By "perfectly designed nozzle", I want to mean that the static pressure of the flow at the exit would be the same as the backpressure. In case of overexpanded nozzles, the static pressure of the exit flow can be much lower than the backpressure. And, before posting this thread, I have searched...
Now, I want to simplify the matter more. As I have already mentioned, that the pressure of the air/gaseous fluid is 5 barA. Now, let's consider that the flow rate 3 cubic meter per minute. Kindly tell the just the radius at the inlet, at the throat and also at the exit of a perfectly designed De...
It's the release pressure, but the pressure of the flow coming out of the nozzle can be less than that. Kindly go through this thread.
Supersonic velocity of the fluid coming out of the nozzle.
So far, what I have found is convergent-divergent nozzles can generate supersonic velocity. But, no idea about design parameters of an overexpanded nozzle.
Suppose there is a steady source of compressed air/gas at 5 barA pressure, and that will be released through an overexpanded convergent-divergent nozzle. I want to know how to design such a nozzle and what would be the parameters for that. And also what would be the minimum pressure at the exit?
I want to start with two scenarios and both contains a blower and a tunnel fitting for the blower and both have almost the same diameter. The tunnel has s little higher so that the blower can be fitted inside. In the first scenario, the blower is fitted at the end of the tunnel i.e. the flow...
This video is just to show that how venturies can drastically increase output of a wind turbine. What you are telling is the next step, that's another matter.
I am not interested about what is the wind turbine capital of US. In the video, you can see two same wind turbines rotating side by...
@ cjl, kindly look at this video and see how the RPM dramatically increased after introduction of a venturi. And here is one more example and see by yourself the difference between having a venturi and not having it. Just search youtube with "venturi turbine" and you can see tons of examples.
I have recently find this video and just curious about the working principle behind this concentrated windmill. Apparently it seems that the basic principle is Venturi effect but the shape of the entry point is more suitable for Coanda effect with its curved surface at the entry. And it's also...
All helices have their axis aligned with the magnetic field. What will differ is the radius of rotation and linear velocity as different atoms have different level of energies.
Not "Various" helical path. All helix have the axis pointed towards the same direction. If the radius of curvature is less than that of the mean free path, there would be little chance of any collision. Collision will occur as the radius of curvature is much much greater than the mean free path...
I have just done some calculations and found that if the Magnetic field is 3 Tesla, then the radius of curvature would be far greater than the mean free path. So there will be collisions. In a little bit different way, but the ultimate goal would be the same as neutral atoms.
But, such collisions are only possible when the diameter of the rotation would be more than the mean free path, right? And at least it's clear now that the randomness of the motion is less in comparison to the scenario without the magnetic field.
Yes, that can be understood. But all axis are...
Ok. Now, I understand. But, the actual movement is along the axis of the helix. So, ultimately it will look like a number of helical motions having same axis buy varying radius, right?
It has the magnetic field. Does that can change its direction of motion? Suppose the the direction of motion isn't perpendicular to the magnetic field but rather at an angle to it.
What will happen if neither of them is zero?
How the motion will look like then?
The helical movement will only occur when the direction of motion is perpendicular to the magnetic field. Will there be any helical motion where the direction of motion isn't perpendicular to the magnetic field?
Magnitude can't be changed because that means because energy can't destroyed. I want to know whether the ionized molecules will still contain the random movements or they will become streamlined perpendicular to the magnetic field?
@Baluncore, I know about MHD Generators. I just want to know whether they, especially the magnetic part is capable of converting the internal energy of the flew gas too or it just converts the velocity of the flow into useful power. To be more precise, whether the magnetic field will force the...
Suppose a molecule from our surrounding air (at ambient temperature) is being selected and is ionized. By some mechanical means, some velocity (say 100 m/s) is added to it and it has been put into a magnetic field perpendicular to its direction of motion. We all know how the molecule will behave...
In case of a flow coming out of a De Laval nozzle, it certainly opposes the backpressure. "The pressure measured in the direction of the motion is called the total pressure and is equal to the sum of the static and dynamic pressureas described by Bernoulli's equation."
From this source. It may...
Reduced effectiveness means it opposes the pressure and that's why the pressure was unable to compress the flow in the direction of flow. Not very complicated.
Ok. No debate now on this point. The pressure is present everywhere. But in the direction of the flow, the dynamic pressure has reduced its effectiveness while in the plane perpendicular to the direction of motion, dynamic pressure has no effect as it's vector (associated with momentum).
Dynamic pressure is associated with momentum. Pressure may be a scalar quantity, but not dynamic pressure because that's associated with momentum. In case of overexpanded nozzles, the force towards the direction of the flow is higher but lower in the plane perpendicular to the direction of motion..
I mean dynamic pressure here. It arises due to velocity that's why should have a direction.
Ok. My mistake. In future, I will mention this as total pressure.
I am talking about the part where the flow encounters a boundary layer i.e. where it started to be compressed instead of being expanded.
From yesterday onwards, I have started to study overexpanded nozzles and I found a curious point. In case of overexpanded nozzles, the flow coming out of the nozzle is compressed by the surrounding atmosphere (or the release fluid) in the plane perpendicular to the direction of motion, not...
Sufficiently means less than half of the release/ambient pressure.
Can you give me a source/reference?
To be precise, I am looking for an example simple compressed air/gas being released through an overexpanded nozzle.
In case of De Laval nozzle, the static pressure at the exit can be as low as 1/3rd of the ambient/release pressure. Kindly look at the conditions of operation part of the page. At present, I want to know any real life example (in as much detail as possible) of a De Laval nozzle where the exit...
In case of De Laval nozzle, the release pressure (mentioned as ambient pressure in the page) can be 2-3 times higher than the pressure at the exit of the nozzle. If that's true, that simply means a high velocity flow of lower pressure can easily enter at higher pressure if the gross i.e. total...
Static pressures are chosen and dynamic velocity calculators available on net. Let's consider the both the fluids at standard temperature. The density can be calculated easily and just put those values into the calculator and you can get the results.
Can't understand what you want to mean.
Why...
In this scenario, we have a compressible flow at 0.99 barA pressure and flowing at 300 m/s velocity and is released into 1 barA pressure. Point is, whether the flow can be released at 1 barA or not at the release pressure is higher than the pressure of the compressible fluid inside the tube. And...
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.
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...
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...
@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...
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...
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...
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...
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...
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...
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...
@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...
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...
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...