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.