---
title: "Conduction through hollow sphere"
subject: "Heat Transfer"
chapter: 2
key: "semester_5/heat_transfer:hollow-sphere-heat-flow"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/hollow-sphere-heat-flow
site: BCT Engineering
---

# Conduction through hollow sphere

Heat Transfer · Chapter 2: Conduction Heat Transfer

- Asked in: year not given, 5 marks
- Video: [Heat Conduction through Hollow Sphere II Heat and Mass Transfer II ](https://www.youtube.com/watch?v=thMrp0xryS4) — Shailendra's Academy
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/hollow-sphere-heat-flow
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-2

## Question

A hollow sphere of inner radius 100 mm and outer radius 200 mm is made of material of thermal conductivity 15 W/m K. The inner surface is kept at 300 °C and the outer surface at 100 °C. Derive the expression you use for the heat flow, and calculate (a) the heat flow rate and (b) the temperature at a radius of 150 mm.

## Short answer

Spherical heat flow: Q = 4πk (T1−T2) / (1/r1 − 1/r2). With r1=0.1 m, r2=0.2 m, k=15 W/m·K, ΔT=200 K: Q = 4π·15·200 / (1/0.1−1/0.2) = 7.54 kW. At r=0.15 m, T = 300 − (7540/(4π·15))(1/0.1−1/0.15) = 166.7 °C.

## Answer

#### Expression for a hollow sphere
Steady radial conduction with no generation: $\dfrac{d}{dr}\left(r^2\dfrac{dT}{dr}\right)=0$, so $T = -C_1/r + C_2$. Using $Q = -k(4\pi r^2)\dfrac{dT}{dr}$ and the boundary temperatures $T_1$ at $r_1$ and $T_2$ at $r_2$:

$$
Q = \frac{4\pi k\,(T_1 - T_2)}{\dfrac{1}{r_1}-\dfrac{1}{r_2}}
$$

and the temperature at any radius:

$$
T = T_1 - \frac{Q}{4\pi k}\left(\frac{1}{r_1}-\frac{1}{r}\right)
$$

#### (a) Heat flow
Given $r_1 = 0.1$ m, $r_2 = 0.2$ m, $k = 15$ W/m K.

$$
\begin{aligned}
\frac{1}{r_1}-\frac{1}{r_2} &= 10 - 5 = 5\ \text{m}^{-1}\\
Q &= \frac{4\pi(15)(300-100)}{5} = 7540\ \text{W}
\end{aligned}
$$

#### (b) Temperature at 150 mm

$$
\begin{aligned}
T &= 300 - \frac{7540}{4\pi(15)}\left(10 - \frac{1}{0.15}\right)\\
&= 300 - 40\,(10 - 6.667) = 300 - 133.3 = 166.7\ ^\circ\text{C}
\end{aligned}
$$

The temperature at r = 0.15 m (166.7 °C) is lower than the arithmetic mean of the surfaces (200 °C), because the area is smaller near the inside and the gradient is steeper there.

**Answer:** $Q \approx 7.54\ \text{kW}$; $T(0.15\ \text{m}) \approx 166.7\ ^\circ\text{C}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which formula gives heat flow through a hollow sphere?
   A. 2 pi k L (T1-T2)/ln(r2/r1)
   B. 4 pi k (T1-T2)/(1/r1 - 1/r2)
   C. k A (T1-T2)/(r2-r1)
   D. 4 pi k (T1-T2)/ln(r2/r1)
   Answer: B. With A = 4 pi r^2 integration gives the 1/r1 - 1/r2 form. (Key point: Sphere formula)

2. Heat flow for the given sphere is about:
   A. 3.77 kW
   B. 1.5 kW
   C. 7.54 kW
   D. 15 kW
   Answer: C. Q = 4 pi(15)(200)/5 = 7540 W. (Key point: Heat flow)

3. Why is T at r = 150 mm (166.7 C) below the 200 C mean?
   A. Conductivity is higher inside
   B. Heat is generated inside
   C. The profile is linear
   D. Area is smaller near the inside so the gradient is steeper there
   Answer: D. Smaller area near the inside needs a steeper gradient, so T drops faster there. (Key point: Mid-radius temperature)

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## Related in other subjects

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