Need a confirmation. Bernoulli.

[Clausius2]

I'm not sure completely of this. Can you confirm if, assuming IDEAL (Bernoulli) flow there is a loss of stagnation(total) pressure in the geometry attached.

I think there is a loss of stagnation pressure indeed.

The flow enters with a stagnation pressure Po1, pass through an orifice at 2 and leaves at section 3 (which area is equal to section 1) in a atmosphere of pressure Pa.

 

[marlon]

The way i see it is that pressure will remain the same when the flow passes through, only the velocity of the flow will be higher when it passes through the orifice...I just used the continuity law : Av = constant where A is the area...

regards
marlon

 

[Clausius2]

What pressure remains constant?? Static or total?

Caution: take the downstream of the tube (3) as a chamber of homogenous pressure Pa. So that, once the jet is discharging at (3) the static pressure just at the exit of the orifice must be Pa in order to be fully expanded. That's part of the enunciated of the problem.

 

[marlon]

Isn't P01 just equal to Pa ?

Isn't this what the continuity law is about ?


regards
marlon

 

[Clausius2]

Maybe my explanation is not enough clear, sorry Marlon and thanks by the way.

Po1=total pressure at section1
Pa=static pressure at section 3 and static pressure at section 2 because the boundary constraint of the jet in ideal flow just at the exit of the orifice must be P2=Pa in order to be fully expanded.

Continuity: U1=U3 but here is the doubt:

For example, when a pipe ends in a large chamber of pressure Pa, I always take the static pressure just at the outlet of the pipe to be Pa. It seems to be a boundary constraint for the flow (it makes sense because eventually the flow jet is surrounded by an atmosphere of pressure Pa).

Thus P2=P3=Pa and Po1 isn't equal to Po3. That was my reasoning. I hope to obtain some opinion about this.

Thanks.

 

[Doc Al]

total pressure remains constant

I don't understand why total pressure (static + dynamic = $P + 1/2\rho v^2$) would change along the streamline. The static pressure in sections 1 & 3 will equal atmospheric pressure. What am I missing?

 

[arildno]

But you CAN'T use Bernoulli in this case!
In particular, between 2&3, the energy loss due to viscous dissipation is simply too large.
(I haven't gone through all the arguments presented as yet, though..)

 

[Clausius2]

Right, I was hoping a confirmation rather than a explanation by myself. Anyway I must acknowledge your effort only by merely trying to understand my poor set up of the problem.

See the geometry attached. There is a pipeline, whose fluid has a total pressure Po. The pipeline discharges into a reservoir of pressure Pa. Which is the lost of total pressure?

$$P_o=P+1/2\rho U^2$$

Into the reservoir total pressure=Pa.

So that the fluid has a loss of total pressure:

$$\Delta P_o=1/2\rho U^2$$

Because just at the pipe outlet P=Pa. What do you think about that? The question here is that in sudden expansions the total pressure is not conserved. Why? See the figure and realize there is a jet flowing into the chamber. The unique boundary constraint a jet can feel while flowing into an atmosphere is a pressure condition---> P=Pa.

Try to assimilate this to the downstream widening of the pipe of the original problem I posted.

Thanks.

 

[Clausius2]

But you CAN'T use Bernoulli in this case!
In particular, between 2&3, the energy loss due to viscous dissipation is simply too large.
(I haven't gone through all the arguments presented as yet, though..)

Yes, I can't use Bernoulli. BUT in my last post I used Bernoulli with a particular boundary constraint (pressure outlet). Moreover:

i) Although I assume ideal flow, there is a loss enhanced by the static pressure boundary condition (P2=Pa). (I always employ the subindex "o" when writting stagnation or total pressures, if not it is a static pressure).

ii) In my analysis: have you seen any experimental coefficient of pressure loss anywhere? Answer: You don't. Such experimental coefficient K measures viscous dissipation and turbulence:

$$\Delta P_o=1/2 \rho K U^2$$ and it is the responsible of stagnation pressure loss.

 

[arildno]

All right, I'll check up on your conundrum after supper!

 

[dextercioby]

Clausius,your notation is very decieving.It can miselead.Mybe it's traditional among engineers,but us physicists tend to take them very seriously and if one does not understand our notations,then it's bad,really bad...
Hopefully mine will e very explicit.
Let's assume that along the pipe the ideal fluid has a laminary stationary flow,which means that the equations of Euler admit the solution due to DANIEL ( ) Bernoulli ("Hydrodynamica",1738).
Let's label the 3 portions of the pipe with "1","2" and "3".

Then considering the gravity contribution as well,we can write
$$P_{1}=P_{2}=P_{3}$$
,which becomes
$$P^{0}_{1}+\frac{\rho_{1}U_{1}^{2}}{2}= P^{0}_{2}+\frac{\rho_{2}U_{2}^{2}}{2}= P^{0}_{3}+\frac{\rho_{3}U_{1}^{3}}{2}$$
,where gravity constribution is canceled/reduces.
$P_{i}^{0}$ is the contribution to total pressure by the pressure field in the absence of movement (velocity field is zero and the gravity does not act) in the portion 'i' of the pipe.
The fluid is assumed incompressible,therefore
$$\rho_{1}=\rho_{2}=\rho_{3}=\rho;S_{1}U_{1}=S_{2}U_{2}=S_{3}U_{3}$$

I found the difference $P_{3}^{0}-P_{2}^{0}$ to be
$$P_{3}^{0}-P_{2}^{0}=\frac{1}{2}\rho U_{2}^{2}(\frac{S_{3}^{2}-S_{2}^{2}}{S_{3}^{2}})$$

Maybe that's not the loss (actually it's a gain,the difference is larger than 0),but at least it may give you ideas to correctly approach the problem.

Daniel.

 

[Q_Goest]

Yes, Bernoulli's is incomplete. Bernoulli's is a conservation of energy equation, and energy is not conserved for a fluid moving through a pipe. Regardless of there being an orifice there or not, there is energy loss as defined by various equations, such as Darcy Weisbach (sp?).

Consider for example a very long pipe. There is a drop in pressure due to the viscous affect/turbulence in the pipe. An orifice can often be viewed as a very long section of pipe. So you're correct that at the outlet of the pipe, the pressure is Pa (assuming incompressible flow). In compressible flow, there could be a shock wave present, just as there could be in an orifice. But neglecting that, the flow is then determined by the dP and the total restriction.

Note that a converging/diverging nozzle may actually provide for very little pressure drop (stagnation pressure drop), but for a flat plate or sharp edged orifice, there is significant loss of energy.

 

[Clausius2]

Clausius,your notation is very decieving.It can miselead.Mybe it's traditional among engineers,but us physicists tend to take them very seriously and if one does not understand our notations,then it's bad,really bad...
Hopefully mine will e very explicit.
Let's assume that along the pipe the ideal fluid has a laminary stationary flow,which means that the equations of Euler admit the solution due to DANIEL ( ) Bernoulli ("Hydrodynamica",1738).
Let's label the 3 portions of the pipe with "1","2" and "3".

Then considering the gravity contribution as well,we can write
$$P_{1}=P_{2}=P_{3}$$
,which becomes
$$P^{0}_{1}+\frac{\rho_{1}U_{1}^{2}}{2}= P^{0}_{2}+\frac{\rho_{2}U_{2}^{2}}{2}= P^{0}_{3}+\frac{\rho_{3}U_{1}^{3}}{2}$$
,where gravity constribution is canceled/reduces.
$P_{i}^{0}$ is the contribution to total pressure by the pressure field in the absence of movement (velocity field is zero and the gravity does not act) in the portion 'i' of the pipe.
The fluid is assumed incompressible,therefore
$$\rho_{1}=\rho_{2}=\rho_{3}=\rho;S_{1}U_{1}=S_{2}U_{2}=S_{3}U_{3}$$

I found the difference $P_{3}^{0}-P_{2}^{0}$ to be
$$P_{3}^{0}-P_{2}^{0}=\frac{1}{2}\rho U_{2}^{2}(\frac{S_{3}^{2}-S_{2}^{2}}{S_{3}^{2}})$$

Maybe that's not the loss (actually it's a gain,the difference is larger than 0),but at least it may give you ideas to correctly approach the problem.

Daniel.

Well, in the last posts I used the notation:
$$P_{oi}$$ as stagnation or total pressure
$$P$$ as static pressure.

Your notation:
$$P_i^0$$ is the static pressure
$$P$$ is the stagnation or total pressure.

You have started by P1=P2=P3 which is wrong. The last formula gives the jump of static pressure between 2 and 3, which must be compensated with the one between 1 and 2 to give zero. The flow is not as an ideal as we could think.

 

[Clausius2]

Yes, Bernoulli's is incomplete. Bernoulli's is a conservation of energy equation, and energy is not conserved for a fluid moving through a pipe. Regardless of there being an orifice there or not, there is energy loss as defined by various equations, such as Darcy Weisbach (sp?).

Consider for example a very long pipe. There is a drop in pressure due to the viscous affect/turbulence in the pipe. An orifice can often be viewed as a very long section of pipe. So you're correct that at the outlet of the pipe, the pressure is Pa (assuming incompressible flow). In compressible flow, there could be a shock wave present, just as there could be in an orifice. But neglecting that, the flow is then determined by the dP and the total restriction.

Note that a converging/diverging nozzle may actually provide for very little pressure drop (stagnation pressure drop), but for a flat plate or sharp edged orifice, there is significant loss of energy.

Thanks for the aid, but here I was not trying to get oneself into turbulence and rugosity (Darcy). My considerations are only involving ¿¿ideal?? flow.

 

[Q_Goest]

The "ideal" flow you're referring to is not real though. There is no conservation of energy. In the first geometry you posted (first post), if you assume a fluid which somehow manages to conserve energy, then you can apply Bernoulli's and P1 = P3 (I'm assuming cross sectional areas are equal) and of course you could use Bernoulli's to determine pressure on the other geometry you've posted. In the case of the second geometry, note that Bernoulli's would also predict no drop it total pressure.

When you say Pa, and "pipe outlet":

For example, when a pipe ends in a large chamber of pressure Pa, I always take the static pressure just at the outlet of the pipe to be Pa. It seems to be a boundary constraint for the flow (it makes sense because eventually the flow jet is surrounded by an atmosphere of pressure Pa).

. . . that seems to indicate you're thinking there's atmospheric pressure on the pipe outlet, and it implies you're trying to find a pressure drop across this orifice. Pressure drop meaning total pressure or stagnation pressure. The only pressure drop is from real, viscious affects. There are standard equations for pressure drop across an orifice. They relate the orifice geometry, fluid properties, and other variables - to flow and pressure drop. These equations give you the pressure drop due to viscous affects. Bernoulli's never predicts a drop in total pressure (ie: the sum of velocity, static, and head pressure).

 

[rcgldr]

Bernoulli's logic for flow in pipes of varying diameters:

The amount of mass flowing past any cross section of the pipe must be constant, otherwise mass will be accumulating. Assuming that this flow has settled into a constant state, no mass is accumulating anywhere. This requires the flow in the smaller pipe to be moving faster, or more compressed, or a combination.

Assuming that the flow is faster in the smaller part of the pipe, then there must be a pressure differential to cause the accleration into the narrower section pipe and deceleration to slow the flow down after it leaves the narrower section of pipe. Based on this logic, the pressure in the narrower section of pipe is lower than the wider sections.

If the fluid or gas is compressible / expandable, the pressure differential is even more, because the gas expands under reduced pressure, requiring even faster flow to maintiain a constant rate of mass flow.

One factor being ignored here is temperature, which resists the expansion of gas, and therefore reduces the increase of speed somewhat, since the temperature is reduced along with any reduction in pressure.

The temperature effect is real, it's not uncommon to see frost accumulate on the surfaces of venturis.

Getting back to reality, because of viscosity, friction, and turbulence, a pressure differential force between the input and output ends of a pipe must exist in order to maintain a flow, or you can turn the pipe vertical and let gravity supply the force to draw a gas or fluid downwards in pipe (assuming that buoyancy issues are eliminated).

 

[Clausius2]

Bernoulli's logic for flow in pipes of varying diameters:

The amount of mass flowing past any cross section of the pipe must be constant, otherwise mass will be accumulating. Assuming that this flow has settled into a constant state, no mass is accumulating anywhere. This requires the flow in the smaller pipe to be moving faster, or more compressed, or a combination.

Assuming that the flow is faster in the smaller part of the pipe, then there must be a pressure differential to cause the accleration into the narrower section pipe and deceleration to slow the flow down after it leaves the narrower section of pipe. Based on this logic, the pressure in the narrower section of pipe is lower than the wider sections.

If the fluid or gas is compressible / expandable, the pressure differential is even more, because the gas expands under reduced pressure, requiring even faster flow to maintiain a constant rate of mass flow.

One factor being ignored here is temperature, which resists the expansion of gas, and therefore reduces the increase of speed somewhat, since the temperature is reduced along with any reduction in pressure.

The temperature effect is real, it's not uncommon to see frost accumulate on the surfaces of venturis.

Getting back to reality, because of viscosity, friction, and turbulence, a pressure differential force between the input and output ends of a pipe must exist in order to maintain a flow, or you can turn the pipe vertical and let gravity supply the force to draw a gas or fluid downwards in pipe (assuming that buoyancy issues are eliminated).

Thanks for the comment, but the case was incompressible for simplicity. See below this post. I'm going to clarify this. It was my fault.

 

[Clausius2]

The "ideal" flow you're referring to is not real though. There is no conservation of energy. In the first geometry you posted (first post), if you assume a fluid which somehow manages to conserve energy, then you can apply Bernoulli's and P1 = P3 (I'm assuming cross sectional areas are equal) and of course you could use Bernoulli's to determine pressure on the other geometry you've posted. In the case of the second geometry, note that Bernoulli's would also predict no drop it total pressure.

When you say Pa, and "pipe outlet":

. . . that seems to indicate you're thinking there's atmospheric pressure on the pipe outlet, and it implies you're trying to find a pressure drop across this orifice. Pressure drop meaning total pressure or stagnation pressure. The only pressure drop is from real, viscious affects. There are standard equations for pressure drop across an orifice. They relate the orifice geometry, fluid properties, and other variables - to flow and pressure drop. These equations give you the pressure drop due to viscous affects. Bernoulli's never predicts a drop in total pressure (ie: the sum of velocity, static, and head pressure).

Thanks QGoest. You're comment is very near of the truth. I'm going to clarify this because I have thinking of it while sleeping.

-First assumption: incompressible flow
-Second assumption: high Reynolds number: Re>>1 but small enough to remain laminar flow.
-Third assumption: I thinking of the original geometry, an orifice inside a duct, the cross area of the orifice is A2 and the cross area of the pipe (recovered symmetrically after the orifice) is A1=A3.
-Fourth assumption: the problem data is a total pressure Po1 at stage 1, and static pressure P3 at 3 where the flow is fully developed.
-Fifth assumption: the total pressure loss between stages 1 and 2 is much smaller than the total pressure loss between stages 2 and 3. It can be demonstrated so. Therefore, I will consider losses only at the sudden expansion.

There are 3 main approach to this problem:

1) Considering entirely ideal flow. Thus, DocAl, Marlon and Daniel are right. Total pressure is conserved: Po1=Po2=Po3 as Bernoulli states. There is no losses of energy. Of course, this behavior is very far of reality, and forecasts a larger mass flow (unreal). There is no possibility of recirculations, the flow remains straight.

2) Considering $$A_2/A_3$$ is of the order 1, and non ideal flow. In fact there are losses. Here I link with Arildno and his explanation about lift loss and vortex implications. Recirculations will be formed due to the violent turn of the flow at the intake and sudden expansion at the outlet. Such recirculations are various orders of magnitude smaller than the characteristic length $$L=\sqrt{A_1}$$, and have a characteristic length $$l$$, where $$L>>l$$. So that, the Reynolds number based on L is much larger than the Reynolds number based on l: $$Re_L>>Re_l$$. Therefore, viscous effects cannot be neglected at recirculation zones. Total Pressure loss will occur (Arildno, this could explain too the loss of lift when boundary layer separation and vortex effects on it, do you realize of that?).

The outflow at stage 2 will be jet-shaped. Maybe its section is stretched. But I don't think the pressure outlet condition (P2=P3) will be satisfied in these conditions. I haven't found any evidence of that. Moreover, the velocity $$U_3\approx U_2$$ (I mean, they are not the same because of continuity, but are of the same order) and surely there will be a small gradient of static pressure between 2 and 3. It can be calculated the mass flow is smaller than in the case 1, and the coefficient of pressure loss can be calculated via integral equations: $$K=(1-A_2/A_3)^2$$

3)Considering $$A_2<<<<A_3$$ so that K-->1 and the total pressure loss is the kinetic energy itself based on the velocity of the orifice, as I predicted in one of my lasts posts, when I was explaining the case of a pipe discharging into a large reservoir. Yes Qgoest, Bernoulli can't be employed, in part because there is none streamline which ends at the center of the reservoir, where the fluid is at rest, without passing through a recirculation zone (where the viscous effects cannot be neglected).

Moreover, I have seen analitycally that as $$A_3\rightarrow \infty$$ then the static pressures $$P_2 \rightarrow P_3$$ so that I recover the result that the total pressure loss through the pipe and orifice is $$\Delta P_o=1/2 \rho U_2^2$$

All of this is curious, isn't it?.

A final question: in which case the mass flow is the largest?, remaining the data problem constant (except the quotient A_2/A_3)

 

[Q_Goest]

Clausius, I think it's worth also pointing out that Bernoulli's equation would predict zero pressure drop for an infinitely long pipe with an infinitely high flow rate. (ie: total pressure drop)

(PS: We must have posted at the same time. I'll have to go back and see what you wrote.)

 

[Q_Goest]

Hi Clausius. You have a pretty good handle on this, but I think there's some significant points worth mentioning.

-Second assumption: high Reynolds number: Re>>1 but small enough to remain laminar flow.

- Bernoulli's is applicable to high or low flow rates. It is applicable to laminar or turbulent flow. It does not distingquish. Real, frictional pressure drop as predicted by various other equations exists regardless of the pressure drop being laminar or turbulent. The pressure drop is slightly higher in proportion for turbulent flow, but still exists for laminar flow.

-Third assumption: I thinking of the original geometry, an orifice inside a duct, the cross area of the orifice is A2 and the cross area of the pipe (recovered symmetrically after the orifice) is A1=A3.

- Bernoulli's is the ONLY equation that predicts complete pressure recovery (ie: no pressure drop) after this type of restriction. If a fluid is flowing through this restriction in real life, there will actually be a pressure drop.

-Fifth assumption: the total pressure loss between stages 1 and 2 is much smaller than the total pressure loss between stages 2 and 3. It can be demonstrated so. Therefore, I will consider losses only at the sudden expansion.

- The restriction coefficient, K, as given by Crane paper #410, for a sudden, sharp edged contraction is 0.5. For a sudden expansion it is 1.0. In other words, the amount of pressure loss (ie: not calculated by Bernoulli's equation) for the expansion is roughly twice the pressure loss for the sudden contraction. The contraction can not be ignored. Regardless, the geometry is of a flat plate orifice, which is usually handled as a single restriction as opposed to two separate restrictions. It is acceptable to do it either way.

There is no losses of energy (predicted by Bernoulli's equation). Of course, this behavior is very far of reality, ...

Yes, this is exactly correct.

You talked a bit in your last post about how the (real) pressure recovers after such a restriction. Of course, Bernoulli's doesn't really say anything about that, but it is an important feature of the flow. Pressure taps used on orifice meters for example, must be placed far enough upstream and downstream of the orifice to ensure the flow is fully developed across the cross section of the pipe. Bernoulli's really doesn't make that distinction. For Bernoulli's equation, we assume the velocity is dependant on the continuity equation, regardless of location to the restriction.

My appologies if any of this sounds like lecturing, I don't mean to be standing on a soap box talking down. You seem to understand this stuff pretty well.

 

[Clausius2]

- Bernoulli's is applicable to high or low flow rates. It is applicable to laminar or turbulent flow. It does not distingquish. Real, frictional pressure drop as predicted by various other equations exists regardless of the pressure drop being laminar or turbulent. The pressure drop is slightly higher in proportion for turbulent flow, but still exists for laminar flow.

.

Well, it doesn't make sense to apply Bernoulli in turbulent flow. That's because when constructing Bernoulli equation you neglect viscous terms. Neglecting viscous terms in turbulent regimen is a barbarity. On the other hand, it can be well established laminar and quasi inviscid flows at enough high Reynolds# and Bernoulli is employable then. To sum up, Bernoulli is employable in a range of Reynolds numbers, not too low for not experimenting viscous effects, and not too high for not experimenting turbulent dissipations.

As a engineer, I am interested now in which configuration the mass flow is greater, in A2/A3 of the order 1, or with A3-->infinity with Po1 and P3 as data problem.

 

[Q_Goest]

Hi again. Let's just clear up one point. Bernoulli's is applicable for both turbulent or laminar flow. And frictional pressure losses, not calculated by Bernoulli's, are also applicable for both turbulent or laminar flow.

***
For example:
Take a long pipe with a "T" or other branch, all branches equal in diameter, with flow going into a T and half the flow going one way, half the other. Regardless of the flow velocity, regardless of whether the flow is turbulent or laminar, there is a change in the static pressure and a change in the pressure due to velocity as the flow splits and goes into each branch. Since the velocity gets cut in half, the dynamic pressure drops by 1/2^2 or 1/4. Since Bernoulli's says Static pressure + Dynamic pressure + Head pressure = constant, we find the Static pressure in this case increases by an amount equal to the drop in Dynamic pressure.

Also, regardless of whether the flow through this pipe system is laminar or turbulent, there is a frictional pressure drop as well. When calculating total actual pressure drop for the REAL fluid, after going through the "T", I will calculate the affects due to Bernoulli's and add the pressure drop due to frictional forces. Bernoulli's says the total pressure is a constant, but that's not true for any real fluid, nor for any time there is any flow whatsoever in the pipe. There are ALWAYS frictional losses in the pipe, regardless of the flow being turbulant or laminar, regardless of Reynold's number. Bernoulli's is always applicable for calculating dynamic, static, and head pressure affects, regardless of Reynold's number, but it is never applicable for calculating frictional pressure drop. Frictional pressure drop calculations are also applicable regardless of Reynold's number.
***

Regarding which mass flow is greater, through A2 or A3 (I think that's what you're getting at) the continuity equation requires all flows to be equal. Mdot2/Mdot3 = 1 (regardless of area) (for the example given)

I think that's what you're asking, but I'm not sure. If I'm missunderstanding you, please rephrase. I hope this is helpful.

 

[Clausius2]

Hi again. Let's just clear up one point. Bernoulli's is applicable for both turbulent or laminar flow. And frictional pressure losses, not calculated by Bernoulli's, are also applicable for both turbulent or laminar flow.

***
For example:
Take a long pipe with a "T" or other branch, all branches equal in diameter, with flow going into a T and half the flow going one way, half the other. Regardless of the flow velocity, regardless of whether the flow is turbulent or laminar, there is a change in the static pressure and a change in the pressure due to velocity as the flow splits and goes into each branch. Since the velocity gets cut in half, the dynamic pressure drops by 1/2^2 or 1/4. Since Bernoulli's says Static pressure + Dynamic pressure + Head pressure = constant, we find the Static pressure in this case increases by an amount equal to the drop in Dynamic pressure.

Also, regardless of whether the flow through this pipe system is laminar or turbulent, there is a frictional pressure drop as well. When calculating total actual pressure drop for the REAL fluid, after going through the "T", I will calculate the affects due to Bernoulli's and add the pressure drop due to frictional forces. Bernoulli's says the total pressure is a constant, but that's not true for any real fluid, nor for any time there is any flow whatsoever in the pipe. There are ALWAYS frictional losses in the pipe, regardless of the flow being turbulant or laminar, regardless of Reynold's number. Bernoulli's is always applicable for calculating dynamic, static, and head pressure affects, regardless of Reynold's number, but it is never applicable for calculating frictional pressure drop. Frictional pressure drop calculations are also applicable regardless of Reynold's number.
***

Regarding which mass flow is greater, through A2 or A3 (I think that's what you're getting at) the continuity equation requires all flows to be equal. Mdot2/Mdot3 = 1 (regardless of area) (for the example given)

I think that's what you're asking, but I'm not sure. If I'm missunderstanding you, please rephrase. I hope this is helpful.

Bernoulli Equation Po=P+1/2rhoV^2+gz=const along a stream line is NOT applicable in turbulent regimen. Again: is NOT applicable:

-If Re<<<1: Viscous effects are predominant: impossible to apply Bernoulli.

-If Re>>1: it can be assumed inviscid flow. There are losses in boundary layers and recirculation vortex. Bernoulli isn't applicable too, but this is the experimental range of actuation of Bernoulli equation, where it gives an acceptable result in spite of not being exact (never can be exact). But viscous terms can be neglected in this range.

-If Re>>>>1: turbulent viscous effects are predominant. It can't be never applied Bernoulli here. The flow is highly dissipative. And viscous terms cannot be neglected at all.

Surely you can apply "Bernoulli with losses" in any regimen, but understanding Bernoulli as stagnation pressure conservation has been only applied at the regimes I posted by the Fluid Mechanics commuinity in the world.

I wasn't referring which section has greater mass flow. I was supposing I was able to vary the geometry, conservating the boundary constraints Po1 and P3 and see how varies the mass flow as I vary the relation A2/A3. Sure mass flow is dependant of the geometry and it changes.

 

[Q_Goest]

From Crane paper #410:

The Bernoulli theorem is a means of expressing the application of the law of conservation of energy to the flow of fluids in a conduit. ... If friction losses are neglected and no energy is added to or taken from the piping system (ie pumps or turbines), the total head, H, in the above equation (Bernoulli's) will be a constant for any point in the fliud. However, in actual practice, losses or energy increases or decreases are encountered and must be included in the Bernoulli eqation. ... All practical formulas for the flow of fluids are derived from Bernoulli's theorem, with modifications to account for losses due to friction.

I've been doing this for over 15 years as a professional engineer. I've always used Bernoulli's to account for energy conservation affects regardless of Reynolds number. This is the only way it's done in industry. For where energy conservation is not applicable (frictional losses) various other equations are used depending on the situation, but most often it's the Darcy Weisbach equation. Crane paper #410 is the bible for fluid flow in industry.

Regarding Reynold's number (dimensionless):

Fluid flows are laminar for Reynolds Numbers up to 2000. Beyond a Reynolds Number of 4000, the flow is completely turbulent. Between 2000 and 4000, the flow is in transition between laminar and turbulent, and it is possible to find subregions of both flow types within a given flow field.

So... Using "Re >>> #" is a bit misleading, since Re is dimensionless.
For Re < 2000 we assume laminar flow
For Re > 4000 we assume turbulent flow
Use your best judgement for the transition.

I wasn't referring which section has greater mass flow. I was supposing I was able to vary the geometry, conservating the boundary constraints Po1 and P3 and see how varies the mass flow as I vary the relation A2/A3. Sure mass flow is dependant of the geometry and it changes.

Ok, I think what you're trying to say is, does Bernoulli's equation predict a mass flow rate through the orifice which changes with the inlet and outlet pressure? In other words, does the mass flow rate through the orifice vary for any given inlet and outlet pressure, depending on the orifice diameter and pipe diameters.

No, Bernoulli's does not. In fact, Bernoulli's does not predict a pressure drop across this orifice. Bernoulli's predicts the inlet and outlet pressure are equal if the areas are equal. It will predict a dynamic pressure increase, and static pressure decrease inside the orifice where the flow area is smallest. And it will predict that dynamic pressure increases and static pressure decreases as the flow rate increases. But it will predict that the static pressure will fully "recover" to the value upstream of the orifice once it has passed through. And it predicts that at all three points, (upstream, orifice, downstream) the total pressure remains constant.

 

[Clausius2]

From Crane paper #410:

I've been doing this for over 15 years as a professional engineer. I've always used Bernoulli's to account for energy conservation affects regardless of Reynolds number. This is the only way it's done in industry. For where energy conservation is not applicable (frictional losses) various other equations are used depending on the situation, but most often it's the Darcy Weisbach equation. Crane paper #410 is the bible for fluid flow in industry.

Yes and I agree with you. But take into account that Bernoulli is an approach, and that approach works well only on the range in which the flow can be considered as approximately inviscid. Notwithstanding engineers use it for everything (very wrong made by the way) it is a misconception to use it in conditions where viscous effects are predominant, and it can lead to numerical error. You always can correct it (via Darcy coefficients) to consider losses, BUT it isn't the Bernoulli equation yet.

Regarding Reynold's number (dimensionless):


So... Using "Re >>> #" is a bit misleading, since Re is dimensionless.
For Re < 2000 we assume laminar flow
For Re > 4000 we assume turbulent flow
Use your best judgement for the transition.

That's vague. The onset of turbulence varies a lot from one geometry to another. See texts like Incropera & DeWitt: "heat transfer fundamentals" where a lot of experiments gives transition Reynolds much different than those you posted. There isn't any general rule as you've posted.

Edit: you must know I'm not an engineer YET. But I have read about this stuff a bit.

 

[Q_Goest]

Loosely speaking, inviscid flows are those for which fluid friction is negligible. But the only time something like that will happen is over very short distances, or extremely low fluid velocities. There are times when this is applicable, and one can largely neglect frictional losses, but they are very rare in practice.

Note, that doesn't mean Bernoulli's is NOT applicable to viscous flow, it is. The only difference is that one must also consider frictional losses which are added into the Bernoulli equation.

Regarding the onset of turbulance, you are correct in saying that the numbers given by typical textbooks (2000 - 4000) are not always correct. One can induce turbulance at much lower Re and one can force laminar flow at much higher Re. But for the vast majority of applications of fluid flow, and without better knowledge of how geometry affects turbulance, these values are generally accepted. Note that for internal fluid flow, the onset of turbulance strongly affects the friction factor which is used directly in the Darcy equation. This is the primary reason we determine Re when calculating fluid flow.

Have a good day!

 

[Clausius2]

Note, that doesn't mean Bernoulli's is NOT applicable to viscous flow, it is. The only difference is that one must also consider frictional losses which are added into the Bernoulli equation.

Defnitely, and as Fluid Mech. community uses it:

$$P_o=P+1/2\rho U^2+\rho g z=constant$$ BERNOULLI

$$P_{o1}=P_{o2}+\Delta P_{o(losses)}$$ NOT BERNOULLI


The last equation, which consider losses is employable over any Reynolds. BUT IT IS NOT THE BERNOULLI EQUATION AT ALL.

 

[Q_Goest]

Hi Clausius,
If you'd like to work on this, please define the variables in the second equation.

Also, don't forget to put rho*g*h in the Bernoulli's equation.

And if you have a system of piping which requires analysis of pressure
1. changes due to elevation
2. changes in flow velocity
3. frictional pressure drops
- regardless of whether your analysis is viscous or not

How would you suggest taking those factors into account without using
- Bernoulli's for 1 and 2
- Some frictional equation such as Darcy for 3.
?

 

[Clausius2]

2nd equation: Pt=total pressure. Delta Pt are the losses of total pressure due to any effect. I consider a straight pipeline z=cte. The last equation IS NOT the Bernoulli equation. Benoulli equation=Ideal Flow. Last equation=Non ideal flow=Not Bernoulli.

You're a bit stubborn, aren't you?

 

[Q_Goest]

If Delta Pt are the losses due to any affect, then how do you calculate changes in elevation and velocity in this portion of the equation? To suggest this is not Bernoulli's equation is only half correct, since although you're not using his equation for frictional losses, you ARE using every bit of it for the conservation of energy affects due to elevation and velocity. Bernoulli's tells you how much pressure change there is due to elevation and velocity in a piping system regardless of Re.

I'm not stubborn, I've been doing this for so long I don't need to think about it any more. My books and tech papers are worn and tattered, marked and dog eared, and have been used to create hundreds of programs. I try not to get caught up in symantics - that is: what to call an equation, but it seems to me the words on this page may confuse someone else who's trying to figure out this issue. I remember having a tough time of it in school, I don't think they really do a very good job of explaining pressure drop in pipe in college. That's something most of us learn after those funny hats with tassels get lost in the attic. How do we calculate pressure drop around bends and elbows? through concentric expanders or reducers? across orificies and through valves? How do you apply thermodynamics and heat transfer to all that... there's a whole lot more to it than can be put into a small message board like this.

If we stay focused on the actual affects of pressure drop in a relatively simple system which contains an incompressible fluid only, with no heat transfer, and neglect thermodynamic changes due to changes in pressure, then we can look at using Bernoulli's and frictional formulas for pressure drop only. In virtually any realistic piping system you will analyze in your career yet to come, you will find that pressure changes due to elevation are calculated using rho*g*h, that changes in velocity result in pressure changes and are calculated by .5*rho*V^2, and these pressure calculations stem directly from Bernoulli's equation because they represent a conservation of energy. I really don't care what you'd like to call equation 2 above, but what I'd call it is Bernoulli's equation with frictional losses added it... assuming no heat transfer, changes of state, ... yadyadayada.

 

[Clausius2]

I'm not stubborn, I've been doing this for so long I don't need to think about it any more.

Your experience is not a seal of guarantee. My father is 61 years and he keeps on being wrong in some things.

I must say the truth, and the truth is that the phrase "Bernoulli with losses" DOESN'T REPRESENT THE BERNOULLI EQUATION, BECAUSE BERNOULLI EQUATION DOES NOT CALCULATE ANY LOSS, BECAUSE BERNOULLI EQUATION STATES THE CONSERVATION OF STAGNATION PRESSURE. The fact of calling Bernoulli to an equation which APPARENTLY is similar to Bernoulli is a MISCONCEPTION, no matter how long have you been calling it Bernoulli.

 

[Clausius2]

For those interested, I have simulated the original problem in Fluent 6.0. I have attached the Total Pressure Contours. Red color means highest total pressure and blue color the lowest one. The computational mesh is axisymmetric, so the lowest line is a symmetry axe.