Ideal-Flow Velocity and Flow Rate Through a Sharp-Edged Nozzle: 04-BS-7, December 2015, Q6

Question 6: Ideal-Flow Velocity and Flow Rate Through a Sharp-Edged Nozzle (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Pipe diameter D₁	40 mm
Jet diameter D₂	10 mm
Pipe gauge pressure p₁	3 MPa
Jet exit pressure (to atmosphere)	0 gauge

Find. V₁ (pipe), V₂ (jet), and Q.

Fig. Q6 — ideal (frictionless) acceleration from the 40 mm pipe through the sharp-edged orifice plate into the 10 mm free jet at atmospheric pressure.

Approach. Combine continuity (A₁V₁=A₂V₂) with Bernoulli between the pipe and the jet (no elevation change, ideal flow) to eliminate V₁ and solve directly for V₂.

Continuity relates the two velocities. $$V_1 = \left(\frac{D_2}{D_1}\right)^2 V_2 = \left(\frac{10}{40}\right)^2 V_2 = 0.0625\,V_2$$
Bernoulli between the pipe and the jet (p₂=0 gauge): $$p_1 = \tfrac{1}{2}\rho\left(V_2^2 - V_1^2\right) = \tfrac{1}{2}\rho V_2^2\left(1-0.0625^2\right)$$
Solve for the jet velocity. $$V_2 = \sqrt{\frac{2p_1}{\rho(1-0.0625^2)}} = \sqrt{\frac{2(3\times10^6)}{1000(0.99609)}} = \boxed{77.61\ \text{m/s}}$$
Back out the pipe velocity and the flow rate. $$V_1 = 0.0625(77.61) = \boxed{4.85\ \text{m/s}}$$ $$Q = \frac{\pi}{4}D_2^2\,V_2 = \frac{\pi}{4}(0.010)^2(77.61) = \boxed{6.10\times10^{-3}\ \text{m}^3/\text{s} = 6.10\ \text{L/s}}$$

Quantity	Result
Pipe velocity V₁	4.85 m/s
Jet velocity V₂	77.61 m/s
Flow rate Q	6.10 L/s

Topic: Ideal-flow nozzle acceleration (continuity + Bernoulli).

Key relations: continuity A₁V₁=A₂V₂; Bernoulli p+½ρV²=const.

Why this works. With no friction and no elevation change, all of the pipe's static pressure converts to kinetic energy at the jet; because the area ratio is so extreme (16:1), the small correction from V₁ in the Bernoulli equation barely changes the answer versus treating the pipe as a reservoir — but it is still the technically correct approach.

Common pitfall. Neglecting the (small) upstream velocity V₁ entirely gives V₂=√(2p₁/ρ)=77.46 m/s, only 0.2% different here because D₂/D₁ is small — but the full continuity + Bernoulli method should always be shown since the correction is not always negligible.

Source: White, Fluid Mechanics, Ch. 3 (Bernoulli equation and jets).

Question 6 of 13: Ideal-Flow Velocity and Flow Rate Through a Sharp-Edged Nozzle

Question 6: Ideal-Flow Velocity and Flow Rate Through a Sharp-Edged Nozzle (5 marks)