11th April 2007 - 02:23 PM
Unfortunately, that is not correct. In fact, Bernoulli's equation is NOT sufficient to determine how the flow splits into the two pipes, because the flow in each pipe also depends on conditions downstream of the split. For instance, if one pipe is partially clogged later on, you will see less flow into it here at the junction.
Here is an analogy which may be useful in thinking about these problems (assuming you are familiar with electric circuits). The Bernoulli formula B gives the energy per unit mass carried by the flow, just as the voltage gives the energy per unit charge carried by electric current. The mass flow rate m' acts just like electric current I, in that both mass and charge are conserved. Power carried by the flow is B*m' and power carried by an electric current is V*I. Current splits when two wires are parallel, and mass flow splits when two pipes are parallel. The voltage drop across two parallel wires is equal, and the drop of B (head loss) across two parallel pipes (i.e. they split and then rejoin later) is also equal. Voltage changes across two resistors or batteries in series sum up, but the currents through them are equal, and head changes across two clogs or pumps in series sum up, but the mass flows through them are equal.
The analogy is quite exact, and all the same kind of reasoning can be applied. In the case of your branching pipes the analogy would say that since the voltages of three wires that are joined at a point are all equal, then B in the three pipes must all be equal at the junction (though they may change later due to clogs or pumps). This means that you DO NOT ADD the B values for the two smaller pipes. Instead, they are EACH equal to the B of the larger pipe. Therefore, your equation should read:
You will also be able to apply the equation of continuity, m'1 = m'2 +m'3, which in terms of velocity is v1*A1*density = v2*A2*density + v3*A3*density, where A is area.
There are 4 unknowns (two velocities and two pressures) but only 3 equations, which is why more information about conditions downstream is needed, just as with electric circuits.
By the way, many people make the mistake of thinking the pressure must be equal in all 3 pipes. This is not true. B must be equal in all 3 pipes, because it represents the energy per unit mass carried by the flow. If B were higher in one exit pipe than in the other, then it would also be higher than B in the inlet pipe (since the average value of B in the inlet and exit flows must be equal by conservation of energy). But that would mean that you could concentrate energy into one flow without doing any mechanical work. If that were true, you could build a system that would violate the second law of thermodynamics, by using heat o drive the flow in the first pipe, then concentrating the energy into one flow, and then converting it back to heat. Since there would be more energy per unit mass, the final temperature would be higher than the initial temperature of the heat source that was driving the flow, so you would have made heat flow from a cold temperature to a hotter one with no applied work. If that were possible, I would already have patented it.
Hope this helps!