---
title: "Laminar oil flow in inclined pipe"
subject: "Hydraulics"
chapter: 1
key: "semester_4/hydraulics:laminar-inclined-oil-upward"
source: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-inclined-oil-upward
site: BCT Engineering
---

# Laminar oil flow in inclined pipe

Hydraulics · Chapter 1: Pipe flow

- Asked in: 2082 Kartik, 2+4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/laminar-inclined-oil-upward
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1

## Question

An oil with density 900 kg/m³ and kinematic viscosity 0.0002 m²/s flows through an inclined pipe as shown in figure. Assuming steady laminar flow, verify that the flow is upward and find the flow rate. [Figure: inclined pipe of diameter d = 6 cm and length 10 m, inclined at 40° to the horizontal; section 1 at the lower end with $p_1$ = 350,000 Pa and $z_1$ = 0; section 2 at the upper end with $p_2$ = 250,000 Pa]

## Short answer

Flow direction is from higher to lower piezometric pressure p+rho g z; since p1* > p2*, flow is upward. Pressure difference is 43,248 Pa. Hagen-Poiseuille equation yields Q = 0.007643 m^3/s (7.64 L/s). Re approx 811 confirms laminar flow.

## Answer

**Given.** $\rho = 900$ kg/m³, $\nu = 0.0002$ m²/s so $\mu = \rho\nu = 0.18$ Pa s; $d = 0.06$ m, $L = 10$ m, inclination 40°. At section 1: $p_1 = 350\,000$ Pa, $z_1 = 0$. At section 2: $p_2 = 250\,000$ Pa, $z_2 = L\sin 40^\circ = 6.428$ m.

#### Direction of flow
Flow goes from high to low **piezometric** pressure $p^* = p + \rho g z$, not from high to low $p$.
$$
p_1^* = 350\,000 + 0 = 350\,000\ \text{Pa}, \qquad p_2^* = 250\,000 + 900(9.81)(6.428) = 306752\ \text{Pa}
$$
Since $p_1^* > p_2^*$, the flow is from 1 to 2, i.e. **upward**, as required. The driving pressure drop is
$$
\Delta p^* = p_1^* - p_2^* = 43248\ \text{Pa}
$$

#### Flow rate (Hagen-Poiseuille, laminar)
$$
Q = \frac{\pi d^4\,\Delta p^*}{128\,\mu L} = \frac{\pi (0.06)^4 (43248)}{128(0.18)(10)} = 0.007643\ \text{m}^3/\text{s} = 7.64\ \text{l/s}
$$

#### Check that the flow is laminar
$$
V = \frac{Q}{\pi d^2/4} = 2.703\ \text{m/s},\qquad Re = \frac{Vd}{\nu} = \frac{2.703(0.06)}{0.0002} = 811 < 2000
$$
So the laminar assumption is valid.

**Answer:** The flow is upward; $Q = 0.007643$ m³/s ($7.64$ l/s), $Re \approx 811$.

## Check yourself

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

1. How is the direction of flow in an inclined pipe decided?
   A. From high p to low p
   B. From high piezometric pressure p + rho g z to low
   C. From low elevation to high elevation always
   D. From high velocity to low velocity
   Answer: B. Fluid moves from high to low piezometric pressure; here p1* = 350000 Pa exceeds p2* = 306752 Pa, so flow is upward. (Key point: Flow direction)

2. What pressure drop is used in the Hagen-Poiseuille formula here?
   A. p1 - p2 = 100000 Pa
   B. Piezometric drop of 43248 Pa
   C. rho g z2 = 56752 Pa
   D. p1 only
   Answer: B. Q = pi d^4 dp* / (128 mu L) uses the piezometric drop, which includes the elevation effect. (Key point: Driving pressure)

3. Why is Re = 811 computed after finding Q?
   A. To confirm Re < 2000 so the laminar assumption is valid
   B. To find the viscosity
   C. To compute the elevation
   D. To get the friction factor for turbulence
   Answer: A. Hagen-Poiseuille holds only for laminar flow, so the result is checked against Re < 2000. (Key point: Laminar check)

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