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Pressure in parallel pipe
Engineering and workplace issues. topic
Posted by Mike on 13 February, 2009 - 10:48 pm
Hi all,
I'd like to ask about the pressure in parallel pipe system:

when a single pipe with diameter D1 and flow rate Q1 branch into two pipe with diameter D2 & D3 and the same length, their flow rate are Q2 & Q3, bellow statement are true or not? :

Q1 = Q2 + Q3; ( i believe this relation is absolutely true)

and how about this?

P1.V1 = P2.V2 = P3.V3 (consider that there are no head loss & pressure loss)

is there any possibility that P1 < P2 + P3?

please response in account of pressure.

Thanks guys..


Posted by Bruce Durdle on 14 February, 2009 - 10:31 pm
Hi Mike,
Your first question is correct - Q1 = Q2 + Q3.

As to the second "It all depends"...

You need to check Mr Bernouilli's fine equation - which relates pressure in a pipe to the flowing velocity.

P/rho + 1/2 v*v = constant.

So if the velocity in all three sections is the same, the pressures will be the same. If the velocity in Q2 in less than that in Q3, then the pressure in Q2 will be higher than that in Q3 (not by much but enough)

If the diameters of all three sections are the same, then the velocity in the inlet section will be higher than that in the other 2, and the pressure will therefore be less. So it is possible for P1 to be less than P2 and P3.

If the velocity in section 1 is 5 m/s and the pressure is 100 kPa, and the velocities in 2 and 3 are the same, the pressure in 2 and 3 will be about 109 kPa.

Cheers,
Bruce


Posted by saravanan on 16 February, 2009 - 11:30 am
Hi Bruce. Your answers is not clear to me. Pls. explain clearly.


Posted by Bruce Durdle on 16 February, 2009 - 2:04 pm
In moving from one section of pipe to another with no intervening pumps or turbines to add or remove energy from the flowing stream, the total energy of the fluid must remain constant (except for losses which we will neglect for now).

There are three forms of energy that are significant. Kinetic energy is given by 1/2 m v^2. v is the velocity of the liquid and m is the mass. Gravitational potential energy is given by mgh where g is the acceleration due to gravity and h is the elevation above some reference. Pressure energy is the energy stored in a volume and is given by P x V where V is the volume involved.

To make things simpler, deal with 1 kg of liquid by looking at the specific energy in J/kg. The equations then are 1/2v^2, gh, and PxV/m = P/rho.

If at one point in a pipe the velocity is 5 m/s and the pressure is 100 kPa, the specific kinetic energy is 25/2 = 12.5 J/kg. The specific pressure energy for water with rho = 1000 kg/m^3 is 100x1000/1000 = 100 J/kg. If we assume that there is no change in elevation we can set gravitational potential energy to 0. So the total energy at this point is 12.5 + 100 = 112.5 J/kg.

If at another point in the same pipeline the velocity is less, the kinetic energy will have fallen and the pressure energy must be increased to make the total constant. In my example I assumed that the feed pipe split into two pipes of the same diameter as the initial section. So the total flow area will be doubled and the velocity will be halved. Kinetic energy will be reduced to 12.5/4 = 3.125 J/kg. The pressure energy must therefore have increased by 12.5 - 3.125 = 9.375 J/kg, so increasing the pressure to 109.375 kPa.

Correction - the spelling of the gentleman's name is Bernoulli. Google this for more info.

Cheers,
Bruce.


Posted by mike on 16 February, 2009 - 10:56 pm
Hi Bruce,

Thanks for the quick response, Mr bernoulli said that:

P/rho + 1/2*v*v = constant,
there is still one thing I still not clear.
In these parallel branching pipe system; based on bernoulli energy equation are you saying that:


P1/rho + 1/2*v1*v1 = P2/rho + 1/2*v2*v2 + P3/rho + 1/2*v3*v3 (eq. 1)

or this one?

P1/rho + 1/2*v1*v1 = P2/rho + 1/2*v2*v2 = P3/rho + 1/2*v3*v3. (eq.2)

In your example it likes that the 2nd equation is the right one, BUT in the other statement that Bernoulli said that "the total energy of the fluid must remain constant (except for losses which we will neglect for now)", based on this statement it likes that the 1st equation is the right one.

I'm rather confused with this, what is the relation between above statement, and finally which equation is the right one for parallel pipe system?

Thanks,
Mike :)


Posted by Bruce Durdle on 17 February, 2009 - 3:54 pm
Hi Mike,

The second equation is the correct one, because we are considering whatv happens to 1 kg of fluid.

If a 1 kg bundle of fluid moves from the inlet to branch 1, its velocity will fall and its pressure will rise so that Bernoulli's equation applies between the first point and the second. If a different bundle of fluid moves from inlet to branch 2, it will do the same. Or if we look at 1 kg entering the inlet and splitting equally between branch 1 and branch 2, we will be dealing with 1/2 kg bundles in each outlet branch.

Another much more simplistic way of viewing it is that if a fluid particle slows down, something must be acting on it to decelerate it. This can only happen if the downstream pressure is greater than the upstream one.

Cheers,
Bruce.


Posted by Mike on 18 February, 2009 - 10:45 pm
Hi Bruce,

so based energy equation :

P1*m1/rho1 + 1/2*m1*v1^2 = P2.m2/rho2 + 1/2*m2*v2^2 = P3*m3/rho3 + 1/2*ms*v3^2

in your example D1=D2=D3, with same length so that:
m2 = m3 = 1/2 m1; Q1= Q2+ Q3 so that v2=v3=1/2 v1;
rho1 = rho2= rho3

112,5 = 1/2*P2 + 1/4 * (2.5)^2

so:
the value for P2=P3 would be 221.8 kPa. The result is different with what your given. Is there something wrong?

Regards,
Mike


Posted by Bruce Durdle on 19 February, 2009 - 2:01 pm
Hi Mike,
Bernoulli's equation applies to energy per unit mass - so you do not include mass terms in the calculation.

P1/rho1 + 1/2*v1^2 = P2/rho2 + 1/2* v2^2 = P3/rho3 + 1/2*v3^2

100,000/1000 + 1/2 * 5^2 = P2/1000 + 1/2 * (1/2 *v1)^2

P2/1000 = 100,000/1000 + 1/2 * 5^2 - 1/2 * (1/2 * 5)^2

P2 = 100,000 + 1000 * (1/2 * 25 - 1/2 * 6.25)

= 100,000 + 1000 * (12.5 - 3.125)

= 100,000 + 1000 * 9.375

= 109,375 Pa

= 109.375 kPa.

If you need any more help, contact me direct,

- bru$ce dot d%urdle at xt#ra dot c(o do)t nz

Without the punctuation.

Bruce


Posted by Mike on 20 February, 2009 - 3:07 am
OK Bruce, Thanks for your help :)

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