SCIENCE · INVESTIGATE AND EXPLAIN

Audit a whole flow-power account before inferring efficiency

The final model uses a hypothetical everyday-style lifting flow to connect fluid mechanics, power and thermal measurements without prescribing any physical apparatus. Liquid density is 1000 kg/m³ and g=10. It rises 2 m from a large reservoir with negligible inlet speed to an outlet speed of 2 m/s. Both endpoint pressures are atmospheric, and a separate prescribed pipe loss is 5000 Pa.

Learning outcome: Retain input, fluid, height, kinetic, pipe and pump powers for contrasting settings. Infer efficiency from an independent hypothetical wattmeter record, then propose a new observation distinguishing an input-power bias from a thermal-capture error. Preserve original prediction, critique and revision in the course project.

Starting recap

Choose recipient, system, time interval and calibration reference before combining equations. Preserve alternative explanations and distinguish predicted, generated and independently observed evidence.

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Try

At Q=0.1 L/s and η=0.5, fluid power is 2.7 W and input 5.4 W. Exported height plus kinetic power is 2.2 W, so receiver heating is 3.2 W and its 60 s temperature rise is 1.92 K.

Open the experiment

Predict

The pump internal loss equals:

Your prediction and reason

Experiment

Retain input, fluid, height, kinetic, pipe and pump powers for contrasting settings. Infer efficiency from an independent hypothetical wattmeter record, then propose a new observation distinguishing an input-power bias from a thermal-capture error. Preserve original prediction, critique and revision in the course project.

Audit a hypothetical lifting-flow power account

Worked teaching model · synthetic values · Fictional incompressible liquid rho=1000 kg/m³, g=10, lift 2 m from large reservoir to outlet speed 2 m/s. Both endpoint pressures atmospheric. Pipe loss 5000 Pa; delivered fluid power/input power is selected efficiency. All pipe/pump waste heats receiver C=100 J/K for 60 s. No physical pump, electricity or fluid handling.

Steady volume flow: 0.2 L/s Pump energy efficiency: 0.6 ratio

Reset keeps your prediction, captures and practice history.

Input and destinations at prescribed efficiency

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: input power (solid).2: height plus exit kinetic power (dashed).3: receiver heating power (dotted). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Input and destinations at prescribed efficiency

input power

volume flow (L/s)	power (W)
0	0
0.2	9

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

height plus exit kinetic power

volume flow (L/s)	power (W)
0	0
0.2	4.4

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

receiver heating power

volume flow (L/s)	power (W)
0	0
0.2	4.6

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Volume flow in SI	0.0002	m³/s
Required fluid mechanical power	5.4	W
Required input power	9	W
Gravitational energy flux increase	4	W
Kinetic energy flux increase	0.4	W
Pipe loss power	1	W
Pump internal loss power	3.6	W
Receiver heating power	4.6	W
Receiver temperature rise after 60 s	2.76	K
Whole power residual	0	W
Efficiency inferred from input and flow	0.6	ratio

Pump efficiency and pipe loss describe different transfers and must not be counted twice.

Per-volume demand is rho g h+rho vout²/2+loss=27000 Pa. Multiply by Q for fluid power, then divide by efficiency for input power. Pump loss is input minus delivered fluid power; pipe loss is already part of that delivered-fluid account. Height and exit kinetic energy leave with the stream, not into the receiver. Varying Q at fixed outlet speed implies a changed outlet area; one unchanged nozzle would not maintain all these settings independently.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Pump efficiency and pipe loss describe different transfers and must not be counted twice.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Freeze the system and measurement contract

Derive the coupled model

The mechanical energy demand per transported volume is ρgh+ρv²/2+loss=20000+2000+5000=27000 J/m³. Multiplying by volume flow Q gives power delivered to the fluid. If pump efficiency η is defined as delivered fluid power divided by input power, input power is 27000Q/η. This efficiency refers to the pump; it does not erase the pipe loss that remains part of the delivered-fluid demand.

Check the numerical energy and power account

At Q=0.2 L/s=0.0002 m³/s and η=0.6, delivered fluid power is 5.4 W and input is 9 W. Of that input, 4 W leaves as height-energy flux, 0.4 W as exit kinetic-energy flux, 1 W becomes pipe heating and 3.6 W becomes internal pump loss. The complete account closes: 9=4+0.4+1+3.6. Counting the 1 W pipe loss again after subtracting all fluid power would double count it.

Reject an unsupported inference

The selected thermal receiver captures both pump and pipe loss, with capacity 100 J/K for 60 s. Its 4.6 W heating gives rise 2.76 K. A different observed rise need not establish a new pump efficiency: flow calibration, heat-capacity error, incomplete capture or external heating are alternatives. Independently measuring input power and flow gives a second conditional estimate. Varying Q while keeping exit speed fixed also implies a corresponding outlet-area change by Q=Av; a single unchanged nozzle would not preserve both. No electrical supply, liquid handling or heat experiment is requested.

Worked original case

At Q=0.1 L/s and η=0.5, fluid power is 2.7 W and input 5.4 W. Exported height plus kinetic power is 2.2 W, so receiver heating is 3.2 W and its 60 s temperature rise is 1.92 K.

Cumulative evidence action

Changed-case numerical practice

Try the changed case before feedback

Use Q=0.1 L/s, fluid demand 27000 J/m³ and pump efficiency 0.5. Find required input power in W.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and completion boundaries

Original synthesis checked 30 October 2026 against OpenStax College Physics 2e and the course’s earlier source-backed force, energy and thermal models. All records are stipulated or generated; no machinery, electrical connection, heating, liquid handling or physical experiment is requested. Completing these authored pages does not establish independent understanding, supervised practical skill or institutional credit. Use the separate retained course assessment and project for actual evidence.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Which observations are needed before a calorimeter reading can support a pump-efficiency claim?

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Reconcile falling-mass rotation with a separate stopping receiver Next: College Physics I — algebra-based measurement and mechanics and next units

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