Water leaving a tank: speed from a height difference
You will be able to: Derive the ideal efflux speed from Bernoulli’s equation and identify the approximations.
Why does a lower hole produce a faster water jet?
An open tank has a small outlet below the water surface. The surface and emerging jet both touch the same atmosphere. A larger vertical drop from surface to outlet makes a faster ideal jet, just as a larger fall height gives more speed to a falling object.
A useful starting point: Bernoulli: track pressure, speed and height together →
Words and symbols before equations
- Efflux speed
- Speed as fluid exits an opening.
- Head h
- Vertical height of the free surface above the outlet center.
- Large-reservoir approximation
- Tank cross-sectional area is much larger than outlet area, so surface speed is negligible.
- Quasi-steady
- Conditions change slowly enough to compare a short interval using a steady-flow model.
What this picture assumes
Large open reservoir, small horizontal outlet area 0.0002 m², g=10 m/s². Surface and emerging jet share atmospheric pressure. Negligible surface speed, viscosity and jet contraction. Each setting is an instantaneous head; water level is not evolved over time. Arrow shows exit velocity, not the downstream path.
Connect the picture to the physics
Apply Bernoulli between the free surface and the exiting jet. Both have atmospheric pressure, so those pressure terms cancel. With negligible surface speed, ρgy_surface=½ρv_exit²+ρgy_exit. Rearranging gives v_exit=√(2gh), called Torricelli’s result.
Density cancels in this ideal speed relation. Outlet area still affects volume flow rate Q=A_out v_exit. A larger hole does not directly appear in the approximate speed formula, but it can make the tank drain faster and invalidate the negligible-surface-speed assumption.
Use the water height above the hole, not the tank height above the floor. As the water level falls, h and exit speed decrease. Real jets can contract and lose energy; this model ignores those effects and does not claim that every real opening delivers the ideal Q.
A worked example, step by step
Water exits a small opening 1.25 m below an open tank’s surface. Use g=10 m/s² and outlet area 0.0002 m². Find ideal exit speed and Q.
- The surface and emerging jet share atmospheric pressure; surface speed is negligible.
- v=√[2(10)(1.25)]=√25=5 m/s.
- Q=A_out v=0.0002(5)=0.001 m³/s.
- That is 1 L/s at that instant. The flow decreases as h falls unless the level is maintained.
Torricelli’s h is the surface-to-outlet drop. Equal atmospheric pressures cancel, but pressure is not zero in absolute units.
Does doubling h double exit speed?
Compare with an explanation
No. Speed scales as √h, so it increases by √2.
Predict. Change one thing. Explain.
Vary the water height above the outlet. Predict speed and instantaneous Q. Compare h with 4h within the slider range; speed should double. The arrow shows exit velocity, not the later projectile path.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Head h=1.25 m. Ideal v=√(2gh)=5 m/s. For outlet area 0.0002 m², Q=0.001 m³/s=1 L/s. This is the instantaneous ideal flow at the selected head.
Large open reservoir, small horizontal outlet area 0.0002 m², g=10 m/s². Surface and emerging jet share atmospheric pressure. Negligible surface speed, viscosity and jet contraction. Each setting is an instantaneous head; water level is not evolved over time. Arrow shows exit velocity, not the downstream path.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant force, motion or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA small outlet is 0.45 m below a large open tank’s surface. Use g=10 m/s² and outlet area 0.0001 m². (a) Find ideal exit speed. (b) Find Q. (c) Find speed when h falls to 0.20 m. (d) Explain why atmospheric pressure cancels.
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Compare with the answer and four-point rubric
- 1 point: v=3 m/s.
- 1 point: Q=0.0003 m³/s=0.3 L/s.
- 1 point: v=2 m/s.
- 1 point: Both the free surface and emerging jet are exposed to the same atmosphere, so equal pressure terms occur on both sides.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Torricelli speed formula?
v=√(2gh) under the stated assumptions.
RECALL 2Which height is h?
Water surface above outlet center.
RECALL 3Does the ideal exit speed depend on fluid density?
Not in this open-reservoir relation; density cancels.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Water leaving a tank: speed from a height difference
- Large open reservoir, small outlet, ideal flow: v_exit=√(2gh).
- Q=A_out v_exit; h is measured to the outlet center.
Remember: Torricelli’s h is the surface-to-outlet drop. Equal atmospheric pressures cancel, but pressure is not zero in absolute units.
Conditions: Large open reservoir, small horizontal outlet area 0.0002 m², g=10 m/s². Surface and emerging jet share atmospheric pressure. Negligible surface speed, viscosity and jet contraction. Each setting is an instantaneous head; water level is not evolved over time. Arrow shows exit velocity, not the downstream path.
Refresh Kid · Unit 8 · Objectives 8.4.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 8.4, objectives 8.4.B. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Fall-2026 corrections also checked. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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