Can kg/m2h Be Converted to m/s for Process Calculations?

  • Thread starter Thread starter chispofish
  • Start date Start date
  • Tags Tags
    Convert
AI Thread Summary
kg/m²h cannot be directly converted to m/s due to their different physical meanings; kg/m²h represents mass flux while m/s indicates linear velocity. To convert kg/m²h to m/s, one must first convert mass flow to volumetric flow by dividing by density. Then, the volumetric flow rate in m³/m²h can be converted to linear velocity in m/h, which can be further adjusted to m/s by dividing by 3600. This process involves calculations rather than a straightforward conversion. Understanding these distinctions is crucial for accurate process calculations.
chispofish
Messages
1
Reaction score
0
Is there a wat to transform kg/m2h to m/s

I'm not sure if kg/m2h is for Evaporation coefficient, Water flux,Drying rate or Mass transfer coefficient. However one of the can be transform to m/s (maybe Impingement velocity)
 
Engineering news on Phys.org
I don't see how you would do that. You are talking about base units in the SI system (except for hours which is easily replaced by seconds). There's no direct conversion.
 
kg/m2h is a mass flux of a substance through an area
m/s is the linear velocity of that substance

Therefore, step 1 is to convert the mass flow to volumetric flow (divide by density).
then you'll need to realize that:
m3/m2h = 1 m/h
so, the volume passing through an area per time equals linear velocity.

Then divide by 3600 to get the /h in stead of /s. (3600 seconds in 1 hour).
 
That's a calculation, not a conversion/transformation. I guess I shouldn't take the posts so literally.
 
FredGarvin said:
That's a calculation, not a conversion/transformation. I guess I shouldn't take the posts so literally.

Hehe... you probably shouldn't. On internet you can never be sure you're talking to a native English speaker (who can be expected to choose exactly the right words) or someone (like me) who is merely likely to choose the right words :D
 
Hi all, I have a question. So from the derivation of the Isentropic process relationship PV^gamma = constant, there is a step dW = PdV, which can only be said for quasi-equilibrium (or reversible) processes. As such I believe PV^gamma = constant (and the family of equations) should not be applicable to just adiabatic processes? Ie, it should be applicable only for adiabatic + reversible = isentropic processes? However, I've seen couple of online notes/books, and...
I have an engine that uses a dry sump oiling system. The oil collection pan has three AN fittings to use for scavenging. Two of the fittings are approximately on the same level, the third is about 1/2 to 3/4 inch higher than the other two. The system ran for years with no problem using a three stage pump (one pressure and two scavenge stages). The two scavenge stages were connected at times to any two of the three AN fittings on the tank. Recently I tried an upgrade to a four stage pump...
Back
Top