---
title: "Model velocity and flow for oil pipeline (0.04 poise)"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:pipeline-oil-model-water-004"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipeline-oil-model-water-004
site: BCT Engineering
---

# Model velocity and flow for oil pipeline (0.04 poise)

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2072 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipeline-oil-model-water-004
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

A pipeline of 2 m diameter is to be designed to carry the oil at the rate 5 m³/s having sp.gr. 0.92 and viscosity $\mu$ = 0.04 poise. Tests were conducted using a pipe of 20 cm diameter and water as a liquid. Find the velocity and rate of flow required for the model pipe. Take $\mu$ (water) = 0.01 poise.

## Short answer

Using Reynolds model law, prototype velocity = 4Qp/(πDp²) = 1.59 m/s. Equating Re gives model velocity Vm = Vp (ρp/ρm)(Dp/Dm)(μm/μp) = 3.66 m/s. Model discharge = (πDm²/4) Vm = 0.115 m³/s.

## Answer

Flow in a pressurised pipe is dominated by viscous forces, so the **Reynolds model law** is used: $Re_m = Re_p$.

#### Given

Prototype: $D_p = 2$ m, $Q_p = 5\ \text{m}^3/\text{s}$, oil of s.g. 0.92, so $\rho_p = 920\ \text{kg/m}^3$, $\mu_p = 0.04$ poise $= 0.004\ \text{N s/m}^2$.

Model: $D_m = 0.2$ m, water $\rho_m = 1000\ \text{kg/m}^3$, $\mu_m = 0.01$ poise $= 0.001\ \text{N s/m}^2$.

#### Prototype velocity

$$
V_p = \frac{Q_p}{\frac\pi4D_p^2} = \frac{5}{\frac\pi4(2)^2} = 1.5915\ \text{m/s}
$$

#### Model velocity

$$
\frac{\rho_mV_mD_m}{\mu_m} = \frac{\rho_pV_pD_p}{\mu_p}
\;\Rightarrow\;
V_m = V_p\,\frac{\rho_p}{\rho_m}\,\frac{D_p}{D_m}\,\frac{\mu_m}{\mu_p}
$$

$$
V_m = 1.5915\times\frac{920}{1000}\times\frac{2}{0.2}\times\frac{0.001}{0.004} = 3.66\ \text{m/s}
$$

#### Model discharge

$$
Q_m = \frac\pi4D_m^2V_m = \frac\pi4(0.2)^2(3.6606) = 0.115\ \text{m}^3/\text{s}
$$

**Answer: model velocity $\approx 3.66$ m/s; model discharge $\approx 0.115\ \text{m}^3/\text{s}$ (115 L/s).**

## Check yourself

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

1. Which law applies?
   A. Froude law
   B. Reynolds law
   C. Weber law
   D. Euler law
   Answer: B. Pressurised pipe flow is viscous dominated. (Key point: Model law)

2. What is rho_p of the oil?
   A. 92 kg/m^3
   B. 1000 kg/m^3
   C. 920 kg/m^3
   D. 9200 kg/m^3
   Answer: C. s.g. 0.92 x 1000. (Key point: Density)

3. What is V_m?
   A. About 1.59 m/s
   B. About 36.6 m/s
   C. About 0.37 m/s
   D. About 3.66 m/s
   Answer: D. 1.5915 x 0.92 x 10 x 0.25 = 3.66. (Key point: Model velocity)

4. What is Q_m?
   A. About 0.115 m^3/s
   B. About 5 m^3/s
   C. About 0.0115 m^3/s
   D. About 1.15 m^3/s
   Answer: A. pi/4 x 0.04 x 3.66 = 0.115. (Key point: Model discharge)

## Related in Fluid Mechanics

- [Model velocity and flow for oil pipeline (0.042 poise)](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipeline-oil-model-water-0042) — Fluid Mechanics, chapter 10
- [Model liquid viscosity, Froude and Reynolds](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Amodel-liquid-viscosity-froude-reynolds) — Fluid Mechanics, chapter 10

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