Question 4 of 13: VDI Orifice Meter — Differential Pressure Reading
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2017-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and capillarity (Ch. 2), integral analysis and buoyancy (Ch. 3), viscous flow in ducts, pipe friction and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), open-channel flow and the hydraulic jump (Ch. 10).
Check — assumptions used across this paper:
Q2's figure shows the vertical leaf of the gate rising above the free surface, so water acts over the full depth x = 2.0 m; both "0.3 m" dimensions locate the centre of gravity (0.3 m right of the vertical leaf, 0.3 m above the arm). Reservoir water also fills the space beneath the 1.2 m arm (the vertical lines below O are dimension extension lines, not a wall), so the arm carries uplift at pressure ρgx, and the arm tip bears up against a lip on the spillway crest — the gate can only open by rotating clockwise, vertical leaf toward the spillway.
Q4's flow coefficient K ≈ 0.69 is read from the attached VDI chart on the Do/D1 = 0.50 curve (third from the top, the one that levels off at 0.62) at the approach Reynolds number R ≈ 2550. The printed curve gives 0.695 at R = 2000 and 0.684 at R = 3000; a reading of 0.68–0.70 moves the answer by only ±0.3 kPa.
Q7 and Q8 friction factors are obtained from the Colebrook–White/Haaland equation the Moody chart itself plots (smooth-wall case, as both problems specify or imply smooth surfaces).
Q9's sphere drag coefficient vs. Reynolds number is obtained from the Schiller–Naumann correlation $C_D=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)$, which reproduces the published sphere-drag curve (the same curve reprinted in the attachment) to within a few percent for $Re<1000$ — used here because the resulting Reynolds number (≈19–20) is well above the range where the simple Stokes'-law formula ($C_D=24/Re$) alone is valid.
Given. Pipe diameter $D_1=50\ \text{mm}$; orifice diameter $D_o=25\ \text{mm}$ ($D_o/D_1=0.50$); oil $SG=0.94$ ($\rho=940\ \text{kg/m}^3$), $\mu=0.015\ \text{Ns/m}^2$; mass flow rate $\dot m=1.5\ \text{kg/s}$; flow coefficient $K\approx0.69$ read from the attached chart on the $D_o/D_1=0.50$ curve at the approach Reynolds number (see check note).
Flange-tap VDI orifice meter: pressure taps immediately upstream and downstream of the orifice plate.
Find. The differential-pressure-gauge reading (kPa) across the orifice.
Approach. Compute the volumetric flow rate and the pipe Reynolds number, use the chart-read discharge coefficient $K$ in the reference-sheet orifice equation to solve for the equivalent head $\Delta h$, then convert to a pressure reading.
Check: K ≈ 0.69 is read from the attached VDI chart on the D₀/D₁ = 0.50 curve at R ≈ 2550 (see paper-level check note). The curve is still falling there, from about 0.70 at R = 2000 to about 0.68 at R = 3000, so the reading must be taken at the computed R and not from the curve's high-R plateau of 0.62.
Reynolds number of approach.
$$R=\frac{D_1V_1\rho}{\mu}=\frac{0.05\times0.8127\times940}{0.015}=\boxed{2546}$$
Read K and solve the orifice equation for $\Delta h$. On the $D_o/D_1=0.50$ curve at $R=2546$ the chart gives $K\approx0.69$. With $A_o=\pi D_o^2/4=4.909\times10^{-4}\ \text{m}^2$ and $Q=K A_o\sqrt{2g\Delta h}$,
$$\Delta h=\frac{1}{2g}\left(\frac{Q}{K A_o}\right)^2=\frac{1}{2\times9.81}\left(\frac{1.596\times10^{-3}}{0.69\times4.909\times10^{-4}}\right)^2=1.132\ \text{m (of oil)}$$
Convert to a pressure-gauge reading.
$$\Delta p=\rho g\Delta h=940\times9.81\times1.132=\boxed{10{,}433\ \text{Pa} = 10.43\ \text{kPa}}$$
A chart reading of $K=0.68$ or $0.70$ gives 10.74 or 10.14 kPa, since $\Delta p\propto1/K^2$.
