Fluid flow - pressure v volume?

[Frangelo]

This a practical problem for the design of a system which uses a water driven hydraulic motor.


Assume a 100 gallon tank is filled with water, constantly filled further pressurized by a pump to 500psi

Assume a hydraulic powered motor which operates with intake required at 200psi, flow of 10 gallons per minute.

How can both pressure stepped down, and flow controlled to the required input rate at the same time? Assume there will be a vave which controls the diameter of the opening at the bottom of the water tank through which the water flows. The pipe connected the two can be of arbitrary volume/diameter, but the pipe connector at the motor end is say 5cm in diameter.

Is this possible, or are pressure and flow equal to some constant which makes fixing both at arbitrary values impossible?

If there are relevant equations can you please point me in the right direction? Thanks.

 

[voko]

Water is incompressible, so the inflow must be equal to the outflow. The inflow, assuming the tank is always full, is controlled solely by the performance of the pressurizing pump - can it match the requirement?

 

[Frangelo]

Thanks but don't think this is quite the full answer. Assume that the pump/water tank can always maintain the water pressure and volume at 500psi no matter what the outflow.

But imagine that the interface between the tank and output water pipe is pin-hole sized, but the hydraulic motor wants a flow rate of 10/gal minute. While the I can believe that the pressure at equilibrium will be the same inside the tank and inside the connector pipe to the motor (assuming no outflow through the motor), I don't think this is going to be true on an instaneous basis (i.e. as fluid is flowing out of the pipe through the motor.) My instinct tells me that pressure will in the pipe will drop very quickly as fluid flows out through the motor if the input flow to the pipe is less than rated 10/gal minute.

Are you saying that if you make the hole big enough to accommodate the flow rate required, there's no way to reduce the pressure inside the connector pipe to a pressure lower than inside the water tank? (i.e. no way to get the pressure down from 500psi to 200psi while maintaining the 10 gal/minute flow rate?)

 

[voko]

I do not believe there is a simple solution to your problem. Only if you can somehow guarantee, which is impossible in practice without some feedback mechanism, that the flow is steady at the given rate and both pressures are also steady at the specified values, and further assuming that the flow is laminar everywhere, you could achieve what you want by setting the cross-section of the tank exit. This is done by applying Bernoulli's principle: $$

\frac {v_{in}^2} {2} + \frac {p_{in}} {\rho} = \frac {v_{out}^2} {2} + \frac {p_{out}} {\rho}

$$ (and I have further assumed that the difference in height is negligible throughout)

The volume of the tank is 100 gallons, or 23100 cubic inches. The volumetric flow rate is 10 gallons per minute, or 2310 cubic inches per minute. Let's assume it is a cube, then its side is 28.5 inches. Then the velocity of the flow in the tank is 2.9 in/minute, or 0.0012 meters per second; which just can be taken as zero for practical purposes. Then the equation simplifies: $$

v_{out} = \sqrt {2 \frac {p_{in} - p_{out}} {\rho}}

$$ The pressure difference is 300 psi, or 2 MPa, and the density of water is 1000 kg per cubic meter, so we have $v_{out} = 63 \ m/s$ (141 mph, not bad). So that the volumetric flow rate be preserved, the product of this speed and the cross-section of the tank exit must be equal to 2310 cubic inches per minute, or 0.00063 cubic meters per second. Thus the cross-section of the exit must be 0.00001 square meters, which is quite tiny. If it is circular, its radius is 1.8 mm. Because we have ignored a lot of possible complications, you should treat that only as a rough estimate, and make the exit bigger than that, and equipped with some adjustable choke, so that you could tune it.

I would like to mention that 500 psi is a lot of pressure. And 100 gallons at this pressure is a lot of energy. Are you sure you are ready to handle that?

 

[Frangelo]

That's great Voko, thanks.

I'm not sure exactly what the exact pressure and flow requirements will end up being (the 500PSI/200PSI 10 gal/min were just placeholder). In practice the pressure inside the tank will be measured constantly as will the flow rate and pressure to the motor. The valve will be computer controlled so there will be a feedback mechanism.

If there is some more conventional way (other than a valve) for stepping down pressure and flow rate to desired target please point me in the right direction.

Many thanks for these formulas and the example, I think I'll be able to work through to what I need.

 

[Frangelo]

One other question -- I don't see a term reflecting viscosity of water in the bernoulli equation you mention. Is that because it has negligible impact on the result? If the flow consisted of more viscous hydraulic fluid would the same equations apply, just changing density value for the proper one for water? Or would viscosity then have to factored in?

 

[voko]

The assumption is that the flow in the tank is "almost no flow", so viscosity cannot affect that much. The extent of the flow at the exit is limited by the length of the exit, which is assumed short, so again not a lot of an effect here.

A more accurate modelling of the flow rate at the exit would use the "flow coefficient", which are determined experimentally for various valves. If I remember correctly, the flow factor for a circular orifice is about 0.6, so you would need to increase the area of the exit inversely proportionally to that, which gives 2.3 mm radius. You could probably use some standardized adjustable valve whose dimensions are in this ballpark (and which is rated for the pressure).

Note also that all this completely ignores whatever happens downstream, which will be significant for the flow velocity you have.