Nozzle Jet Velocity and Discharge: 04-BS-7, May 2014, Q6

Question 6 — Nozzle Jet Velocity and Discharge (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
Jet (nozzle exit) diameter, D₂	75 mm
Pipe diameter, D₁	200 mm
Pipe pressure, p₁	80 kPa gauge
Coefficient of velocity, Cₜ	0.96
Coefficient of contraction, C_c	1.00 (no contraction stated)

Find. The actual jet velocity V₂ and discharge Q.

Fig. Q6 — nozzle reducing from a 200 mm supply pipe to a 75 mm free jet; the coefficient of velocity Cₜ accounts for the small friction loss in the nozzle.

Approach. Bernoulli between the pipe and the (ideal) jet gives the ideal velocity including the pipe's own approach velocity (found from continuity in terms of the actual jet velocity); the real velocity is then Cₜ times the ideal value.

Set up simultaneous continuity + Bernoulli. Continuity (real flow, C_c=1): V₁ = (D₂/D₁)² V₂_actual = r V₂, r=(75/200)²=0.1406. Bernoulli, pipe→jet (same elevation, p_jet=0 gauge): $$\frac{{p_1}}{{\rho}}+\frac{{V_1^2}}{{2}} = \frac{{V_{{2,ideal}}^2}}{{2}},\qquad V_{{2,ideal}}=\frac{{V_2}}{{C_v}}$$
Solve for the actual jet velocity. Substituting V₁=rV₂ and V_2,ideal=V₂/Cₜ: $$\frac{{2p_1}}{{\rho}} = V_2^2\left(\frac{{1}}{{C_v^2}}-r^2\right)$$ $$V_2 = \sqrt{{\frac{{2(80,000)/1000}}{{1/0.96^2 - 0.1406^2}}}} = \sqrt{{\frac{{160}}{{1.0851-0.01977}}}} = \sqrt{{150.2}}$$ $$V_2 = \boxed{{12.25\text{{ m/s}}}}$$
Discharge. $$A_2 = \frac{{\pi}}{{4}}(0.075)^2 = 4.418\times10^{{-3}}\text{{ m}}^2$$ $$Q = A_2 V_2 = 4.418\times10^{{-3}}\times12.25 = \boxed{{0.0541\text{{ m}}^3/\text{{s}}}}$$

Quantity	Result
Actual jet velocity, V₂	12.25 m/s
Discharge, Q	0.0541 m³/s (54.1 L/s)

Topic: Real nozzle flow — coefficients of velocity and contraction.

Key relations: Bernoulli energy equation; C_v=V_actual/V_ideal; continuity through an area change.

Why this works: the nozzle's small internal friction loss is folded entirely into Cₜ (an empirical correction on the ideal Torricelli-type velocity), while the approach velocity in the 200 mm pipe still has to be carried through continuity because the area ratio, though small, is not negligible.

Common pitfall: squaring the diameter ratio twice by mistake (using r⁴ instead of r² in the Bernoulli substitution) — here r²=0.0198 is already small enough that neglecting the approach-velocity term entirely changes the answer by only about 1%, but it should still be carried through the algebra correctly.

Source: White, Fluid Mechanics, 8th ed., §3.7 & §6.11 (nozzle/orifice discharge coefficients).

Question 6 of 13: Nozzle Jet Velocity and Discharge

Question 6 — Nozzle Jet Velocity and Discharge (5 marks)