Im having trouble with a scenario where a flow 𝑄𝑆𝑡𝑎𝑟𝑡 splits into two parallel pipes 𝑄𝐴 and 𝑄𝐵 then rejoins 𝑄𝐸𝑛𝑑 before exiting the control volume
what makes this scenario difficult is the parallel pipes are of varying diameter
At the diffluence Pipe A has a diameter 2D and Pipe 2 has a diameter of 1D by the time they reach the confluence they have reversed diameter now have Pipe A has a dimeter of 1D and Pipe B has a dimeter of 1D
Start - assume a flow velocity 5ms static pressure SP 100
$2A.5ms = 1A.10ms$
$1A.5ms^-1 = 2A.2.5ms^-1$
Head Loss What we know
All elements of flow converging at WILL have the same head loss. The flow will adjust automatically so that the head loss in each branch pipe WILL BE THE SAME
According to resistance coefficient tables the divergent pipe has a K value of 0.46 and the convergent pipe has a K value of 0.1
As these are Losses are proportional to – velocity of flow, this suggests that the expansion pipe will decrease its flow (to decrease its losses) while the convergent pipe must increase its flow to maintain continuity This means that the flow rates have diverged not come together but we know that they must be the same at the exit of the control volume examined
Continuity also tells us that the total flow rate must be the same at all points in the pipe
$𝑄_𝑆=𝑄𝑎_1+𝑄𝑏_1 =𝑄a_2+𝑄b_2 =𝑄_𝐸$
$𝑣_𝑆.𝐴_𝑆=𝑣_a. 𝐴_a+𝑣_𝑏.𝐴_𝑏=𝑣_𝑎.𝐴_𝑎++𝑣_𝑏 𝐴_𝑏 =𝑣_𝑒.A_𝑒$
So on total head/ stagnation value we will have the same value at the convergence as both paths have experienced the same head loss but Bernoullis tells us that we have very different velocities and static pressures at this point .
My question is how at the confluence does this follow that we do not require the same value of pressure and velocity at the confluence for both streams?
If this can occur we must then have a mechanism to achieve the expected uniform velocity and pressure (Not considering head losses) at the exit $𝑣_𝑆.3 =𝑣_𝑒.3𝑒$
What would this mechanism be ?