---
title: "Streamline, $\\psi$ and $\\phi$ for given field"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:streamline-psi-phi-2x3-6x2y"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/streamline-psi-phi-2x3-6x2y
site: BCT Engineering
---

# Streamline, $\psi$ and $\phi$ for given field

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2073 Shrawan, 3+3 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_3/fluid_mechanics/streamline-psi-phi-2x3-6x2y
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

Velocity vector of flow field is given by $\vec V = 2x^3\hat i - 6x^2 y\hat j$. Determine the equation of stream line. Also determine expression of $\psi$ and $\phi$.

## Short answer

Streamline equation is x^3 y = C. Stream function ψ = 2x^3 y. Velocity potential φ does not exist because the flow is rotational (vorticity ω_z = -6xy).

## Answer

**Given:** $u = 2x^3$, $v = -6x^2y$.

**Check of continuity**

$$
\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} = 6x^2 - 6x^2 = 0
$$

so the flow is incompressible and a stream function exists.

#### Equation of streamline

$$
\frac{dy}{dx} = \frac{v}{u} = \frac{-6x^2y}{2x^3} = -\frac{3y}{x}
$$

$$
\frac{dy}{y} = -3\frac{dx}{x}\;\Rightarrow\; \ln y = -3\ln x + \ln C\;\Rightarrow\; x^3y = C
$$

#### Stream function ψ

$u = \partial\psi/\partial y = 2x^3\Rightarrow\psi = 2x^3y + f(x)$.

$v = -\partial\psi/\partial x = -6x^2y - f'(x)$. Comparing with $v = -6x^2y$ gives $f'(x) = 0$.

$$
\psi = 2x^3y\ (+\text{constant})
$$

This agrees with the streamline $x^3y = C$.

#### Velocity potential φ

A velocity potential exists only for irrotational flow, i.e. when $\partial v/\partial x = \partial u/\partial y$.

$$
\frac{\partial v}{\partial x} = -12xy,\qquad \frac{\partial u}{\partial y} = 0
$$

The vorticity $\omega_z = \tfrac12(\partial v/\partial x - \partial u/\partial y) = -6xy \ne 0$, so the flow is rotational.

**Answer: streamline $x^3y = C$; $\psi = 2x^3y$; $\phi$ does not exist (flow is rotational, $\omega_z = -6xy$).**

## Check yourself

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

1. Why check du/dx + dv/dy = 0 first?
   A. It shows the flow is incompressible so psi exists
   B. It shows the flow is irrotational
   C. It gives phi
   D. It gives the pressure
   Answer: A. A stream function exists only for continuity-satisfying flow; 6x^2 - 6x^2 = 0. (Key point: Continuity check)

2. What is the streamline equation?
   A. x y^3 = C
   B. x^3 y = C
   C. x^2 y = C
   D. y = Cx^3
   Answer: B. dy/dx = -3y/x integrates to ln y = -3 ln x + ln C. (Key point: Streamline)

3. What is psi?
   A. x^3 y^2
   B. -6x^2 y
   C. 2x^3 y
   D. 2x^3 + y
   Answer: C. u = dpsi/dy = 2x^3 gives psi = 2x^3 y + f(x), f' = 0. (Key point: Stream function)

4. Why does phi not exist?
   A. The flow is compressible
   B. The flow is unsteady
   C. The velocity is too large
   D. The flow is rotational (omega_z = -6xy is not zero)
   Answer: D. A velocity potential requires irrotational flow, dv/dx = du/dy; here -12xy is not 0. (Key point: Potential)

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