Bernoulli Principle and fluid velocity

[gloo]

I have been trying to fully understand this concept of Bernoulli's Principle , the Venturi effect and fluid velocity and energy. I need to grasp this to develop an idea that I have been working on.

1. As water is forced down a narrowing pipe, as the pipe gets more narrow, it means it losses more pressure and it's velocity increases. When it comes out of the narrow section (nozzle) it will have a higher velocity versus just a regular orifice (no narrowing). The higher velocity means a higher force or further distance traveled ?

2. The trade off of the higher force (velocity of water...assuming above is correct), is less flow, thus the conservation of energy principle is relevant here??

3. For a given static pressure of water in the pipe, we can aim for a higher force, or velocity by :
- making the final diameter more narrow
- creating a longer narrowing length (versus a short wide funnel) because this accelerates the water to faster speeds for a longer span
- smoother wall material

 

[boneh3ad]

Above all else, mass has to be conserved, so the rate of mass coming out of one end of your pipe or nozzle has to be the same as the rate of mass going into the pipe unless the flow is both compressible and unsteady. In this case (and many cases), it is neither. So, regarding your third question, it makes no difference how long the nozzle is unless you consider the effect of viscosity. Also, the smoothness of the wall has a lot less of an effect than you might imagine. The primary consideration is simply the ratio of the entrance and exit areas. More specifically, since mass must be conserved,
$$\rho_{in} V_{in} A_{in} = \rho_{out} V_{out} A_{out}.$$
The two values of $\rho$ are the same since you are considering water and water is effectively incompressible, so
$$V_{out} = \dfrac{A_{in}}{A_{out}}V_{in}.$$

So then, given your second question, the answer is no, there is not less flow when you speed up. The mass flow (and in the case of water, volumetric flow) rate is the same either way. However, conservation of energy is still relevant in that Bernoulli's equation is a statement of energy conservation. The units of pressure are the same as energy per unit volume, so Bernoulli's equation is essentially a conservation statement about the energy per volume.

 

[gloo]

Above all else, mass has to be conserved, so the rate of mass coming out of one end of your pipe or nozzle has to be the same as the rate of mass going into the pipe unless the flow is both compressible and unsteady. In this case (and many cases), it is neither. So, regarding your third question, it makes no difference how long the nozzle is unless you consider the effect of viscosity. Also, the smoothness of the wall has a lot less of an effect than you might imagine. The primary consideration is simply the ratio of the entrance and exit areas. More specifically, since mass must be conserved,
$$\rho_{in} V_{in} A_{in} = \rho_{out} V_{out} A_{out}.$$
The two values of $\rho$ are the same since you are considering water and water is effectively incompressible, so
$$V_{out} = \dfrac{A_{in}}{A_{out}}V_{in}.$$

So then, given your second question, the answer is no, there is not less flow when you speed up. The mass flow (and in the case of water, volumetric flow) rate is the same either way. However, conservation of energy is still relevant in that Bernoulli's equation is a statement of energy conservation. The units of pressure are the same as energy per unit volume, so Bernoulli's equation is essentially a conservation statement about the energy per volume.

Thanks boneh3ad.

1. Ok so what about Statement number 1? The faster the stream of water, means a greater force felt if i stood in front of it and it hit me it would hurt more? Or if it is unimpeded it will travel further in the air correct?

2. Yes, I messed up my statement on # 2. I meant to say smaller diameter stream but faster thus making up for total flow correct?

3. Can you tell me what you mean by ratio of Area into Area out? do you mean the area before it starts tappering, to the final exit point area (talking diameters i guess)

 

[boneh3ad]

1. Ok so what about Statement number 1? The faster the stream of water, means a greater force felt if i stood in front of it and it hit me it would hurt more? Or if it is unimpeded it will travel further in the air correct?

That is correct. The force imparted by a fluid is equal to the change in momentum, so a fluid with greater velocity but an otherwise equal mass flow rate will have greater momentum and exert more force on an object. It would probably hurt more.

2. Yes, I messed up my statement on # 2. I meant to say smaller diameter stream but faster thus making up for total flow correct?

Yes.

3. Can you tell me what you mean by ratio of Area into Area out? do you mean the area before it starts tappering, to the final exit point area (talking diameters i guess)

If you have two ends of your pipe or nozzle or whatever out, one will be the inlet (water entering) and the other will be the outlet (water leaving), hence in and out.

 

[gloo]

That is correct. The force imparted by a fluid is equal to the change in momentum, so a fluid with greater velocity but an otherwise equal mass flow rate will have greater momentum and exert more force on an object. It would probably hurt more.
Yes.
If you have two ends of your pipe or nozzle or whatever out, one will be the inlet (water entering) and the other will be the outlet (water leaving), hence in and out.

1. Your equation: Vout=Area in/Area out * Vin - The "V" stands for Velocity right?? Not volume?!

2. If this is V is for velocity, isn't it the same equation as to a regular pipe with no narrowing? For instance, a 5 meter diameter pipe with no narrowing has a 30 centimeter orifice versus a 5 meter pipe narrowing to 30 centimeter exit gradually. You are saying that the latter will have a 500/30 =16.67 faster speed exit velocity versus input velocity? But the first one will not have a 16.67 faster speed exit velocity than the input?

 

[russ_watters]

1. As water is forced down a narrowing pipe, as the pipe gets more narrow, it means it losses more pressure and it's velocity increases.

The first thing to learn about Bernoulli's principle is that there is more than one kind of pressure, so that statement is incomplete. In that scenario, static pressure decreases and velocity pressure (and velocity) increases. If you just say "pressure", by Bernoulli's principle you must be talking about all types of pressure and the whole point of Bernoulli's principle is that the sum of all types of pressure is constant.

When it comes out of the narrow section (nozzle) it will have a higher velocity versus just a regular orifice (no narrowing). The higher velocity means a higher force or further distance traveled ?

This depends on the specifics of the system. What you are saying applies to putting your thumb on the output of a water hose, for example, but the key difference in that case is actually that the piping pressure losses become concentrated at the outlet instead of lost throughout the pipe. So in that case you actually end up with more pressure in the pipe than in the scenario where the pipe is open.

2. The trade off of the higher force (velocity of water...assuming above is correct), is less flow, thus the conservation of energy principle is relevant here??

One important aspect of Bernoulli's principle is that it describes one flow situation at a time. Describing a before and after you add an obstruction is a completely separate analysis that may or may not even involve Bernoulli's principle. In the case of putting your thumb over a garden hose outlet, energy is not conserved between the two scenarios: the stream of water with your thumb over the outlet has less total energy than without your thumb over the end because you've added a restriction to the pipe (an energy loss).

3. For a given static pressure of water in the pipe, we can aim for a higher force, or velocity by :
- making the final diameter more narrow
- creating a longer narrowing length (versus a short wide funnel) because this accelerates the water to faster speeds for a longer span
- smoother wall material

Again, changing things about the flow scenario means Bernoulli's principle doesn't necessarily apply from one to the next. What happens when you change the system depends more on the structure and dynamics of the system. In general though, no, adding a long, narrow section will add loses to the system, which may or may not result in a higher outlet velocity...and it doesn't accelerate throughout the length of the pipe (since that would violate conservation of mass).

 

[boneh3ad]

1. Your equation: Vout=Area in/Area out * Vin - The "V" stands for Velocity right?? Not volume?!

Yes.

2. If this is V is for velocity, isn't it the same equation as to a regular pipe with no narrowing? For instance, a 5 meter diameter pipe with no narrowing has a 30 centimeter orifice versus a 5 meter pipe narrowing to 30 centimeter exit gradually. You are saying that the latter will have a 500/30 =16.67 faster speed exit velocity versus input velocity? But the first one will not have a 16.67 faster speed exit velocity than the input?

I never said there was anything different about those two situations. Insofar as conservation of mass is concerned, there is no difference. For a constant mass flow, if the inlet area is 16.67 times larger than the outlet area, then the velocity of the outlet flow has to be 16.67 times that of the inlet flow. Also note that you just used length instead of area. If those were diameters, the factor would actually be 277.78. Also note that as @russ_watters has mentioned, the very act of closing off the end or changing shapes can change the upstream flow and the two situations cannot be necessarily considered equivalent, but conservation of mass does not care.

 

[gloo]

Yes.
I never said there was anything different about those two situations. Insofar as conservation of mass is concerned, there is no difference. For a constant mass flow, if the inlet area is 16.67 times larger than the outlet area, then the velocity of the outlet flow has to be 16.67 times that of the inlet flow. Also note that you just used length instead of area. If those were diameters, the factor would actually be 277.78. Also note that as @russ_watters has mentioned, the very act of closing off the end or changing shapes can change the upstream flow and the two situations cannot be necessarily considered equivalent, but conservation of mass does not care.

Oh crap...my bad...forgot to do the area calculation before dividing. Thanks for that!

So taking the 277.78 number... if the Vin is 10m/s, then the Vout is 2777.80 m/s ? This is just with a regular 500 centimeter diameter pipe and a cap with a 30 centimeter hold in the cap?

If we used a gradual tapering nozzle instead of just a cap ending with an orifice plate, we get close to the 2777.80 (discharge co-efficient is close to 1) . If we use the orifice plate we get energy loss and only get about 0.6 * 2777.80 = 1666 m/s?

 

[gloo]

The first thing to learn about Bernoulli's principle is that there is more than one kind of pressure, so that statement is incomplete. In that scenario, static pressure decreases and velocity pressure (and velocity) increases. If you just say "pressure", by Bernoulli's principle you must be talking about all types of pressure and the whole point of Bernoulli's principle is that the sum of all types of pressure is constant.

This depends on the specifics of the system. What you are saying applies to putting your thumb on the output of a water hose, for example, but the key difference in that case is actually that the piping pressure losses become concentrated at the outlet instead of lost throughout the pipe. So in that case you actually end up with more pressure in the pipe than in the scenario where the pipe is open.

One important aspect of Bernoulli's principle is that it describes one flow situation at a time. Describing a before and after you add an obstruction is a completely separate analysis that may or may not even involve Bernoulli's principle. In the case of putting your thumb over a garden hose outlet, energy is not conserved between the two scenarios: the stream of water with your thumb over the outlet has less total energy than without your thumb over the end because you've added a restriction to the pipe (an energy loss).

Again, changing things about the flow scenario means Bernoulli's principle doesn't necessarily apply from one to the next. What happens when you change the system depends more on the structure and dynamics of the system. In general though, no, adding a long, narrow section will add loses to the system, which may or may not result in a higher outlet velocity...and it doesn't accelerate throughout the length of the pipe (since that would violate conservation of mass).

Thanks Russ... you are very technical and exact and I need to have an awake mind to absorb half of what you teach. Let me digest this tomorrow.

 

[russ_watters]

Oh crap...my bad...forgot to do the area calculation before dividing. Thanks for that!

So taking the 277.78 number... if the Vin is 10m/s, then the Vout is 2777.80 m/s ? This is just with a regular 500 centimeter diameter pipe and a cap with a 30 centimeter hold in the cap?

If we used a gradual tapering nozzle instead of just a cap ending with an orifice plate, we get close to the 2777.80 (discharge co-efficient is close to 1) . If we use the orifice plate we get energy loss and only get about 0.6 * 2777.80 = 1666 m/s?

No, the ratio has to be constant as a matter of geometry and conservation of volumetric flow (otherwise your pipe would be accumulating water!). If you use a less efficient nozzle, you get less flow out the outlet and less flow everywhere else in the pipe, keeping the ratios the same.

 

[gloo]

No, the ratio has to be constant as a matter of geometry and conservation of volumetric flow (otherwise your pipe would be accumulating water!). If you use a less efficient nozzle, you get less flow out the outlet and less flow everywhere else in the pipe, keeping the ratios the same.

Ok...but the 277.78 ratio number is theoretically correct? It is the same for both the orifice opening and the tapered nozzle opening?

 

[russ_watters]

Yes.

 

[gloo]

Ok...but the 277.78 ratio number is theoretically correct? It is the same for both the orifice opening and the tapered nozzle opening?

So...I am lost now and i think it's party the semantics of static pressure, total pressure , velocity pressure etc...

How is it that using a narrowing of a pipe, which accelerates the water (convert static pressure to velocity pressure) more efficiently effect this 277.78 number? You said that just having an orifice plate opening means energy loss because water has to make 90 degree turns and loses energy.

Is this how it works? Assuming water moves 10m/s with no nozzle, gets discharged out the opening at 2777.8?

But with a nozzle the 10m/s speeds up to 13 m/s, and then gets discharted at 13m/s* 277.78 = 3611.m/s?

 

[russ_watters]

Important piece of the puzzle from post #6:

One important aspect of Bernoulli's principle is that it describes one flow situation at a time. Describing a before and after you add an obstruction is a completely separate analysis that may or may not even involve Bernoulli's principle. In the case of putting your thumb over a garden hose outlet, energy is not conserved between the two scenarios: the stream of water with your thumb over the outlet has less total energy than without your thumb over the end because you've added a restriction to the pipe (an energy loss).

You are trying to compare two completely different flow scenarios with one application of Bernoulli's principle. You can't do that.

 

[boneh3ad]

Is this how it works? Assuming water moves 10m/s with no nozzle, gets discharged out the opening at 2777.8?

But with a nozzle the 10m/s speeds up to 13 m/s, and then gets discharted at 13m/s* 277.78 = 3611.m/s?

No, the key thing to remember is that you can't have more or less fluid leaving the pipe than what is entering the pipe. Mass must be conserved. If a pipe has a 1 m² area at its inlet and the water is moving 1 m/s, then the mass flow entering the pipe is $\rho V A$, or about 1,000 kg/s (since the density of water is about 1,000 kg/m³. That means that no matter how the area changes, exactly that much mass must be leaving the pipe at any given time. So if the area through which water is leaving the pipe is 0.5 m², then no matter what happens between the inlet and the exit, so long as water is not entering or leaving the system anywhere else, the exit velocity must be 2 m/s so that you still have 1,000 kg/s leaving the pipe and mass is conserved. This is the same whether there is just a cap on the exit end with a 0.5 m² hole in it or if there is some smoothly converging section that ends with a 0.5 m² area. Mass must be conserved.

The efficiency of the nozzle comes into play when you start trying to determine how much pressure you need to maintain a given flow scenario.

 

[gloo]

No, the key thing to remember is that you can't have more or less fluid leaving the pipe than what is entering the pipe. Mass must be conserved. If a pipe has a 1 m² area at its inlet and the water is moving 1 m/s, then the mass flow entering the pipe is $\rho V A$, or about 1,000 kg/s (since the density of water is about 1,000 kg/m³. That means that no matter how the area changes, exactly that much mass must be leaving the pipe at any given time. So if the area through which water is leaving the pipe is 0.5 m², then no matter what happens between the inlet and the exit, so long as water is not entering or leaving the system anywhere else, the exit velocity must be 2 m/s so that you still have 1,000 kg/s leaving the pipe and mass is conserved. This is the same whether there is just a cap on the exit end with a 0.5 m² hole in it or if there is some smoothly converging section that ends with a 0.5 m² area. Mass must be conserved.

The efficiency of the nozzle comes into play when you start trying to determine how much pressure you need to maintain a given flow scenario.

I get your conservation of flow numbers. I just don't see how nozzle even matters anymore. I don't understand what you guys mean by efficiency?? That seems like loosely used term now in light of how specific you guys want to be when words like "pressure" is used (static pressure, velocity pressure, total pressure, sum of all pressure ...but velocity pressure not the same as static pressure).

 

[russ_watters]

I get your conservation of flow numbers. I just don't see how nozzle even matters anymore. I don't understand what you guys mean by efficiency?? That seems like loosely used term now in light of how specific you guys want to be when words like "pressure" is used (static pressure, velocity pressure, total pressure, sum of all pressure ...but velocity pressure not the same as static pressure).

Bernoulli's principle is a conservation of energy statement, with energy converted to pressure. Efficiency in a fluid system is measured by pressure loss. So in a Bernoulli's equation description of the system, that manifests as the sum of the pressures on one side of the equation being lower than on the other -- and the difference is the pressure or energy loss. So the more efficient the system, the less pressure is lost when the fluid flows from one end to the other. So if the nozzle in one flow scenario is less efficient than in the other flow scenario and the pipe sizes are the same, the flow rate through the entire second system will be lower.

Say you have a large tank of water of a certain volume and height. The static pressure at the bottom is due to the heigh. Let's say that pressure is 50 psi (gauge). You attach a 1" pipe of a certain length to the bottom of the tank. The inlet (static) pressure of the water is 50 psi and the outlet static pressure is 0 psi. Let's say at the efficiency of this pipe, you lose 20 psi of pressure to friction. So the outlet velocity pressure is 50-20=30 psi. This system is 30/50=60% efficient at converting static pressure to velocity pressure.

Now you add a second pipe to this large tank of water: a 2" pipe instead of a 1" pipe. But the efficiency (pressure loss rate) is exactly the same. What is the inlet and outlet pressure and pressure loss? Is the flow rate the same, more or less than in the first pipe and by how much?

 

[gloo]

Bernoulli's principle is a conservation of energy statement, with energy converted to pressure. Efficiency in a fluid system is measured by pressure loss. So in a Bernoulli's equation description of the system, that manifests as the sum of the pressures on one side of the equation being lower than on the other -- and the difference is the pressure or energy loss. So the more efficient the system, the less pressure is lost when the fluid flows from one end to the other. So if the nozzle in one flow scenario is less efficient than in the other flow scenario and the pipe sizes are the same, the flow rate through the entire second system will be lower.

Say you have a large tank of water of a certain volume and height. The static pressure at the bottom is due to the heigh. Let's say that pressure is 50 psi (gauge). You attach a 1" pipe of a certain length to the bottom of the tank. The inlet (static) pressure of the water is 50 psi and the outlet static pressure is 0 psi. Let's say at the efficiency of this pipe, you lose 20 psi of pressure to friction. So the outlet velocity pressure is 50-20=30 psi. This system is 30/50=60% efficient at converting static pressure to velocity pressure.

Now you add a second pipe to this large tank of water: a 2" pipe instead of a 1" pipe. But the efficiency (pressure loss rate) is exactly the same. What is the inlet and outlet pressure and pressure loss? Is the flow rate the same, more or less than in the first pipe and by how much?

Wouldn't the flow rate be the same? Bigger opening, less speed, but greater amount of mass per time unit?

 

[russ_watters]

Wouldn't the flow rate be the same? Bigger opening, less speed, but greater amount of mass per time unit?

That's both wrong and self-contradictory.

Velocity pressure is a function of velocity, right? So by following that you should see that I already said that the velocity is the same. And if the velocity is the same but the second pipe is bigger, the mass flow rate has to be higher in the bigger pipe, right?

Again, you are erroneously trying to apply Bernoulli's principle to two separate scenarios at the same time. Bigger only means less speed at different locations in the same pipe. Conversely, if you have two different pipes of different sizes, their sizes don't tell us much of anything about their relative velocities.

And "flow rate" and "mass per time" are the same thing -- it can't be both the same and greater. That's the contradiction.

 

[gloo]

That's both wrong and self-contradictory.

Velocity pressure is a function of velocity, right? So by following that you should see that I already said that the velocity is the same. And if the velocity is the same but the second pipe is bigger, the mass flow rate has to be higher in the bigger pipe, right?

Again, you are erroneously trying to apply Bernoulli's principle to two separate scenarios at the same time. Bigger only means less speed at different locations in the same pipe. Conversely, if you have two different pipes of different sizes, their sizes don't tell us much of anything about their relative velocities.

And "flow rate" and "mass per time" are the same thing -- it can't be both the same and greater. That's the contradiction.

Russ (and boneh3ad), I apologize for my oversimplistic mind. Physics was not my strong point in school (actually nothing was...much to your surprise I am sure). I can't express my gratitude in getting help here from experts like you guys. I just really need help to explain an idea I am trying to put on a power point presentation to sell an idea...but I need to get it right.

What my obsession seems to be is that "geyser guy" water cannon toy argument (the cone shaped pulling toy). I am trying to apply this phenomena in creating something.

http://www.geyserguys.com/science.php

I am trying hard to vet out what is happening. First I am imagine a kid with a cylinder , (not a cone) that has a small hole and him pulling on the handles at the end with a force of "x" Then I imagine the same kid with that cone shape with same size opening , and pulling on the handle with force "x". Intuitively, it makes sense to me why the cone will fire the water further. Russ, you pointed out that the advertisement got the explanation wrong. But you did attribute the fact that the water cone employs the venturi effect in using the narrowing of the cone; this results in increased velocity of water exciting the cone and thus can travel further.

The cylinder, I intuitively guess (wrongly or rightly) that it can't fire the water as far when pulling the handle. I imagine all that water hitting the end of the cylinder, having to make a turn, losing energy (as you explained), then exiting the hole. Thus the pulling force gets wasted a bit (less efficient).

I am assuming that using a cone shape, versus just the cylinder will significantly increase the speed or force, at which the the water comes out -- everything else the same.

If there is some kind of percentage that can be used (please throw me a bone), what are we talking about in efficiency improvement.. If some young kid can use a cone to fire water 20 feet, I can't even imagine him pulling on a cylinder with a hole and getting it to fire 10 feet.

I am trying not to get tripped up and using the wrong words in explaining my query.

 

[boneh3ad]

The cylinder, I intuitively guess (wrongly or rightly) that it can't fire the water as far when pulling the handle. I imagine all that water hitting the end of the cylinder, having to make a turn, losing energy (as you explained), then exiting the hole. Thus the pulling force gets wasted a bit (less efficient).

I am assuming that using a cone shape, versus just the cylinder will significantly increase the speed or force, at which the the water comes out -- everything else the same.

If there is some kind of percentage that can be used (please throw me a bone), what are we talking about in efficiency improvement.. If some young kid can use a cone to fire water 20 feet, I can't even imagine him pulling on a cylinder with a hole and getting it to fire 10 feet.

Your guess is essentially correct. By pulling on the device, the person is providing energy to the system. Essentially, he or she is supplying the pressure difference between the inlet and outlet.

Think of it this way: the inlet velocity for the device is simply the velocity with which it is dragged through the water. For an identical velocity between your two cases, the exit velocity will be identical. However, the cylinder will be harder to pull through the water at that velocity than the cone will, thus the effect of the efficiency of the nozzle. This essentially means the cylinder is losing more of the pressure energy you are providing with the pull force to produce the same velocity.

If instead you held the force constant, then yes, the cylinder would fire the water a shorter distance. After all, that same force will not pull the cylinder as fast, and the mass flow into and out of it will be lower as compared to the cone.

 

[Malverin]

The mass flow is always the same (when you have incompressible fluid. For low speeds, they all are).
So when the cross section changes, the speed changes too. And
Q_m=ρ⋅A⋅V=const

In the smaller cross section, you have more speed, and so you have more kinetic energy.
But potential energy (correlated with static pressure) is lower.
So the energy of the flow is constant for any cross section.

E = E_k + E_p= const

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html

 

[gloo]

Your guess is essentially correct. By pulling on the device, the person is providing energy to the system. Essentially, he or she is supplying the pressure difference between the inlet and outlet.

Think of it this way: the inlet velocity for the device is simply the velocity with which it is dragged through the water. For an identical velocity between your two cases, the exit velocity will be identical. However, the cylinder will be harder to pull through the water at that velocity than the cone will, thus the effect of the efficiency of the nozzle. This essentially means the cylinder is losing more of the pressure energy you are providing with the pull force to produce the same velocity.

If instead you held the force constant, then yes, the cylinder would fire the water a shorter distance. After all, that same force will not pull the cylinder as fast, and the mass flow into and out of it will be lower as compared to the cone.

Boneh3ad - thanks for expanding on that layman's explanation. I have just a few more queries...

1. So the velocity is multiplied for the outstream by the ratio of the Area of the instream/ Area outstream. Thus if ratio is 10, then in velocity of 5m/s means an outflow velocity of 50. This multiplication also applies to the force of the stream correct? if force in is x...then force out is 10x?

2. Is there some kind of ball park figure in terms of efficiency using the cone versus the cylinder? Like between 20 to 50 percent or higher? This is assuming same pulling force, same size outlet hole, same size original diameter and optimal cone shape (Russ mentioned more curved wall etc..). I know this is very open ended and unscientific, but if you could humor me.

Thanks

 

[gloo]

Boneh3ad - thanks for expanding on that layman's explanation. I have just a few more queries...

1. So the velocity is multiplied for the outstream by the ratio of the Area of the instream/ Area outstream. Thus if ratio is 10, then in velocity of 5m/s means an outflow velocity of 50. This multiplication also applies to the force of the stream correct? if force in is x...then force out is 10x?

2. Is there some kind of ball park figure in terms of efficiency using the cone versus the cylinder? Like between 20 to 50 percent or higher? This is assuming same pulling force, same size outlet hole, same size original diameter and optimal cone shape (Russ mentioned more curved wall etc..). I know this is very open ended and unscientific, but if you could humor me.

Thanks

ok ...i will try to use my crappy logic to answer myself. If someone can tell me how i am wrong..(which i probably will be). Since the velocity out is greater and using the equation E=0.5mv^2, if the velocity goes up, then energy goes up. And, if energy or work=f x d, then either distance is greater (if fired through air, or force could be greater if the stream of water hits something and doesn't travel as far??

For ballpark figure efficiency?? I have no idea...but can it be ballparked by measuring the distanced travel if fired by the cone, versus distance traveled by the cylinder?

 

[boneh3ad]

Sorry, @gloo, I've been pretty busy lately. Anyway...

1. So the velocity is multiplied for the outstream by the ratio of the Area of the instream/ Area outstream. Thus if ratio is 10, then in velocity of 5m/s means an outflow velocity of 50. This multiplication also applies to the force of the stream correct? if force in is x...then force out is 10x?

Yes. Mathematically, since the force is the rate of change of momentum, you are looking at $\dot{m}v$ to get force. Since $\dot{m}$ is constant through the tube, it means the force that could be generated by the stream increases proportionally to $v$. That said, this is just the potential to exert a force with the exit stream compared to the conditions at the inlet. If you increased your pulling force, that wouldn't necessarily follow the same pattern since force and velocity are not necessarily linear (and a very likely to not be).

2. Is there some kind of ball park figure in terms of efficiency using the cone versus the cylinder? Like between 20 to 50 percent or higher? This is assuming same pulling force, same size outlet hole, same size original diameter and optimal cone shape (Russ mentioned more curved wall etc..). I know this is very open ended and unscientific, but if you could humor me.

I couldn't really tell you off the top of my head. That's an enormously complex problem, to be honest. You could measure it experimentally to get some empirical answer, but solving it analytically would be difficult without CFD software. You may be able to get a decent answer by adapting the concept of head loss for pipe constrictions. The reason here is that Bernoulli's equation is not really applicable in situations like this and only gives you an approximate answer in situations like this where viscosity is likely to be quite important. In order to use a Bernoulli-like analysis, you need to incorporate the concept of head loss to correct the Bernoulli equation. You could therefore likely come up with a relative efficiency be comparing the head loss associated with a smooth taper compared to that of an abrupt area change.

ok ...i will try to use my crappy logic to answer myself. If someone can tell me how i am wrong..(which i probably will be). Since the velocity out is greater and using the equation E=0.5mv^2, if the velocity goes up, then energy goes up. And, if energy or work=f x d, then either distance is greater (if fired through air, or force could be greater if the stream of water hits something and doesn't travel as far??

For ballpark figure efficiency?? I have no idea...but can it be ballparked by measuring the distanced travel if fired by the cone, versus distance traveled by the cylinder?

Keep in mind that kinetic energy in a moving fluid is typically represented by dynamics pressure, $q = \frac{1}{2}\rho v^2$. The stream with the higher dynamic pressure (kinetic energy) will certainly travel farther, but keep in mind that a design with a closed end with a hole in it can still produce the same outlet velocity as the tapered design, albeit with a greater pull force required. So for the same velocity, both designs will travel the same distance and be capable of the same impact force.

So, your goal in determining efficiency is to determine either the the difference in force required to achieve the same distance or else apply the same force and compare the distance. The head loss method approach is effectively doing the first of these options. Doing either option experimentally in a pool would be nearly impossible, however, as you would need a way to make sure you are pulling with the same force both times and/or find a way to measure the force accurately, neither of which you could do manually.

 

[gloo]

Sorry, @gloo, I've been pretty busy lately. Anyway...
Yes. Mathematically, since the force is the rate of change of momentum, you are looking at $\dot{m}v$ to get force. Since $\dot{m}$ is constant through the tube, it means the force that could be generated by the stream increases proportionally to $v$. That said, this is just the potential to exert a force with the exit stream compared to the conditions at the inlet. If you increased your pulling force, that wouldn't necessarily follow the same pattern since force and velocity are not necessarily linear (and a very likely to not be).
I couldn't really tell you off the top of my head. That's an enormously complex problem, to be honest. You could measure it experimentally to get some empirical answer, but solving it analytically would be difficult without CFD software. You may be able to get a decent answer by adapting the concept of head loss for pipe constrictions. The reason here is that Bernoulli's equation is not really applicable in situations like this and only gives you an approximate answer in situations like this where viscosity is likely to be quite important. In order to use a Bernoulli-like analysis, you need to incorporate the concept of head loss to correct the Bernoulli equation. You could therefore likely come up with a relative efficiency be comparing the head loss associated with a smooth taper compared to that of an abrupt area change.
Keep in mind that kinetic energy in a moving fluid is typically represented by dynamics pressure, $q = \frac{1}{2}\rho v^2$. The stream with the higher dynamic pressure (kinetic energy) will certainly travel farther, but keep in mind that a design with a closed end with a hole in it can still produce the same outlet velocity as the tapered design, albeit with a greater pull force required. So for the same velocity, both designs will travel the same distance and be capable of the same impact force.

So, your goal in determining efficiency is to determine either the the difference in force required to achieve the same distance or else apply the same force and compare the distance. The head loss method approach is effectively doing the first of these options. Doing either option experimentally in a pool would be nearly impossible, however, as you would need a way to make sure you are pulling with the same force both times and/or find a way to measure the force accurately, neither of which you could do manually.

Again, thank you for taking the time and explaining things to me with my layman physics knowledge. Ultimately, my aim is searching for the ability to drive a stream of water upward (head) as hard, or high as possible with a given input force (like the Geyser guys cannon toy). My end goal is for the stream of water to be able to have a large enough upward impact force, or to be able to push through a body of water of a certain depth. But my confusion started with the explanation of venturi effect, and applying Bernoulli (still not strong on it) principle.

You said force and velocity increase is not linear, but is the force increase greater, or less in proportion to the velocity?

 

[boneh3ad]

Generally, the pressure (head) loss due to the shape goes up with the square of the velocity, so the force required is going to rise faster than the velocity being output. If I remember my undergraduate course well enough, the head loss is something like
$$h_L = K\dfrac{V^2}{2g}$$
where $h_L$ is the head loss (has units of length), $K$ is a discharge coefficient that depends on the shape, $V$ is the velocity and $g$ is the acceleration due to gravity. Generally speaking, and abrupt area change like a hole in the end of a capped tube is going to have a higher value of $K$ than a smooth taper will. You could further reduce the value of $K$ if you designed the taper with certain curves such that the boundary layer did not separate at any point.

 

[gloo]

Generally, the pressure (head) loss due to the shape goes up with the square of the velocity, so the force required is going to rise faster than the velocity being output. If I remember my undergraduate course well enough, the head loss is something like
$$h_L = K\dfrac{V^2}{2g}$$
where $h_L$ is the head loss (has units of length), $K$ is a discharge coefficient that depends on the shape, $V$ is the velocity and $g$ is the acceleration due to gravity. Generally speaking, and abrupt area change like a hole in the end of a capped tube is going to have a higher value of $K$ than a smooth taper will. You could further reduce the value of $K$ if you designed the taper with certain curves such that the boundary layer did not separate at any point.

Ok...so interpreting... you are saying (will probably get it wrong)

with the less efficient (bigger pressure loss) shape , the force required to achieve the target velocity or head will be higher by some power of 2?