---
title: "Model velocity and flow for oil pipeline (0.042 poise)"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:pipeline-oil-model-water-0042"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pipeline-oil-model-water-0042
site: BCT Engineering
---

# Model velocity and flow for oil pipeline (0.042 poise)

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2073 Shrawan, 3 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-0042
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

A pipe line of 2 m diameter is to be designed to carry the oil at the rate of 5 m³/s with specific gravity 0.8 and viscosity of 0.042 poise. Test were conducted using a pipe of 20 cm diameter with water having viscosity of 0.01 poise. Calculate the velocity and rate of flow required for model.

## Short answer

Reynolds similarity applied. Model velocity ≈ 3.03 m/s, model discharge ≈ 0.0952 m³/s (95.2 L/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 with $\rho_p = 800\ \text{kg/m}^3$ (s.g. 0.8), $\mu_p = 0.042$ poise $= 0.0042\ \text{N s/m}^2$.

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

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

$$
V_m = 1.5915\times\frac{800}{1000}\times\frac{2}{0.2}\times\frac{0.001}{0.0042} = 3.03\ \text{m/s}
$$

#### Model discharge

$$
Q_m = \frac\pi4D_m^2V_m = \frac\pi4(0.2)^2(3.0315) = 0.0952\ \text{m}^3/\text{s}
$$

**Answer: model velocity $\approx 3.03$ m/s; model discharge $\approx 0.0952\ \text{m}^3/\text{s}$ (95.2 L/s).**

## Check yourself

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

1. Which law is used for pressurised pipe flow?
   A. Froude law
   B. Weber law
   C. Reynolds law
   D. Mach law
   Answer: C. Viscous forces dominate. (Key point: Model law)

2. How is V_p found?
   A. Q x D
   B. Q/D
   C. Q/D^2
   D. Q/(pi D^2/4) = 1.59 m/s
   Answer: D. 5/(pi/4 x 4) = 1.5915 m/s. (Key point: Prototype velocity)

3. What is V_m?
   A. About 3.03 m/s
   B. About 1.59 m/s
   C. About 30.3 m/s
   D. About 0.3 m/s
   Answer: A. Reynolds equality gives 1.5915 x 0.8 x 10 x 0.2381 = 3.03. (Key point: Model velocity)

## Related in Fluid Mechanics

- [Model velocity and flow for oil pipeline (0.04 poise)](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Apipeline-oil-model-water-004) — Fluid Mechanics, chapter 10

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