---
title: "Rate of level change in standpipes"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:standpipes-rate-of-level-change"
source: https://www.bctengineer.com/q/semester_4/hydraulics/standpipes-rate-of-level-change
site: BCT Engineering
---

# Rate of level change in standpipes

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2072 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/standpipes-rate-of-level-change
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Water flows by gravity in two open stand pipes shown in figure. Estimate the rate of change of water level in left standpipe. [Figure: two open standpipes each of diameter D = 0.75 m connected at the bottom by a pipe of d = 75 mm, e = 0.3 mm, L = 4 m; the water level in the left standpipe is 2.5 m above that in the right]

## Short answer

Head loss equation (K_entrance+ fL/d+1) V²/2g=2.5 m gives V=4.02 m/s, Q=0.01776 m³/s. With standpipe area 0.4418 m², left level falls at 0.0402 m/s (≈40 mm/s).

## Answer

**Data.** Two open standpipes, $D = 0.75$ m, joined at the bottom by a pipe $d = 75$ mm, $L = 4$ m, $e = 0.3$ mm. At the instant considered the water in the left standpipe is $H = 2.5$ m above that in the right. Water at 20 °C ($\nu = 10^{-6}$ m²/s).

#### Velocity in the connecting pipe
The head difference is used up in the entrance loss (0.5), friction, and the exit loss (1.0) at the other standpipe:
$$
H = \left(0.5 + f\frac{L}{d} + 1.0\right)\frac{V^2}{2g}, \qquad \frac{e}{d} = 0.0040
$$
$f$ depends on $Re$ (Colebrook-White), so iterate: assume $f \approx 0.029$, $V \approx 4$ m/s, $Re \approx 3\times10^5$; the Colebrook equation then gives
$$
f = 0.0288,\qquad f\frac{L}{d} = 1.535,\qquad 2.5 = 3.035\,\frac{V^2}{2g} \;\Rightarrow\; V = 4.020\ \text{m/s},\quad Re = 301490
$$

#### Discharge
$$
Q = \frac{\pi}{4}d^2V = \frac{\pi}{4}(0.075)^2(4.020) = 0.01776\ \text{m}^3/\text{s} = 17.76\ \text{l/s}
$$

#### Rate of change of level
Area of the standpipe $A_s = \dfrac{\pi}{4}(0.75)^2 = 0.4418$ m². The left standpipe loses water at the rate $Q$, so
$$
\frac{dh}{dt} = -\frac{Q}{A_s} = -\frac{0.01776}{0.4418} = -0.0402\ \text{m/s} = -40.2\ \text{mm/s}
$$

**Answer:** The water level in the left standpipe falls at about 0.040 m/s ($\approx 40$ mm/s) at this instant, while the right one rises at the same rate; the head difference therefore decreases by twice this rate, 0.080 m/s, and the flow slows down gradually.

## Check yourself

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

1. What does the 2.5 m head difference supply in the connecting pipe?
   A. Friction only
   B. Entrance loss 0.5, friction and exit loss 1.0, all times V^2/2g
   C. Exit loss only
   D. Static lift
   Answer: B. H = (0.5 + fL/d + 1.0) V^2/2g. (Key point: Losses balancing head)

2. How is f obtained?
   A. Colebrook-White with e/d = 0.004 and Re by iteration
   B. f = 64/Re
   C. f = 0.02
   D. Not needed
   Answer: A. f depends on Re (which depends on V), so it is iterated; f is about 0.0288. (Key point: Friction factor)

3. How is the rate of fall in the left standpipe found?
   A. dh/dt = -Q/A_s with A_s the standpipe area
   B. dh/dt = -V
   C. dh/dt = -Q x A_s
   D. dh/dt = -Q/a pipe area
   Answer: A. The pipe discharge Q leaves a tank of area A_s, so dh/dt = -Q/A_s. (Key point: Level change rate)

4. How fast does the head difference between the standpipes change?
   A. At the same rate as the left level
   B. Twice the rate (left falls, right rises)
   C. Zero
   D. Half the rate
   Answer: B. The left level falls and the right rises at the same rate, so the difference changes at double the rate. (Key point: Head difference change)

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