Why a Plastic Container Leaks Only Above a Critical Level
Lecture notes for a physicist new to fluids
We have all seen a takeout container that holds water fine when half full, but begins to drip from a spot on the bottom once you fill it past a certain mark. The drip is not a jet. It starts at about one drop per second, then it slows, and then it stops with water still in the container. The rate of one drop per second is an estimate, not a timed measurement. This note explains why, with enough detail to check the numbers for your container.
We will assume you know mechanics and thermodynamics, but not fluid interfacial physics. We will define every fluid symbol when it first appears.
0. What we are looking at — a visual dictionary
Before any formula, look at this diagram. It shows the same container you measured — 9.4 in long, 4.2 in wide, 3.1 in tall — and every quantity we will use, drawn where it lives.
Figure 0 — Symbols and their values: The container is drawn to scale: 9.4 in (23.9 cm) long, with full depth $H_{max}=3.10$ in $=7.874$ cm. The dashed line marks $H_{crit}=3.97$ cm for the fitted radius, and $H$ is any depth above the hole. The first magnified circle shows the meniscus in the hole mouth, with contact angle $\theta$, surface tension $\gamma$ and hole radius $r$. The second circle shows the wall thickness $L$, the wetted spot of radius $r_{out}$, and the drop of volume $V_{drop}$ fed by the flow rate $Q$. The key lists each symbol with its value and says whether the value is assumed, fitted or a target. The container is drawn as a straight-walled box in side view, with rounded bottom corners. The 4.2 in width is perpendicular to the view.
With that picture in mind, we ask three questions: why threshold, why slow, why discrete drops.
1. Why a threshold? Hydrostatic vs capillary
Water at depth $H$ presses on the bottom with a pressure $P=\rho g H$, where $\rho=1000$ kg/m³ and $g=9.81$ m/s². The full depth $H_{max}=0.07874$ m gives $P_{full}=772$ Pa.
A pinhole does not leak automatically, because the water-air interface can curve and support a pressure difference. Surface tension $\gamma=0.072$ N/m is energy per unit area, which is the same as force per unit length. The contact angle $\theta$ is the angle between the water surface and the wall, measured at the line where water, air, and plastic meet. Bulk polyethylene is hydrophobic, with a contact angle of 95-110°. This note quotes the range without a source. We also do not state the temperature for $\gamma=0.072$ N/m and $\mu=0.001$ Pa·s, and both depend on it. For $\theta>90°$ the wall repels water, and the interface resists the entry of water into the hole. The resisting pressure depends on $|\cos\theta|$. We choose the upper end of the range, $\theta=110°$, so $|\cos\theta|=0.34$. This is the largest value in the range. At $\theta=95°$, $|\cos\theta|=0.087$, and every $H_{crit}$ below would be 3.9 times lower. We did not measure the contact angle on your container.
A curved interface supports a pressure jump $\Delta P_{cap}$ given by the Young-Laplace equation $\Delta P_{cap}=2\gamma/R$. In a cylindrical hole the radius of curvature is $R=r/|\cos\theta|$. The maximum pressure that the interface can hold is therefore
$$P_{cap}=\frac{2\gamma|\cos\theta|}{r}$$
A leak starts only when $\rho g H > P_{cap}$. This gives the critical height:
$$H_{crit}=\frac{2\gamma|\cos\theta|}{\rho g r}$$
For $r=0.08$ mm, $H_{crit}=6.24$ cm. For $r=0.10$ mm, $H_{crit}=4.99$ cm. For $r=0.12$ mm, $H_{crit}=4.16$ cm. All three values are below the full height $H_{max}=7.87$ cm, so the container leaks above these levels. The smallest radius that leaks at the full level is $2\gamma|\cos\theta|/(\rho g H_{max})=0.063$ mm. This note considers radii of 0.08-0.15 mm. It cites no measured size distribution for real defects.
Figure 1 — Capillary threshold: This figure answers the question: why does a threshold exist? Panels (a) and (b) show the container at $H=3.0$ cm and $H=7.0$ cm for a hole of radius $r=0.10$ mm, with the hole magnified. For $H < H_{crit}$ there is no leak. For $H > H_{crit}$ there is a leak. The chart gives $H_{crit}$ against $r$ and marks the smallest radius that leaks at the full level, 0.063 mm. The figure shows $P_{cap}=2\gamma|\cos\theta|/r$ and $H_{crit}=P_{cap}/\rho g$. Without these two relations, we cannot explain why the container does not leak below a certain level.
2. Why slow? A small hole, viscous loss, and exit kinetic energy
Once the meniscus breaks, water flows through a channel of length $L≈0.8$ mm and radius $r≈0.1$ mm. Suppose the container drips at one drop per second. Section 3 shows that each drop has a volume $V_{drop}=27.7$ µL, so the flow rate is $Q=27.7$ µL/s. The mean velocity is $v=Q/(\pi r^2)$. At the fitted radius $r=0.1256$ mm this gives $v=0.56$ m/s and a Reynolds number $Re=\rho v D/\mu=140$, where $\mu=0.001$ Pa·s and $D=2r$. This is below the pipe-flow transition value of about 2000, so the flow is laminar. The drip is slow because the hole area $\pi r^2=5.0×10^{-8}$ m² is small. It is not slow because the velocity is small. The Bernoulli velocity $\sqrt{2gH}=1.24$ m/s through the same area would give 62 µL/s, which is 2.2 drops per second. The viscous part of the flow follows from the Navier-Stokes equations with the no-slip condition $u=0$ at the wall, where $u$ is the fluid velocity. It gives the Hagen-Poiseuille law:
Here $\Delta P=\rho g(H-H_{crit})$ is the pressure difference that drives the flow. The $r^4$ dependence is the key point for small holes. Halving the radius cuts the viscous flow by a factor of 16, so a small change in $r$ changes the drip rate by a large factor.
The Hagen-Poiseuille law omits the kinetic energy of the water that leaves the hole. With that term included, the driving pressure splits into a kinetic part and a viscous part:
$$\rho g (H-H_{crit}) = \tfrac12\rho v^2 + \frac{8\mu L v}{r^2},\qquad v=\frac{Q}{\pi r^2}$$
We solve this quadratic equation for the positive root $Q$. The kinetic term assumes a uniform exit velocity. A parabolic exit profile would double it, and then 1 Hz needs $r=0.135$ mm. At $r=0.1256$ mm the kinetic term is 156 Pa and the viscous term is 227 Pa, out of 383 Pa. For small $r$ the viscous term dominates and $Q$ follows $r^4$. For large $r$ the kinetic term takes over and $Q$ grows more slowly. The channel is also shorter than the entrance length $0.06\,Re\,D≈2$ mm, a textbook coefficient that we did not check against a source, so the viscous term is itself approximate. Two assumptions also enter this balance. First, the capillary back pressure $2\gamma|\cos\theta|/r$ that sets $H_{crit}$ keeps its static value while the water flows, so the driving pressure is $\rho g(H-H_{crit})$. A moving contact line can have a different angle, and we did not model it. Second, the kinetic term assumes that the kinetic energy of the exit flow is lost. If the growing drop recovers part of it as pressure, the fitted radius lies between 0.113 mm (no kinetic term) and 0.1256 mm.
Figure 2 — Flow through the hole: This figure answers the question: why is the flow slow? The magnified circle shows the 0.8 mm channel of radius $r=0.1256$ mm, drawn with true proportions ($L/D=3.2$), with the parabolic velocity profile, $v=0.56$ m/s and $Re=140$. The bar splits the driving pressure $\Delta P=383$ Pa into 156 Pa kinetic and 227 Pa viscous. The chart gives $Q$ against $r$ at the full level. The viscous term alone (Hagen-Poiseuille) falls as $r^4$ and gives 1 drop/s at $r=0.1134$ mm. With the exit kinetic term, 1 drop/s needs $r=0.1256$ mm. At $r=0.10$ mm the model with the kinetic term gives 0.39 Hz, and at $r=0.15$ mm it gives 1.79 Hz.
3. Why discrete drops? Tate's law
Outside the container, the water hangs as a pendant drop. The drop is attached to the wetted spot around the hole, a circle of radius $r_{out}$. The drop grows until its weight $mg$ exceeds the surface tension force $2\pi r_{out}\gamma f$, where $m$ is the drop mass. We assume the Harkins-Brown correction factor $f≈0.6$. For these inputs $r_{out}/V_{drop}^{1/3}=0.33$, and the Harkins-Brown factor depends on this ratio. We did not check $f≈0.6$ against the Harkins-Brown table. We use $r_{out}≈1$ mm:
The target flow rate for one drop per second is $Q_{target}=1×27.7$ µL/s $=1.66$ mL/min.
We chose $r_{out}≈1$ mm as an assumed value. We did not measure it and we cite no source for it. The geometry requires only that $r_{out}$ is larger than the inside radius $r$. Here it is 8.0 times larger than $r=0.1256$ mm. You can measure it on your container: it is the radius of the wetted spot where the drop hangs. The drop volume is proportional to $r_{out}$. With the other parameters unchanged, $r_{out}=0.5$ mm gives $V_{drop}=13.8$ µL and a fitted radius $r=0.105$ mm for 1 Hz. The value $r_{out}=2$ mm gives $V_{drop}=55.3$ µL and $r=0.156$ mm. A factor of 2 in $r_{out}$ therefore changes the fitted $r$ by −16% and +24%, against $r=0.1256$ mm for $r_{out}=1$ mm.
The model also sets the pressure outside the hole to zero. A pendant drop of radius $r_{out}=1$ mm holds a Laplace pressure $2\gamma/r_{out}=144$ Pa, which is 1.47 cm of water. This is up to 38% of the driving pressure of 383 Pa at the fitted radius. A model that includes it needs a larger $r$ for 1 Hz. We did not compute that value.
Figure 3 — Tate's law: This figure answers the question: why are the drops discrete? The upper panel shows the whole container, full to $H_{max}$. The magnified circle shows the drop hanging from the wetted spot of radius $r_{out}$. The weight $mg=\rho V g$ and the surface tension force $2\pi r_{out}\gamma f$ are both 271 µN. The chart gives $V_{drop}$ against $r_{out}$. Without the force balance, we cannot convert the 1 Hz target into the flow rate $Q=27.7$ µL/s.
4. Numbers for your container
$r$
$H_{crit}$
Excess
$Q$
$f_{drip}$
0.10 mm
4.99 cm
2.88 cm
10.9 µL/s
0.39 Hz
0.12 mm
4.16 cm
3.71 cm
23.4 µL/s
0.85 Hz
0.1256 mm
3.97 cm
3.90 cm
27.7 µL/s
1.00 Hz
0.15 mm
3.33 cm
4.55 cm
49.6 µL/s
1.79 Hz
The table uses the model with the exit kinetic term. A radius $r=0.1256$ mm $=125.6$ µm gives 1 Hz at the full level. This note does not give a source for the size of real defects.
The level falls according to $dH/dt=-Q/A_c$, where $A_c$ is the area of the water surface and $Q$ comes from the model with the exit kinetic term. This equation is not exponential, so we integrate it numerically. We take $A_c=9.4×4.2$ in $=254.7$ cm², the footprint of the container outline. We did not measure the water surface, and rounded corners would reduce it. The full volume is then $A_c H_{max}=2006$ mL, which is the volume of the 9.4 × 4.2 × 3.1 in box. For $r=0.12$ mm, the level falls from $H_0=H_{max}$ toward $H_{crit}$. The volume that leaks before the flow stops is $A_c(H_0-H_{crit})≈946$ mL, and about 1059 mL stays in the container. The leak reaches 63.2% of this volume after 9.98 h and 99% after 37.6 h. An earlier draft used $A_c=127$ cm² with no recorded source. That value gives 472 mL, 4.97 h and 18.7 h.
Verdict: The model reproduces the estimated 1 drop/s, but one parameter is fitted and four are chosen. The radius $r=0.1256$ mm is solved from the 1 Hz target, so it is not an independent prediction. The values $L=0.8$ mm, $r_{out}=1$ mm and $f=0.6$ are chosen, not measured. The value $|\cos\theta|=0.34$ is the upper end of the 95-110° range for bulk polyethylene. It was not measured on this container. The model includes the exit kinetic term with a uniform exit velocity. With a parabolic exit profile, 1 Hz needs $r=0.135$ mm. It uses the Hagen-Poiseuille viscous term although the channel is shorter than the entrance length (Section 2). It also sets the pressure outside the hole to zero (Section 3). The area $A_c=254.7$ cm² (the 9.4 × 4.2 in outline) affects only the level-fall times and volumes. A threshold of 3.3-6.2 cm for $r=0.08$-0.15 mm explains why there is no leak below a certain level.
With these inputs, you can compute $H_{crit}$, $Q$ and the drip rate for any hole radius, and you can fit $r$ to a measured drip rate.
5. Glossary
$\rho$
1000 kg/m³
$g$
9.81 m/s²
$H$, $H_{max}$, $H_{crit}$
Depth, full 7.874 cm, critical
$\gamma$
0.072 N/m
$\theta$
Contact angle, 110° ($|\cos\theta|=0.34$)
$r$, $r_{out}$
Inside ~0.1256 mm, outside ~1 mm
$L$
Wall 0.8 mm
$\mu$
0.001 Pa·s
$Q$
Root of $\rho g (H-H_{crit}) = \tfrac12\rho v^2 + 8\mu L v/r^2$