Are Total and Static Temperatures Equivalent for Incompressible Fluids?

[defunc]

Am I correct if I say the total and static temperatures of an incompressible fluid are the same?

Thanks.

 

[minger]

Generally, the ratio between static and total temperature is:
<br /> \frac{T_0}{T} = 1 + \frac{\gamma -1}{2}M^2<br />
So if one is assuming low Mach incompressible flow then it's a reasonable assumption. It depends on the accuracy you're looking for though.

 

[defunc]

Yep. But from my understanding, for incompressible flow the total and static temperatures are exactly the same. Regardless the velocity. Just want to confirm this.

Regards.

 

[FredGarvin]

How can they be the same given what Minger pointed out to you? Imagine hydraulic fluid or water (both incompressible) at a high Ma. The definition of a total or stagnation property is in bringing a moving fluid to rest isentropically. Granted, most places you see total properties are in compressible flow areas, but there is nothing (at least that I can think of) that doesn't say you can't use them for incompressible.

What is it in your understanding makes them the same based on compressibility?

 

[defunc]

Mingers equations are based on perfect gas laws. For incompressible flow entropy is only a function of temperature. From there my conclusion...

 

[minger]

Incompressible is an assumption.

Everything is compressible. Water is compressible. It may be essentially incompressible but its not. That's what you're not fully understanding.

 

[defunc]

The equations you gave is also based on an assumption: perfect gas behaviour. So its not applicable to any fluids. So my question still remains, under the assumption of incompressible flow, what's the relation between total and static temperatures? Is it the same?

 

[minger]

Static and total temperature are no for equal at any condition than static and total pressure are. They are both simply functions of Mach. If you want a closed-form solution assuming a real gas, then I'm sorry, I don't think any exist. Run a CFD using real gas propreties...which you typically can't either, unless you hardwire it yourself.

Sure suuuuree, under your assumption that for incompressible flow, Mach = 0, they are the same. How same are they actually? I don't know, plug and chug. You'll have orders of magnitude less error assuming ideal gas than assuming imcompressible.