---
title: "Head loss and flow direction in tapering oil pipe"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:tapering-pipe-oil-flow"
source: https://www.bctengineer.com/q/semester_6/hydro_power/tapering-pipe-oil-flow
site: BCT Engineering
---

# Head loss and flow direction in tapering oil pipe

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2070 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/tapering-pipe-oil-flow
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A pipe line (AB) conveying oil (specific gravity = 0.90) has diameter [?] ... [?] and the discharge is [?] litres per second. Find the loss of head and the direction of flow. [Figure: tapering pipe AB rising from the smaller end A to the larger end B; the centre line at B is 5 m above A.] (Most of the data in this question is unreadable in the scan.)

## Short answer

Flow from A to B since total energy at A (13.177 m) exceeds B (11.720 m). Head loss = 1.457 m of oil.

## Answer

Most of the data are unreadable in the scan, so the standard textbook values for this problem are assumed: $d_A = 200$ mm, $d_B = 500$ mm, $p_A = 9.81$ N/cm² (98.1 kPa), $p_B = 5.886$ N/cm² (58.86 kPa), $Q = 200$ L/s, oil S = 0.90, B is 5 m above A. The method is the same for any values.

**Principle:** flow always goes from the section of higher total energy to the section of lower total energy, and the difference is the head loss.

```text
                         B  (d = 500 mm, z = 5 m)
                      ___/
          ________---
 A ___---            flow ?
 (d = 200 mm, z = 0)
```

#### Velocities
$$
\begin{aligned}
V_A &= \frac{0.2}{\frac{\pi}{4}(0.2)^2} = 6.366\ \text{m/s} \\
V_B &= \frac{0.2}{\frac{\pi}{4}(0.5)^2} = 1.019\ \text{m/s}
\end{aligned}
$$

#### Total energy (take datum through A, $\rho g = 0.9 \times 9810 = 8829$ N/m³)
$$
\begin{aligned}
E_A &= \frac{p_A}{\rho g} + \frac{V_A^2}{2g} + z_A = \frac{98100}{8829} + \frac{6.366^2}{19.62} + 0 \\
&= 11.111 + 2.066 + 0 = 13.177\ \text{m} \\
E_B &= \frac{58860}{8829} + \frac{1.019^2}{19.62} + 5 \\
&= 6.667 + 0.053 + 5 = 11.720\ \text{m}
\end{aligned}
$$

#### Result
Since $E_A > E_B$, flow is **from A to B** (upwards, from the smaller to the larger end).
$$
h_L = E_A - E_B = 13.177 - 11.720 = 1.457\ \text{m of oil}
$$

**Answer (with assumed data):** Flow from A to B; loss of head ≈ 1.46 m of oil. If $E_B$ had come out larger, the flow would be from B to A.

## Check yourself

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

1. How is the direction of flow decided in a pipe between A and B?
   A. Flow goes from the smaller diameter to the larger
   B. Flow goes from the section with higher total energy to lower total energy
   C. Flow goes from the lower elevation to the higher
   D. Flow goes from higher velocity to lower velocity
   Answer: B. Flow is always from higher total energy to lower; the difference is the head loss. (Key point: Direction of flow)

2. Which terms make up total energy at a section (in head of oil)?
   A. Pressure head + elevation only
   B. Velocity head + elevation only
   C. Pressure head + velocity head + elevation head
   D. Pressure + velocity + discharge
   Answer: C. E = p/rho g + V^2/2g + z, with the pressure head expressed using the oil's density (0.9 x 9810). (Key point: Total energy)

3. How are VA and VB found from Q = 200 L/s?
   A. V = Q x area
   B. V = Q / (pi/4 d^2) at each section, giving 6.37 and 1.02 m/s
   C. V = sqrt(2gh) at each section
   D. Both velocities are equal because Q is the same
   Answer: B. Continuity gives V = Q/A, larger at the small end A (6.366 m/s) and smaller at B (1.019 m/s). (Key point: Velocities)

4. With EA = 13.177 m and EB = 11.720 m, what is the result?
   A. Flow B to A with loss 1.46 m
   B. Flow A to B with loss 1.46 m of oil
   C. Flow A to B with loss 24.9 m
   D. No flow, as B is higher
   Answer: B. EA > EB so flow is A to B, and hL = EA - EB = 1.457 m of oil, even though B is 5 m higher. (Key point: Head loss)

## Related in Hydro Power

- [Discharge and head loss in horizontal pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-discharge-head-loss) — Hydro Power, chapter 1
- [Head loss with enlarged central pipe length](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-replaced-central-length) — Hydro Power, chapter 1
- [Flow between reservoirs (20 cm and 30 cm pipe)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Areservoirs-pipeline-20-30cm) — Hydro Power, chapter 1
- [Head loss at sudden enlargement](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asudden-enlargement-loss) — Hydro Power, chapter 1
- [Specific gravity, weight and Newtonian fluids](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Anewtonian-non-newtonian-fluids) — Hydro Power, chapter 1
- [Differential manometer between CCl4 and oil pipes](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Amanometer-ccl4-oil-pipes) — Hydro Power, chapter 1

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- [Viscosity of oil from laminar pipe flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Ahorizontal-oil-pipe-laminar-viscosity) — Hydraulics, chapter 1
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