Venturi Effect based thought experiment

In summary: The mixture will be at equilibrium and the temperature and pressure will be the same as the surrounding atmosphere.
  • #1
T C
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TL;DR Summary
This is a thought experiment where pressure over water has been lowered by using venturi effect and I want to know whether the temperature will remain the same or differ from the initial state as the LHV of vapour is added to the airflow.

Though Experiment.jpg
The drawing above is regarding a thought experiment of mine. Air at ambient pressure is being blown through a convergent-divergent nozzle and in the throat section, a water filled enclosed container is attached through a tube. The water container has sufficient surface area for supplying necessary LHV for the water evaporated. It's very much clear that when the air starts to flow,, the pressure in the throat section will be lower than before and that means vaporisation from the water inside the container. Now, I want to know what will be the temperature of the air after exiting the divergent section as the LHV of the steam from the vaporised air will be added to the flow. Will it be lower or higher than the initial/entry temperature? Let's assume that that entry and exit area of the nozzle is almost same.
This thought came to my mind when I am studying cold steam generation and some video on youtube regarding rotating a Tesla turbine by using cold steam in combination with compressed air flow.​
 
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  • #3
T C said:
Summary:: This is a thought experiment where pressure over water has been lowered by using venturi effect and I want to know whether the temperature will remain the same or differ from the initial state as the LHV of vapour is added to the airflow.

Will it be lower or higher than the initial/entry temperature?
What will happen if no water is in the tank? Doesn't that answer your question?
 
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  • #4
hutchphd said:
What will happen if no water is in the tank? Doesn't that answer your question?
No. My point of concern is if there is water in the chamber and vaporisation occurs from it. The vapour will mix with the flow and in the divergent section, the velocity will decrease and the pressure will increase again.
 
  • #5
Lnewqban said:
Just as a reference to a practical problem:
https://en.wikipedia.org/wiki/Carburetor_icing
Not exactly. There is no doubt about fall in pressure and temperature due to venturi effect. I want to know what will happen when the pressure will rise again and this time with extra vapour mixed in it. Remember that the surface area of the water container is enough to supply necessary heat from surrounding atmosphere.
 
  • #6
When you pass the throat, you will have a gas (air + water vapor) with different characteristics. Use those new characteristics to determine the changes in pressure, temperature, density, and enthalpy along the path.

To find the composition of the gas - that might vary along the path - you will need to find out first how much water is pulled at the throat and then use partial pressures.
 
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  • #7
I don't get it; is the water tank vented to external pressure?
 
  • #8
hutchphd said:
is the water tank vented to external pressure?
No. Due to venturi effect, the pressure over the water is lowered and it starts to evaporate. This newly formed vapour will be mixed with flow and in the divergent section, where the velocity will be lowered and the temperature and pressure will rise again, will the temperature back to the initial/entry level or not, that's my question.
 
  • #9
And the tank water is maintained at inlet temperature?
 
  • #10
hutchphd said:
And the tank water is maintained at inlet temperature?
The inlet temperature is same as the surrounding temperature. The tank temperature may be a little lower (for the sake of heat transfer, I have already stated that the heat transfer area is sufficient to maintain the flow of LHV from outside to the water inside). But that temperature difference isn't that significant, it's negligible.
 
  • #11
Your scenario is not completely described. Please state all the relevant conditions, known and unknown, at the inlet, throat and outlet of the venturi: temperature, pressure and humidity at least.

[edit]
Some thoughts though:
  • If the entering air is at less than 100% RH you will get evaporation and a temperature drop without a Venturi. That's what a cooling tower/evaporative cooler does.
  • A strong Venturi will cause both a temperature and pressure drop at the throat, which can cause condensation/icing (as @Lnewqban pointed out). It's the opposite of what you are trying to do.
 
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  • #12
T C said:
No. My point of concern is if there is water in the chamber and vaporisation occurs from it. The vapour will mix with the flow and in the divergent section, the velocity will decrease and the pressure will increase again.
So what would happen with, rather than water vapour being injected into the venturi, it was some other gas in the tank such as N2, O2, CO2, Ar, ..., air. The air blown through the venturi and gas from the tank would mix adiabatically would they not? The air and water vapor are at the same temperature <-- we could make that assumption
hutchphd said:
What will happen if no water is in the tank? Doesn't that answer your question?

Your problem would be if the air/water vapour mixture becomes super saturated. Then there would be condensation in the divergent section. ( or in the throat, also due to the slight temperature drop of a gas passing through the throat )
russ_watters said:
Your scenario is not completely described. Please state all the relevant conditions, known and unknown, at the inlet, throat and outlet of the venturi: temperature, pressure and humidity at least.
 
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  • #13
256bits said:
Your problem would be if the air/water vapour mixture becomes super saturated. Then there would be condensation in the divergent section. ( or in the throat, also due to the slight temperature drop of a gas passing through the throat )
Condensation at the throat can be logical. But how can there be condensation at the divergent section? At the divergent section, there is increase in pressure and if it's adiabatic, that means increase in temperature too. How can condensation occur at that point. Though it may be possible that if the exhaust from the divergent section comes in contact with a cooler surface, condensation can occur there.
russ_watters said:
A strong Venturi will cause both a temperature and pressure drop at the throat, which can cause condensation/icing (as @Lnewqban pointed out). It's the opposite of what you are trying to do
It may cause condensation but at the same time pressure drop over the stored water inside the container. The container has sufficient surface area for supplying the necessary LHV. Why evaporation wouldn't occur?
 
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  • #14
T C said:
Condensation at the throat can be logical. But how can there be condensation at the divergent section? At the divergent section, there is increase in pressure and if it's adiabatic, that means increase in temperature too. How can condensation occur at that point.
If it so happens that the temperature of the mixture remains below dew point there should be condensation.
The exiting stream may be cloudy.
 
  • #15
256bits said:
If it so happens that the temperature of the mixture remains below dew point there should be condensation.
The exiting stream may be cloudy.
As I have already told that inlet and exit have almost the same size and that means the pressure at the inlet and exit would be almost the same. If the temperature at the entry is above dew points, how can the almost same temperature become below dew point at the exit?
 
  • #16
T C said:
As I have already told that inlet and exit have almost the same size and that means the pressure at the inlet and exit would be almost the same. If the temperature at the entry is above dew points, how can the almost same temperature become below dew point at the exit?
over 100% relative humidity
 
  • #17
256bits said:
over 100% relative humidity
Ok. But what will happen to the LHV of the condensed water?
 
  • #18
Good question.
Let me think about that for a while and get back if I can.
The LHV though does go into the air as the water condenses.
I am just wondering what would happen to the temp of the air.
Need a psychometric chart which I haven't used in while.
 
  • #19
That was my point from the very beginning. LHV means large amount of heat.
 
  • #20
Let's omit the water container and consider both input and the exhaust air to be totally dry i.e. devoid of any kind of humidity. And also let's consider the tube at the throat is made up of good conducting material and heat from outside can enter into the flow inside to raise its temperature. In such a scenario, the temperature at the exit will certainly rise as the divergent section will convert the velocity into pressure and if the pressure level will rise to the same as the inlet, the temperature will certainly rise due to the inflow of heat from outside to the flow.
The similarity between this situation and the initial thought experiment is the inflow of enthalpy from outside to the flow. In case of the thought experiment, it's in the form of LHV and while in this later scenario, as simple heat. Point is, can the outcome be the same?
 
  • #21
T C said:
It may cause condensation but at the same time pressure drop over the stored water inside the container. The container has sufficient surface area for supplying the necessary LHV. Why evaporation wouldn't occur?
Because the air is already saturated with water. It can't hold any more.
But what will happen to the LHV of the condensed water?
Condensation is exothermic. It heats the air (or rather, resists further cooling).
 
  • #22
T C said:
Let's omit the water container and consider both input and the exhaust air to be totally dry i.e. devoid of any kind of humidity. And also let's consider the tube at the throat is made up of good conducting material and heat from outside can enter into the flow inside to raise its temperature. In such a scenario, the temperature at the exit will certainly rise as the divergent section will convert the velocity into pressure and if the pressure level will rise to the same as the inlet, the temperature will certainly rise due to the inflow of heat from outside to the flow.
The similarity between this situation and the initial thought experiment is the inflow of enthalpy from outside to the flow. In case of the thought experiment, it's in the form of LHV and while in this later scenario, as simple heat. Point is, can the outcome be the same?
If you heat the air so that it doesn't cool below the dew point, you can get evaporation.
 
  • #23
In steady state, why will there be appreciable flow of water vapor? The pressure in water chamber will be equilibrate at the venturi pressure and then no more exchange. What am I missing? I suppose you could make ambient T just below 100C but then you have simply reinvented the heat pump as the water boils in the chamber and condenses outside. So is this just a very bad heat pump ?
 
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  • #24
hutchphd said:
In steady state, why will there be appreciable flow of water vapor? The pressure in water chamber will be equilibrate at the venturi pressure and then no more exchange. What am I missing?
If the air at the throat is <100% RH (saturated), there will be evaporation, just like with a puddle of water on the floor.
 
  • #25
The more I look at your drawing, the more I think that the water reservoir will not have the effect you expect on the relative humidity of the airflow.

If your drawing is proportionally correct (the tube connecting to the reservoir is as big as the throat diameter), then the pressure will not be as low as you might expect. The huge area change will be seen as a divergent section by the airflow and pressure will increase dramatically. The larger the tube, the more it will look like this (the airflow will probably be more turbulent over the reservoir):

swimming_pool_evaporation.png

If your reservoir tube is made smaller such that it doesn't disrupt the throat flow and does create a low pressure to create more water vapor, then the tube may be small enough that the water vapor flow will so small that it will not change the airflow humidity level enough to make a difference. It's basically the same effect as using a pitot tube like so:

?u=https%3A%2F%2Ftse3.mm.bing.net%2Fth%3Fid%3DOIP.jpg
 
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  • #26
hutchphd said:
In steady state, why will there be appreciable flow of water vapor? The pressure in water chamber will be equilibrate at the venturi pressure and then no more exchange. What am I missing? I suppose you could make ambient T just below 100C but then you have simply reinvented the heat pump as the water boils in the chamber and condenses outside. So is this just a very bad heat pump ?
I was going to mention something like that. The vapour pressure in the tank would have to be above that in the venturi to obtain some flow ( in regards to your question if the tank is vented to external pressure )
 
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  • #27
256bits said:
I was going to mention something like that. The vapour pressure in the tank would have to be above that in the venturi to obtain some flow ( in regards to your question if the tank is vented to external pressure )
Vapor pressure is related to temperature. If the pressure inside the container is the same as the venturi, that means the temperature inside is lower than the surrounding and in that case, heat from outside will start to flow in and that means increase in temperature and pressure of the container. I have already stated that the container has sufficient surface area for heat transfer. In short, the pressure inside the container would be higher than that of the throat but lower than that of the surrounding.
russ_watters said:
Because the air is already saturated with water. It can't hold any more.
But it's at a lower pressure than the pressure inside the container.
 
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  • #28
My conclusion now is like that. As the pressure at the throat is lower than that of the container, some vapor from container will enter into it. After entering there, some condensation will occur as the pressure and temperature there is lower. But, that means release of heat and that will increase the temperature and pressure at the throat a bit. And in the divergent section, where the pressure (and temperature) will rise again, it will be at a higher level than the inlet due to this released LHV.
 
  • #29
T C said:
After entering there, some condensation will occur as the pressure and temperature there is lower. But, that means release of heat and that will increase the temperature and pressure at the throat a bit.
You cannot say something like that. You are basically saying "Because the pressure & temperature will go down, the pressure & temperature will increase."

You have to take the problem as a whole by using a control volume. I would set it as the inside of the convergent-divergent nozzle, including the reservoir up to the water level. (But I would best trust @Chestermiller to select the better approach)

From there, you have two inlets (convergent part of the nozzle and the reservoir) & one outlet (divergent part of the nozzle). Whatever comes in must come out - momentum, energy & mass.

Energy-wise, there will necessarily be more out from the divergent nozzle compare to what is going in the convergent nozzle, because you say yourself energy is allowed to come in the reservoir and thus added to flow.

How that added energy will show up depends on the design conditions. It could show up as a temperature rise, a higher pressure, or a faster-moving fluid. I don't think you can get an obvious answer without solving the equations and maybe even having numbers put into those design conditions (dimensions & initial conditions). For example, will the flow reach supersonic speeds somewhere in the nozzle?
 
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  • #30
Basically everything you're saying now is wrong and contradicts what we've been telling you:
T C said:
As the pressure at the throat is lower than that of the container
The dry air pressure at the throat is equal to the dry air pressure in the container. The [water] vapor pressure may be higher, lower, or the same based on conditions you haven't specified.

T C said:
...some vapor from container will enter into it.
Maybe, depending on the conditions you haven't specified.
T C said:
After entering there, some condensation will occur as the pressure and temperature there is lower.
No, and no. The pressure drop happens before the container, so you get one or the other (evaporation or condensation), but not both.

To put a finer point on it, I see three scenarios, with the second being the boundary between the other two:

1. Low humidity entering air results in evaporation from the container.
2. Moderate humidity entering air results in saturated air in the Venturi and no evaporation from the container.
3. High humidity entering air results in condensation/icing into the container.

[Note: I have a vision of what the boundaries of the problem are, and if someone pictures something outside that there may be differences. For example, if we get into supersonic flow.]

You need to start trying to absorb/learn what we're trying to teach you and not just cling to your initial idea, accepting only validation in response.
 
  • #31
@jack action. i haven't said it. First the vapor from the container will enter throat. The pressure and temperature there is lower than that of the injected vapor. That's why the a section of the injected vapor will condense there and that means release of LHV there. This LHV will increase the pressure and temperature. The air at the venturi has lower pressure and temperature while the vapor coming from the container will have higher pressure and temperature. The pressure and temperature of the vapor will go down and the pressure and temperature of the air at the venturi will go up. That's simple and as far as I can understand doesn't violate any law of physics.
russ_watters said:
2. Moderate humidity entering air results in saturated air in the Venturi and no evaporation from the container.
3. High humidity entering air results in condensation/icing into the container.
In case of moderate and high humidity air, the pressure and temperature at the venturi will be lower than that of the container and that will be enough to suck vapor from the water inside the container. And after that, what will happen is described above.
Kindly think of this scenario. Two containers, one have saturated air at lower temperature and pressure while other have vapor at higher temperature and pressure, If the two are connected through a tube, will steam from the higher pressure and temperature will enter the saturated air container or not? That's a pretty simple question. And in case of carburetors, icing occurs around the throat part. And I have already told at the very beginning that the container has sufficient heat transfer area. Why icing will occur inside the container despite having sufficient heat transfer surface area?
 
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  • #32
hutchphd said:
In steady state, why will there be appreciable flow of water vapor? The pressure in water chamber will be equilibrate at the venturi pressure and then no more exchange. What am I missing? I suppose you could make ambient T just below 100C but then you have simply reinvented the heat pump as the water boils in the chamber and condenses outside. So is this just a very bad heat pump ?
This was my immediate reaction. At equilibrium, the volume of water will remain at a fixed level. To maintain a flow of water, you need to maintain a high. temperature in the container.
However, air is not an ideal gas, so that will involve energy loss in the Venturi process and that energy could maintain some evaporation of water in the reservoir.
 
  • #33
sophiecentaur said:
This was my immediate reaction. At equilibrium, the volume of water will remain at a fixed level. To maintain a flow of water, you need to maintain a high. temperature in the container.
Higher than that of venturi you mean. And, as this is a running process, at equilibrium the vapor from the chamber will be steady.
sophiecentaur said:
However, air is not an ideal gas, so that will involve energy loss in the Venturi process and that energy could maintain some evaporation of water in the reservoir
Energy loss means?
 
  • #34
T C said:
Energy loss means?
Well . . . . er. . . . . Something will get warm or change state. I didn't think this through very carefully. Isn't the problem assuming ideal conditions and gases?
 
  • #35
russ_watters said:
If the air at the throat is <100% RH (saturated), there will be evaporation, just like with a puddle of water on the floor.
T C said:
Higher than that of venturi you mean. And, as this is a running process, at equilibrium the vapor from the chamber will be steady.

But the rate will be diffusion limited. Just to beat the horse, as @jack action notes this will not create a venturi but rather a swampcooler. For me this ge-danken-experiment is ge-done.
 
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