Water flow from pressurized tank - versus temperature

[Anachronist]

Bernoulli's equation gives us this as the flow rate from a pressurized tank:
$$v = C_v \sqrt{2gh + \frac{p}{\rho}}$$
where $C_v$ is the velocity coefficient, $g$ is the acceleration due to gravity, $h$ is the height of the water above the exit hole, $p$ is the excess pressure above ambient, and $\rho$ is the fluid density (1000 kg/m³ for water).

My problem is, the viscosity of water changes significantly with temperature... like, a factor of 2 difference between cold water and warm water.

Wouldn't that mean hot water would exit the tank faster than cold water, for the same pressure?

I would think, with such a large difference in viscosity between cold and hot water, I'd see a large difference in flow rate also. If I wanted to increase the mass flow rate without increasing pressure or hole size, it would seem that increasing the temperature of the water would accomplish that.

How would I calculate the flow velocity to account for temperature?

The velocity coefficient $C_v$ seems like a fudge factor that would account for viscosity. The Engineering Toolbox gives a value of 0.97 for water, but I suspect that this is for a specific temperature. And I haven't been able to find anything that helps me account for viscosity when calculating the flow from a pressurized tank.

 

[gmax137]

First, you are mixing the terms "flow rate" and "velocity." These seem related - and they are - but to relate them you need the flow area. The flow area that "goes with" the velocity is not the area of the hole, rather the flow stream contracts as it leaves the hole. See in your Toolbox link further on down the page, look for the contraction coefficient. The overall coefficient Cd = 0.6.

This is as you suspect a function of viscosity, but it is a weak function. Normally these coefficients are given as functions of Reynolds number (which includes viscosity). But what is seen is, for large Reynolds number, the function is flat (coefficient Cd is essentially constant). As long as the flow is fully turbulent (large Re) you can use a Cd = 0.6.

Also note that "hole in tank" implies the area of the hole is small in comparison to the tank diameter -- in other words the velocity in the tank in negligible and there is "no" friction at the tank walls. The only flow resistance is that associated with the flow through the hole.

Hope that helps some.

 

[Chestermiller]

The equation you wrote assumes inviscid flow.

 

[Anachronist]

First, you are mixing the terms "flow rate" and "velocity." These seem related - and they are - but to relate them you need the flow area. The flow area that "goes with" the velocity is not the area of the hole, rather the flow stream contracts as it leaves the hole. See in your Toolbox link further on down the page, look for the contraction coefficient. The overall coefficient Cd = 0.6.

Yes, you're right.

In my case, the 'tank' is a soda bottle, the 'hole' is the neck. This is for a water rocket application, in which I'm trying to validate a physics-based model I developed for the thrust. I'd say the taper toward the neck qualifies as a "well rounded aperture" which would have a contraction coefficient of 0.97 according to that Engineering Toolbox page.

What I'm after is to maximize the mass flow rate of water through the aperture, given that I cannot change the aperture size or the initial internal pressure. So I was wondering if using warm water would make any difference, and if so, how to account for it.

Also note that "hole in tank" implies the area of the hole is small in comparison to the tank diameter -- in other words the velocity in the tank in negligible and there is "no" friction at the tank walls. The only flow resistance is that associated with the flow through the hole.

Hmm, in this case the velocity in the tank wouldn't be negligible. The diameter of the hole is about 20% of the diameter of the bottle. At the extreme, a cylindrical tank with a hole equal to the diameter of the cylinder is equivalent to water flowing out of the end of a short pipe. So I wonder if my problem is closer to a pipe flow than a tank leak?

The equation you wrote assumes inviscid flow.

OK. In that case, warming up the water would make the real-world situation approach the inviscid ideal, would it not? Or would the difference be negligible?

 

[Chestermiller]

Yes, you're right.

In my case, the 'tank' is a soda bottle, the 'hole' is the neck. This is for a water rocket application, in which I'm trying to validate a physics-based model I developed for the thrust. I'd say the taper toward the neck qualifies as a "well rounded aperture" which would have a discharge coefficient of 0.97 according to that Engineering Toolbox page.

What I'm after is to maximize the mass flow rate of water through the aperture, give that I cannot change the aperture size or the initial internal pressure. So I was wondering if using warm water would make any difference, and if so, how to account for it.Hmm, in this case the velocity in the tank wouldn't be negligible. The diameter of the hole is about 20% of the diameter of the bottle. At the extreme, a cylindrical tank with a hole equal to the diameter of the cylinder is equivalent to water flowing out of the end of a short pipe. So I wonder if my problem is closer to a pipe flow than a tank leak?OK. In that case, warming up the water would make the real-world situation approach the inviscid ideal, would it not? Or would the difference be negligible?

It would probably be negligible. But heating the water would increase the vapor pressure of the water. But, be careful, as this might not be safe.

 

[Anachronist]

It would probably be negligible. But heating the water would increase the vapor pressure of the water. But, be careful, as this might not be safe.

Well, the tank is pumped up to 100 psi after it is filled with water, so the vapor pressure would be part of the initial internal pressure. And because the tank is a soda bottle, the water can't be too hot, just warm enough that dipping your finger in it may be uncomfortable but not burning. Any warmer than that, and the plastic starts to soften. That's where the safety issue comes in, not from vapor pressure, but weakening the tank walls. A PET soda bottle can take about 160 psi before rupturing, but I suspect that safety margin is drastically reduced if the water temperature is too hot... which is why I bought 50 feet of pressure hose, and safety goggles, for pumping it up.