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SCIENCE · INVESTIGATE AND EXPLAIN

Terminate a draining solution at its physical endpoint

For a large tank with surface area Atank and a small outlet, a quasisteady Torricelli model gives outflow Cd a sqrt(2gH). Mass continuity then gives Atank dH/dt = −Cd a sqrt(2gH). The discharge coefficient is stipulated, and surface kinetic energy is neglected under the disclosed small outlet-to-tank area ratio.

Learning outcome: Derive the separated solution, identify its endpoint and compare Euler event times on two grids without hiding the signed error.

Starting recap

Derive equations with their initial or boundary conditions, integrate distributed profiles, and check model domains independently of numerical outputs.

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Try

At outlet area 0.001 m², exact emptying time is about 74.5356 s. Refining the Euler grid changes its numerical event error, not this analytic target. A changed discharge coefficient would instead change the model itself.

Open the experiment

Predict

Why must the squared exact-height expression stop at emptying?

Experiment

Derive the separated solution, identify its endpoint and compare Euler event times on two grids without hiding the signed error.

Integrate a draining-height law without stepping below the tank bottom

Worked teaching model · synthetic values · Large tank area.1 m²,initial height1 m,orifice discharge coefficient.6,g=10. Quasisteady Torricelli outflow with inlet-surface kinetic energy neglected. Small orifice/tank area ratio.01–.03; emptying ends the model. No physical vessel or plumbing.

Reset keeps your prediction, captures and practice history.

Exact draining event and bounded numerical steps

liquid height (m)time (s)037.267810

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: exact until empty (solid).2: Euler with within-step zero event (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Exact draining event and bounded numerical steps

exact until empty

time (s)liquid height (m)
01
0.46584750.9751563
0.9316950.950625
1.3975420.9264063
1.863390.9025
2.3292370.8789063
2.7950850.855625
3.2609320.8326562
3.726780.81
4.1926270.7876562
4.6584750.765625
5.1243220.7439063
5.590170.7225
6.0560170.7014063
6.5218650.680625
6.9877120.6601563
7.453560.64
7.9194070.6201562
8.3852550.600625
8.8511020.5814062
9.316950.5625
9.7827970.5439063
10.248640.525625
10.714490.5076563
11.180340.49
11.646190.4726563
12.112030.455625
12.577880.4389062
13.043730.4225
13.509580.4064062
13.975420.390625
14.441270.3751563
14.907120.36
15.372970.3451562
15.838810.330625
16.304660.3164063
16.770510.3025
17.236360.2889062
17.70220.275625
18.168050.2626562
18.63390.25
19.099750.2376562
19.565590.225625
20.031440.2139063
20.497290.2025
20.963140.1914062
21.428980.180625
21.894830.1701562
22.360680.16
22.826530.1501562
23.292370.140625
23.758220.1314063
24.224070.1225
24.689920.1139062
25.155760.105625
25.621610.09765625
26.087460.09
26.553310.08265625
27.019150.075625
27.4850.06890625
27.950850.0625
28.41670.05640625
28.882540.050625
29.348390.04515625
29.814240.04
30.280090.03515625
30.745930.030625
31.211780.02640625
31.677630.0225
32.143480.01890625
32.609320.015625
33.075170.01265625
33.541020.01
34.006870.00765625
34.472710.005625
34.938560.00390625
35.404410.0025
35.870260.00140625
36.33610.000625
36.801950.00015625
37.26780

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.

Euler with within-step zero event

time (s)liquid height (m)
01
1.863390.9
3.726780.8051317
5.590170.7154025
7.453560.6308211
9.316950.5513968
11.180340.4771407
13.043730.4080654
14.907120.3441854
16.770510.285518
18.63390.2320842
20.497290.183909
22.360680.1410244
24.224070.1034712
26.087460.07130427
27.950850.04460141
29.814240.02348236
31.677630.008158406
33.360720

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
QuantityValueUnit
Exact emptying time37.2678s
Euler detected emptying time33.36072s
Euler emptying-time error-3.907084s
Euler time step1.86339s
Orifice to tank area ratio0.02ratio

Emptying is a domain endpoint, not permission to square a negative height root and refill the tank.

Continuity gives Atank dH/dt=−Cd a sqrt(2gH). Integrating yields sqrtH=sqrtH0−kt/2 only until it reaches zero, at2sqrtH0/k. Euler predicts decreasing positive heights; if its next step would be negative, interpolate to zero within that step and terminate. That numerical event time is approximate and is compared with the exact analytic event, not hidden by replacing it.

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

Emptying is a domain endpoint, not permission to square a negative height root and refill the tank.

Continuity gives Atank dH/dt=−Cd a sqrt(2gH). Integrating yields sqrtH=sqrtH0−kt/2 only until it reaches zero, at2sqrtH0/k. Euler predicts decreasing positive heights; if its next step would be negative, interpolate to zero within that step and terminate. That numerical event time is approximate and is compared with the exact analytic event, not hidden by replacing it.

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

Understand

Terminate a draining solution at its physical endpoint

For a large tank with surface area Atank and a small outlet, a quasisteady Torricelli model gives outflow Cd a sqrt(2gH). Mass continuity then gives Atank dH/dt = −Cd a sqrt(2gH). The discharge coefficient is stipulated, and surface kinetic energy is neglected under the disclosed small outlet-to-tank area ratio.

Write k = Cd a sqrt(2g)/Atank. Separating variables gives sqrt(H) = sqrt(H₀) − kt/2 until the right side reaches zero. Thus H(t) is the square of that expression only before emptying, at time 2sqrt(H₀)/k. Continuing the square beyond that time would produce a fictitious refilling solution that no longer satisfies the original sign of dH/dt.

At H₀ = 1 m, Atank = 0.1 m², outlet area 0.002 m², Cd = 0.6 and g = 10, k is about 0.0536656 in compatible units and exact emptying time is about 37.2678 s. Halving outlet area doubles the emptying time in this bounded model, not in every real drain geometry.

Forward Euler advances H using its current derivative. When the next step would become negative, the implementation locates a zero within that step by linear interpolation and terminates. The resulting event time is still approximate and is compared with the exact event. It is not replaced by the analytic answer or hidden by letting the tank height become negative.

Changed-case check

Predict before reading the comparison. At outlet area 0.001 m², exact emptying time is about 74.5356 s. Refining the Euler grid changes its numerical event error, not this analytic target. A changed discharge coefficient would instead change the model itself.

Retained investigation

Derive the separated solution, identify its endpoint and compare Euler event times on two grids without hiding the signed error. Which observation would test the assumed discharge model rather than the numerical step size?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All prose, functions and evidence packets are original. Bounded mathematical models do not establish measured apparatus, supervised practical skill, independent learner mastery or institutional credit. No physical loading, liquids, heat or sound exposure is requested.

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Why must the squared exact-height expression stop at emptying?

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Which observation would test the assumed discharge model rather than the numerical step size?

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