"Y" Fitting and Energy Equation

[Mechwill]

Hi Engineers,

I have searching an analysis for this "Y" fitting configuration. Consider two inlets and one outlet. Assuming fluid is water and the fitting has the same diameter on all three ports. I can simply write and equation Q_outlet = Q_inlet1 + Q_inlet2. Can I use energy equation to analysis this "Y" fitting? What I mean by that is this: E_in = E_out, which means E_inlet1 + E_inlet2 = E_outlet? Somehow I feel that I am way off the track

 

[LittleInch]

Why do you feel that way? sounds good to me on the conservation of energy principle.

My motto: Learn something new every day

Also: There's usually a good reason why everyone does it that way

 

[zdas04]

There has to be more to it than you've said. Total energy is conserved. Pressure drop (potential energy) will be converted (reversibly) to velocity and (irreversibly) to heat due to friction. The interaction of the two streams can create rotational flows that have to be accounted for. If you keep all your energy conversions straight then energy in will equal energy out. I'm not sure what you can do with that factoid, but it is true.

David Simpson, PE
MuleShoe Engineering

Law is the common force organized to act as an obstacle of injustice Frédéric Bastiat

 

[Mechwill]

My mistake for not including rest of energy. Allow me to re-write this equation.

E_in = E_out ----->

delta_E_system = E_in - E_out
= delta_Internal Energy + delta_Kinetic Energy + delta_potential Energy (Conservation of Energy)
= delta_heat transfer + delta_work transfer (Energy Transfer)


The intention here is to use this equation and find out the relationship of pressure of inlets and pressure of outlet. However, I am making myself complicated by introducing equations into this application. I guess what I have to do is to put 3 pressure gauges and flow meters on each port and play around with it. Just trying to get some ideas of how this "Y" fitting works (efficiency, losses, etc)