EngiMetric Arena / Design Challenge
Specify a pump efficiency that keeps the duty inside its budget
What you should take away: Separate a hydraulic requirement from an efficiency assumption, and see that every efficiency above a floor satisfies the same stated constraints.
The situation
A transfer pump must raise the liquid through a known static lift, pipe friction and fittings, and deliver a stated residual pressure at the discharge. The project caps the motor power at a stated figure. Pump and motor efficiencies are assumptions you choose. Specify the pump efficiency you would design to.
Given
| Quantity | Value | Unit | Note |
|---|---|---|---|
| Flow | 60 | m³/h | Design flow. |
| Static lift | 28 | m | Geometric rise. Fixed by the layout, not by the pump. |
| Pipe friction | 12 | m | Friction loss in the pipework. |
| Minor losses | 3 | m | Fittings and bends. |
| Residual pressure | 4 | bar | Pressure required at the discharge. |
| Motor efficiency | 95 | % | Assumed motor efficiency. |
| Approved motor power | 18.5 | kW | The project power cap the duty must fit inside. |
Constraints
- Membrane elements are supplied whole. A fraction of an element cannot be ordered.
- Pressure vessels are loaded with a whole number of elements — 6 per vessel here.
- Every vessel in the array is filled to the same element count.
Limits this decision must satisfy
| Limit | Requirement | If it is breached |
|---|---|---|
| Motor power | ≤ 18.5 kW | Above the cap the motor duty exceeds what the project has approved for this duty, so the pump has to be oversized or the head re-examined. |
More than one recovery or efficiency can satisfy every limit at once. There is no single right number here — find a value that holds all of them.
Your answer
Hints
- The head requirement and the power requirement are different questions, and the head does not depend on efficiency at all.
- Compute the static, friction and minor losses, then add the residual pressure as a head. That total is fixed regardless of which pump you choose.
- Power is where efficiency enters. A more efficient pump needs less shaft power for the same head — so ask what the lowest acceptable efficiency is, not what the best one is.
Where that comes from
Each limit, checked
Why
- The pump head engine adds the three stated height components to the residual pressure expressed as head, giving a total dynamic head of 83.79 m. That figure is a property of the system, not of the pump, and no efficiency choice changes it — which is why it is stated as information rather than as a constraint you can satisfy or fail.
- The pump power engine then converts that head and the flow into hydraulic, shaft and motor power at the stated efficiencies. Hydraulic power is a fixed 13.69 kW; everything above it moves with the efficiency assumption: 20.59 kW at 70 %, 18.25 kW at 79 %, 18.02 kW at 80 %, 16.96 kW at 85 %.
- So the design question is a floor, not a target. Motor power equals 18.5 kW at exactly 77.9 % pump efficiency, so any efficiency at or above that holds the duty inside the cap — and each of those choices is a defensible design.
- Choosing above the floor buys margin against fouling, wear and off-design flow — a commercial judgement, not a correctness question.
- Note what the head engine contributed: it fixed the duty at 83.79 m regardless of pump choice, and it is the power engine plus the efficiency assumption that decides whether the project can afford it.
What this challenge does not establish
- No pump curve is modelled. This challenge does not consider the best-efficiency point, so it cannot tell you where a real pump should be selected.
- Efficiencies are stated assumptions, not vendor data. A real selection needs manufacturer curves at the design flow.
- Static lift, friction and minor losses are taken as fixed; none is re-evaluated as the design changes.
- No allowance for suction-side conditions, NPSH, or start-up and shut-down transients.
- Static friction and velocity head are not modelled separately from the stated losses.