---
title: "Air jet on curved vane with pitot"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:pitot-air-jet-curved-vane-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pitot-air-jet-curved-vane-force
site: BCT Engineering
---

# Air jet on curved vane with pitot

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2078 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/pitot-air-jet-curved-vane-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

A stream of air at standard condition from 2 cm dia. nozzle strikes a curved vane as shown in figure. A stagnation pitot tube connected to water-filled U-tube manometer is located in the nozzle exit plane. Calculate the speed of the air leaving the nozzle. Estimate the horizontal component of force exerted on the vane by the jet. [Figure: air flowing through a 2 cm diameter nozzle; stagnation tube connected to a water U-tube with 7 cm reading; free air jet deflected by a fixed vane through 30° at exit]

## Short answer

Air speed V = √(2ρ_w g h / ρ_a) = √(2×1000×9.81×0.07/1.225) ≈ 33.5 m/s. Horizontal force on the vane is ρ_a A V² (1 − cos30°) = 0.058 N in the jet direction.

## Answer

**Given:** nozzle $d = 2$ cm, standard air $\rho_a = 1.225$ kg/m³, water manometer reading $h = 7$ cm (water $\rho_w = 1000$ kg/m³). The jet is deflected by the fixed vane through $30^\circ$ and leaves at the same speed (no friction).

#### Speed of the air from the pitot reading

A stagnation pitot tube in the nozzle exit plane (where static pressure is atmospheric) reads the dynamic pressure:

$$
p_0 - p = \frac{1}{2}\rho_aV^2 = \rho_wgh
$$

$$
V = \sqrt{\frac{2\rho_wgh}{\rho_a}} = \sqrt{\frac{2(1000)(9.81)(0.07)}{1.225}} = \sqrt{1121.1} = 33.5\ \text{m/s}
$$

#### Horizontal force on the vane

$$
A = \frac{\pi}{4}(0.02)^2 = 3.1416\times 10^{-4}\ \text{m}^2
$$

$$
\dot m = \rho_aAV = 1.225\times 3.1416\times 10^{-4}\times 33.48 = 0.01288\ \text{kg/s}
$$

The jet enters horizontally with velocity $V$ and leaves at $30^\circ$ to the horizontal with the same speed (pressure atmospheric everywhere, so only momentum flux matters). Taking $x$ along the incoming jet:

$$
\begin{aligned}
-R_x &= \dot m\,(V\cos30^\circ - V)\\
R_x &= \dot mV(1 - \cos30^\circ) = 0.01288\times 33.48\times 0.13397\\
&= 0.0578\ \text{N}
\end{aligned}
$$

where $R_x$ is the force on the vane in the direction of the incoming jet.

**Answer: speed of air $V \approx 33.5$ m/s; horizontal force on the vane $\approx 0.058$ N in the direction of the jet.** (Equivalently $R_x = \rho_aAV^2(1-\cos30^\circ) = 0.4315 \times 0.134$.)

## Check yourself

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

1. What is the air speed?
   A. About 335 m/s
   B. About 33.5 m/s
   C. About 1121 m/s
   D. About 11.2 m/s
   Answer: B. V = sqrt(2 x 1000 x 9.81 x 0.07/1.225) = 33.5 m/s. (Key point: Speed)

2. Why does only momentum flux determine the force on the vane?
   A. Pressure is atmospheric everywhere on the jet
   B. The vane is moving
   C. Air is incompressible so no force exists
   D. Gravity is dominant
   Answer: A. With atmospheric pressure all around, the force equals the change of momentum flux. (Key point: Force reason)

3. What is the horizontal force on the vane?
   A. m V (1 - cos 30 degrees)
   B. m V cos 30 degrees
   C. m V (1 + cos 30 degrees)
   D. m V sin 30 degrees
   Answer: A. The jet loses x-momentum from V to V cos30, so R = m V (1 - cos30) = 0.058 N. (Key point: Force expression)

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