---
title: "Laminar oil flow in a pipe"
subject: "Fluid Mechanics"
chapter: 6
key: "semester_4/fluid_mechanics_mechanical:laminar-oil-pipe-pressure-drop"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/laminar-oil-pipe-pressure-drop
site: BCT Engineering
---

# Laminar oil flow in a pipe

Fluid Mechanics · Chapter 6: Viscous Effects

- Asked in: year not given, 8 marks
- Video: [Fluid Mechanics example on oil flow in pipes](https://www.youtube.com/watch?v=_b08n9Shh-c) — AKMAL DANIAL BIN CHE MAHMUD STUDENT
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/laminar-oil-pipe-pressure-drop
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-6

## Question

Oil of specific gravity 0.85 and dynamic viscosity 0.08 Pa s flows through a 50 mm diameter, 100 m long horizontal pipe at a rate of 1.5 litres per second. Show that the flow is laminar and find (a) the maximum velocity, (b) the pressure drop and head loss, (c) the wall shear stress, (d) the friction factor, (e) the radius at which the local velocity equals the mean velocity, (f) the velocity at 15 mm from the axis, and (g) the power needed to overcome the friction.

## Short answer

Re = 406 < 2000, laminar. u_max = 2V = 1.53 m/s. Δp = 32μVL/D^2 = 78.2 kPa, h_f = 9.38 m. τ_w = Δp D/(4L) = 9.78 Pa. f = 64/Re = 0.158. r where u=V: r = R/√2 = 17.7 mm. u(15 mm) = 0.978 m/s. Power = QΔp = 117 W.

## Answer

#### Given data
$\rho = 850\ \text{kg/m}^3$, $\mu = 0.08\ \text{Pa s}$, $D = 0.05\ \text{m}$, $R = 0.025\ \text{m}$, $L = 100\ \text{m}$, $Q = 1.5\times10^{-3}\ \text{m}^3/\text{s}$.

#### Mean velocity and Reynolds number
$$
\begin{aligned}
A &= \frac{\pi}{4}(0.05)^2 = 1.9635\times10^{-3}\ \text{m}^2 \\
\bar V &= \frac{Q}{A} = \frac{1.5\times10^{-3}}{1.9635\times10^{-3}} = 0.764\ \text{m/s} \\
Re &= \frac{\rho \bar V D}{\mu} = \frac{850 \times 0.764 \times 0.05}{0.08} = 406
\end{aligned}
$$

$Re = 406 < 2000$, so the flow is **laminar**.

#### (a) Maximum velocity
$$
u_{max} = 2\bar V = 2 \times 0.764 = 1.528\ \text{m/s}
$$

#### (b) Pressure drop and head loss
From Hagen-Poiseuille, $\Delta p = \dfrac{32\mu \bar V L}{D^2}$:

$$
\Delta p = \frac{32 \times 0.08 \times 0.764 \times 100}{(0.05)^2} = 78\,228\ \text{Pa} \approx 78.2\ \text{kPa}
$$

$$
h_f = \frac{\Delta p}{\rho g} = \frac{78\,228}{850 \times 9.81} = 9.38\ \text{m of oil}
$$

#### (c) Wall shear stress
$$
\tau_w = \frac{\Delta p D}{4L} = \frac{78\,228 \times 0.05}{4 \times 100} = 9.78\ \text{Pa}
$$

#### (d) Friction factor
$$
f = \frac{64}{Re} = \frac{64}{406} = 0.158
$$

Check: $h_f = f\dfrac{L}{D}\dfrac{\bar V^2}{2g} = 0.158 \times 2000 \times 0.02975 = 9.38\ \text{m}$. This agrees.

#### (e) Radius where $u = \bar V$
Since $u = u_{max}(1 - r^2/R^2) = \bar V = u_{max}/2$:

$$
r = \frac{R}{\sqrt2} = \frac{25}{1.414} = 17.7\ \text{mm}
$$

#### (f) Velocity at $r = 15\ \text{mm}$
$$
u = 1.528\left[1 - \left(\frac{15}{25}\right)^2\right] = 1.528 \times 0.64 = 0.978\ \text{m/s}
$$

#### (g) Power to overcome friction
$$
P = Q\,\Delta p = 1.5\times10^{-3} \times 78\,228 = 117.3\ \text{W}
$$

**Answer: $Re = 406$ (laminar); $u_{max} = 1.53\ \text{m/s}$; $\Delta p = 78.2\ \text{kPa}$ ($h_f = 9.38\ \text{m}$); $\tau_w = 9.78\ \text{Pa}$; $f = 0.158$; $r = 17.7\ \text{mm}$; $u = 0.978\ \text{m/s}$; $P = 117\ \text{W}$**

## Check yourself

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

1. Q = 1.5 L/s in a 50 mm pipe gives V = 0.764 m/s. How is laminar flow shown?
   A. Re = rho V D/mu = 406, below 2000
   B. V is below 1 m/s
   C. Re = 4000 exactly
   D. f is below 0.2
   Answer: A. Re = 850 x 0.764 x 0.05 / 0.08 = 406 < 2000, so the flow is laminar. (Key point: Laminar flow check)

2. Which formula gives the pressure drop for the laminar pipe flow?
   A. Delta p = 128 mu V L / D^2
   B. Delta p = 8 mu V L / D^2
   C. Delta p = 32 mu V L / D^2
   D. Delta p = 64 mu V L / D
   Answer: C. Delta p = 32 mu V_bar L/D^2 = 78.2 kPa, giving h_f = 9.38 m of oil. (Key point: Pressure drop)

3. At what radius does u equal the mean velocity?
   A. R/2
   B. R/sqrt(2) = 17.7 mm
   C. R/4
   D. R = 25 mm
   Answer: B. u = u_max(1 - r^2/R^2) = u_max/2 gives r^2 = R^2/2, so r = R/sqrt2. (Key point: Radius where local velocity equals mean)

## Related in Fluid Mechanics

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