---
title: "Diameter to thickness ratio for stress limit"
subject: "Hydraulics"
chapter: 4
key: "semester_4/hydraulics:pipe-diameter-thickness-ratio-hoop-stress"
source: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-diameter-thickness-ratio-hoop-stress
site: BCT Engineering
---

# Diameter to thickness ratio for stress limit

Hydraulics · Chapter 4: Unsteady flow in pipes

- Asked in: 2074 Bhadra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/pipe-diameter-thickness-ratio-hoop-stress
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4

## Question

In a pipe of length 500 m and uniform circular cross-section, water flows at a steady velocity of 2 m/s and discharges to atmosphere through a valve. Under steady conditions the static head just before the valve is 300 m. Calculate the ratio of internal diameter to wall thickness of the pipe so that, when the valve is completely and instantaneously closed, the increase in circumferential stress is limited to 20 MPa, and determine the maximum time for which the closure could be described as rapid. The bulk modulus of water = 2 GPa, and the elastic modulus of the pipe material = 200 GPa.

## Short answer

Solving for D/t from hoop stress limit of 20 MPa yields D/t ≈ 15.2. Wave speed is 1318 m/s, so closure is rapid for t_c ≤ 0.76 s.

## Answer

**Data.** $L = 500$ m, $V_0 = 2$ m/s, static head 300 m, instantaneous valve closure, increase of circumferential stress $\Delta\sigma \le 20$ MPa, $K = 2$ GPa, $E = 200$ GPa. Let $x = D/t$.

#### Relations
Wave speed: $c = \dfrac{\sqrt{K/\rho}}{\sqrt{1 + \dfrac{K}{E}\dfrac{D}{t}}} = \dfrac{1414.2}{\sqrt{1 + 0.01x}}$ (since $K/E = 0.01$).

Pressure rise: $\Delta p = \rho cV_0 = 1000(2)c = 2000c$.

Hoop stress increase (thin-walled pipe): $\Delta\sigma = \dfrac{\Delta p\,D}{2t} = \dfrac{\Delta p\,x}{2}$.

#### Solve for $D/t$
$$
\Delta\sigma = \frac{x}{2}\cdot\frac{2000(1414.2)}{\sqrt{1 + 0.01x}} = 20\times10^6 \;\Rightarrow\; \frac{x}{\sqrt{1 + 0.01x}} = \frac{40\times10^6}{2000(1414.2)} = 14.142
$$
Square both sides: $x^2 = 200(1 + 0.01x) = 200 + 2x$, so
$$
x^2 - 2x - 200 = 0 \;\Rightarrow\; x = 1 + \sqrt{201} = 15.18
$$
$$
\boxed{\frac{D}{t} \approx 15.2}
$$

#### Maximum time for rapid closure
Wave speed with this thickness:
$$
c = \frac{1414.2}{\sqrt{1 + 0.01(15.18)}} = 1317.7\ \text{m/s}
$$
(then $\Delta p = 1000(1317.7)(2) = 2.64$ MPa, and $\Delta\sigma = 20.0$ MPa, as required; the maximum pressure is $2.943 + 2.64 = 5.58$ MPa.)

Closure is rapid if its time $t_c \le \dfrac{2L}{c}$:
$$
t_c \le \frac{2(500)}{1317.7} = 0.759\ \text{s}
$$

**Answer:** $D/t \approx 15.2$ (wall thickness is about 1/15 of the diameter); closure is rapid for $t_c \le 0.76$ s.

## Check yourself

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

1. With K/E = 0.01 and x = D/t, how is c expressed?
   A. c = 1414.2 / sqrt(1 + 0.01x)
   B. c = 1414.2 x sqrt(1 + 0.01x)
   C. c = 1414.2 / (1 + 0.01x)
   D. c = 1414.2 x 0.01x
   Answer: A. Substituting K/E = 0.01 into c = sqrt(K/rho)/sqrt(1 + (K/E)(D/t)). (Key point: Wave speed in terms of D/t)

2. How is the hoop stress increase related to the pressure rise?
   A. delta sigma = delta p D / (2t) = delta p x / 2
   B. delta sigma = delta p t / D
   C. delta sigma = 2 delta p
   D. delta sigma = delta p x
   Answer: A. Thin-wall hoop stress is pD/2t, so with x = D/t it equals delta p x /2. (Key point: Hoop stress increase)

3. Solving x^2 - 2x - 200 = 0 gives which D/t?
   A. About 15.2
   B. About 14.1
   C. About 200
   D. About 7.6
   Answer: A. x = 1 + sqrt(201) = 15.18. (Key point: Resulting D/t)

4. What is the maximum rapid-closure time with c = 1317.7 m/s and L = 500 m?
   A. 0.76 s
   B. 0.38 s
   C. 1.5 s
   D. 7.6 s
   Answer: A. t_c <= 2L/c = 1000/1317.7 = 0.759 s. (Key point: Rapid closure limit)

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