# Fluid dynamics: Initial velocity of milk outflow from carton

## Homework Statement

A hole is punched in a full milk carton, 10cm below the top. What's the initial velocity of outflow?

## Homework Equations

P2 = P1 + ρgh
Conservation of energy of fluids: P2 + 1/2ρv22 + ρgy2 = P1 + 1/2ρv12 + ρgy1

## The Attempt at a Solution

Set the top of the carton as P1, and the point 10cm below the surface as P2.
Assume that milk has the same density as water: 1000kg/m3

P1 = atmospheric pressure = 101325Pa
P2 = 101325 + 1000(9.8)(0.1) = 102305Pa

P2 + 1/2ρv22 + ρgh2 = P1 + 1/2ρv12 + ρgy1
The milk doesn't have kinetic energy at the top of the carton, so K=1/2ρv22 =0
At 10cm below the surface, y=0 (although this is still elevated above the ground, we're trying to find the change in energy between the top of the carton and hole punched in it); therefore U = ρgy1 = 0

P2 + ρgh2 = P1 + 1/2ρv12
102305 + 1000(9.8)(0.1) = 101325 + 1/2(1000)v12
103285 - 101325 = 500v12
V1 = 1.98m/s

Is my method correct? The actual answer is V=1.4m/s, but perhaps it's because I assumed that the density of milk is the same as water....

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gneill
Mentor
I think you'll find that for the fluid stream coming out of the hole the pressure is atmospheric pressure. That's the environment it is in as soon as it passes through the wall of the carton, and that's where you want to determine its speed. So the pressure terms should cancel, and you're left with the energy change due to the change in elevation.

Essentially this is the derivation of Torricelli's Law. I think a web search on that will be helpful

vetgirl1990
SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

A hole is punched in a full milk carton, 10cm below the top. What's the initial velocity of outflow?

## Homework Equations

P2 = P1 + ρgh
Conservation of energy of fluids: P2 + 1/2ρv22 + ρgy2 = P1 + 1/2ρv12 + ρgy1

## The Attempt at a Solution

Set the top of the carton as P1, and the point 10cm below the surface as P2.
Assume that milk has the same density as water: 1000kg/m3

P1 = atmospheric pressure = 101325Pa
P2 = 101325 + 1000(9.8)(0.1) = 102305Pa

P2 + 1/2ρv22 + ρgh2 = P1 + 1/2ρv12 + ρgy1
The milk doesn't have kinetic energy at the top of the carton, so K=1/2ρv22 =0
At 10cm below the surface, y=0 (although this is still elevated above the ground, we're trying to find the change in energy between the top of the carton and hole punched in it); therefore U = ρgy1 = 0

P2 + ρgh2 = P1 + 1/2ρv12
102305 + 1000(9.8)(0.1) = 101325 + 1/2(1000)v12
103285 - 101325 = 500v12
V1 = 1.98m/s

Is my method correct? The actual answer is V=1.4m/s, but perhaps it's because I assumed that the density of milk is the same as water....
The density of milk is not your problem here.

Review the derivation of Torricelli's Law from the Bernoulli equation:

https://en.wikipedia.org/wiki/Torricelli's_law

vetgirl1990
I think you'll find that for the fluid stream coming out of the hole the pressure is atmospheric pressure. That's the environment it is in as soon as it passes through the wall of the carton, and that's where you want to determine its speed. So the pressure terms should cancel, and you're left with the energy change due to the change in elevation.

Essentially this is the derivation of Torricelli's Law. I think a web search on that will be helpful
Extremely useful, thank you so much!

My logic (although wrong) was that the pressure at the hole would be affected by the volume of milk above it, which would have a direct effect on speed of the milk leaving the carton.
I suppose using P2 = P1 + ρgh then, would only be applicable [if for some reason], they were asking what the pressure was at the bottom of the carton vs. the top, in a closed container with no holes.

Chestermiller
Mentor
Extremely useful, thank you so much!

My logic (although wrong) was that the pressure at the hole would be affected by the volume of milk above it, which would have a direct effect on speed of the milk leaving the carton.
I suppose using P2 = P1 + ρgh then, would only be applicable [if for some reason], they were asking what the pressure was at the bottom of the carton vs. the top, in a closed container with no holes.
Yes. That's why it's called the hydrostatic equation.

haruspex
Homework Helper
Gold Member
2020 Award
What's the initial velocity of outflow?
This is rather a tricky question, too tricky for the question setter, it seems.
Torricelli's Law does not apply to the flow immediately after creating the hole. It applies only once a steady flow has been established.
The difference is observable. If you pull a plug out near the bottom of a barrel you will see the flow strengthen for the first second, or thereabouts.
The immediate flow can be estimated by considering rate of change of momentum. The water inside the tank, near the hole, is going from a standing start. If it exits at speed v through a hole area A, the mass flow rate is ##\rho A v##, the rate of gain of momentum is ##\rho A v^2##, which therefore equals the force ##\rho h g A##. This gives ##v=\sqrt{gh}##.
To get the ##\sqrt{2gh}## answer, it would be better to state that after some seconds the water is 10cm above the hole, and ask what the velocity is then.

Chestermiller
Mentor
This is rather a tricky question, too tricky for the question setter, it seems.
Torricelli's Law does not apply to the flow immediately after creating the hole. It applies only once a steady flow has been established.
The difference is observable. If you pull a plug out near the bottom of a barrel you will see the flow strengthen for the first second, or thereabouts.
The immediate flow can be estimated by considering rate of change of momentum. The water inside the tank, near the hole, is going from a standing start. If it exits at speed v through a hole area A, the mass flow rate is ##\rho A v##, the rate of gain of momentum is ##\rho A v^2##, which therefore equals the force ##\rho h g A##. This gives ##v=\sqrt{gh}##.
To get the ##\sqrt{2gh}## answer, it would be better to state that after some seconds the water is 10cm above the hole, and ask what the velocity is then.
Hi haruspex,

I'm a little confused. Shouldn't the velocity be zero when you first pull the plug since there have been no impulsive forces applied? I mean, doesn't it take time for the fluid to accelerate?

haruspex
Homework Helper
Gold Member
2020 Award
Hi haruspex,

I'm a little confused. Shouldn't the velocity be zero when you first pull the plug since there have been no impulsive forces applied? I mean, doesn't it take time for the fluid to accelerate?
Yes, but consider the first 'plug' of water to emerge. It will accelerate from nothing as it emerges, but by the time it is clear of the hole and falling freely it will have reached roughly the velocity calculated from force. An experiment is in order.

Chestermiller
Mentor
Yes, but consider the first 'plug' of water to emerge. It will accelerate from nothing as it emerges, but by the time it is clear of the hole and falling freely it will have reached roughly the velocity calculated from force. An experiment is in order.
In my judgement, only if the compressibility of the water were taken into account could this happen. Otherwise, to satisfy continuity, any initial velocity of any part of the water would mean that all the water would have to be moving initially. If compressibility were included, then an infinitecimal amount of water could start emerging at finite velocity without all the water it being affected.

haruspex