HYD Flow

Yes, use this formula, Flow rate = Cross Sectional Area of Pipe x Flow velocity

or Q = AxV; Q-flow rate, A-Area cross section, V- flow velocity

Q_supply = Q_discharge = Q_(5/8)+Q_(1/4) is it necessary to solve it all for you?

I could simply do it for 10 bucks. or by a bottle of this

Thank you very much! I got it from here.

One last question, does Ft/sec (supply) = Ft/sec (5/8") + Ft/sec (1/4")?

Thank you so much again.

No.

Flow (volume per time) is approximately proportional to the 2.5 power of pipe diameter. Then velocity (distance per time) can be computed. It will be less in the smaller pipe.

I am afraid you did not understand the problem. I shall make a similitude with an electric circuit may be il will be easier for you to get the point:

- imagine you have a current source (this is an electric equivalent of a hydraulic pump which delivers flow) and you connect to this source 2 resistors in parallel (the valves cross sections).

Can you tell me how much current will flow through each one ?

As the question was written the answer is "NO" since the flow through a valve will depend on the opening of it and on the pressure drop at this opening as for the resistance you must know the resistance values and the voltage drop at each one.

The section of pipes or hoses will define the velocity for the given flows but not the opposite.

Have a look at hydraulics how it works.

Thank you very much, after at your comment I looked at the valve spec and on of them only allows 1 GPM through, thus if both valves were open 100% this one valve will allow 1 GPM therefore 9 GPM will flow through the other one back into the Charge port of the piston pumps.

My answer is undone and true to the assumption that loses and static heads are the same both discharge branch and no pressure head. Of course it will matter if both ends of the tee have different lenghts and fittings plus if both ends are to supply a close air water tank.

It will depend on what you've got downstream of those two pipes, Murphy, because if one of them were open-ended and the other were closed, then all the flow would go down the first branch and none the second. As you haven't told me anything about those tbings, then that's about all anyone could say.

you're one of a kind Murphy, LOL.

In the conditions you have outlined you cannot predict a hydraulic flow with any accuracy. Even if you have all the information about where the flow is going and the amount of work that the flow will perform, the slightest obstruction in one of the pipes or a small amount of sticktion on a valve will disrupt the calculation. Something as insignificant as a burr on the inside of one of the pipe fittings will produce disproportionately large distortions in the flow.

You would get better results by working backwards from how much flow you want to achieve in each leg of the circuit. Then put a flow splitter (i.e. two positive displacement pumps attached to a common shaft and driven by the flow itself, rather like a flow meter) into the line instead of a T junction. Specially shaped spacers can be fitted to each of the rotors in the flow splitter to adjust the proportion of flow directed to the two outputs, but I should caution you that most flow splitters are only accurate to 1-2%.

You can do a theoretical calculation only. Just like water, hydraulic oil will flow in the pass of least resistance. You will need to know what "load" you have on each side of the T.

If everything is equal in both branches except the pipe diameter, you could estimate the relative flows using a ratio of their cross-sectional areas.

In this case the ratio of the cross-sectional areas is (5/8)² ÷ (1/4)² = 6.25/1

If everything else is exactly equal, 10 gpm would split into 2 streams of approximately 8.6 gpm (5/8" ID) and 1.4 gpm (1/4" ID).