How do pumped flows combine in a Y connector?

Question

I'm curious about pumping a fluid through two different tubes (coming from two different pumps) into a Y connector. The third side of the connector runs into a collector without any resistance (other than that of the length of tubing).

1. If each pump runs at a different rate, say one is twice as strong as the other, do I run any risk of any backing up into the weaker pump? What if one is three times stronger than the other, etc.?

2. Does the third side of my Y connector (the side the fluid leaves through) now pump with the equivalent of pump a + pump b total power? Are they averaged together? Is it diminished?

3. Does the type of fluid matter for parts 1 and 2 above?

This question assumes that all tubing is the same thickness, but answers regarding different thicknesses could also be interesting (and worth extra credit!)

Answer 1

General rule when dealing with fluids

The fluids will travel the path of least resistance. Treat them as dumb and lazy and you'll understand the process better; fluids will only travel where you make them and if you allow them to take a shorter path they will take it.

1 If the smaller pump puts less pressure than the third side orifice then the small pump will likely cavitate and back flow. The relative pressure of the two pumps will play into where they operate on the pump curve however if they both push higher pressure than the orifice there will not be back flow.

2 The resulting pressure of the orifice is a complicated answer solved by Bernoulli's Equation (https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=bernoulli%20equation). In real life the texture, size, fullness, route and shape of the pipe opening play major roles in determining if the pressure is higher, lower or nil.

3 They type of fluid does not change how physics work, unless you're talking about pumping different fluids in each pump and making the Y into a mixing chamber (asking for trouble).

If you do run into an issue that the pressure differential in the system is too great and the smaller pump would back flow are you able to create an air gap in the (liquid) system? Rather than directly piping the two pumps into each other have them pour into a common container which gravity discharges through the final orifice.

4 Bonus based on comment The same answers would apply to liquids or gasses. you would still need the same type of information. In a previous life, I designed piping systems which covered hundreds of miles and thousands of junctions in each system. There were 3-8 fans passing 600k-1,000k cfm into different points in the system. We had to determine with a fairly high reliability (90%+-) what the pressure and air flow at each junction would be or faced shutdown by regulators.

Answer 2

Pumps output based on a pump curve graph that has been determined empirically by the manufacturer for that specific pump.

In your question, "Twice as strong as the other", is not meaningful because it is a complex relationship of output pressure(head) and output flow. If you have the pump curve for both pumps and the minor and major losses you can solve for a flow direction, total flow, pressures, etc.

Answer 3

Pumps are complicated

Many questions regarding fluid machinery start with these two assumptions:

1. A pump or gravity is needed to make the fluid go faster
2. High pressure means faster flow

All of these are correct in principle, but none of them are accurate. The first is inaccurate. Due to conservation of mass, fluid flowing into a pipe on one end will be equal to the flow out the other end. While the pump will make the fluid flow faster, the reality is the pump imparts energy on the flow, until the added pressure loss from the higher flow rate is equal to the added energy.

Because of how complicated the energy loss in a pipe can become (Reynolds numbers, Darcy friction factor, hydraulic diameter, etc.), since most engineering problems deal with water and piping, Hazen Williams equations are typically used. In addition to simplifying the pipe, we simplify the pump. Pumps have a curve, which is always convex. By taking the max flow rate (at no pressure), and the maximum pressure (at no flow), a linear equation can be developed - just use the intercept form. This simplifies the pump. Now, a set of equations can be developed called the Hardy Cross Method. My textbook shows a typical setup:

A source of unknown pressure is feeding the system 3 cfs of liquid - via valves, pumps, or who knows what. The exits have 1 and 2 cfs - matching the conservation of mass. Again, pressure at these points is unknown, but not unlike pressure loss across a resistor, the energy difference between each point means that energy is conserved across the loop. This is why Hardy-Cross Method can be seen to be the fluid version of Kirchoff's Voltage Law

As can be seen, the top line (only 6" and a full 3000' + 4000') will get minimal flow, while the bottom line, which is shorter and has larger diameter pipes, should be expected to hold most of the flow. When adding various pumps, it can be seen that even with sufficiently large pumps, circulation starts to become inevitable:

The Y-Connector

Your Y-Connector method is interesting because you now have an unknown flow rate - but a known pressure. In this case, you need to use the air in the atmosphere to connect the pressures in a loop (See for example, page 15 of this pdf, showing how this is done. Also, don't forget to add in all of the minor loss coefficients

However, all of this is a model. To prevent backflow, always add a check valve to your pumps. Because you never know what could happen, and backflow can be very expensive.