Can convergent nozzles convert heat into motion?

[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.

 

[boneh3ad]

The sum history of all of human experience with compressible flows is proof of what I am saying. If a gas accelerates to a higher velocity, it must move from high to low pressure. That's Newton's second law. If there is an increase in momentum, there must be a force.

I'm not honestly how else to discus this because it seems like you are pretty set in your assumptions and aren't willing to listen. I even provided equations to back up what I've been saying.

 

[jack action]

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.

You are saying that the increase in velocity comes at the expense of a reduction in internal energy AND pressure. Then you ask us how much is the pressure increase. You just said it: there is none. Because of the isentropic nature of the process, the temperature cannot go down alone, pressure will decrease as well.

To feel a «pressure increase», you will have to reverse the process and slow down the fluid (ex.: hitting the blades of turbine). At that point the temperature AND pressure can only go back to their original value (stagnation point). There will not be any magical gain in pressure.

 

[pranj5]

To feel a «pressure increase», you will have to reverse the process and slow down the fluid (ex.: hitting the blades of turbine). At that point the temperature AND pressure can only go back to their original value (stagnation point). There will not be any magical gain in pressure.

I hope you can understand that the fluid will hit the blades of the turbine harder in comparison to the scenario where there is no nozzle. If the turbine is attached to generator, then a part of energy will be converted into electricity and that means the pressure and temperature can't go back to the original value. To me what matters is the force with which the fluid will struck the blades and what I can understand is that in case of a nozzle, it will be harder. Question is how much harder and that's everything for me.

The sum history of all of human experience with compressible flows is proof of what I am saying. If a gas accelerates to a higher velocity, it must move from high to low pressure. That's Newton's second law. If there is an increase in momentum, there must be a force.

Force is not energy that force can be exerted by the walls of the nozzle. This force will be a reaction to striking of the walls of the nozzle by the molecules of fluid. There is no doubt that when a gas accelerates, it moves from higher to lower pressure. But in case of a nozzle, the acceleration will be higher due to the geometry of the nozzle that converts internal energy of the fluid into velocity along with the pressure. And that's why the temperature of the fluid falls when it accelerates. Just try to think where the energy that has been liberated by the fall in temperature can go.

I'm not honestly how else to discus this because it seems like you are pretty set in your assumptions and aren't willing to listen. I even provided equations to back up what I've been saying.

Your equation is about a scenario where the fluid enters and leaves the nozzle and I am focused on how much internal energy has been converted into velocity at the throat. They are different.

 

[Mech_Engineer]

So @pranj5, when are you going to start actually accepting the advice of the multiple experts providing input in this thread? I feel like this thread is doomed to run around in circles forever at this point!

 

[pranj5]

I will accept everything when I get the explanation of temperature drop at the throat. Initially all have agreed on one point that convergent and/or c/d nozzles can convert internal heat into motion inside it and the increase in velocity comes at the cost of both pressure and internal heat.

So the kinetic energy comes initially from heat (temperature). It is the isentropic condition that makes the pressure going down as well.

That's what Jack Action had said in post no. 2 of this thread. I just want to understand how much extra thrust can be gained by conversion of this internal heat into motion.

 

[jack action]

I hope you can understand that the fluid will hit the blades of the turbine harder in comparison to the scenario where there is no nozzle.

No, it won't.

You look at the force this way:
$$F = \left(p + \frac{1}{2}\rho v^2\right)A$$
You look at this and think: "the nozzle increases $v$, therefore the force $F$ will be greater." But you are forgetting that the pressure $p$ and $\rho$ are also decreasing with the nozzle. Converting to the Mach number $M$, you get:
$$p + \frac{1}{2}\rho v^2 = p + \frac{1}{2}\rho \gamma RTM^2$$
$$p + \frac{1}{2}\rho v^2 = p + \frac{1}{2}p \gamma M^2$$
$$p + \frac{1}{2}\rho v^2 = p \left(1 + \frac{\gamma}{2}M^2\right)$$
$$p + \frac{1}{2}\rho v^2 = p_t \frac{\left(1 + \frac{\gamma}{2}M^2\right)}{\left(1 + \frac{\gamma-1}{2}M^2\right)^\frac{\gamma}{\gamma-1}}$$

Where $p_t$ is the total pressure. Plotting the equation $\frac{F}{p_t A}$ for $\gamma = 1.4$, you will notice that it is at its maximum ($=1$) at $M=0$. It actually decreases as $M$ increases.

I'm pretty sure the previous analysis is wrong because it doesn't take into account the compressibility effect.

The thing is that as the speed increases, the fluid decompresses (lower density) which releases the internal energy and the temperature drops. But when you slow down the fluid (hitting the blade), you have to compressed it again, so you cannot actually used that energy to push on the blade. Everything comes back to the initial state.

It's like if the moving fluid was a moving spring. When you accelerate one end you see the spring decompressing and you're thinking: "Look, I released some internal energy. Great! I'll use it to do some work!" But when the first end of the spring hits the blade, it slows down. Then, the other end that follows still arrive at a higher speed (It has accelerated too), thus the spring re-compresses itself and you loose the energy you gained earlier. The only difference with a moving fluid is that it happens in a continuous fashion. Maybe my comparison is questionable, but it might help further the discussion from more knowledgeable people than me.

 

[boneh3ad]

Yes, internal energy can be used as an additional "pool of energy" to feed a flow's acceleration, but even that can be related back to pressure as I showed previously. You just factor in the ratio of specific heats. If you would like more, allow me to show you exactly how to arrive at the isentropic pressure relationship for any compressible flow. Again, start with the energy equation in terms of enthalpy, which is $h = e + p/\rho$.
$$d\left( h + \dfrac{V^2}{2}\right) = 0$$
Let's treat this as being a statement about the relationship between two states: one at rest and one at any arbitrary velocity, $V$:
$$h_0 = h +\dfrac{V^2}{2}$$
We can assume the gas is calorically perfect (we aren't talking about super hot or cold things here so it's a good assumption), so $h = c_p T$:
$$c_p T_0 = c_p T + \dfrac{V^2}{2}$$
From the ideal gas law, we know that $p = \rho R T$, so
$$\dfrac{c_p p_0}{\rho_0 R} = \dfrac{c_p p}{\rho R} + \dfrac{V^2}{2}.$$
I already showed that $c_p/R$ can be cast in terms of $\gamma$ for the gas, so
$$\left( \dfrac{\gamma}{\gamma -1} \right)\dfrac{p_0}{\rho_0} = \left( \dfrac{\gamma}{\gamma -1} \right)\dfrac{p}{\rho} + \dfrac{V^2}{2}$$
or
$$\dfrac{p_0}{\rho_0} = \dfrac{p}{\rho} + \left( \dfrac{\gamma - 1}{\gamma} \right)\dfrac{V^2}{2}$$
If you divide through by $RT$, you end up with a term that is $a^2 = \gamma R T$, which, when combined with $V^2$ gives the square of the Mach number:
$$\dfrac{p_0}{\rho_0 RT} = \dfrac{p}{\rho RT} + \left( \dfrac{\gamma - 1}{2} \right)M^2$$
Also note that $p/(\rho R T) = 1$ from the ideal gas law, so
$$\dfrac{p_0}{\rho_0 RT} = 1 + \left( \dfrac{\gamma - 1}{2} \right)M^2$$
So far that is getting more useful than what we had before, but ideally, we'd like to keep this in terms of only pressure and Mach number, which means it would be nice to cast the $\rho_0$ term in terms of $\rho$ (so that it can be converted into a pressure term due to the ideal gas assumption and the $RT$ term) and pressures. Luckily, isentropic gas relations give us this. We know that
$$\dfrac{\rho}{\rho_0} = \left( \dfrac{p}{p_0} \right)^{1/\gamma}$$
so if you solve that for $\rho_0$, the resulting $\rho RT$ term becomes a $p$ and, and after some algebra the final relationship is
$$\boxed{ \dfrac{p}{p_0} = \left[ 1 + \left( \dfrac{\gamma-1}{2} \right)M^2 \right]^{-\frac{\gamma}{\gamma-1}} }$$

This is a very classical result, and hopefully the derivation has proven to you that it already takes into account the change in $e$ associated with acceleration of the gas to a higher speed. This is isentropic, so it also applies to slowing a flow back down. It's an energy conservation statement. It also has a maximum at $M=0$ (see plot below that includes the equivalent relations for $T$ and $\rho$) so no matter what process you use to speed it up and no matter what it then impacts, the maximum stagnation pressure it can recover is still the same stagnation pressure that occurred upstream when it was at rest. That's true whether you are talking about bringing it up to sonic velocity at a throat or supersonic velocity downstream. Of course, if it is supersonic, that means that slowing it down likely incurs a shock, which is not an isentropic process and dissipates some of the available energy, meaning you recover less total pressure even when slowing to zero. The bottom line is that you will never increase total pressure without adding energy to the system. This is dictated by the second law of thermodynamics.

I'll even do you one further. Using a similar analysis to that which I did above, you can get similar relations for $rho$ and $T$, which are plotted above. You can also pick a Mach number you'd like to use and use its relationship with velocity, namely $V = Ma$ to derive a relationship for a dimensionless momentum flux ($\rho u^2$ in a one-dimensional flow) as a function of the $M$ and $\gamma$.

So, the end result is that if you assume you have expanded your flow to be moving one dimensionally, then the plot of momentum flux versus Mach number looks like the following:

So, there actually is a peak in momentum flux that occurs as $M\approx 1.41$. Momentum flux is probably the closest related quantity to force on an object it eventually hits. If you imagine an object redirecting all of that momentum flux, then it requires a force in order to do that. However, this has nothing (directly) to do with pressure (or "effective pressure", whatever that may be), and you can never recover more pressure than exists in the form of total or stagnation pressure.

In case you were curious, I also included similar plots for mass flux, $\rho u$ (the maximum is unsurprisingly at $M=1$) and the kinetic energy flux, $\rho u^3$, which has a maximum at $M\approx 1.73$.

 

[pranj5]

The thing is that as the speed increases, the fluid decompresses (lower density) which releases the internal energy and the temperature drops. But when you slow down the fluid (hitting the blade), you have to compressed it again, so you cannot actually used that energy to push on the blade. Everything comes back to the initial state.

The fluid decompresses that means it's doing some work. That's why both internal energy and pressure are decreasing. Both of you are pointing to the fact that when the fluid will come out of the nozzle and then what happens while I am focusing on what is happening at the throat.

It's like if the moving fluid was a moving spring. When you accelerate one end you see the spring decompressing and you're thinking: "Look, I released some internal energy. Great! I'll use it to do some work!" But when the first end of the spring hits the blade, it slows down. Then, the other end that follows still arrive at a higher speed (It has accelerated too), thus the spring re-compresses itself and you loose the energy you gained earlier. The only difference with a moving fluid is that it happens in a continuous fashion. Maybe my comparison is questionable, but it might help further the discussion from more knowledgeable people than me

The analogy here is incorrect on two points. First, spring is solid while the medium here is gas. Their properties are different. I can accept your analogy if the blade is fixed, but in reality the blade is moving and how can the particles that are hitting the blade can slower the particles coming after?

Yes, internal energy can be used as an additional "pool of energy" to feed a flow's acceleration, but even that can be related back to pressure as I showed previously. You just factor in the ratio of specific heats. If you would like more, allow me to show you exactly how to arrive at the isentropic pressure relationship for any compressible flow

If by "isentropic", you want to mean that the entropy of the gas will be same as before and after, that's not the case here. As the gas will hit the blades of the turbine while at higher speed, the turbine will rotate and produce power by the generator attached to it. So, the entropy of the fluid can't be the same as before and after. That's what I want to mean repeatedly.

 

[boneh3ad]

The fluid decompresses that means it's doing some work. That's why both internal energy and pressure are decreasing. Both of you are pointing to the fact that when the fluid will come out of the nozzle and then what happens while I am focusing on what is happening at the throat.

...

If by "isentropic", you want to mean that the entropy of the gas will be same as before and after, that's not the case here. As the gas will hit the blades of the turbine while at higher speed, the turbine will rotate and produce power by the generator attached to it. So, the entropy of the fluid can't be the same as before and after. That's what I want to mean repeatedly.

You keep switching foci. You talk about flow at the throat or after a diverging section and then we discuss that. Then when we provide the answers there, you move the goal posts on us and so that no, you are talking about interacting with a turbine. At any rate, it is important to discuss the nature of the flow prior to its interaction with any sort of turbine because it illustrates that there is still a fundamental upper limit to the amount of work that can be done by the fluid in order to turn the turbine, and that fundamental limit is related to the upstream conditions. Specifically, it is related to the total enthalpy, which includes both pressure and internal energy in its definition, and for most gases, this can be expressed solely in terms of total (or stagnation) pressure. That's what I've illustrated above. You seem to have ignored that, however.

Addtionally, until the fluid interacts with the turbine, it is still isentropic unless there is heat being added to the system. The fact that it eventually hits a turbine doesn't change the fact that the expansion of flow through a nozzle (whether it is converging or converging-diverging) is very, very nearly isentropic. That is what we are discussing because you seem to to be having problem with that part of the problem well before you consider the turbine.

So which is it? Are you talking about what happens at the throat of the nozzle or are you talking about what happens after the nozzle when it interacts with the turbine? You can't keep changing on us. At either rate, the above statement still applies. The incoming flow is still important, and the flow at the throat is fundamentally related to the flow everywhere else.

 

[pranj5]

So which is it? Are you talking about what happens at the throat of the nozzle or are you talking about what happens after the nozzle when it interacts with the turbine? You can't keep changing on us. At either rate, the above statement still applies. The incoming flow is still important, and the flow at the throat is fundamentally related to the flow everywhere else.

What I want to mean is that if the nozzle is sufficiently large sized so that we put a turbine near the throat at the exit end. Let's forget about increasing pressure by using a nozzle, I accept that pressure of a fluid can't be increased by using a nozzle. But, if we put a turbine inside the divergent section of the nozzle so that the fluid can interact with it while its speed is still increasing, then what?

 

[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):
$$\begin{matrix} & \underline{1} & \underline{2i} & \underline{2c} \\ P\ (Pa) & 300\ 000 & 278\ 223 & 273\ 581 \\ T\ (K) & 300 & 300 & 292.2 \\ \rho\ (kg/m^3) & 3.484 & 3.484 & 3.26228 \\ v\ (m/s) & 100 & 150 & 160.2 \\ M & 0.288 & 0.432 & 0.467565 \\ m\ (kg) & 1 & 1 & 1 \\ V\ (m^3) & 0.287 & 0.287 & 0.306534 \\ A\ (m^2) & 2.87 & 1.913 & 1.913 \\ L\ (m) & 0.1 & 0.15 & 0.1602 \\ A*\ (m^2) & 1.36 & & 1.36 \\ \frac{A}{A*} & 2.111 & & 1.407248 \end{matrix}$$
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:
$$\begin{matrix} & \underline{equation} & \underline{2i} & \underline{2c} \\ work & P_2V_2 - P_1V_1 & -6250 & -2238.154 \\ internal\ energy & \frac{R}{\gamma - 1}\left(m_2 T_2 - m_1 T_1\right) & 0 & -5595.385 \\ kinetic\ energy & \frac{1}{2}\left(m_2 v_2^2 - m_1 v_1^2\right) & 6250 & 7833.540 \\ total & & 0 & 0 \end{matrix}$$
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.