Venturi Effect based thought experiment

[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.