Can convergent nozzles convert heat into motion?

[boneh3ad]

Well that would depend on the geometry, but at the very least the flow leading up to it would still effectively be isentropic and the acceleration up to that point would follow the pattern of everything I've posted so far. That provides the inflow condition for the turbine section.

How the turbine functions would be a pretty complicated function of its geometry and the inflow conditions. I'm not sure how to be more descriptive there because it would be a pretty complicated flow field.

It would also be an immensely bad idea to place a turbine in a supersonic flow. The flow approaching the turbine would have to slow down, rather abruptly, and a shock would form. This dissipates energy, meaning the system is less efficient, and the problem gets worse as the Mach number increases.

 

[pranj5]

Can you give any example of slowing of supersonic flow for injected into a turbine? And I haven't said that that the flow exiting the nozzle would be supersonic. I have said that the speed will increase at the throat. How you have concluded that the speed will be supersonic after exiting the throat?

 

[jack action]

OK, I've put numbers to do some comparisons.

I've taken a fluid - with $R = 287\ \frac{J}{kg.K}$ and $\gamma = 1.4$ - at state $1$ where I arbitrarily set the pressure $P$ at $300\ kPa$, the temperature $T$ at $300\ K$, the velocity $v$ at $100\ m/s$, the mass $m$ at $1\ kg$ and the area $A$ at $2.87\ m^2$. I calculated the density $\rho$, the volume $V$, the length of the duct $L$, the Mach number $M$ and the area needed for choked flow $A*$ based on these conditions.

Then I asked myself how would the fluid be affected if I divided the area by $1.5$. I considered 2 cases: incompressible and compressible flows, that I labeled states $2i$ and $2c$, respectively.

The following is the comparison of the 3 states (the values are rounded, but I did all calculations with the true values):
\begin{matrix}<br /> &amp; \underline{1} &amp; \underline{2i} &amp; \underline{2c} \\<br /> P\ (Pa) &amp; 300\ 000 &amp; 278\ 223 &amp; 273\ 581 \\<br /> T\ (K) &amp; 300 &amp; 300 &amp; 292.2 \\<br /> \rho\ (kg/m^3) &amp; 3.484 &amp; 3.484 &amp; 3.26228 \\<br /> v\ (m/s) &amp; 100 &amp; 150 &amp; 160.2 \\<br /> M &amp; 0.288 &amp; 0.432 &amp; 0.467565 \\<br /> m\ (kg) &amp; 1 &amp; 1 &amp; 1 \\<br /> V\ (m^3) &amp; 0.287 &amp; 0.287 &amp; 0.306534 \\<br /> A\ (m^2) &amp; 2.87 &amp; 1.913 &amp; 1.913 \\<br /> L\ (m) &amp; 0.1 &amp; 0.15 &amp; 0.1602 \\<br /> A*\ (m^2) &amp; 1.36 &amp; &amp; 1.36 \\<br /> \frac{A}{A*} &amp; 2.111 &amp; &amp; 1.407248<br /> \end{matrix}
The first thing you notice is that the volume has increase in $2c$ and is fixed in $2i$. Because the particles have to cover a greater distance, it's also normal to have a greater velocity. The ratio $\frac{v}{L}$ stay constant in both cases.

What is more interesting is the energy analysis. Comparing the changes in work, internal energy and kinetic energy, we get:
\begin{matrix}<br /> &amp; \underline{equation} &amp; \underline{2i} &amp; \underline{2c} \\<br /> work &amp; P_2V_2 - P_1V_1 &amp; -6250 &amp; -2238.154 \\<br /> internal\ energy &amp; \frac{R}{\gamma - 1}\left(m_2 T_2 - m_1 T_1\right) &amp; 0 &amp; -5595.385 \\<br /> kinetic\ energy &amp; \frac{1}{2}\left(m_2 v_2^2 - m_1 v_1^2\right) &amp; 6250 &amp; 7833.540 \\<br /> total &amp; &amp; 0 &amp; 0<br /> \end{matrix}
First, it's clear that everything balances. There is a larger increase in kinetic energy in state $2c$ since the fluid goes faster. But you can see that as all the kinetic energy comes from a change in work in state $2i$, there is a lot less in state $2c$.

The "compressible" internal energy is not added to the "incompressible" work, it actually almost replaces it. It's just the fact that it is an ideal gas and this means that some work must be done when the internal energy changes, so a little bit or work is added to the internal energy change, both contributing to the increase in kinetic energy.

 

[pranj5]

The "compressible" internal energy is not added to the "incompressible" work, it actually almost replaces it. It's just the fact that it is an ideal gas and this means that some work must be done when the internal energy changes, so a little bit or work is added to the internal energy change, both contributing to the increase in kinetic energy.

Wonderful details! Just prove what I have thought. Actually, it's the initial pressure difference that's the source of work that will extract this internal energy.

 

[pranj5]

Let's forget about the nozzle temporarily and go for another calculation.
If 300 kPa pressure can give rise to 100 m/s speed, how much it would take to give rise 160.2 m/s speed. Everything is without the nozzle.
Exit velocity is directly proportional to square root of pressure i..e. if the pressure rises four times then the exit velocity will be twice as before.
That means the required pressure would be (300*(160.2/100)²) i.e. 769.9212 or almost 770 kPa pressure, right?
A big factor here is the extra increase in velocity in case of a compressible fluid. That's the factor here that's converting the internal heat into motion.

 

[boneh3ad]

If 300 kPa pressure can give rise to 100 m/s speed, how much it would take to give rise 160.2 m/s speed. Everything is without the nozzle.

This question doesn't even make sense. Pressure doesn't create flow. Pressure difference creates flow. This will also be geometry-dependent. What sort of situation are you now asking about? I get the impression you have something in mind that all of these different questions pertain to and are just not asking us the real problem at this point.

Exit velocity is directly proportional to square of velocity i..e. if the pressure rises four times then the exit velocity will be twice as before.

What does this mean? How can velocity be proportional to the square of velocity? Was there a type somewhere. Did you mean that pressure is proportional to the square of velocity? I think you are going back to your erroneous equation that you proposed in another thread where you claimed that $pV = m v^2/2$. That equation is not correct and does not reflect reality, so you can't use it here.

A big factor here is the extra increase in velocity in case of a compressible fluid. That's the factor here that's converting the internal heat into motion.

Again, this seems like word soup to me. The reason internal energy is converted into motion in the compressible case is precisely because of the fact that it is compressible. Since $\rho$ and $T$ are no longer constants, then $e=c_v T$ is no longer a constant.

 

[pranj5]

This question doesn't even make sense. Pressure doesn't create flow. Pressure difference creates flow. This will also be geometry-dependent. What sort of situation are you now asking about? I get the impression you have something in mind that all of these different questions pertain to and are just not asking us the real problem at this point.

I have asked the same question to Jack and haven't got any reply yet. I have arbitrarily consider that at the other end there is vacuum. I can understand that it's the "pressure difference" that will give rise to the flow.

What does this mean? How can velocity be proportional to the square of velocity? Was there a type somewhere. Did you mean that pressure is proportional to the square of velocity? I think you are going back to your erroneous equation that you proposed in another thread where you claimed that pV=mv2/2pV=mv2/2pV = m v^2/2. That equation is not correct and does not reflect reality, so you can't use it here.

Sorry for the mistake and corrected it.

Again, this seems like word soup to me. The reason internal energy is converted into motion in the compressible case is precisely because of the fact that it is compressible. Since ρρ\rho and TTT are no longer constants, then e=cvTe=cvTe=c_v T is no longer a constant.

Instead of T, use ΔT and all will be clear. Just calculate the decrease in internal energy and add that as motion and you can see that it will end up in the same velocity at the throat. Jack has clearly shown that in his last calculations.

 

[boneh3ad]

I'll respond to the PM that I got from @pranj5 here.

1. Do you agree Jack's calculation of the increase in velocity at the throat from 100 m/s to 160.2 m/s for a compressible fluid?
2. Do you agree to Jack's calculation regarding fall in temperature of the compressible gas at the throat?

I do know know his methodology, but the numbers I got are the same. I don't necessarily agree with using extensive properties like mass and volume here; though it seems to check out, it's not intuitive for a continuous medium, in my opinion. I think it makes more sense to use intensive properties like density in a fluid flow, but the overall conclusion is the same. It still shows that the temperature change is a larger factor in the kinetic energy change than is the pressure.

I have asked the same question to Jack and haven't got any reply yet. I have arbitrarily consider that at the other end there is vacuum. I can understand that it's the "pressure difference" that will give rise to the flow.

Your question still doesn't make sense, though. You are essentially just picking some arbitrary pressure and assigning it an arbitrary velocity based on essentially nothing. I think you just picked the initial conditions given by @jack action and then somehow misunderstood the process. At a given point, the pressure and velocity don't necessarily have anything to do with each other without consideration of how those quantities are changing in space.

300 kPa did not give rise to 100 m/s in his example. Those were simply arbitrarily chosen initial conditions for the sake of illustrating. Any change in velocity from 100 m/s to 160 m/s in his example corresponded to a decrease in pressure, which equates to a force. You got an answer that was somehow higher than the original pressure which makes no sense. Perhaps you have just specified your parameters incorrectly.

Instead of T, use ΔT and all will be clear. Just calculate the decrease in internal energy and add that as motion and you can see that it will end up in the same velocity at the throat. Jack has clearly shown that in his last calculations.

I understand the role temperature plays here. In fact, temperature is the limiting factor for how fast you can make a flow. You will hit absolute zero long before you hit zero pressure. At issue is that the statement you made to which I was replying was wrong. It's not simply an increase in velocity that causes internal energy to be converted. That's kind of a non-statement. If you look at the two cases, in fact, what you should notice is that the contribution to the total energy change as a result of pressure has decreased substantially when you consider compressibility. Temperature changes are therefore not simply responsible for the velocity difference between the two cases. There's a lot more depth to it than that.

The internal energy changes because of the fact that the density changes in a compressible flow, which is intrinsically coupled to the temperature and the pressure. That temperature change corresponds to changes in internal energy, and pressure changes are obviously already a form of energy change. The relative changes in each of those quantities are related and determined by the properties of the gas.

This is why I've derived all these equations in earlier posts. I discussed all of that. You have conveniently ignored all of those discussions, though.

 

[jack action]

I do know know his methodology, but the numbers I got are different.

Here are the equations on my Excel sheet (reference):

State $1$:
$\rho_1 = \frac{P_1}{RT_1}$
$M_1 = \frac{v_1}{\sqrt{\gamma RT_1}}$
$\frac{A_1}{A^*} = \frac{1}{M_1}\left(\frac{2+(\gamma-1)M_1^2}{\gamma+1}\right)^\frac{\gamma+1}{2(\gamma-1)}$
$A^* = \frac{A_1}{^{A_1}/_{A^*}}$
$V_1 = \frac{m_1}{\rho_1}$
$L_1 = \frac{V_1}{A_1}$

State $2i$:
$A_2 = \frac{A_1}{1.5}$ (by definition)
$\frac{A_2}{A^*} = \frac{1}{M_2}\left(\frac{2+(\gamma-1)M_2^2}{\gamma+1}\right)^\frac{\gamma+1}{2(\gamma-1)}$ (find $M_2$ by trial and error)
$T_2 = T_1\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}$
$P_2 = P_1\left(\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\right)^\frac{\gamma}{\gamma-1}$
$\rho_2 = \frac{P_2}{RT_2}$
$v_2 = M_2\sqrt{\gamma RT_2}$
$m_2 = m_1$
$V_2 = \frac{m_2}{\rho_2}$
$L_2 = \frac{V_2}{A_2}$

 

[boneh3ad]

@jack action I meant to change that. I actually had an error in my code and fixed that and got the same compressible numbers as you. Either way, all my other points stand. They were in light of my corrected numbers. I edited the post to reflect that. It made no sense as written.

 

[jack action]

I've been thinking about the initial question:

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?

The work of a turbine is the difference between inlet and outlet enthalpy, $w_t = h_{out} - h_{in}$.

If you put a nozzle before the turbine inlet, it won't change anything for the for the turbine, because for a nozzle $h_{out} = h_{in}$. So the same enthalpy will be available at the turbine inlet.

Enthalpy can be calculated based on total temperature, i.e. $h = C_p T_0 = C_p \left(T + \frac{v^2}{2C_p}\right)$. Thus, for a nozzle, whatever increase in velocity you get, it will be at the expense of a temperature decrease. Enthalpy wise, you gain nothing, you loose nothing and the turbine sees the same thing.

The only reason to put a nozzle at the inlet of a turbine would be to adjust the flow conditions such that the turbine is doing its job as efficiently as possible.

 

[pranj5]

Enthalpy can be calculated based on total temperature, i.e. h=CpT0=Cp(T+v22Cp)h = C_p T_0 = C_p \left(T + \frac{v^2}{2C_p}\right). Thus, for a nozzle, whatever increase in velocity you get, it will be at the expense of a temperature decrease.

That's exactly I want to know. Enthalpy gain is simply impossible because that's against 1st law of thermodynamics. But, by converting internal heat into motion, we can convert a little more of the gross enthalpy into power IMO.
No nozzle (actually nothing) can change the gross enthalpy of a fluid stream (in fact anything) without some kind of energy input or extraction. That's violation of 1st law of thermodynamics. But, what matters here is the question of exergy. Motion can be more useful in extracting energy than internal heat alone.

 

[jack action]

But, by converting internal heat into motion, we can convert a little more of the gross enthalpy into power IMO.

You just said that you agree that enthalpy is the same, why do you insist it will be converted to something else?

Power is the mass flow rate times the enthalpy ($\dot{m}h$). The nozzle doesn't change the mass flow rate nor the enthalpy, thus potential power cannot be affected.

Like I said earlier, any machine is designed to perform based on certain inlet & outlet conditions. In a case of turbine, it works best with a high velocity flow and a low pressure pressure differential. But if you replace your turbine by a piston engine, it will be better to have a low velocity flow and high pressure differential. Both will produce the same power under correct flow conditions, but the torque-RPM output will also be different. A gearbox can be used on the shaft to adapt the torque and RPM to what you need. In any case the power is conserved (minus some minor friction losses).

 

[pranj5]

You just said that you agree that enthalpy is the same, why do you insist it will be converted to something else?

I want to mean that the gross amount of energy stored in the fluid will be the same while the form may change. I have repeatedly mentioned 1st law of thermodynamics.

Power is the mass flow rate times the enthalpy (˙mhm˙h\dot{m}h). The nozzle doesn't change the mass flow rate nor the enthalpy, thus potential power cannot be affected.

What you want to mean is the gross amount of power embedded in the flow, point is how much can be converted into useful power.

Like I said earlier, any machine is designed to perform based on certain inlet & outlet conditions. In a case of turbine, it works best with a high velocity flow and a low pressure pressure differential. But if you replace your turbine by a piston engine, it will be better to have a low velocity flow and high pressure differential. Both will produce the same power under correct flow conditions, but the torque-RPM output will also be different. A gearbox can be used on the shaft to adapt the torque and RPM to what you need. In any case the power is conserved (minus some minor friction losses).

There is no doubt about that because that will violate 1st law of thermodynamics. Again, question is how much of the embedded power can be converted into useful power.

 

[jack action]

Again, question is how much of the embedded power can be converted into useful power.

Again, it depends on the design of the machine itself. It must be adapted to the given flow to produce as much power as possible. The capacity of a machine to produce useful work is called isentropic efficiency. No flow is better than the others. No flow has the potential of producing more work than another. They all have the same potential.

 

[pranj5]

The nozzle itself here is a part of the machine design.

 

[jack action]

The nozzle itself here is a part of the machine design.

It doesn't change anything to my statement.

For example, if you add a nozzle to accelerate the flow feeding a piston engine, you will most likely see a decrease in performance. Even with a turbine, you will see a drop in efficiency if the velocity gets too high.

You cannot state that an increase in velocity necessarily correspond to an increase in isentropic efficiency.

 

[pranj5]

Just give a look at the http://twisterbv.com/products-services/twister-supersonic-separator/how-it-works/ below that works well with supersonic input.

In short, such machinery is available that can perform with supersonic velocity.

 

[boneh3ad]

From what I can tell, the device you have pictured has absolutely nothing to do with the questions you've been asking. For example:

• It does not feature a turbine.
• It's inlet is subsonic; it only generates supersonic flow downstream of its vortex generator using a Laval nozzle.
• The primary purpose is not power generation, but instead the use of supersonic expansion to cause a temperature drop that allows water and hydrocarbons to condense and be centrifugally separated from the gas flow.

This really is not germane to the previous pages of discussion.

 

[RogueOne]

Just give a look at the http://twisterbv.com/products-services/twister-supersonic-separator/how-it-works/ below that works well with supersonic input.

In short, such machinery is available that can perform with supersonic velocity.

I am not sure why you think that this machine represents a nozzle that converts heat into motion and enables air to move from a low pressure zone to a higher one. The input to the system had both a higher temp AND pressure than the outlet.

 

[pranj5]

I am not sure why you think that this machine represents a nozzle that converts heat into motion and enables air to move from a low pressure zone to a higher one. The input to the system had both a higher temp AND pressure than the outlet.

I am not. It's just an example that there is machinery that can work well with supersonic flow.

The above graph of the impulse turbine shows that it's the pressure, that will be converted into velocity first and then this velocity will produce power. Now, anybody can calculate that whenever the pressure difference is 3 bar and above, the flow coming out will be supersonic. And I am sure that impulse turbines in use at present are working with much more than the mentioned pressure difference.

 

[pranj5]

The photo given above is from a home made experiment. I have made a simple homemade device where I have put a turbine like structure inside a tube and placed a convergent nozzle like structure at the inlet and placed the structure before a table fan. Thanks to one of my friend who temporarily handed me over the infrared camera to take pictures. I have take and few pictures with varying speed, but every time I have found that the air coming out of the structure is colder than the inlet like the picture given above. Though the slower the input speed, the lesser is the difference.
Now, what can be proved by it. At least I can say that the nozzle like structure have converted internal heat of the input flow into velocity and that has been converted into power. That's why the exhaust is colder.

 

[boneh3ad]

I have no idea what I'm looking at.

 

[russ_watters]

View attachment 113309
The photo given above is from a home made experiment. I have made a simple homemade device where I have put a turbine like structure inside a tube and placed a convergent nozzle like structure at the inlet and placed the structure before a table fan. Thanks to one of my friend who temporarily handed me over the infrared camera to take pictures. I have take and few pictures with varying speed, but every time I have found that the air coming out of the structure is colder than the inlet like the picture given above. Though the slower the input speed, the lesser is the difference.
Now, what can be proved by it. At least I can say that the nozzle like structure have converted internal heat of the input flow into velocity and that has been converted into power. That's why the exhaust is colder.

This is tough to interpret, but it seems you might be circling back to what you were told in post 2: during expansion out of a nozzle, internal thermal energy is converted to work. You were claiming the opposite: that during compression, you could convert thermal energy to work. That is still wrong.

 

[boneh3ad]

I am not. It's just an example that there is machinery that can work well with supersonic flow.

Except that isn't what you showed. That image had a series of vanes that interacted with subsonic flow, then expanded the flow to supersonic speed in order to utilize the resulting temperature drop. At no point was any energy extraction occurring. In fact, in general, supersonic flow is a huge negative when it comes to flows interacting with machinery since it must necessarily involve shock waves, which increase entropy, thereby decreasing the available pool of energy for extraction.

The above graph of the impulse turbine shows that it's the pressure, that will be converted into velocity first and then this velocity will produce power.

You didn't source the image so I can't comment on exactly what it is showing, but it isn't showing what you claim it is. This (1) is not dealing with compressible flows, (2) is not dealing with supersonic flows, (3) is not quantitative and therefore not a definitive source on exactly what the pressure and velocity are doing simultaneously with respect to energy extraction, and (4) appears to be simply showing some sort of comparison between these two methods of spinning a turbine, not any of the things you are trying to extract from it.

Now, anybody can calculate that whenever the pressure difference is 3 bar and above, the flow coming out will be supersonic.

Really? Show me how to prove this. Show it using just raw variables if possible, but if you really prefer to use actual values, I choose a reservoir pressure $p_0 = 10\mathrm{\; bar}$ and a downstream pressure of $p = 7\mathrm{\; bar}$.

 

[pranj5]

I have no idea what I'm looking at.

An infrared photograph of the device during its work.

 

[pranj5]

This is tough to interpret, but it seems you might be circling back to what you were told in post 2: during expansion out of a nozzle, internal thermal energy is converted to work. You were claiming the opposite: that during compression, you could convert thermal energy to work. That is still wrong.

The turbine like structure is inside the tube and the nozzle like structure is at the inlet. Therefore, if the turbine doing some work here, it's out of the nozzle. How you are interpreting something real is beyond my knowledge.
Can you tell me which factor is reducing the temperature of the input air?

 

[boneh3ad]

An infrared photograph of the device during its work.

But you've made no indication of scale or where the walls or devices are located or pressures or anything. The photo is essentially useless without more information.

 

[russ_watters]

The turbine like structure is inside the tube and the nozzle like structure is at the inlet.

What device from which post are you referring to? If you are referring to Post #82, the only thing I know for sure about the setup (a regular photo would help...) is you powered the device with a table fan, which means it is way, way below the velocity required for the effects of pressurization to manifest.

Also, air is transparent to the infrared, so you can't take pictures of it with an infrared camera. So there's that too...

 

[pranj5]

But you've made no indication of scale or where the walls or devices are located or pressures or anything. The photo is essentially useless without more information.

It's a homemade experiment and laboratory like perfection can't be expected.

Also, air is transparent to the infrared, so you can't take pictures of it with an infrared camera. So there's that too...

As far as I know, we can study the effect of temperature to air by adjusting the infrared. Actually, I can't say much about that as my friend is the expert. He has adjusted the camera to detect whether the air coming out is colder or not. But I can say that the picture isn't something made with animation. It's real!