Venturi Effect based thought experiment

[T C]

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.

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.​

 

[Lnewqban]

Just as a reference to a practical problem:
https://en.wikipedia.org/wiki/Carburetor_icing

 

[hutchphd]

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?

 

[T C]

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.

 

[T C]

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.

 

[jack action]

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.

 

[hutchphd]

I don't get it; is the water tank vented to external pressure?

 

[T C]

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.

 

[hutchphd]

And the tank water is maintained at inlet temperature?

 

[T C]

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.

 

[russ_watters]

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.

 

[256bits]

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

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 )

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.

 

[T C]

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.

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?

 

[256bits]

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.

 

[T C]

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?

 

[256bits]

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

 

[T C]

over 100% relative humidity

Ok. But what will happen to the LHV of the condensed water?

 

[256bits]

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.

 

[T C]

That was my point from the very beginning. LHV means large amount of heat.

 

[T C]

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?

 

[russ_watters]

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).

 

[russ_watters]

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.

 

[hutchphd]

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 ?

 

[russ_watters]

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.

 

[jack action]

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):

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:

 

[256bits]

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 )

 

[T C]

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.

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.

 

[T C]

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.

 

[jack action]

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?

 

[russ_watters]

Basically everything you're saying now is wrong and contradicts what we've been telling you:

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.

...some vapor from container will enter into it.

Maybe, depending on the conditions you haven't specified.

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.

 

[T C]

@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.

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?

 

[sophiecentaur]

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.

 

[T C]

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.

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?

 

[sophiecentaur]

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?

 

[hutchphd]

If the air at the throat is <100% RH (saturated), there will be evaporation, just like with a puddle of water on the floor.

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.

 

[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):