Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations — 16-Mec-A6 Fluid Machinery, May 2019. Closed-book, three hours. Section A is calculative (5 questions) and Section B is descriptive (3 questions); candidates answer four questions from Section A and two from Section B (six questions, 60 marks, 10 marks each). All eight questions are solved as a study resource.
Reference texts. Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Fox & McDonald, Introduction to Fluid Mechanics (10th ed.); F. M. White, Fluid Mechanics (8th ed.); Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.).
Paper. This is the 16-Mec-A6, May 2019 Fluid Machinery examination.
Given. Model form $H=a-bQ-cQ^2$; specified points (L/s, m): $(0,5.5),(0.7,4.6),(1.7,3.1),(2.4,1.5)$.
Find. The coefficients from the first three points, the physical meaning of $a,b,c$, the predicted head at 2.4 L/s versus the specified value, and an improvement.
Approach. Substitute the first three points to get three linear equations in $a,b,c$ and solve; then evaluate at the fourth point and compare.
Fitted quadratic $H=5.5-1.20Q-0.126Q^2$ (blue) through the first three specified points (red). At 2.4 L/s the fit predicts 1.90 m against the specified 1.50 m — a 0.40 m over-prediction because the fourth point was not used.
Set up from the first three points. At $Q=0$, $H=5.5$ gives $a=5.5$ directly. The other two points give
$$0.7\,b+0.49\,c=5.5-4.6=0.9,\qquad 1.7\,b+2.89\,c=5.5-3.1=2.4.$$
Solve for $b$ and $c$. Eliminating $b$,
$$c=0.126,\qquad b=1.20.$$
so the pump characteristic is
$$\boxed{H=5.5-1.20\,Q-0.126\,Q^{2}}\quad(H\text{ in m},\;Q\text{ in L/s}).$$
Meaning of the constants (b). $a=5.5\;\text{m}$ is the shut-off head (the H-intercept at $Q=0$); $b$ is the initial slope of the falling curve (the near-linear droop at low flow); $c$ is the curvature that steepens the drop at high flow, reflecting the growing hydraulic losses. Both $b$ and $c$ are positive so head falls monotonically with flow.
Predicted vs specified at 2.4 L/s (c).
$$H(2.4)=5.5-1.20(2.4)-0.126(2.4)^2=1.90\;\text{m}.$$
The specified head is 1.50 m, so the fit over-predicts by $\boxed{0.40\;\text{m}}$ (shown as the orange gap on the sketch).
Improvement (d). Because the curve was forced exactly through only the first three points, it drifts at high flow. Fit all four points instead — either by least-squares regression of the quadratic over the full data set, or by adding a higher-order ($Q^3$) term so four coefficients can pass through all four points. Either spreads the small residual error and greatly reduces the 0.40 m mismatch at the top of the range.
