Relation between Inlet Diameter and Cv for Check Valve

[rajcoep88]

Hello all,

This is my first post here. I do not know whether my question is repetition. But, I have searched all posts first and then I am posting a question, as I could not find satisfactory answer to my doubt.

I have to design pneumatic check valve. From what I read or researched, I could understand that one must calculate Cv and select valve from catalogue. But, the thing is, I have to design my own valve. I could not use already available valves in the market. Inlet condition of check valve for compressed air is 6 Nl/min @50hPa at room temperature. I can assume pressure drop. So, I am assuming 10hPa pressure drop. I know its very small value. But, the requirement is such that there should be minimal pressure drop. Using this much data, I can calculate Cv. But, how should I use this Cv to determine Inlet diameter of check valve or other dimesions for sizing? Because, I searched a lot regarding relation between Cv and diameter of check valve. But, after Cv calculation everyone goes for selection through Catalogues, which in my case not possible. For more understanding, I have attached PPT and drawn simple ball valve with spring. Lets also assume, that spring can crack open at 20hPa (I am not sure whether at such low pressure ball valve with spring can open or not, so making assumption). My main focus is relating Cv to calculate check valve dimensions.

Any help or input is welcome and Thanks in advance!

Best Regards,
Rajdeep Rajput

 

[rajcoep88]

Please guys reply if you know anything.

Regards,
Rajdeep Rajput

 

[gerhardl]

We are talking very small values here, and my experience are based on larger values.

Is there already given a pipeline diameter? How and where is the pressure controlled in the pipeline? What type of checkvalve are you constructing? (I could not open your attachment in ppt at my available PC).

Have you any way of making a prototype and experiment with one or several variations, er check the one constructed as your best guess?

One rule for all checkvalves is to construct or choose it so small in regard to the flow, that it does not gulp (let in a large amount of fluid and immediately closes because the pressure in front drops, and then repeats).

One rough estimation is to calculate roughly that the speed of flow through the orifice (smallest opening), has a reasonable value.

Check also commercial available similar data for flow compared to CV, or compared to actual opening area of the valve. Note: not connecting dimension but smallest orifice.

 

[rajcoep88]

Hi gerhardl,

First of all thanks for the reply. I will try to clarify your doubts as far as I can. Due to confidentiality, that I have signed, I could not give you perfect details. But, I can let you know the mechanism.

There is motor to provide compressed air and it is of very small size. I am sharing one link for more details.

http://www.smartproducts.com/docs/Series_Air_Specs.pdf

We are not using the motor shown in the link. But, it is of same size and nearly same specifications. Outlet inside diameter is 3mm for motor. Then this outlet is connected using plastic tube (inner dia 4mm x 1.5mm thickness) to valve housing, whose inlet has inside diameter of 3mm and outside 5mm. Valve allows flow through it in one direction and air is filled in air cushion or balloon like structure which has volume 500 ml. Existing valve has problems so I have to design new one with new design of my own. Like I can decide housing size, inlet and outlet dimensions, valve mechanism or techniques. But, remaining system parts like motor or balloon structure will remain same.

Now, have a look at this link:

http://www.alibaba.com/product-deta...69453540.html?spm=a2700.7724838.30.322.E9ROOp

The one I want to design is of such type. Now, you can understand there are valves for such low values of pressure. But, I cannot use these valves directly. I can use the same mechanism or rubber valves (see attached image) like shown on that website. Why? Because, the valve housing has to be small in dimension and rectangular (max L 25mm x W 15mm x h 10mm). It has to be fitted in another part and there I have maximum this much space. That is the reason, I cannot just do calculations and select from manufacturer’s catalogue.
Prototyping and Experimentation: I thought about it at the very first I started working on it. But, right now my department does not have budget. I am Master Student there. So, next year may be if renewed budget allows then we can go for it. But for now I have to do simulation. And obviously, in either cases practical experimentation or simulation I have to have some design ready. It means I should freeze the dimensions or valve sizing and valve mechanism. For that, I should have some reliable calculated data or reasons why I chose/decided few parameters likewise. So, for me I am stuck at the point how Cv can be used or related to find inlet diameter of valve and other dimensions. I hope, I explained enough points. If you have doubts you can ask me and I will try to clarify it.

The way you said in last point, that I should look commercial data. I did it. But, for such small values there is not enough data. And again more can be found in hydraulics and very less in pneumatics. Or maybe I am not looking at correct place. Also, in Alibaba site you can see, they have given cracking and back pressure but no one talks about Cv. I am uploading image of the PPT slide also.
Eagerly waiting for your reply.

Regards,
Rajdeep

 

[gerhardl]

Based on your illustrations I observe following:

a. You need to have as small cracking pressure as possible. This is a result of pressure times exposed area of the closing member of the valve. Eg as large area as possible to have smallest cracking pressure. (Increased area contributes by square factor, increased pressure by linear factor.)

b. In your second picture this is illustrated by the ball closing the whole pipeline area. But the picture shows also a larger tube (inner body of checkvalve). With the rubber part pictured the closing member can be moved somewhat into the larger tube, the troughlet area and can be enlarged, checkvalve closing member enlarged, and opening pressure will work on a larger area because of the construction of the closing member.

c. Immediately behind the closing member you have shown a disc with smaller hole as a throttling device. We can assume that the opening of the closing device is near or quite much larger than the pipeline. This is not important if the room after the valve is small, and a throttling device controls how much of the air is led further into the system. The CV you are seeking is then not the CV of the valve, but actually given by how small in area the holes need to be (all summarized together either with one or several holes). The area must be small enough so the amount of air in front of the valve (eg. air pressure) maintains a pressure large enough to keep the valve open. This area is not the full area of the front tube (inner body), but necessarily equal or probably somewhat less than the the tube area, and the throttling device need to be as close to the closing member sa possible.

d. Difficult to calculate, easy to experiment: If you mount the closing member in a correct sized tube and experiment either with different throttling discs, or a needle valve and measure the flow or opening or cv when flow and behavior is ok.

Good luck, soory I cannot help you further!

 

[rajcoep88]

Hi again and Thanks for the reply!

Please see the attached images (simple concept representation) first so that you can understand the terms I am going to use in explanation.

From what you are saying, I understand that inlet diameter (inlet pipe area before large rectangular chamber area) is not of much concern. I should take care about the opening area of orifice plate (which fits in chamber). The opening area in this case will be total sum of areas of all small hole openings. So, basically I should calculate area which should be enough to give me very low cracking pressure which should keep the valve open and as well as give the required minimal flow with the same cracking pressure. In this case, rubber valve circular thin wall should open up all orifices which it is covering. Then, the elasticity of rubber should be such that it should help it to come back as soon as flow is stopped and again back flow could help it. Correct me if I have understood you wrong.

So, it means I should not worry about Cv calculations. I should consider cracking pressure and flow to calculate required area. Then, I should find out how many orifices I should make and their areas so that it should be equal to total required area. Then, I can finalize orifice plate dimensions and chamber dimensions. Please comment on this and tell me if I am thinking in right direction or not.

Best Regards,
Rajdeep Rajput

 

[gerhardl]

Yes. To specify a bit further on details: The cracking pressure and maintained pressure in main flow direction must be kept as large as necessary to continuously keep the checkvalve open as long as you wish to supply to the system.

The valve will close when back pressure is larger than front pressure (when you disconnect or turn down the flow.

The rest as summarized by you.

In my mind I had a construction with a larger rubber valve in front with a large opening and an separate orifice plate immidiately downstream from the valve in the chamber. Your construction is simpler, but I am a bit unsure on how it would actually would behave in the process.

Good luck!

 

[rajcoep88]

Thanks for good luck as I need it.

Normally things should be like you mentioned. But, the way I have shown in the picture, inlet and outlet areas in existing system are smaller and in between there is rectangular chamber with more space. Due to required small size of valve or space restriction, I should design something compact.

Regards,
Rajdeep Rajput

 

[gerhardl]

I woould appreciate a short final report when you finally have tested the product!

Even more good luck!

 

[rajcoep88]

Once I am done with it I will surely post my approach. But, it will take few months. As, it is my thesis topic and I have to thoroughly go through it. I have to defend with actual facts and data.

 

[rajcoep88]

Hi again,

I have one more question. I did the calculations for C[sub]v[/sub]. Input conditions are Volumetric Flow Rate 6Nl/min at Gage Pressure 500hPa (I had wrong input pressure earlier) at room temperature(I am taking 20ºC). The permissible ΔP is between 50-100 hPa. C[sub]v[/sub](US) comes out to be 0.06 for ΔP=50 hPa and 0.04 for ΔP=100 hPa. Then, I converted it to equivalent valve area S[sub]e[/sub]. It comes out to be 1.08 mm[sup]2[/sup] and 0.72 mm[sup]2[/sup] respectively, when I multiply by factor 18. (Have a look on the page number 69-70 of PDF available on the following link to understand the calculations). So, if I am right till this point, does it mean, when I have to design check valve with only two conditions open and close, then in open condition the valve open area should be equal to calculated equivalent area? Obviously, I have to also consider other losses, back pressure, pressure required to open up valve. But, I want to make sure am I right till this point in line with the theoretical concepts?

Link for the one of the Pneumatic Handbook :

http://www.adi-sales.com/docs/engineeringtools/British-Pneumatic-Handbook.pdf

If I am right, then such small area would suffice in practice or there is any thumb rule to take certain minimal area?

Best Regards,
Rajdeep Rajput

 

[gerhardl]

Sorry, I have already commented as far as I am able to do, the rest is up to you!