Can convergent nozzles convert heat into motion?

[pranj5]

It's a well known fact that convergent and/or convergent-divergent nozzles convert internal enthalpy into forward motion i.e. dynamic pressure. But, enthalpy means both internal heat and the pressure-volume. I want to know whether the internal heat of a fluid can also be converted into forward motion by a convergent and/or convergent-divergent nozzle or it's the pressure part only that increases the dynamic pressure of fluid?

 

[jack action]

Well I see a temperature decrease as well as a pressure decrease in this diagram from Wikipedia:

The basic equation is:
c_pT_0 = \frac{V^2}{2}+c_pT
where the following relation comes from (the one plotted on the graph):
\frac{T_0}{T} = 1+\frac{k-1}{2}M^2
So the kinetic energy comes initially from heat (temperature). It is the isentropic condition that makes the pressure going down as well.

 

[Jaxv]

Jacks reply above should help you get the gist of it. Another way of looking at it could be

Vexit=a*Mexit -->a is speed of sound and M is exit mach number

a=sqrt(gamma*R*T) --> where gamma is specific heat ratio of fluid, R is gas constant and T is static temperature.

therefore Vexit=sqrt(gamma*R*T)*Mexit , or exit velocity is proportional to square root of temperature.

you could use a cycle deck and see that increasing turbine inlet temperature alone increases thrust.

 

[pranj5]

Thanks to all for your responses. It seems that as a convergent or c/d nozzle can convert heat into velocity, there it can be used for other purposes too. As for example, we all know that pressure is nothing but result of random motion of molecules. Now, if a convergent and/or c/d nozzle can increase speed in a specific direction, can we use that to reduce power consumption to compress compressible fluid from lower pressure to higher pressure.
I mean if we put a convergent and/or c/d nozzle at the exit of a compressor, does it help to reduce power consumption by the compressor? It apparently seems it can be because nozzles can convert heat into forward motion and therefore dynamic pressure in a particular direction increases. It can be said in this way:
H = U + PV; when a compressible fluid has been injected to a higher pressure by using a nozzle, its PV part becomes bigger at the expense of U as the nozzle converted part of U into PV. By this process, the first law of thermodynamics isn't violated. Am I right?

 

[jack action]

can we use that to reduce power consumption to compress compressible fluid from lower pressure to higher pressure.

Short answer: No.

You have to look at this as the equivalent of the mechanical advantage in mechanics. Power is torque times angular velocity and, for example, a gear set can transform torque into rpm and vice-versa, but the power remains constant throughout the process. In fluid mechanics, power is pressure times velocity. A c-d nozzle will increase one at the expense of the other, but it doesn't add anything either.

So your power source have to produce the power you need and your c-d nozzle will «adjust» the flow to the conditions you need (pressure and velocity).

 

[pranj5]

If by "power" you want to mean gross energy, I agree with you. But, I just want to point out your attention to another point. The nozzle converts internal energy i.e. heat into motion and you yourself have admitted that in your very first reply. Kindly read my post above, where I want to say that the gross enthalpy of the fluid will remain constant but the PV part will increase at the expense of U by the nozzle. At least, this don't have any contradiction with first law of thermodynamics.
I want to mean dynamic pressure here. As the nozzle can convert heat into velocity and can increase the speed of molecules in a particular direction, then if the mean velocity of molecules there (at the high pressure zone) would less than the velocity of molecules attained at the throat, then certainly it's possible. At least theoretically. In fact, the own heat of a fluid will be used to compress it.

 

[jack action]

In fact, the own heat of a fluid will be used to compress it.

If you have a «free» heat source, then you are right, you will be able to reduce power consumption. If not, you will have to add some energy yourself by either heating the fluid or compressing it.

Note also that there is a limit on converting the heat into velocity. The lowest temperature feasible is zero and you would end up with an infinite Mach number. At the very least, we know that it cannot be larger than the speed of light.

 

[pranj5]

Well, that too is an advantage as heat can be considered as a lesser form of energy in comparison to electricity. Whatsoever, can you explain what you want to mean by "The lowest temperature feasible is zero and you would end up with an infinite Mach number"? And if possible, kindly explain that mathematically.

 

[Chestermiller]

If by "power" you want to mean gross energy, I agree with you. But, I just want to point out your attention to another point. The nozzle converts internal energy i.e. heat into motion and you yourself have admitted that in your very first reply. Kindly read my post above, where I want to say that the gross enthalpy of the fluid will remain constant but the PV part will increase at the expense of U by the nozzle. At least, this don't have any contradiction with first law of thermodynamics.

Actually, this does contradict the first law of thermodynamics. From the open system steady state flow version of the first law of thermodynamics, the enthalpy of the gas decreases as the flow velocity increases, and vice versa. In particular:
$$\Delta \left(h+\frac{v^2}{2}\right)=0$$

Getting back to basics, for inviscid adiabatic reversible flow of an ideal gas through a nozzle, the enthalpy actually has to satisfy 3 separate equations:

1. The differential form of the open system steady state flow version of the first law of thermodynamics:$$dh+vdv=0$$where v is the velocity.
2. The enthalpy property relationship for an ideal gas: $$dh=C_pdT$$
3. The constraint that the process is adiabatic and reversible (constant entropy):$$dh=TdS+VdP=VdP$$where V is the specific volume.
So, depending on how you want to interpret it, dh changes because the velocity changes, because the pressure changes (your P-V effect), or because the temperature changes. Actually, it is a combination of all three of these things. dh can be eliminated by combining any two of these equations, which, together with the ideal gas law and the continuity equation, then yields a pair of equations for determining the changes in T and P with cross sectional area variation.

 

[jack action]

Whatsoever, can you explain what you want to mean by "The lowest temperature feasible is zero and you would end up with an infinite Mach number"? And if possible, kindly explain that mathematically.

If you remove all internal energy from the fluid, you will reach the minimum possible absolute temperature you can achieve (i.e. 0 K), which is $T$ in the following equation:
\frac{T_0}{T} = 1+\frac{k-1}{2}M^2
or:
\frac{T_0}{(0)} = \infty = 1+\frac{k-1}{2}M^2
which can only be true if $M = \infty$.

Of course, you'll never reach that (for one, your fluid will turn to solid at some point ), but it is just to show that the more heat you try to remove, the harder it is. You can see on the previous graph that as the divergent part of the nozzle increases, the temperature line flattens (as well as all the other properties too).

 

[boneh3ad]

Also note that the velocity cannot be increased to infinity, as it is limited by the first law of thermodynamics (even though the Mach number can go to infinity in theory). If your flow starts from rest, then it's total enthalpy is
h_0 = c_p T_0.
For an isentropic system (as is typically the case for nozzles), the first law of thermodynamics states, as previously mentioned by @Chestermiller, that
\Delta h = 0,
so between the initial rest state and any point in a nozzle, you have
c_p T_0 = c_p T + \dfrac{u^2}{2}.
In the case where the flow has been expanded so far as to reach absolute zero ($T=0$), which is the limiting factor on velocity, this gives
c_p T_0 = \dfrac{u_{\textrm{max}}^2}{2},
or
u_{\textrm{max}} = \sqrt{2 c_p T_0}.
So, the maximum velocity in a nozzle is limited by your starting temperature, $T_0$, and the specific heat at constant pressure, $c_p$. In that sense, yes, a nozzle converts the internal energy into velocity, and the maximum attainable velocity scales as the square root of these two variables and shows marked diminishing returns.

Of course, you also need a pressure difference to make a nozzle operate, and this pressure ratio, $p_0/p_e$, grows very rapidly as you increase the Mach number of a nozzle. So, a nozzle is theoretically limited by the total temperature (total enthalpy) of the working gas, but in practice is generally limited to the pressure you can obtain (safely) in your reservoir.

Either way, though, you aren't going to be able to use a converging-diverging nozzle to decrease the power consumption of a compressor located upstream in any useful sense. For one, a converging-diverging nozzle is going to undo the compression obtained by the compressor, as the purpose of such a nozzle is to expand the flow to a higher velocity/Mach number and lower pressure, temperature, and density.

Second, once the nozzle "starts", the compressor can receive no information about what is occurring downstream of the throat of the nozzle. If the throat is sonic and the expansion is supersonic, then information about that downstream flow, which can only travel at the speed of sound, cannot travel upstream faster than the air is coming out of the nozzle. Essentially, a nozzle operates only in one direction, so whatever outlet pressure is supplied by the compressor is going to drive the nozzle.

Now, for a given reservoir pressure, a nozzle may operate in one of several modes depending on the downstream conditions. Assuming the compressor supplies a pressure high enough to choke the throat, then the mass flow rate through the nozzle is no longer sensitive to downstream conditions and depends only on the reservoir temperature and pressure. If the compressor is not capable of keeping up with this mass flow rate, then the nozzle won't function properly because it will try to draw gas away faster than the compressor can supply it and you will end up with subsonic flow throughout. If the compressor supplies gas faster than the nozzle can expend it, then you might see a reduction in power consumption since the compressor doesn't have to supply gas as fast as it possibly can, but I don't see the use here.

So, I guess then the operative question is this: what exactly are you hoping to use this compressor for? You might reduce the power consumption slightly, but all of that gas is going to end up just moving through the nozzle and being expanded back to ambient pressure, so I am not sure that you will get any use out of it unless that use is simply making a supersonic wind tunnel.

 

[pranj5]

Either way, though, you aren't going to be able to use a converging-diverging nozzle to decrease the power consumption of a compressor located upstream in any useful sense. For one, a converging-diverging nozzle is going to undo the compression obtained by the compressor, as the purpose of such a nozzle is to expand the flow to a higher velocity/Mach number and lower pressure, temperature, and density.

I just want to point out that portion. Velocity itself can be termed as dynamic pressure and means forcing the molecules to a specific direction. What you have said that means static pressure may decrease but dynamic pressure on a particular direction increase and why not we be able to force molecules into higher pressure zone by using the velocity/dynamic pressure?
I am writing some formulation about this matter and may probably submit that tomorrow.

 

[boneh3ad]

Dynamic pressure is essentially another name for kinetic energy. At the expense of kinetic energy, you can force a flow into a higher pressure region. This happens all the time. If you have a flow moving with some velocity, $u_1$, and it moves into a downstream region with velocity, $u_2$, then one of several things have occurred. The pressure can decrease ($p_2 < p_1$), and that pressure difference, which can be viewed as a force, accelerates the flow such that $u_2 > u_1$. The pressure could stay the same, and in that case, disregarding viscosity, the velocities are the same. Finally, if $p_2 > p_1$, that is equivalent to saying the flow is moving in a direction against a force, so $u_2 < u_1$.

In other words, you can certainly have a flow with a high dynamic pressure (proportional to $u^2$, after all) that moves into a region of higher pressure. You will just lose some of your kinetic energy in the process. However, you will never be able to accelerate that flow to a point where it can travel against a pressure gradient far enough to reach a pressure higher than what the compressor originally supplied in the first place. That would violate energy conservation.

 

[pranj5]

That simply means first heat has been converted into kinetic energy and the kinetic energy into higher pressure. In short, low quality energy (heat) is converted into higher quality energy. Right?
In that case, you are agreeing with me that a convergent and/or c/d nozzle can be used to inject low pressure fluid into higher pressure.

 

[boneh3ad]

But not against the first law of thermodynamics. Kinetic energy is generated, but that results in a lower pressure. Pressure is essentially a form of potential energy per unit volume, and if you are increasing kinetic energy, that energy has to come from somewhere. In an incompressible flow, this means a lower pressure and/or a lower gravitational potential energy (this is essentially the gist of Bernoulli's equation). For a compressible flow, things are a bit more complicated, as there are more variables involved (e.g. temperature and density), but still, if you increase the dynamic pressure, the static pressure is going to decrease, not increase. You still need a force to accelerate the flow, and that comes in the form of a negative pressure gradient.

 

[pranj5]

That has already been discussed. The nozzle converts internal energy i.e. heat into kinetic energy and that kinetic energy will be converted into pressure. I can't understand how the violation of first law of thermodynamics can arise here.

 

[boneh3ad]

A nozzle converts enthalpy to velocity, if you will. Not all of that comes from internal energy. Essentially, it pulls that energy from the internal energy and the pressure energy. If you accelerate an ideal gas through a nozzle, you cause it to expand, which means the velocity goes up while the density, pressure, and temperature go down. It must have a high-pressure reservoir upstream as a source (supplied by the compressor in your example) and it ends up at a much lower pressure.

If you want to bring that accelerated flow back into a higher pressure region (isentropically), you aren't going to get it up beyond the pressure at which it started without adding energy, because it only has exactly as much kinetic energy as the enthalpy it already lost during the acceleration. It gained a certain amount of energy in the acceleration, so bringing it back to zero will cause it to lose exactly that much energy again back to enthalpy. That means that, at best, your nozzle expands the flow, and then in allowing it to isentropically recompress, you have zero net energy gained or loss and end up exactly where you started at the outlet of the compressor. This is essentially the first law of thermodynamics: you can't win.

Of course, isentropically slowing a supersonic gas is nearly impossible, and what will actually happen is a shock will form, causing a massive increase in entropy, and you will never actually recover the full compressor outlet pressure. This is the second law of thermodynamics: you can't break even.

 

[pranj5]

A nozzle converts enthalpy to velocity, if you will. Not all of that comes from internal energy. Essentially, it pulls that energy from the internal energy and the pressure energy. If you accelerate an ideal gas through a nozzle, you cause it to expand, which means the velocity goes up while the density, pressure, and temperature go down. It must have a high-pressure reservoir upstream as a source (supplied by the compressor in your example) and it ends up at a much lower pressure.

Question is, how much of it comes from it and how much from pressure. What you are saying above is about the process by which nozzles are used so far. But, kindly note that I am discussing about the using the "dynamic pressure" i.e. the velocity itself to inject low pressure fluid into higher pressure zone.

If you want to bring that accelerated flow back into a higher pressure region (isentropically), you aren't going to get it up beyond the pressure at which it started without adding energy, because it only has exactly as much kinetic energy as the enthalpy it already lost during the acceleration. It gained a certain amount of energy in the acceleration, so bringing it back to zero will cause it to lose exactly that much energy again back to enthalpy. That means that, at best, your nozzle expands the flow, and then in allowing it to isentropically recompress, you have zero net energy gained or loss and end up exactly where you started at the outlet of the compressor. This is essentially the first law of thermodynamics: you can't win.

If it can be done by just adding heat, that too can be considered as some kind of advantage. Low temperature heat sources are abundant around and if that can be used for producing pressurised fluid, that too can be considered as advantage.
And, by the way, I just want to add that by heating the fluid before entering the nozzle means less chance of choking because speed of sound through a fluid is directly proportional to square root of its temperature In short, M α https://www.physicsforums.com/file:///C:/Users/Payel/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif (in Kelvin). That simply means the higher the temperature, the higher will be velocity of sound through it and the nozzles will choke at higher velocity.

Of course, isentropically slowing a supersonic gas is nearly impossible, and what will actually happen is a shock will form, causing a massive increase in entropy, and you will never actually recover the full compressor outlet pressure. This is the second law of thermodynamics: you can't break even.

2nd law of thermodynamic simply says that no process can be isentropic (all ideal/reversible processes are imaginary and can't be achieved in real world). But, it doesn't say that how much entropy increase will occur in every process and actually that will depend on our own creativity. As for example, we all are mortal but what medical science is trying and doing is to lengthen the span. Second law doesn't dictate that there will be a "massive increase" in entropy and if it happens, that means faulty design. The better the design, the lesser will be the increase in entropy.

 

[pranj5]

From the discussion, it's now clear that c/d nozzles can give rise to supersonic velocity of the fluid, but can anybody tell me what can be the maximum limit of the speed?

 

[boneh3ad]

From the discussion, it's now clear that c/d nozzles can give rise to supersonic velocity of the fluid, but can anybody tell me what can be the maximum limit of the speed?

See post #11 where I answered exactly that question.

 

[pranj5]

Thanks! Can you elaborate the equation with one example at least. I mean just show me the maximum attainable velocity at a given temperature like 20C.

 

[boneh3ad]

Just plug the numbers into the formula. Make sure to use an absolute scale like Kelvin.

 

[pranj5]

Well, I just have some different thought. We all know that heat is nothing but expression of random motion of molecules and hotter a thing is, higher is the random motion of molecules. Therefore, heat of gaseous fluids are half of the result of multiplication of mass of molecules and square of root mean square velocity. That simply means the root mean square velocity can be considered as umax.
But, as per your formula, it's much less than the root mean square velocity. In short, after extracting all heat, there will be still huge amount of velocity left for the fluid i.e. there will be heat inside after extracting all heat.

 

[boneh3ad]

That's not at all what my formula says. It comes directly from the first law of thermodynamics and represents conversion of all internal energy into velocity.

 

[pranj5]

I just want to show that as per your formula, there will be still velocity left and that means that gas molecules will still have energy even at 0K. Do you think it's possible?

 

[boneh3ad]

It is about as possible as bringing anything truly to absolute zero. If you could somehow do it in this situation, then yes, you could have velocity of the bulk gas while its temperature was zero. It would mean that the thermal motion of all of the particles has stopped and all that is left is their bulk motion. Of course, the bigger problem is that if you got a gas that cold, it would no longer be a gas. Still, that's why it is called the theoretical limit.

 

[cjl]

That has already been discussed. The nozzle converts internal energy i.e. heat into kinetic energy and that kinetic energy will be converted into pressure. I can't understand how the violation of first law of thermodynamics can arise here.

Yes, the gas cools and accelerates through the nozzle, but when you recompress it, it heats up again. You'll never end up with the gas at a higher pressure than the reservoir supplying the nozzle.

 

[pranj5]

Yes, the gas cools and accelerates through the nozzle, but when you recompress it, it heats up again. You'll never end up with the gas at a higher pressure than the reservoir supplying the nozzle.

When it will cool, if that area is at a pressure higher than before, that means a part of its internal energy will be converted into pressure. In short, some U will be converted into PV and by this process the enthalpy will remain same as per first law of thermodynamics.

 

[boneh3ad]

No, when the gas cools due to expansion like this, the pressure gets lower, not higher.

 

[Chestermiller]

When it will cool, if that area is at a pressure higher than before, that means a part of its internal energy will be converted into pressure. In short, some U will be converted into PV and by this process the enthalpy will remain same as per first law of thermodynamics.

As I said in post #9, as per the first law, the enthalpy does not remain constant.

 

[pranj5]

No, when the gas cools due to expansion like this, the pressure gets lower, not higher.

Sorry as I myself have misused words. What I want to say is that when the fluid will be slowed instead of being "cooled".

 

[cjl]

When it will cool, if that area is at a pressure higher than before, that means a part of its internal energy will be converted into pressure. In short, some U will be converted into PV and by this process the enthalpy will remain same as per first law of thermodynamics.

What do you mean "when it will cool, if that area is at a pressure higher than before"? It cannot flow into an area with a higher pressure than the reservoir pressure. The maximum possible pressure recovery would result in the same conditions that existed in the reservoir upstream of the nozzle.

 

[pranj5]

Now, a new thought comes to my mind and I am sharing it others here.
If a pressurised gas/fluid will pass through a convergent and/or c/d nozzle, then it's velocity increases in a specific direction inside the nozzle. We both agreed to one fact that a part of the velocity comes at the price of internal heat of the gas/fluid.
In case of Nitrogen at 4 barA pressure and 27°C, if a turbine is used to release the Nitrogen at 1 barA with a flowrate of 1 kg/sec; then the output is around 95 kW. Now, if the pressurised Nitrogen is released through a convergent or c/d nozzle shaped structure before the turbine, it's velocity will be higher than the previous case. Does that means effective rise in the pressure? If yes, then how much?
As per my thoughts, if the cross section at the throat is half that of the inlet, then the velocity at the throat is twice. That means the effective pressure will be 4 times than the inlet. Am I right?

 

[jack action]

Now, if the pressurised Nitrogen is released through a convergent or c/d nozzle shaped structure before the turbine, it's velocity will be higher than the previous case. Does that means effective rise in the pressure? If yes, then how much?

No, because the static pressure will drop. The power is (static pressure) X (velocity). When you increase dynamic pressure, it only relates to velocity. The word 'pressure' in 'dynamic pressure' only means «if we stop the flow, this is the pressure we would get». But if the flow doesn't slow down, it doesn't exist. As @boneh3ad mentioned in one of his posts, dynamic pressure represents kinetic energy, i.e. energy stored in motion.

 

[pranj5]

Do you want to mean that the velocity inside a nozzle comes only at the expense of pressure only? In this thread, it's you who have said that a part of the internal energy too will convert into motion.
You are right, if all the velocity inside the nozzle will come at the expense of pressure only, but that's not true. The temperature too will fall and that means part of internal energy will be converted into motion. In short, the random motion of molecules converted to the velocity towards a specific direction.

 

[jack action]

Do you want to mean that the velocity inside a nozzle comes only at the expense of pressure only?

I did not say that at all.

Maybe I misunderstood what you said. The way you wrote your post, I understood that you think that by increasing dynamic pressure, you actually increase the pressure of the fluid. You don't. That is what I said. Sorry if I misunderstood your post.

 

[pranj5]

What I want to mean is that whether convergent or c/d nozzle can increase the effective pressure of fluid going through it or not. If yes, then how much?

 

[boneh3ad]

Now, if the pressurised Nitrogen is released through a convergent or c/d nozzle shaped structure before the turbine, it's velocity will be higher than the previous case. Does that means effective rise in the pressure? If yes, then how much?
As per my thoughts, if the cross section at the throat is half that of the inlet, then the velocity at the throat is twice. That means the effective pressure will be 4 times than the inlet. Am I right?

Your thoughts appear to be incorrect. For one, the velocity at a throat that is half the area of its inlet will not be double that of the incoming flow. This is a compressible flow and simple relations like $V_1 A_1 = V_2 A_2$ do not hold. The real relationship (in a 1D sense) is $\rho_1 V_1 A_1 = \rho_2 V_2 A_2$ and you have to account for how $\rho$ will be changing (typically by assuming an ideal gas, $p = \rho RT$).

Whether or not your thoughts on increasing effective pressure are correct depends on what you are describing as "effective pressure". I am not familiar with this term. What exactly do you mean?

 

[jack action]

As shown in the graph of my first post the c/d nozzle reduces the pressure, if that is what you mean by 'effective' pressure. It also increases the velocity which can be expressed in another way as 'dynamic pressure'. But dynamic pressure is 'potential' pressure, not actual pressure.

It's like considering an object at an height of 10 meters and a speed of 31.32 m/s going upward. The object has the potential to reach an height of 60 meters $\left(= \frac{31.32^2}{2g} + 10\right)$.

What is the effective height of the object? It's still 10 meters. Even if you double or halve the velocity, it will always be 10 meters. The 60 meters is not an actual property of the object (just like dynamic pressure), it's a potential result that can be achieved if you set the conditions right. One could refer to the 50 meters as 'dynamic height', which would only express the amount of kinetic energy stored in the object.

 

[boneh3ad]

As shown in the graph of my first post the c/d nozzle reduces the pressure, if that is what you mean by 'effective' pressure. It also increases the velocity which can be expressed in another way as 'dynamic pressure'. But dynamic pressure is 'potential' pressure, not actual pressure.

I am not a big fan of this description, to be honest. Dynamic pressure is really not a pressure nor a potential in any sense, but a kinetic energy per volume of the flow at a given point due to its bulk fluid motion. The static pressure is better described as a potential energy in the sense that it is the stored energy per volume at a point in the flow due to the random motion of its molecules. The total pressure (the static pressure if you isentropically slowed the flow velocity to zero) essentially represents a total energy pool from which the flow can draw.

It should also be noted that, in a compressible flow, the sum of the static and dynamic pressures does not, in general, equal the total pressure as it does in incompressible flows.

For example, if you look at the energy equation for a steady flow with no heat transfer in a control volume, you get
\iint\limits_{S}\rho\left( e + \dfrac{|\vec{V}|^2}{2} + \dfrac{p}{\rho}\right)\vec{V}\cdot d\vec{A} = 0

If we assume that the flow is quasi-one-dimensional here, then you get
d\left( e + \dfrac{V^2}{2} + \dfrac{p}{\rho} \right) = 0

If this looks familiar, it is because, for an incompressible flow, #e#, the internal energy, is constant, so this reduces to Bernoulli's equation and states that the sum of static and dynamic pressures are the total pressure, which is constant.
d\left(\dfrac{V^2}{2} + \dfrac{p}{\rho}\right) = 0

However, in a compressible flow, $e$ is not constant and is effectively a measure of temperature, so static and dynamic pressure do not add up to total pressure anymore.

 

[pranj5]

Whether or not your thoughts on increasing effective pressure are correct depends on what you are describing as "effective pressure". I am not familiar with this term. What exactly do you mean?

By "effective pressure", I want to mean that the force per unit area with which the fluid will come out of the nozzle. As for example, if the pressure at the inlet is 5 barA and the fluid is vented at atmospheric pressure, then the effective pressure is (5-1)barA i.e. 4 barA. While with a nozzle, a part of the internal energy is also being converted into motion and the speed at the throat will be higher in comparison to the case when it will be just vented.
This increased pressure that has been gained by conversion of internal energy plus the original pressure is called the "effective pressure". I hope I have explained enough.

 

[jack action]

This increased pressure that has been gained by conversion of internal energy

There is no gain in pressure.

The speed has increased, but the pressure (as well as the temperature) has decreased. The pressure will only increase back to its original value IF the flow slows down again.

With a nozzle, the pressure might be 3 bar and the "effective" pressure will be (3-1) bar, i.e. 2 bar.

This is for a force balance. For an energy balance or a momentum balance, you will have to consider the fluid velocity as well.

 

[pranj5]

There is no gain in pressure.
The speed has increased, but the pressure (as well as the temperature) has decreased. The pressure will only increase back to its original value IF the flow slows down again.

If there is air at 5 barA pressure and if that's connected to a turbine and exhausted at 1 barA, then the air will come out with a velocity. Now, when a convergent and/or c/d nozzle will be fitted in between, speed at the throat will be higher than before and that means it will struck the blades of turbine with increased velocity. By "effective pressure" I want to mean the pressure level that can give rise to the speed at the throat without a nozzle, whether convergent and/or c/d.

 

[pranj5]

Thanks to boneh3ad, it's the "de" I am more concerned here.
As, jack action has said in this thread about an example of throwing a stone by standing at a height of 10 meters and that stone will reach 50 meter more, by "effective pressure" I want to mean here (50+10) or 60 meters because when that object will hit ground, it will struck the ground with a force that is equal to force of another object of same mass fallen from 60 meters. I hope I am able to understand what I want to mean.
[PLAIN]https://www.physicsforums.com/members/boneh3ad.268837/[/PLAIN]

 

[jack action]

That effective pressure your talking about will be the total pressure, i.e. the theoretical pressure achieved if the flow velocity was reduced to zero. If your fluid comes from a tank where velocity is zero and its pressure is 5 bar, then the total pressure is 5 bar and you won't achieved a greater pressure than that. If you move the fluid from tank A to tank B, the pressure will decrease in the line connecting the two tanks - the faster it will go, the lower the pressure - and it will go back up to 5 bar in tank B where the fluid will be slowed back to zero.

Like I told you earlier - when talking about mechanical advantage - it's like torque and rpm with a gear set. You can increase torque at the expense or rpm or increase rpm at the expense of torque. But you cannot increase power unless you input some new energy somehow. You cannot take energy stored internally and expect a bigger impact from your fluid, because it already has an impact.

If you take a slow moving, high pressure fluid you can use a slow-revving high-torque turbine. Or you can accelerate the fluid with a nozzle to get a high speed, low pressure fluid that will be able to feed a high-revving, low-torque turbine. No matter what, the power output will be the same. The torque-rpm relationship you will need will depend on what you want to do.

 

[pranj5]

That effective pressure your talking about will be the total pressure, i.e. the theoretical pressure achieved if the flow velocity was reduced to zero. If your fluid comes from a tank where velocity is zero and its pressure is 5 bar, then the total pressure is 5 bar and you won't achieved a greater pressure than that. If you move the fluid from tank A to tank B, the pressure will decrease in the line connecting the two tanks - the faster it will go, the lower the pressure - and it will go back up to 5 bar in tank B where the fluid will be slowed back to zero.

I have repeatedly want to tell that I want to know what will be the actual pressure at the throat. And what you are telling contradicts Joule-Thompson effect. If no internal energy will be converted into motion, then why temperature at the throat decreases?

 

[boneh3ad]

I have repeatedly want to tell that I want to know what will be the actual pressure at the throat. And what you are telling contradicts Joule-Thompson effect. If no internal energy will be converted into motion, then why temperature at the throat decreases?

If you start with a reservoir with stagnant gas at some pressure, then it starts at some total enthalpy, $h = e +\frac{p}{\rho}$. This, in a certain sense, represents the total pool of energy available to expand a flow through a nozzle. In an incompressible flow, since $e$ effectively doesn't change, then the total pressure plays the same role. In the absence of heat transfer into or out of the flow, the total pool of $h$ remains constant regardless of how fast the flow starts moving. In equation form,
d\left( h + \dfrac{V^2}{2} \right) = 0.
So, let's assume for a moment that we are dealing with a calorically perfect gas (specific heats are constant). Then $h=c_p T$ and the above can be changed to
d\left[\dfrac{p}{\rho}\left( \dfrac{\gamma}{\gamma - 1} \right) + \dfrac{V^2}{2} \right] = 0
where $\gamma = c_p/c_v$ is the ratio of specific heats.

In essence, you are still pulling from the same total pressure reservoir (with an important density factor built in as well). Using a nozzle to accelerate the flow still leaves you with the same total pressure at best. In practice, when you slow it back down from a supersonic condition, there is a shock that forms and that decreases total pressure.

 

[pranj5]

Do you agree that at the throat, a part of e will be converted into forward motion. I am not interested in what happens when the fluid leaves the nozzle, but solely on the condition at the throat. At the throat, the velocity increases at the expense of e and I just want to know how much pressure increase can be achieved by that. To be precise, I again want to "at the throat" and/or at the point where the velocity is maximum.

 

[boneh3ad]

At the throat, the velocity increases at the expense of e and I just want to know how much pressure increase can be achieved by that.

At any point, if the velocity has increased, then the pressure has decreased. There is no increase in pressure. I just don't know how much more plainly that can be stated.

 

[pranj5]

Just one problem! You are continuously denying the fact that internal energy too can be converted into motion by a convergent and/or c/d nozzle. What you are saying can only be true if the increase in velocity comes at the expense of pressure only and that's not the case. Joule-Thompson effect is the proof of what I am saying.