---
title: "Viscosity of oil from laminar pipe flow"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:horizontal-oil-pipe-laminar-viscosity"
source: https://www.bctengineer.com/q/semester_4/hydraulics/horizontal-oil-pipe-laminar-viscosity
site: BCT Engineering
---

# Viscosity of oil from laminar pipe flow

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2072 Asoj, 4+2+2 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/horizontal-oil-pipe-laminar-viscosity
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

A horizontal pipe 60 mm in diameter carries oil of specific gravity 0.8. The pressure difference between two sections 5 km apart is found to be 200 kPa. The oil flowing through the pipe is collected in a tank. It is found that 1962 N of oil is collected in 4 minutes. Compute the dynamic viscosity of the oil. Assume the flow to be laminar and verify it. Also, find the velocity at a distance of 20 mm from the pipe wall.

## Short answer

Applying Hagen-Poiseuille equation from the measured pressure drop (200 kPa over 5 km) and flow rate (0.001042 m³/s) gives dynamic viscosity μ=0.0122 Pa·s. The Reynolds number is 1448, confirming laminar flow. The velocity 20 mm from the wall is 0.655 m/s.

## Answer

**Data.** $D = 0.06$ m ($R = 0.03$ m), s.g. 0.8 so $\rho = 800$ kg/m³, $\Delta p = 200$ kPa over $L = 5$ km. Weight collected: 1962 N in 4 min = 240 s.

#### Discharge
Weight flow rate $= 1962/240 = 8.175$ N/s, so
$$
Q = \frac{W}{\rho g} = \frac{8.175}{800(9.81)} = 0.001042\ \text{m}^3/\text{s} = 1.042\ \text{l/s}
$$

#### Dynamic viscosity (laminar flow, Hagen-Poiseuille)
$$
\Delta p = \frac{128\mu L Q}{\pi D^4} \;\Rightarrow\; \mu = \frac{\pi D^4\,\Delta p}{128 L Q} = \frac{\pi(0.06)^4(200\,000)}{128(5000)(0.001042)} = 0.01221\ \text{Pa s}
$$

#### Verification that the flow is laminar
$$
V = \frac{Q}{\pi D^2/4} = 0.3684\ \text{m/s},\qquad Re = \frac{\rho VD}{\mu} = \frac{800(0.3684)(0.06)}{0.01221} = 1448 < 2000
$$
So the flow is laminar and the assumption is correct.

#### Velocity at 20 mm from the wall
Radius at that point: $r = R - 0.020 = 0.010$ m. For laminar flow
$$
u = \frac{1}{4\mu}\frac{\Delta p}{L}\left(R^2 - r^2\right) = \frac{1}{4(0.01221)}\left(\frac{200\,000}{5000}\right)\left(0.03^2 - 0.01^2\right) = 0.655\ \text{m/s}
$$
(The centre-line velocity is 0.737 m/s, twice the mean velocity.)

**Answer:** $\mu = 0.0122$ Pa s; $Re \approx 1448$ (laminar); $u$ at 20 mm from the wall $= 0.655$ m/s.

## Check yourself

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

1. How is the discharge found from 1962 N of oil collected in 4 minutes?
   A. Q = 1962/(4 x 60) directly in m3/s
   B. Q = W/(rho g) where W = 1962/240 N/s
   C. Q = 1962 x 240 / (rho g)
   D. Q = 1962/(rho x 4) using 4 minutes as seconds
   Answer: B. Weight rate 8.175 N/s divided by the specific weight rho g = 800 x 9.81 gives 0.001042 m3/s. (Key point: Discharge from weight collected)

2. Which equation is used to obtain mu for laminar flow?
   A. mu = rho V D / Re with Re = 2000
   B. mu = tau0 / (du/dy) using a turbulent profile
   C. mu = pi D^4 dp / (128 L Q)
   D. mu = 64 / Re
   Answer: C. For laminar flow dp = 128 mu L Q/(pi D^4), which rearranges to mu = pi D^4 dp/(128 L Q) = 0.0122 Pa s. (Key point: Hagen-Poiseuille for viscosity)

3. How is the laminar assumption verified after finding mu?
   A. Check that the velocity is under 1 m/s
   B. Check that the pressure drop is below 200 kPa
   C. Check that mu is below 0.1 Pa s
   D. Compute Re = rho V D / mu = 1448 and confirm it is below 2000
   Answer: D. Laminar flow in a pipe requires Re below about 2000, and 1448 satisfies this. (Key point: Verifying laminar flow)

4. Why is the velocity at 20 mm from the wall found using r = 0.010 m?
   A. Because the profile u = (dp/4 mu L)(R^2 - r^2) uses the distance r from the axis, r = R - 0.020
   B. Because the distance from the wall is used directly as r
   C. Because the velocity is uniform across the section
   D. Because r is the diameter minus 20 mm
   Answer: A. The parabolic profile is written in terms of the radius r from the axis, so r = 0.03 - 0.02 = 0.01 m, giving 0.655 m/s. (Key point: Laminar velocity profile)

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