---
title: "Stream function and velocity potential problem"
subject: "Fluid Mechanics"
chapter: 3
key: "semester_4/fluid_mechanics_mechanical:stream-function-potential-check"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/stream-function-potential-check
site: BCT Engineering
---

# Stream function and velocity potential problem

Fluid Mechanics · Chapter 3: Kinematics of Fluid Flow

- Asked in: year not given, 8 marks
- Video: [Problem No. 1 on Velocity Potential Function and Streamline Function - Fluid Mechanics 1](https://www.youtube.com/watch?v=yLb9tjUXifk) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/stream-function-potential-check
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-3

## Question

The stream function of a two-dimensional incompressible flow is $\psi = 3x^2y - y^3$ (in m$^2$/s, with $x$, $y$ in m). (a) Derive the velocity components and verify that the flow satisfies continuity. (b) Determine whether the flow is irrotational. (c) Find the velocity potential function. (d) Find the velocity and resultant speed at the point (1 m, 2 m) and the discharge between the points (1, 1) and (2, 1).

## Short answer

Velocity components: u = 3x² – 3y², v = –6xy, satisfying continuity. Vorticity is zero, so flow is irrotational. Velocity potential φ = x³ – 3xy². At (1 m, 2 m) resultant speed is 15 m/s, and discharge between (1,1) and (2,1) is ψ₂–ψ₁ = 9 m²/s per unit depth.

## Answer

#### Velocity components
Using $u = \dfrac{\partial\psi}{\partial y}$ and $v = -\dfrac{\partial\psi}{\partial x}$:

$$
\begin{aligned}
u &= 3x^2 - 3y^2 \\
v &= -6xy
\end{aligned}
$$

#### (a) Continuity
$$
\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x - 6x = 0
$$

Continuity is satisfied (as it always is when a stream function exists).

#### (b) Irrotationality
$$
\zeta_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = -6y - (-6y) = 0
$$

The vorticity is zero, so the flow is **irrotational**.

#### (c) Velocity potential
Since $u = \partial\phi/\partial x$:

$$
\phi = \int (3x^2 - 3y^2)\,dx = x^3 - 3xy^2 + f(y)
$$

Then $\partial\phi/\partial y = -6xy + f'(y)$. This must equal $v = -6xy$, so $f'(y) = 0$. Taking the constant as zero,

$$
\phi = x^3 - 3xy^2
$$

#### (d) At the point (1, 2)
$$
\begin{aligned}
u &= 3(1) - 3(4) = -9\ \text{m/s} \\
v &= -6(1)(2) = -12\ \text{m/s} \\
V &= \sqrt{9^2 + 12^2} = 15\ \text{m/s}
\end{aligned}
$$

#### Discharge between (1, 1) and (2, 1)
$$
\begin{aligned}
\psi_1 &= 3(1)(1) - 1 = 2 \\
\psi_2 &= 3(4)(1) - 1 = 11 \\
q &= \psi_2 - \psi_1 = 9\ \text{m}^2/\text{s per unit depth}
\end{aligned}
$$

**Answer: $u = 3x^2 - 3y^2$, $v = -6xy$; irrotational; $\phi = x^3 - 3xy^2$; at (1, 2) $V = 15\ \text{m/s}$; $q = 9\ \text{m}^2/\text{s}$**

## Check yourself

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

1. For psi = 3x^2 y - y^3, with u = dpsi/dy and v = -dpsi/dx, the velocity components are:
   A. u = 6xy, v = 3x^2 - 3y^2
   B. u = 3x^2 - 3y^2, v = -6xy
   C. u = -6xy, v = 3x^2 - 3y^2
   D. u = 3x^2 + 3y^2, v = 6xy
   Answer: B. dpsi/dy = 3x^2 - 3y^2 and -dpsi/dx = -6xy. (Key point: Velocity from stream function)

2. The flow satisfies continuity and is irrotational because:
   A. du/dx + dv/dy = 6x - 6x = 0 and dv/dx - du/dy = -6y + 6y = 0
   B. du/dx = dv/dy
   C. u and v are both zero at the origin
   D. du/dx - dv/dy = 12x
   Answer: A. du/dx = 6x and dv/dy = -6x sum to zero; dv/dx = -6y equals du/dy = -6y. (Key point: Continuity and irrotationality checks)

3. The velocity potential phi (u = dphi/dx, v = dphi/dy) is:
   A. x^3 + 3xy^2
   B. x^3 - 3xy^2
   C. 3x^2 y - y^3
   D. 3x^2 - 3y^2
   Answer: B. Integrating u gives x^3 - 3xy^2 + f(y); matching dphi/dy = -6xy gives f'(y) = 0. (Key point: Velocity potential)

4. The discharge per unit depth between (1,1) and (2,1) is found by:
   A. adding psi at the two points
   B. the difference of psi values, 11 - 2 = 9 m^2/s
   C. multiplying u by distance
   D. the difference of phi values
   Answer: B. Flow between two points equals the difference in stream function: psi2 - psi1 = 11 - 2 = 9 m^2/s. (Key point: Discharge from stream function)

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