---
title: "Water jet on moving blade, 2 in diameter"
subject: "Applied Mechanics (Dynamics)"
chapter: 3
key: "semester_3/applied_mechanics_dynamics:water-jet-moving-blade-2in"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/water-jet-moving-blade-2in
site: BCT Engineering
---

# Water jet on moving blade, 2 in diameter

Applied Mechanics (Dynamics) · Chapter 3: System of particles

- Asked in: 2078 Kartik, 6 marks
- Video: [FORCE EXERTED BY JET ON A MOVING CURVED SYMMETRICAL PLATE](https://www.youtube.com/watch?v=MrJzqrdmhr4) — Mech PM
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/water-jet-moving-blade-2in
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-3

## Question

A 2-in diameter water jet having a velocity of 25 ft/s impinges upon a single moving blade as shown in figure. If the blade moves with a constant velocity of 5 ft/s away from the jet, determine the horizontal and vertical components of force which the blade is exerting on the water. What power does the water generate on the blade? Water has a specific weight of 62.4 lb/ft$^3$. [Figure: jet $v_w = 25$ ft/s from the left hitting curved blade AB moving to the right at $v_{bl} = 5$ ft/s.]

## Short answer

Use relative velocity 20 ft/s and mass flow 0.8456 slug/s. Force components on water: horizontal -8.46 lb, vertical +14.65 lb. Power developed is 42.3 ft·lb/s (0.077 hp).

## Answer

**Assumptions (figure not shown):** the blade turns the water so that, relative to the blade, it leaves at $60^\circ$ above the horizontal with unchanged relative speed (no friction). The blade moves away from the jet at $5$ ft/s. Pound-force units.

#### Data
$$
\rho=\frac{62.4}{32.2}=1.938\ \text{slug/ft}^3,\qquad A=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}=0.02182\ \text{ft}^2
$$

$$
v_{rel}=25-5=20\ \text{ft/s}
$$

#### Mass flow rate hitting the moving blade
$$
\dot m=\rho A\,v_{rel}=1.938(0.02182)(20)=0.8456\ \text{slug/s}
$$

#### Force of the blade on the water
In the frame moving with the blade, the water enters at $20$ ft/s horizontally and leaves at $20$ ft/s at $60^\circ$:

$$
\begin{aligned}
F_x&=\dot m(20\cos60^\circ-20)=0.8456(-10)=-8.46\ \text{lb}\\
F_y&=\dot m(20\sin60^\circ)=0.8456(17.32)=14.65\ \text{lb}
\end{aligned}
$$

The blade pushes on the water with $8.46$ lb opposite to the jet direction (horizontal component) and $14.65$ lb upward (vertical component); resultant $16.9$ lb.

#### Power on the blade
The water pushes the blade forward with $8.46$ lb at speed $5$ ft/s:

$$
P=8.456(5)=42.3\ \text{ft lb/s}=\frac{42.3}{550}=0.0769\ \text{hp}
$$

**Answer: $F_x=-8.46$ lb, $F_y=+14.65$ lb (on the water); power developed $\approx42.3$ ft lb/s ($0.077$ hp).**

## Check yourself

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

1. What relative speed of water to the blade is used?
   A. 30 ft/s
   B. 25 ft/s
   C. 5 ft/s
   D. 20 ft/s
   Answer: D. The blade moves away from the jet, so v_rel = 25 - 5 = 20 ft/s. (Key point: Relative speed)

2. Which expression gives the mass flow rate onto the blade?
   A. rho A v_rel with rho = 62.4/32.2 slug/ft^3
   B. (62.4) A (25)
   C. rho A (25 + 5)
   D. 62.4 A v_rel
   Answer: A. Use mass density in slugs (weight density/g) and the relative speed: 1.938(0.02182)(20) = 0.8456 slug/s. (Key point: Mass flow)

3. What are the force components on the water?
   A. F_x = +8.46 lb, F_y = 14.65 lb
   B. F_x = -8.46 lb, F_y = +14.65 lb
   C. F_x = -14.65 lb, F_y = 8.46 lb
   D. F_x = 0, F_y = 16.9 lb
   Answer: B. F_x = m dot(20 cos60 - 20) = -8.46 lb and F_y = m dot(20 sin60) = 14.65 lb. (Key point: Force components)

4. How is the power on the blade found?
   A. Resultant force times jet speed
   B. Vertical force times blade speed
   C. Horizontal force 8.46 lb times blade speed 5 ft/s
   D. Weight times velocity
   Answer: C. The forward force of the water on the blade times the blade speed gives 42.3 ft lb/s, about 0.077 hp. (Key point: Power on blade)

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