imagine replacing the entire composite cylinder (both the core and the shell) with a solid cylinder of the exact same total size (radius $2R$, length $l$). The effective thermal conductivity is the conductivity this new, uniform cylinder must have so that it conducts the exact same total amount of heat per second as the original composite cylinder.

Question

A cylinder of radius $R$ made of a material of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2R$ made of a material of thermal conductivity $K_2$. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

Options:

(A) $K_1 + K_2$

(B) $\frac{K_1 K_2}{K_1 + K_2}$

(C) $\frac{K_1 + 3K_2}{4}$

(D) $\frac{3K_1 + K_2}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at the cross-section of a composite cylinder. Inside, we have a solid core—a cylinder of radius $R$ with a thermal conductivity of $K_1$. Surrounding this core is a cylindrical shell extending from radius $R$ to an outer radius of $2R$, made of a different material with thermal conductivity $K_2$.

The problem states that the two ends of this combined system are maintained at two different temperatures. This is the crucial piece of information! Because the ends are at the same two temperatures, the temperature difference $\Delta T$ across the inner cylinder is exactly the same as the temperature difference across the outer shell.

When two components in a circuit—whether it's an electrical circuit with resistors or a thermal circuit with heat conductors—experience the same potential difference, they are connected in **parallel**.

### The Master Equation

Since the inner cylinder and the outer shell act as thermal resistances in parallel, we can use the parallel combination formula. The reciprocal of the equivalent thermal resistance $R_{eq}$ is simply the sum of the reciprocals of the individual thermal resistances:

$$ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} $$

To use this master equation, we need to find the thermal resistance of each part. Remember the formula for thermal resistance? It is given by the length of the conductor divided by the product of its thermal conductivity and its cross-sectional area:

$$ R = \frac{l}{KA} $$

### Calculating Individual Resistances

Let's start with the inner cylinder. Its cross-sectional area is simply the area of a circle with radius $R$.

$$ A_1 = \pi R^2 $$

So, its thermal resistance $R_1$ will be:

$$ R_1 = \frac{l}{K_1 (\pi R^2)} $$

Now, let's look at the outer shell. This is where mistakes often happen! The cross-sectional area of the shell is not just $\pi (2R)^2$. It is the area of the larger circle minus the area of the inner circle.

$$ A_2 = \pi (2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2 $$

With the correct area, the thermal resistance of the outer shell $R_2$ is:

$$ R_2 = \frac{l}{K_2 (3\pi R^2)} $$

### Final Calculation

Now, let's treat the entire composite cylinder as a single, uniform material with an effective thermal conductivity $K_{eq}$. The total cross-sectional area of this equivalent cylinder is the area of the outer circle:

$$ A_{eq} = \pi (2R)^2 = 4\pi R^2 $$

So, the equivalent thermal resistance $R_{eq}$ is:

$$ R_{eq} = \frac{l}{K_{eq} (4\pi R^2)} $$

Let's substitute all these resistance expressions back into our parallel combination equation:

$$ \frac{K_{eq} (4\pi R^2)}{l} = \frac{K_1 (\pi R^2)}{l} + \frac{K_2 (3\pi R^2)}{l} $$

Notice how beautifully the math simplifies! The common term $\frac{\pi R^2}{l}$ is present in every single term. We can cancel it out from both sides of the equation.

$$ 4 K_{eq} = K_1 + 3K_2 $$

Dividing by 4, we arrive at our final elegant result for the effective thermal conductivity:

$$ K_{eq} = \frac{K_1 + 3K_2}{4} $$

This result tells us that the effective conductivity is a weighted average of the individual conductivities, weighted by their respective cross-sectional areas. Since the outer shell has three times the area of the inner core, its conductivity $K_2$ contributes three times as much to the overall effective conductivity!

Select Answer:

Visualized Solution

Visualizing the Setup

Identifying the Combination

Parallel Resistance Formula

Thermal Resistance of Inner Cylinder

Thermal Resistance of Outer Shell

Equivalent Thermal Resistance

Substituting into Parallel Formula

Final Calculation

The Way Forward

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Calculating Individual Resistances

Final Calculation

Similar Questions

One end of a thermally insulated rod is kept at a temperature $T_{1}$ and the other at $T_{2}$ . The rod is composed of two sections of lengths $l_{1}$ and $l_{2}$ and thermal conductivities $K_{1}$ and $K_{2}$ respectively. The temperature at the interface of the two sections is

Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively are connected in series. The effective thermal conductivity of the combination is

The figure shows a system of two concentric spheres of radii $r_{1}$ and $r_{2}$ and kept at temperatures $T_{1}$ and $T_{2}$ , respectively. The radial rate of flow of heat in a substance between the two concentric spheres, is proportional to

Select Answer:

Visualized Solution

Visualizing the Setup

Identifying the Combination

Parallel Resistance Formula

Thermal Resistance of Inner Cylinder

Thermal Resistance of Outer Shell

Equivalent Thermal Resistance

Substituting into Parallel Formula

Final Calculation

The Way Forward

The Sigma Insight: Heat Transfer

Analyzing the Setup

The Master Equation

Calculating Individual Resistances

Final Calculation

Similar Questions

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Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively are connected in series. The effective thermal conductivity of the combination is