Hello, hope my answer will be of your help. If you have two inlets and one outlet, the pressures of both the inlets must be equal, otherwise the flow will always flow from the inlet with the higher pressure to the other two which have lower pressures.

So, P1 = P2 is mandatory for fluid to flow to 3, with P3

Hi, Logically it seems correct. But the issue is I am simulating such a flow in fluent and I am unable to decide which pressure to consider as total pressure required for the flow.

Moreover, the values of pressure are not the same at both faces as evaluated from FLUENT. This may be due to the fact that I have imposed a boundary condition that there is flow inlet at faces mentioned, but how to evaluate the net pressure drop

You have three parts of the boundary there fluid go in or out. In this case flow depends on two pressure drops. Look for example the paper in open access journal. There is correct statements of boundary value problems in your case.

Moshkin, N. P. On numerical solution of the incompressible Navier – Stokes equations with static or total pressure specified on boundaries / N. P. Moshkin, D. Yambangwai // Mathematical Problems in Engineering. — 2009. — Vol. 2009. — P. 1–27.

Pressure at the main inlet should be higher or equal to that at the port inlet. such that the pressure immediately before the inlet will be (P1-(DP/Dx)*(l/2)) which resembels the available pressure at that point. This pressure should be such that the incoming flow from the inlet get align to the direction of main flow or else again there will be loses due cross momentum. Now lets say if P1 = P2( port inlet pressure) then net pressure in the flow direction will be PP= (P1- (DP/Dx)*(l/2) - (cross flow losses)) where the second term is pressure losses per unit length, therefor PP gives the net pressure just ahead of the port. If you have two such port the effect can be additive and the total pressure loss in the duct over the entire length will become something like PPF= (P1- (DP/Dx)*(l/2) - 2* (cross flow losses) -Pexit) if the fluid is entering from the both the port inlets at velocity slightly greater than that in the main stream.

Hi

Thanks for your reply. But I have certain concerns regarding your reply. As you mentioned the net pressure in the flow direction will be PP= (P1- (DP/Dx)*(l/2) - (cross flow losses)). But my concern is since my flow arte has changed ahead of port 2, then how can I have same DP/Dx. Wont it be different before and after the port.

Further as per your explanation, pressure up to port 2 is P1-(dp/dx)*(l/2). If I am not wrong, then this pressure has to be less than inlet pressure at port 2 else the flow may also start moving out or at least some reverse flow may be observed. But since it is assumed that flow occurs without any such issue then this means, the pressure at the inlet is certainly higher than value attained after drop from port 1.

I agree there will again be certain mixing etc. closes, again there may be development of boundary layer, but then what should I take the total pressure drop for the flow.

I simulated such condition in fluent for a simple case where instead of port being at l/2 was at l/4 and these are some of the typical values (average total pressure) (in Pa) Port 1 9156; Port 2 71000 and outlet 6300 approx.

Now can someone suggest what the pressure drop for the flow is?

Since friction s always a loss, flow only goes from higher pressure to lower pressure. So if 1 and 2 are inlets and 3 is an outlet, the pressure at 1 is highest and at 3 is lowest. Because the flow rate will be higher downstream of the second inlet, the rate of pressure drop will (likely - depending on the shape of the duct) be faster from 2 to 3. Without knowing the geometry, the fluid, and the flow rates, it's impossible to say what the pressure difference will be. You need to know boundary conditions somewhere - if you don't know the inlet pressures, do you know the mass flow rates into the inlets? If not, you may need to model further upstream.

@Lawrence

Yes I know the inlet velocity (hence flow rate and mass flow rate too) at inlet 1 and also at inlet 2.. The values mentioned above are obtained after simulating such a case. But to conclude something from these I need to have pressure drop which I am not clear at.

The issue is I am solving this using CFD where I have imposed the boundary conditions. Hence the pressure I suppose i adjusted to at both the inlets to as to satisfy the boundary conditions and continuity equation. Now I have the values but I am looking for conclusive justification to consider any pressure drop as the required for aforementioned flow conditions.

Hi

As far as I understand your problem is rather a definition of the pressure drop in the case of multiple inlets. In fact, in the present case there is not one pressure drop, you can define three : 1->2, 2->3 and 1->3

To understand correctly, you must go back to the kinetic energy theorem : if you consider a stream tube with one inlet section and one outlet section, mechanical energy balance states that (I assume incompressible flow) :

(p+1/2 rho v^2 + rho g z)_out - (p+1/2 rho v^2 + rho g z)_in = -Phiv / Q

where all quantities p, v, z are averages over inlet (resp outlet) section, Q the volume flow rate in the stream tube, and Phiv the total power dissipated in the tube by viscous friction. Applying this to a constant section horizontal tube, you get the classical pressure drop p_out - p_in = - Phiv / Q.

Now in your case you can apply this for THAT PART OF THE FLUID FLOWING OUT IN 3 COMING FROM 1, with volume flow rate Q1 (assume horizontal tube to simplify)

(p+1/2 rho v^2)_3 - (p+1/2 rho v^2)_1 = -Phiv_13 / Q1

where Phiv13 is the mechanical energy dissipation rate for the fluid coming from 1.

Same thing between 2 and 3 :

(p+1/2 rho v^2)_3 - (p+1/2 rho v^2)_2 = -Phiv_23 / Q2

where Phiv23 is the mechanical energy dissipation rate for the fluid coming from 2.

Add both equations, knowing that mass conservation imposes Q3 = Q1 + Q2

Q3 (p+1/2 rho v^2)_3 - Q2 (p+1/2 rho v^2)_2 - Q1 (p+1/2 rho v^2)_1 = -Phiv

where Phiv = Phiv_13 + Phiv_23, total mechanical energy dissipated in the whole flow.

Not that the LHS terms are mechanical energy flow rates.

You see that this is difficult to speak here about a total pressure drop. However, Phiv represents the total energy dissipated in the flow, and Phiv/Q3 (homogenous to a pressure) is therefore the energy dissipated per unit volume of fluid crossing 3.

Note that you can generalize the above equation to an arbitrary number of inlets/ outlets, suming over the formers / latters. Clearly it's just a balance and does not tell you haw to calculate the Phiv's. To do that, you must use some correlation, for regular losses, but note that the singularity provoked by your middle port will also add some energy dissipation, this is a singlar loss. You may find such informations here :

http://www.lavoisier.fr/livre/notice.asp?ouvrage=1996434 (don't know if translated in English)

I hope that this is readable... If not I can send you the formulas in LaTeX. If you have some typical figures for your flow, send it to me.

Regards,

Olivier

Deewaker,

I'm sorry if I'm getting off track, but I'm curious. Please help me understand the inlet conditions. Let's forget the CFD for a moment. Can you tell me how you know the inlet velocities? Are they being individually supplied by some sort of positive displacement pumps that can provide constant flow rate at varying back pressure? Otherwise, it seems pressure in the manifold can feed back, and the flow rates at both inlets (and the outlet) would be coupled.

Regards,

Larry

I Can refer a book which can help you out for your question The Book’s name is “Fluid Mechanics “ by Yunus A.Cengel and John M. Cimbala

I feel the argument is being a little carried away. Forgetting CFD, Without considering mixing losses, the problem can be simply dealt using the concept of networks/pipes, which is the electrical analogy for such problems. where in the pressure drop is only due to friction along the pipe. Oliver Louisnard has given the correct mathematical explanation.

I agree above, but since I am simulating the above case and have pressure values at aforementioned points.I need to know what is total pressure drop for the system.

The analogy i suppose can dealt with when we have some length of pipe 2. But as mentioned earlier I am directly giving 2 as inlet surface for CFD. Then the mathematical relation for Q1+Q2=Q3 and then substituting the relevant pressure head in these to get he over all pressure drop doesn't fit here.

ideally we take as delP1j and del P2j where j is junction. But here I don't have the relevant pipe length in case of 2j to apply H-P equations (deltta P =(flQ^2/3d^5) and proceed.

Hope I was clear

I do not understand discussion on this problem.

If you try to create formula for engineers you need used some additional information.

If you try to developed a mathematical model you have to use Navier-Stoks equations with correct boundary conditions. These conditions are the flowing:

1. Solid boundary - no-slip BC;

2. Inlet or outlet parts:

a) three component of velocity vector;

b) pressure /or head + tangent components of velocity vector.

All other types of conditions are some simplifications of real flow and needs some additional assumptions.

If you know Q1 and Q2 the pressure drops \del P depends on domain configuration.

So, first you have to find out what do you need from your research and start from correct boundary problem for NS equations.

@Nikolay

Yeah you got it correct, I am solving using NVS and my domain is fluid domain as can be seen in attached file. I am hypothesizing that total pressure drop will be (static+dynamic=total at 1) +total at 2 - total at 3. I am not concerned with what goes at junction or mixing. I am simply concerned with the pressure drop across the system which causes the described flow (i.e inlet at 1 and 2 and total outlet at 3).

Kindly correct me if I am wring in my hypothesize. In order to ease and simplify computation.I have specified only a surface as inlet and 1 and inlet 2 domains. SO the total pressure at both - total pressure at outlet should give me total drop. I hypothesize total pressure at 2 (and not just static or dynamic) because, there has to be some pressure in statics form to cause the flow rate (or flow inlet velocity given at 2) and further dynamic head is accounted as the flow is not develop but will be developing.

Kindly correct me if I am wrong.

https://lh6.googleusercontent.com/-HMKp2zSgoGQ/UnDeVEepn7I/AAAAAAAAAEs/Oas43V___ck/s1440/2%2520inlets.jpg

Deevakar,

So far in the discussion, I could not find whether the fluid you considered is compressible or noncompressible because the considerations may vary perticularly when flow is in z direction. To have the rough values of pressure drop you can use on-line calculator from veb (for example, www.pressuredrop.com).

I agree with Lawrence Matta on how you arrange to inject required flow rates from two sources.

Velocities at section1-2 and section 2-3 of the duct are different say V1 and V2 which are directly proportional to Q1 and Q2. Here, you encounter with three distinct pressure drops; 1) pressure drop across duct length 1-2 depending on velocity V1, 2) At point 2 due to conflugurance of two streams, and 3) pressure drop across duct length 2-3 depending on velocity V2. Toal pressure drop shall be sum of these three.