Tube bundle head loss
Tube bundle head loss
(OP)
Hi,
It was told me in the Piping forum, to post my question here so that someone may see it and answer...
I'm calculating a superheater, and i was faced with the following problem.
when we calculate the external (gas) head loss, we have the following expression: dP=2*f*G^2*N/density
In this equation, we have the value G, that represents: "mass velocity at minimum flow area" (extracted from Holman). This "G" value is = mass flowrate/minimum area. My problem is related with this minimum area. Not many books refer what is this "minimum flow are", not eaven Steam (Babcock). A bigger problem, is when we find several books with different calculations for this area.
Because of this now i'm starting to wonder if my idea of this "minimum area" is correct.
can anyone help me on this?
Thanks!
It was told me in the Piping forum, to post my question here so that someone may see it and answer...
I'm calculating a superheater, and i was faced with the following problem.
when we calculate the external (gas) head loss, we have the following expression: dP=2*f*G^2*N/density
In this equation, we have the value G, that represents: "mass velocity at minimum flow area" (extracted from Holman). This "G" value is = mass flowrate/minimum area. My problem is related with this minimum area. Not many books refer what is this "minimum flow are", not eaven Steam (Babcock). A bigger problem, is when we find several books with different calculations for this area.
Because of this now i'm starting to wonder if my idea of this "minimum area" is correct.
can anyone help me on this?
Thanks!





RE: Tube bundle head loss
Yes, that's the way the gas-side mass velocity is calculated in exchangers. Basically it's calculated through the net flow area, which is the total area minus the obstruction (tubes, fins, supports, etc.)as taken in a cross-section.
Regards,
Speco (www.stoneprocess.com)
RE: Tube bundle head loss
So if i have for example a superheater with bare tubes, to be installed vertically, the ares of gas pass, is the conduit section area, minus the area of all tubes as if we look frontally to them, following the gas flow. For instance, to calculate this tubes area, the area to be calculated for rach one is: Diameter * Vertical tube length.
CORRECT??
thanks!
RE: Tube bundle head loss
I believe you are correct. It's the mass velocity through the minimum flow area in the direction of the gas flow. I presume that the tubes are transverse to the gas flow. And, yes, the obstruction per tube would be the diameter X length. Be sure that the diameter and length are converted to the same units. That is, you need to be consistent with feet and inches, or whatever units you are working with.
Regards,
Speco
RE: Tube bundle head loss