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LabView inline temperature integration

Hello, 

I would like to know if it is possible to have in-line temperature integration while measuring data. Right now we are using LabView to measure the horse power, RPM, integrated power, and temperature on a lab scale Banbury rubber mixer. The application file is called bodytemp. The power is integrated in-line in units of kilowatt hours so I was hoping integrating temperature the same way wouldn't be too difficult or strange. For certain products that we use it is preferable to use temperature integration rather than power integration. If this topic is discussed already, let me know where since it is difficult to locate this topic when there are a couple meanings for the word integration and the meaning I am using is mathematical (area under curve), not programming related.

 

Thanks.

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Message 1 of 6
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Sure you can.  The same way you do any other integration.  The math is the same, you are just using values that have different meanings in term of engineering units.

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So is it possible to have this reading while it’s running? I don’t think I can rewrite the code myself. 

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@Maiq wrote:

So is it possible to have this reading while it’s running? I don’t think I can rewrite the code myself. 


I would recommend looking at the online LabVIEW tutorials
LabVIEW Introduction Course - Three Hours
Learn LabVIEW

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Okay, this is helpful but how do open the application that will allow you to program the VI? I don't know what it would be called.

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OK, I'm not sure that I'm following you so let me ask a few questions.

 

1. Are you running from LabVIEW or an executable created in LabVIEW?

2. If an executable created in LabVIEW do you have the source code?

3. If you have the source code is there a file with the extension .lvproj?

4. Do you have a copy of LabVIEW in which to open the source code?

 

These are the first steps to determining how to proceed. Do you know who originally wrote the code? Contracting with them to add this feature might be your best option as this may be more than you want to do on your own. 

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