NivaarExam PrepOfficial exam papers ↗

04-BS-7 · Undated paper

Question 5 of 13: Ideal-Flow Orifice Jet from a Pressurized Pipe

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — undated sitting, identified as May 2019 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs “do seven”; Section B (Analytical) offers 4 questions and instructs “do three.” A complete paper is any 10 of the 13, each worth 5 marks. Every question is answered below (13 of 13) so the set can be used as a full study resource. Constants used throughout (from the paper's own Constants page, p.12): g = 9.81 m/s², patm = 100 kPa, ρwater = 1000 kg/m³, SGbenzene = 0.90, SGmercury = 13.56, SGcarbon tetrachloride = 1.59, ρair = 1.19 kg/m³ (20°C), μwater = 1.0×10⁻³ N·s/m², μair = 1.8×10⁻⁵ N·s/m².

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), hydrostatic forces on plane surfaces (Ch. 2), the Bernoulli/continuity pair and orifice flow (Ch. 3), the linear-momentum equation for moving vanes (Ch. 3), pipe friction and the Moody/Colebrook relation (Ch. 6), boundary layers and drag (Ch. 7), capillary rise (Ch. 1), high-lift devices and aircraft wing aerodynamics (J. D. Anderson, Fundamentals of Aerodynamics, Ch. 4–5).

Question 5: Ideal-Flow Orifice Jet from a Pressurized Pipe (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.

QuantityValue
Pipe diameter, d140 mm
Jet (orifice) diameter, d210 mm
Gauge pressure in pipe, P13 MPa
Discharge conditionjet exits to atmosphere, P2 = 0 gauge

Find. V1 (pipe), V2 (jet), and Q.

d1 = 40 mm, V1 d2 = 10 mm, V2
Sharp-edged orifice plate at the pipe outlet contracts the flow to a 10 mm jet.

Approach. Combine continuity (constant volumetric flow through pipe and jet) with the ideal Bernoulli equation (no elevation change, no friction) between the pipe and the jet exit.

  1. Continuity relates V1 and V2. $$ A_1 V_1 = A_2 V_2 \quad\Rightarrow\quad V_1 = V_2\left(\frac{d_2}{d_1}\right)^2 = V_2(0.0625) $$
  2. Bernoulli, pipe to jet exit (P2=0 gauge, no elevation change). $$ P_1 = \tfrac12\rho\left(V_2^2 - V_1^2\right) = \tfrac12\rho V_2^2\left[1-\left(\frac{A_2}{A_1}\right)^2\right] $$
  3. Solve for the jet velocity. $$ V_2 = \sqrt{\frac{P_1}{\tfrac12\rho\left[1-(0.0625)^2\right]}} = \sqrt{\frac{3\,000\,000}{500(0.99609)}} $$ $$ \boxed{V_2 \approx 77.6\ \text{m/s}} $$
  4. Pipe velocity and flow rate. $$ V_1 = 0.0625(77.6) \approx 4.85\ \text{m/s} $$ $$ Q = A_2 V_2 = \tfrac{\pi}{4}(0.010)^2(77.6) $$ $$ \boxed{Q \approx 6.10\ \text{L/s}} $$
QuantityResult
Pipe velocity, V14.85 m/s
Jet velocity, V277.6 m/s
Flow rate, Q6.10 L/s