---
title: "Streamlines from stream function"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:stream-function-6ax-8ay"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stream-function-6ax-8ay
site: BCT Engineering
---

# Streamlines from stream function

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2076 Chaitra, 2+2+2 marks
- Video: [Introductory Fluid Mechanics L13 p1 - Stream Function - 2D Incompressible Flow](https://www.youtube.com/watch?v=kDO3EcXblwg) — Ron Hugo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stream-function-6ax-8ay
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

An incompressible, frictionless flow specified by $\psi = -6Ax - 8Ay$; x, y in meters, A = 1 m/s. Find (i) sketch streamlines $\psi = 0$ and $\psi = 8$ m²/s (ii) velocity vector at (0, 0) and its direction. (iii) flow rate between streamlines passing through points (1, 1) and (4, 1).

## Short answer

Streamlines: ψ=0: y=-0.75x; ψ=8: y=-0.75x−1. Velocity V = -8i+6j m/s, 10 m/s at 143.1°. Flow rate between (1,1) and (4,1) is 18 m²/s per unit width.

## Answer

**Given:** $\psi = -6Ax - 8Ay$ with $A = 1$ m/s, so $\psi = -6x - 8y$. Using $u = \partial\psi/\partial y$ and $v = -\partial\psi/\partial x$.

#### (i) Streamlines

- $\psi = 0$: $-6x - 8y = 0\Rightarrow y = -0.75x$
- $\psi = 8$: $-6x - 8y = 8\Rightarrow y = -0.75x - 1$

Both are parallel straight lines of slope $-0.75$ ($-36.9^\circ$ to the x-axis). The second line cuts the y-axis at $y = -1$ m and the x-axis at $x = -1.333$ m.

```text
 y
   |   \  
   |    \ psi=0  (y = -0.75x)
   |     \
---+------\------------ x
   |\      \
   | \ psi=8  (y = -0.75x - 1)
```

#### (ii) Velocity at (0, 0)

$$
u = \frac{\partial\psi}{\partial y} = -8A = -8\ \text{m/s},\qquad v = -\frac{\partial\psi}{\partial x} = 6A = 6\ \text{m/s}
$$

$$
|V| = \sqrt{8^2 + 6^2} = 10\ \text{m/s},\qquad \theta = \tan^{-1}\frac{6}{-8} = 143.1^\circ\ \text{from +x}
$$

The velocity is uniform everywhere (the same at every point) and parallel to the streamlines.

#### (iii) Flow rate between (1, 1) and (4, 1)

$$
\psi_{(1,1)} = -6 - 8 = -14\ \text{m}^2/\text{s},\qquad \psi_{(4,1)} = -24 - 8 = -32\ \text{m}^2/\text{s}
$$

$$
q = |\psi_2 - \psi_1| = |-32 - (-14)| = 18\ \text{m}^2/\text{s per metre depth}
$$

**Answer: (i) $y = -0.75x$ and $y = -0.75x - 1$; (ii) $\vec V = (-8\hat i + 6\hat j)$ m/s, $10$ m/s at $143.1^\circ$ to +x; (iii) $q = 18$ m²/s per unit width.**

## Check yourself

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

1. What is the velocity at (0,0) for psi = -6x - 8y?
   A. (-6, -8) m/s, 10 m/s
   B. (-8, 6) m/s, 10 m/s at 143.1 degrees
   C. (8, -6) m/s, 10 m/s
   D. (6, 8) m/s
   Answer: B. u = dpsi/dy = -8, v = -dpsi/dx = 6; magnitude 10, angle atan2(6,-8) = 143.1 degrees. (Key point: Velocity at origin)

2. What do the streamlines psi = 0 and psi = 8 look like?
   A. Concentric circles
   B. Parabolas
   C. Parallel straight lines of slope -0.75
   D. Lines through the origin with different slopes
   Answer: C. psi = const gives y = -0.75x - psi/8, parallel lines. (Key point: Streamline shape)

3. How is the flow rate between two points found, and its value between (1,1) and (4,1)?
   A. psi1 + psi2 = 46 m^2/s
   B. psi2 x psi1 = 448 m^2/s
   C. |u| x distance = 30 m^2/s
   D. |psi2 - psi1| = |-32 - (-14)| = 18 m^2/s
   Answer: D. Flow between streamlines equals the difference of stream function values. (Key point: Flow rate)

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