Mixing streams at different pressures 2

[mbac123]

Hello,

I am mixing two streams of the same fluid at different temperatures and pressures in a pipe. I am trying to calculate the pressure at the new point where the two streams mix. I know that I must use conservation of mass and energy to determine the enthalpy at the new point, but, I do not know how to determine the pressure.

 

[ione]

Usually mass and energy conservation are used in psicrometry, when dealing with adiabatic mixing of two air streams (but at the same pressure):

MA + MB = MC
MA xA + MB xB = MC xC
hA MA + hB MB = hC MC

Where:

MA = mass of dry air (stream A)
MB = mass of dry air (stream B)
Mc = mass of o dry air (resulting stream C)

xA= dryness fraction (stream A)
xB= dryness fraction (stream A)
xC = dryness fraction (resulting stream C)

hA= hentalpy (stream A)
hB= hentalpy (stream B)
hC = hentalpy (resulting stream C)

I am wondering if it is really possible to do what you’ve described? If so I’ll go for the ideal gas law

Ptot = R*T*(nA + nB)/V

Ptot = total pressure of stream C
R = ideal gas constant
T = temperature of stream C
nA = moles of gas A
nB = moles of gas B
V = control volume

 

[mjpetrag]

mixing streams at different pressures may cause backflow of the higher pressure stream into the lower pressure stream...in which case the streams will not mix the way you want them to and the pressure downstream would just be the higher static pressure.

can you provide a detail sketch of the problem?

-Mike

 

[mbac123]

Sorry, for the confusion. I should have provided more information. The fluid is not air so psychrometry cannot be used. Fluid at point (1), (2) and (3) in the schematic below are all the same, but at different temperatures and pressures. The fluid enters at points (1) and (2) and exits at point (3).

----------------------------------
(1) (3)
---------| |------------------
| |
| (2) |
| |
| |

The pressure at point (2) is higher than the pressure at point (1). As stated before, all properties at point (3) are unknown, however, the mass flow rate and enthalpy at point (3) can be determined, but I do not know what the pressure should be at point (3).

 

[MrBTU]

As stated above, won't stream 2 flow both into 3 and 1? How will stream 1 flow out to a higher pressure? You could look at an eductor, and use the pressure from stream 2 to pump the stream 1, and mix and combine into 3. An eductor converts pressure to velocity through a venturi and creates an area of low pressure used to pump a second stream, combining both at the outlet.

 

[MikeClay]

Assuming this is a flowing, adiabatic system and the flow is not choked, P1 and P2 are set upstream before some restriction, otherwise the backflow situation will exist. The pressure at the mix point will be at most the same as the lower pressure.

You need to know the final destination pressure (where it exits the pipe). This temperature can be found by mixing the streams and doing an adiabatic flash to this pressure .

You now have the ending temperature and pressure. Work the stream backwards to the mixing point using the piping pressure losses.

If for any reason P1 or P2 is measured just upstream of the mixing point while flowing, then the lower of these two pressures will have to increase due to the increased backpressure.

--Mike--

 

[mbac123]

I should mention that I am not interested in the pressure at the mixing point, but rather the pressure at point (3) should be sufficiently downstream from the mixing point.
I also forgot to mention that the amount of fluid entering point (2) is relatively small in comparison to point (1). The system is also a closed loop (no fluid is exiting the system).

If there is a backflow situation as many of you suggest, what will be the pressure at point (3)? Will it be the same as point (2) (The high pressure stream)? Also if there is a backflow how do I know how much mass flow is being forced into pipe (1) and pipe (3)?

 

[mjpetrag]

Can you give us some numbers for pressures? Is this water? What are the fluid velocities?

I'm trying to get an idea of the whole system. Why is P2 greater than P1? What is the motive force in leg 2? A centrifugal pump? The reason I ask is that usually when you have a piping system like this, leg 1 and leg 2 will balance their pressures naturally so that at the mixing point they have the same pressure.

In another case, this could be a design problem. I would expect that whoever designed this would put a restriction orifice or something to adjust the pressure so that P2 matched P1.

If the amount of fluid you are introducing is a very very small amount in relation to Leg 1, I would expect the pressure to DROP downstream from P1 only due to the friction loss of a pipe tee. This would be a small pressure drop.

If you are looking for exact numbers and you have a significant amount of flow from P2, then you might want to look into a solution using CFD.

-Mike

 

[zekeman]

This not a real problem.
In order to get a flow process, you have to supply the downstream conditions at 3.
If you do, you have your answer.
So, as stated the problem is untenable

 

[mbac123]

Alright, I think I should provide more information and just define the entire problem. The fluid is carbon dioxide. Point (1) represents fluid exiting a turbine at about 80 bar and 800 K, 36 m/s. At normal conditions, point (2) will be shut off from this system by a valve at both ends of the pipe. Point (3) will therefore have the same flow conditions as point (1).

However, the valve at point (2) will open, allowing flow to enter this pipe when it is desired to have flow bypass the turbine. Therefore, as the valve is opened, initially point (1) and (3) have the same flow conditions. Point (2) will be at about 220 bar and 1000 K (the amount of fluid passing through this pipe has not yet been decided, but should not exceed 50% of the mass flow at point 1). But, I want to know once point (2) sends fluid into the pipe tee, what the new conditions will be at point (3) as this fluid will continue on to another component in my system.

 

[zekeman]

Since you are talking increaasing the flow rate by 50%, it would help if you gave us the main duct size and the distance to its discharge point. Now I think you have a doable problem.

 

[mjpetrag]

not sure if I understand the question correctly but ill take a stab at it...

you need to know your outlet pressure. Let's say it is venting to atmosphere at 0 barg

the flow will take the path of least resistance. so you will have to do an iterative calculation of flow rate to find where the pressure drop across the turbine equals the pressure drop through the bypass for each of your conditions depending on how much the bypass is open.

-Mike

 

[mbac123]

Perhaps I should just attach a figure to help explain the problem.

Zekeman I am not increasing the flow rate by 50%, I am simply rerouting it through pipe 2.

My outlet pressure from pipe 3 is known at the initial point in time before the valve at point 2 is opened (80 bar). I understand that the flow will take the path of least resistance, but won't that mean initially, all flow that can enter pipe (2) will enter it (since initially valve is closed, pressure is zero) and that fluid will then mix with the fluid from pipe (1) (and perhaps with some backpressure situation)?

 

[dcasto]

high pressure always flows to low pressure, its that simple. You open a bypass valve on your system and the fluid will just bypass and the system will stop as you have it drawn. This looks like an ORC, to me.

 

[mjpetrag]

in addition to dcasto...

assuming flow rate is constant, as long as the bypass line is large enough to handle the pressure drop. if not the bypass line will reach its maximum flow and the rest will go through the turbine since the turbine will now have less flow resistance than the fully loaded bypass.

-Mike

 

[mjpetrag]

and if im not mistaken...opening the bypass line will raise the pressure in the discharge piping. if your inlet pressure and flow are constant, and more flow is going through the bypass, less flow is going through the turbine = smaller dP across turbine = higher discharge pressure.

-Mike

 

[mbac123]

Your second point mjpetrag is referring to the fact that the turbine pressure ratio will change when opening the bypass line and thus change the turbine exit pressure? Correct?
This is useful information, however, I still do not know what the pressure at point (3) should be (my original question).
I need to know how to model what the mixing pressure (point 3) is after the bypass line is opened. (The pressure does not need to be directly at the mixing point, simply the pressure before entering the next component of my system).

 

[zekeman]

The momentum equation is
p1*A1_+A1/v1*V1^2=(A2/v2*V2+A1/v1*V1)*V3+p3*A3
or using M as the mass transfer rate
p1*A1+M1*V1=(M1+M2)*V3+p3*A3 eq (1)
conserving horizontal momentum
The energy equation is
M1*(p1*v1+u1+V1^2/2)+M2*(p2*v2+u2+V2^2/2)=(M1+M2)*(p3*v3+u3+V3^2/2)
or using enthalpy,h for u +pv
M1*(h1+V1^2/2)+M2*(h2+V2^2/2)=(M1+M2)*(h3+V3^2/2) eq(2)
M3=*A3*V3/v3 , continuity eq(3)
Knowns
M1,M2,p1,h1,T1,p2,h2,T2

Unknowns
V3,,p3,h3,u3,v3,T3
u3=u3(T) = cv*T3 internal energy eq(4)
h3=p3v3+u3=h3(T)= cp*T3 enthalpy eq(5)

6 unknowns, 5 equations so far
The final equation is the flow equation from 3 to the entrance of the compressor, point 4 (my point) which is
(p3*v3+u3+V3^2/2g)= p4*v4+u4+V4^2/2g)+integral(fl/D*V^2/2g) +Q
which is usually handled numerically between the heat sinks.But p4 and V4 are negligible as well as the friction term,integral(fl/D*V^2/2g) vs the heat rejected, so I think, the flow equation is approximately, although the friction term could be included with some labor.
p3*v3+u3+V3^2/2g = Q eq(6)
where Q is the heat rejected to the regenerator and the precooler.
6 equations, 6 unknowns should be solvable.
Now if you know p3 when there is no bypass,M2=0, then
p’3*v’3+u’3+V’3^2/2g = Q ‘ using primes for this case, then
p3 for the case M2=0 should be the same as p’3 determined above.Of course you need to know Q and Q'.

 

[mbac123]

Thank you zekeman. I think the answer I was looking for is applying the momentum equation and solve for pressure at (3). However, I do not understand why your equation is as follows:
p1*A1+M1*V1=(M1+M2)*V3+p3*A3

Shouldn't the equation be:
p1*A1+M1*V1 + p2*A2 +M2*V2 =(M1+M2)*V3+p3*A3

Since the fluid is entering the pipe at point 1+2 and is exiting at point 3. (At least if there is no back pressure effect) What if there is a backflow (and how do I implement that into the equations)?

 

[zekeman]

"p1*A1+M1*V1 + p2*A2 +M2*V2 =(M1+M2)*V3+p3*A3 "


No, we are conserving horizontal momentum, The two terms you added are not horizontal.

zeke