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. Specified curve (converting $Q$ to L/s): $(Q,H)=(0,\,5.5),(0.7,\,4.6),(1.7,\,3.1),(2.4,\,1.5)$ in L/s and m. Input power $P=186\;\text{W}$. Field test: $Q=1.6\;\text{L/s}$, static head difference $2.10\;\text{m}$, pipe friction loss $1.03\;\text{m}$.
Specified characteristic
$Q$ (L/min)
0
42
102
144
$Q$ (L/s)
0
0.7
1.7
2.4
$H$ (m)
5.5
4.6
3.1
1.5
Find. The H–Q and efficiency curves, the best-efficiency point, and the actual operating point and efficiency from the field test.
Approach. Convert the flows, compute the pump (water) efficiency $\eta=\rho g Q H/P$ at each point to locate the peak, then get the actual operating head as static lift plus friction loss and evaluate its efficiency.
Specified H–Q curve (solid, blue) and efficiency curve (dashed, green) versus flow in L/s. The best-efficiency point sits near 1.7 L/s at about 28%; the field-test operating point (1.6 L/s, 3.13 m) falls just below it.
Convert flows and tabulate. Divide L/min by 60: $42\rightarrow0.7$, $102\rightarrow1.7$, $144\rightarrow2.4\;\text{L/s}$. The H–Q curve is the blue line above.
Efficiency at each point. Using $\eta=\dfrac{\rho g Q H}{P}$ with $P=186\;\text{W}$:
$$\eta(0.7)=\frac{9810(0.0007)(4.6)}{186}=17.0\%,\;\;\eta(1.7)=\frac{9810(0.0017)(3.1)}{186}=27.8\%,\;\;\eta(2.4)=\frac{9810(0.0024)(1.5)}{186}=19.0\%.$$
Best operating point. The efficiency rises to a maximum near mid-flow and falls again, so the best-efficiency point is at $Q\approx1.7\;\text{L/s}$, $H\approx3.1\;\text{m}$, $\eta\approx\boxed{27.8\%}$ — the pump should be run near this flow.
Actual operating point. The pump must supply the static lift plus the pipe friction, so its head at the test flow is
$$H_{act}=\Delta z+h_f=2.10+1.03=\boxed{3.13\;\text{m at }1.6\;\text{L/s}}.$$
The corresponding efficiency is
$$\eta_{act}=\frac{9810(0.0016)(3.13)}{186}=\boxed{26.4\%}.$$
Comparison (e). At 1.6 L/s the specified curve reads about 3.2 m, so the measured 3.13 m agrees within a few percent — the pump is performing to specification. The operating point sits just left of the best-efficiency point at essentially peak efficiency, which is exactly where a well-matched system curve should intersect the pump curve.
