Venturi Effect based thought experiment

[T C]

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?

Most probably you want to mean frictional and other parasitic losses. If the venturi is properly designed and made, then that would be negligible. Kindly search net and youtube with venturi effect and you can see tons of videos and materials. It's actually based on venturi effect and that is applicable for any kind of fluid; whether gaseous (compressible) or liquid (incompressible). It has nothing to do with ideal gas. Venturi effect can be observed in reality and that has been done by real gases, not ideal gases or something like that.

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

I have clearly mentioned at the very beginning that it's gedanken (not ge-danken). And swamp cooler means evaporation, how can the water inside the container will evaporate without having a lower pressure zone? And if you agree that it's something like a swamp cooler, that means some evaporation will be there. And only the venturi can create this low pressure zone necessary for evaporation. And, by the way, kindly look at this. And most probably you will get more by searching google with "venturi evaporator".

 

[jack action]

how can the water inside the container will evaporate without having a lower pressure zone? [...] And only the venturi can create this low pressure zone necessary for evaporation.

A stagnant bowl of water (actually any liquid) will evaporate no matter what. Lowering the air pressure on top of the bowl just means faster evaporation.

Leave the apparatus in the OP sitting there, with no forced air flow, no matter what is the atmospheric pressure, and the water will eventually evaporate.

Kindly search net and youtube with venturi effect and you can see tons of videos and materials.

Here's a youtube video with a simple experiment to show my point:

​

 

[russ_watters]

And, by the way, kindly look at this.

That's nothing like the system you are describing.

What you are saying is still wrong. It would help you understand why if you picked real conditions and calculated what happens.

I think the basic problem here is that you think the pool of liquid can evaporate due to its own temperature regardless of the temperature/humidity of the air it is evaporating into*. This is false. It is the air conditions that dictate whether you can have evaporation until boiling temp is reached.

*Unless due to mixing or other heat transfer you can get a small temperature increase in the air. But this will be very small/inefficient.

 

[russ_watters]

A stagnant bowl of water (actually any liquid) will evaporate no matter what. Lowering the air pressure on top of the bowl just means faster evaporation.

As long as the humidity is below 100%/saturation in both cases.

 

[T C]

I have searched google with "venturi evaporation" and found a few technology where venturi effect is used to lower pressure over liquid (maple syrup, water) to evaporate water out of it. That simply means venturi effect can be used (actually being used) for evaporation of water. At present, my only concern is what will happen at the divergent section.

Leave the apparatus in the OP sitting there, with no forced air flow, no matter what is the atmospheric pressure, and the water will eventually evaporate.

My point is using venturi effect for faster evaporation and what will happen when this mixed flow will pass through the divergent section.

 

[russ_watters]

I have searched google with "venturi evaporation" and found a few technology where venturi effect is used to lower pressure over liquid (maple syrup, water) to evaporate water out of it. That simply means venturi effect can be used (actually being used) for evaporation of water. At present, my only concern is what will happen at the divergent section.

You are mis-applying the example you provided and we cannot move on until you correctly analyze your scenario (please note: at this point, this approach is getting dangerously close to misinformation/crackpottery). Since you have not specified a scenario, I'm going to specify it for you:

• Entering Conditions: Room temperature and 50% RH
• Venturi Performance: 50% pressure drop
• Water temperature: Room temperature

Please answer yes, or specify your own conditions.

 

[jack action]

Since you have not specified a scenario, I'm going to specify it for you:

• Entering Conditions: Room temperature and 50% RH
• Venturi Performance: 50% pressure drop
• Water temperature: Room temperature

Please answer yes, or specify your own conditions.

And the ratio of the cross-sectional area of the pipe connected to the reservoir vs the one of the throat is also important.

 

[russ_watters]

And the ratio of the cross-sectional area of the pipe connected to the reservoir vs the one of the throat is also important.

I was going to make a simplifying assumption about that or use a different case...

 

[russ_watters]

I'm running short on time, so I'll just give an answer regarding my scenario, skipping the methodology:

If we have 21C air entering a Venturi at 50% RH and the pressure drops by half at the throat of the Venturi, the temperature drops to -126C. At -126C roughly 99.8% of the water will condense (frost) out of the air. There can be no evaporation from the water container, only condensation into it, given these conditions. The following should also be obvious:

1. Evaporation from the tank is only possible if you have exceptionally dry entering air.
2. Because the moisture carrying capacity of air decreases with a decrease in temperature, evaporation works better at higher air temperatures than lower air temperatures.

Note: @jack action I'm assuming little to no mixing/turbulence of air at the intersection of the Venturi and pipe/container, and a small pipe. This makes it a strictly thermodynamic/diffusion process with little or no heat transfer. I'm more concerned with the "if" than the rate of heat transfer/evaporation. And I'm trying to emphasize that the conditions of the air matter by far the most. If you went entirely in the other direction you could spray water across the throat of the Venturi and get the air back up to room temperature. But even still, the drop in pressure causes the air to lose moisture not gain it, even as it goes from 50% to 100% RH.

Note also: the pressure drop I picked is pretty large. I was going to make it less severe, but didn't get a chance before it was responded to. Achieving this pressure drop isn't easy and the system driving the Venturi would need to be designed. This isn't just a table fan blowing air into a Venturi. I think I may have also gotten into a choked flow situation with that large a drop.

 

[T C]

• Entering Conditions: Room temperature and 50% RH
• Venturi Performance: 50% pressure drop
• Water temperature: Room temperature

Condition 1 ok. And for the venturi, the velocity at the throat is Mach 1 or close to it. Choose the inlet as per your choice to fulfil just that condition. Main point is the temperature at the venturi section should be lower than that of the saturation temperature of the water inside. That is certainly possible. In case carburettor icing, the temperature at the throat can go below zero while the atmospheric temperature can be 100 F. Water temperature would be less than room temperature and choose both as per your will. I suggest 30 C and 25 C respectively.

 

[russ_watters]

Condition 1 ok. And for the venturi, the velocity at the throat is Mach 1 or close to it.

I believe a pressure ratio of 50% gets us there.

Main point is the temperature at the venturi section should be lower than that of the saturation temperature of the water inside. That is certainly possible.

Absolutely. And if the temperature inside the venturi is lower than the saturation temperature of the water, then clearly you can only have condensation/frost, not evaporation. That's my entire point here.

Water temperature would be less than room temperature and choose both as per your will. I suggest 30 C and 25 C respectively.

I'm not going to re-calculate, but that just makes things a little worse.

 

[T C]

Absolutely. And if the temperature inside the venturi is lower than the saturation temperature, then clearly you can only have condensation/frost, not evaporation. That's my entire point here.

Condensation will occur inside the venturi as the evaporated water from the tank will enter into it and become cold and condensed. You are too much focused on just RH and ignored the pressure factor. If the pressure inside the venturi will be less than that of the saturation pressure at that specific water temperature, evaporation from water will certainly occur even in case of 100% RH inside the venturi. Just like, as I stated before, steam power plants work perfectly in a rainy day.
Just google with "venturi evaporator" and see how they works.

 

[russ_watters]

Condensation will occur inside the venturi as the evaporated water from the tank will enter into it and become cold and condensed.

No. That's not how it works. The air can only hold as much water as it can hold. You can't simultaneously be evaporating water from the tank into the air and condensing it out of the air. Those are opposing forces.

You are too much focused on just RH and ignored the pressure factor.

You are simply guessing here and not actually paying attention to any of the thermodynamics. Or worse, just rejecting and making it up when it doesn't say what you want. There is no thermodynamic process for what you describe.

If the pressure inside the venturi will be less than that of the saturation pressure at that specific water temperature, evaporation from water will certainly occur even in case of 100% RH inside the venturi.

That's the opposite of how it works.

Just like, as I stated before, steam power plants work perfectly in a rainy day.

That's not how they work.

 

[T C]

No. That's not how it works. The air can only hold as much water as it can hold. You can't simultaneously be evaporating water from the tank into the air and condensing it out of the air. Those are opposing forces.

When the evaporated vapor will come in contact with the saturated air inside the venturi, it just simply don't mix with ad but rather add some heat/enthalpy to it as it's a higher pressure and temperature than the saturated air inside. It condenses and lowers its temperature by this process and the temperature, pressure of the saturated air inside will increase. That's pretty simple.

That's the opposite of how it works

You want to mean no evaporation will occur even in case the pressure over the water is lower than that of the saturation pressure of steam at this specific water temperature? That's direct violation of physics and Moliere's diagram.

 

[jack action]

I will attempt my solution to the problem. As I mentioned previously in post #29, I would consider this as a fluid dynamic problem for compressible flow and use conservation of mass, energy, and momentum. I would consider a three-pipe branch pipe junction, consisting of the venturi throat inlet and outlet, and the pipe coming from the reservoir. For the conservation of momentum, the angle between the pipes should matter, but I'll keep it simple. I will also not consider condensation within the venturi nozzle (RH always below 100%).

For the following, subscripts represent the conditions at the following locations:
$in$ : throat inlet
$out$ : throat outlet
$wv$ : water vapor pipe

Conservation of mass

$$\dot{m}_{in}+\dot{m}_{wv}=\dot{m}_{out}$$
$$\rho_{in}v_{in}A_{in}+\rho_{wv}\color{red}{v_{wv}}A_{wv}=\rho_{out}\color{red}{v_{out}}A_{out}$$
Where:
$\dot{m}$: mass flow rate
$\rho$: fluid density
$v$: fluid velocity
$A$: cross-sectional area of pipe ($A_{in}=A_{out}$)

Conservation of energy
$$\dot{m}_{in}h_{in}+\dot{m}_{wv}\Delta H_{vap}=\dot{m}_{out}h_{out}$$
$$\rho_{in}v_{in}A_{in}\left(T_{in} + \frac{v^2_{in}}{2C_p}\right)+\rho_{wv}\color{red}{v_{wv}}A_{wv}\Delta H_{vap}=\rho_{out}\color{red}{v_{out}}A_{out}\left(\color{red}{T_{out}} + \frac{\color{red}{v^2_{out}}}{2C_p}\right)$$
Where:
$T$: temperature
$C_p$: specific heat capacity
$\Delta H_{vap}$: enthalpy of vaporization (value for given water temperature)

Conservation of momentum
$$p_{in}A_{in} + \dot{m}_{in}v_{in} + p_{wv}A_{wv} + \dot{m}_{wv}v_{wv}=p_{out}A_{out} + \dot{m}_{out}v_{out}$$
$$p_{in}A_{in} + \rho_{in}v^2_{in}A_{in} + \color{red}{p_{wv}}A_{wv} + \rho_{wv}\color{red}{v^2_{wv}}A_{wv} = \color{red}{p_{out}}A_{out} + \rho_{out}\color{red}{v^2_{out}}A_{out}$$

Where:
$p$: pressure

Stagnation conditions
$$\color{red}{T_{0\ out}} = \color{red}{T_{out}} + \frac{\color{red}{v^2_{out}}}{2C_p}$$
$$T_{0\ wv} = \color{red}{T_{wv}} + \frac{\color{red}{v^2_{wv}}}{2C_p}$$
$$\frac{p_{0\ out}}{\color{red}{p_{out}}} = \left(\frac{\color{red}{T_{0\ out}}}{\color{red}{T_{out}}}\right)^{\frac{k}{k-1}}$$
$$\frac{p_{0\ wv}}{\color{red}{p_{wv}}} = \left(\frac{T_{0\ wv}}{\color{red}{T_{wv}}}\right)^{\frac{k}{k-1}}$$
Where:
$T_0$: stagnation temperature
$p_0$: stagnation pressure
$k$: ratio of specific heats

We get seven unknowns (in red) with seven equations to solve (densities can be found with pressures and temperatures). And then the flow out of the throat get into the divergent section where we can get the final values to determine what temperature we get. Again, there would be the Mach numbers to watch out for as well.

I really don't know how to give a simple answer without solving the system of equations with actual numbers given. Throwing the possibility of condensation in the mix complicates the system even further.

Edit (In response to next @hutchphd 's post):

 

[hutchphd]

We get seven unknowns (in red)

The LaTeX for the color rendering didn't work for me the \colorred command appears inline. Very hard to read.

 

[hutchphd]

The LaTeX for the color rendering didn't work for me the \colorred command appears inline. Very hard to read.

Now its better! Magic

 

[T C]

Actually my question wasn't that complicated. I just want to know whether the temperature of the mixture will be at higher temperature or not. The pressure at the venturi will be lower than that of the saturated vapour pressure of water at that temperature inside the container. That simply means water will evaporate and will enter the venturi. Inside the venturi, it will come in contact with the lower pressure saturated air inside. It will release a part of its enthalpy to the saturated air and after some time, both will be at the same temperature. The basic principle behind venturi evaporators is this. The only point to be discussed now is whether the mixture coming out of the divergent section will be at higher temperature than the inlet or not.

 

[jack action]

I just want to know whether the temperature of the mixture will be at higher temperature or not.

Considering the dry air case, the only thing for sure is that the total temperature $T_{0\ out}$ will increase. Because the energy coming out $C_pT_{0\ out}$ will be equal to sum of the energy coming in $C_pT_{0\ in}$ (where $T_{0\ in} = T_{atm}$) and whatever you put into maintain the water temperature ($\dot{m}_{wv}\Delta H_{evap}$).

Your problem now is that the total temperature can be split into 2 components:
$$T_{0\ out} = T_{out} + \frac{v^2_{out}}{2C_p}$$
You say "I'm sure $T_{out}$ will increase". Well, maybe not. The fluid might come out at a faster speed instead. Or it could be a mix of both.

What will it depend on? Basically, the cross-sectional area ratio of your pipes. That is what will determine the ratio of the flow rates and forces they will create internally. If the flow becomes hypersonic, more fun is expected.

If the air is not dry and condensation appears, it will remove energy from the flow. Let's consider only liquid condensation (no icing). A few scenarios come to mind:

• The water vaporizes back in the flow in the divergent section. In this case, the energy is back in the system, so the scenario is similar to the one with dry air.
• The water accumulates and drips out of the apparatus. In this case, the energy is removed from the system. If it is equivalent to what you added to heat your water, then $T_{0\ out} = T_{0\ in}$ and $v_{out}$ is still not necessarily zero. But if it is more, the total temperature will drop, and this guarantees a drop in temperature.
• The [cold] water drips into the water reservoir. In this case, the water cools down the water in the reservoir, which must be heated up to maintain the original water temperature. It will take extra energy, but this energy will not go into the airflow. It is thus the same scenario as if the water dripped out of the apparatus.

 

[T C]

@jack action And about the scenario, where the input air is highly humid. As the air will enter venturi, its velocity will increase at the expense of its internal enthalpy i.e. it will become 100% humid as the temperature will decrease and part of its vapour will condense. But evaporation will occur as the pressure inside the venturi is lower than that of the water inside the container. That vapour will enter the venturi and will act like a compressor to the saturated air inside the venturi. Previously at the entry of the venturi, the enthalpy of the air entering the venturi was converted into velocity and the vapour coming from the container will supply some of its enthalpy to the air inside and its temperature will rise without decreasing the velocity. And when this flow will enter the divergent section, its temperature and pressure will start to rise again. But, in this case, the temperature and pressure will be higher than that is at the entry of the venturi because a part of the enthalpy of the input steam is added to the enthalpy.
And, I want to be very very clear about one point. Vapour will certainly enter the venturi even in case of airflow to be totally saturated inside the venturi because the pressure inside the venturi is lower than that of the saturated vapour pressure of water at that specific temperature inside the container. No evaporation will occur when the pressure and temperature inside the venturi will be the same as the container and the air inside venturi will be 100% saturated. Vaporisation will certainly occur when the pressure and temperature inside the venturi will lower than that of the saturated vapour pressure inside the container. Too much emphasis has been given to RH while no attention has been given to the pressure factor. A very easy way to check what I am saying is doing this experiment in a rainy day and by replacing the heir dryer with a blower. It would be better if the blower is fitted with a convergent nozzle.

 

[jack action]

That vapour will enter the venturi and will act like a compressor to the saturated air inside the venturi.

I agree. But that doesn't mean the newly added water will not begin to condensate as soon it enters the main airflow stream
.

and the vapour coming from the container will supply some of its enthalpy to the air inside and its temperature will rise without decreasing the velocity.

That is a statement that is not exactly true. What you can only state as true is "its total temperature will rise". You cannot state that the velocity will not decrease without specifying the design conditions.

Just imagine that the pipe from the water reservoir is placed such that it opposes the airflow in the throat of the venturi. Don't you think the dynamic force of the incoming flow has no effect on pushing against the water vapor, even if the throat pressure is lower? Don't you think this can affect the mass flow rate of how much water vapor gets out of the reservoir?

And when this flow will enter the divergent section, its temperature and pressure will start to rise again. But, in this case, the temperature and pressure will be higher than that is at the entry of the venturi because a part of the enthalpy of the input steam is added to the enthalpy.

The pressure at the outlet of the divergent section will be the same - by design - as the one from the inlet of the convergent section, i.e. atmospheric pressure.

Again, the total temperature will increase, which is a combination of the internal energy (temperature) and kinetic energy (velocity) of the airflow. How the total energy will be split between those two will depend on design conditions.

 

[T C]

I agree. But that doesn't mean the newly added water will not begin to condensate as soon it enters the main airflow stream.

It will and I have clearly clarified that. Enthalpy from this vapor will enter the air inside the venturi. So the temperature and pressure of the air inside venturi will increase and the temperature and pressure of the vapour will decrease and enthalpy transfer will stop when both are at the same temperature and pressure. That's pretty simple. They will mix and such transfer of enthalpy can happen quickly.

That is a statement that is not exactly true. What you can only state as true is "its total temperature will rise". You cannot state that the velocity will not decrease without specifying the design conditions.

Just imagine that the pipe from the water reservoir is placed such that it opposes the airflow in the throat of the venturi. Don't you think the dynamic force of the incoming flow has no effect on pushing against the water vapor, even if the throat pressure is lower? Don't you think this can affect the mass flow rate of how much water vapor gets out of the reservoir?

By "the velocity will not decrease", I want to mean that the velocity decrease wouldn't occur at the expense of enthalpy of the air that it container before entering the system. I hope I have clarified my point. And I wouldn't discuss the 2nd scenario you have said, because that will be matter of totally different thread/scenario.

The pressure at the outlet of the divergent section will be the same - by design - as the one from the inlet of the convergent section, i.e. atmospheric pressure.

Again, the total temperature will increase, which is a combination of the internal energy (temperature) and kinetic energy (velocity) of the airflow. How the total energy will be split between those two will depend on design conditions.

At the divergent section, the temperature and pressure will be higher because the temperature and pressure of the flow will be higher than that of the temperature and pressure of the flow when it entered venturi and that's due to the addition of enthalpy from the vapour coming from the container.

 

[jack action]

and I have clearly clarified that.

No, you haven't. you are just repeating a scenario you imagined in your head without showing how this will happen, without doing the math.

Enthalpy from this vapor will enter the air inside the venturi. So the temperature and pressure of the air inside venturi will increase and the temperature and pressure of the vapour will decrease and enthalpy transfer will stop when both are at the same temperature and pressure.

While all this can be true, it doesn't mean the newly added water vapor will not condensate at these new temperature & pressure and just come out of the airflow, resting on the venturi wall, together with this added enthalpy.

By "the velocity will not decrease", I want to mean that the velocity decrease wouldn't occur at the expense of enthalpy of the air that it container before entering the system. I hope I have clarified my point.

This is absolutely not clear to me. Are you talking about the velocity of the air stream or the vapor stream?

And I wouldn't discuss the 2nd scenario you have said, because that will be matter of totally different thread/scenario.

So the velocity at which the vapor gets into the airflow is irrelevant? So is the airflow velocity? Given the same enthalpy, big-hole/slow-velocity or small-hole/large-velocity is all the same to you?

At the divergent section, the temperature and pressure will be higher because the temperature and pressure of the flow will be higher than that of the temperature and pressure of the flow when it entered venturi and that's due to the addition of enthalpy from the vapour coming from the container.

Show me that mathematically while respecting the conservation of mass, energy, and momentum.

That's pretty simple.

Do the math and you will see that it is not.

 

[T C]

No, you haven't. you are just repeating a scenario you imagined in your head without showing how this will happen, without doing the math.

It's a complex scenario and doing the math will be extremely tough. And this is the basic principle behind venturi evaporators, not just my own imagination.

While all this can be true, it doesn't mean the newly added water vapor will not condensate at these new temperature & pressure and just come out of the airflow, resting on the venturi wall, together with this added enthalpy

It wouldn't and I have never said so. A part of the vapor will condense and enthalpy can be transferred from the vapour to the air only by this process.

This is absolutely not clear to me. Are you talking about the velocity of the air stream or the vapor stream?

Velocity of the flow of course because that velocity is the result of conversion of enthalpy of the flow.

So the velocity at which the vapor gets into the airflow is irrelevant? So is the airflow velocity? Given the same enthalpy, big-hole/slow-velocity or small-hole/large-velocity is all the same to you?

The amount of vapour is minuscule in comparison to the amount of air. Addition of this small mass of vapour wouldn't alter the flow velocity much. And, BTW, the amount of vapour may be small but the enthalpy is sufficient in comparison to the enthalpy of the air inside the venturi.