---
title: "Stagnation points in 2D velocity field"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_3/fluid_mechanics:stagnation-points-velocity-field"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stagnation-points-velocity-field
site: BCT Engineering
---

# Stagnation points in 2D velocity field

Fluid Mechanics · Chapter 4: Hydrokinematics

- Asked in: 2072 Chaitra, 3 marks
- Video: [UD MEEG331 Ch3 17: Streamline Ex 2 Stagnation Flow](https://www.youtube.com/watch?v=T08vdvLfRO8) — Joseph Feser
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/stagnation-points-velocity-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4

## Question

A steady, incompressible, two dimensional velocity field is given by $\vec V = (1 + 2.5x + y)\hat i + (-0.5 - 3x - 2.5y)\hat j$ where x and y are in m and magnitude of velocity in m/s. Determine, if there are any stagnation points in this flow field and if so, where they are.

## Short answer

Yes, one stagnation point exists at x = -0.615 m, y = 0.538 m where u and v both vanish.

## Answer

A **stagnation point** is a point where the velocity is zero, i.e. both $u = 0$ and $v = 0$ at the same time.

**Given:** $u = 1 + 2.5x + y$, $v = -0.5 - 3x - 2.5y$.

Set both components equal to zero:

$$
\begin{aligned}
1 + 2.5x + y &= 0 \quad (1)\\
-0.5 - 3x - 2.5y &= 0 \quad (2)
\end{aligned}
$$

From (1): $y = -1 - 2.5x$. Substituting in (2):

$$
\begin{aligned}
-0.5 - 3x - 2.5(-1 - 2.5x) &= 0\\
-0.5 - 3x + 2.5 + 6.25x &= 0\\
2 + 3.25x &= 0\\
x &= -0.6154\ \text{m}
\end{aligned}
$$

$$
y = -1 - 2.5(-0.6154) = 0.5385\ \text{m}
$$

Check: $u = 1 - 1.5385 + 0.5385 = 0$ and $v = -0.5 + 1.8462 - 1.3462 = 0$.

**Answer: yes, there is one stagnation point, at $(x, y) = (-0.615\ \text{m},\ 0.538\ \text{m})$.**

## Check yourself

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

1. What defines a stagnation point?
   A. Pressure is zero
   B. Acceleration is zero
   C. Only u = 0
   D. u = 0 and v = 0 simultaneously
   Answer: D. Velocity vector vanishes. (Key point: Definition)

2. How are stagnation points found?
   A. Solve u=0 and v=0 as simultaneous equations
   B. Integrate u
   C. Differentiate v
   D. Set du/dx = 0
   Answer: A. Two linear equations give a unique point. (Key point: Method)

3. Where is the stagnation point?
   A. (0.615, -0.538) m
   B. (-0.615, 0.538) m
   C. (0,0)
   D. None exists
   Answer: B. From y = -1-2.5x: 2 + 3.25x = 0, x = -0.6154, y = 0.5385. (Key point: Result)

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