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July 16th, 2018, 06:48 AM   #1
Joined: Jul 2018
From: UK

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HVAC finding the value of BETA help

Good afternoon from a very sunny UK,

I’m not particularly strong at maths but I’m hoping that there is a mathtastic wizz who can spare a minute and help me out, I would be eternally grateful.

What I am doing is HVAC related. Presently there are a number of calculations where you can calculate a capture distance for an extraction system welding hood based upon quantitative measurements taken on site:
(Equation 1) ɑ=(x/√A)*(W/L)^-β
(Equation 2) β=0.2*(x/√A)^-1
(Equation 3) Vh=Vc(0.93+8.58a²)

I can do Equations 1-3 fine but what I’m trying to find out what the capture distance would be based upon limited information that I would have at the design stage. I’ve got as far as rearranging the equations:

(Equation 1a) x=a*√A*(W/L)^β
(Equation 2a) β=0.2*(x/√A)^-1
(Equation 3a) a = (√100Vh-93Vc)/( √858*√Vc)

x is the distance the hood is positioned from the process (m)
W Width of the hood face (m)
L Length of hood face (m)
A Area of the hood face (m2)
α Distance from the hood face for satisfactory capture (m)
β Beta
Vh Hood entry velocity (m/s)
Vc Hood capture velocity (m/s)

I’ve been doing the maths both ways to check that the answers tally and they do not. It maybe because it’s stressing me out but I just can’t see how to move forward.

For small circular hoods the answers are near enough. However, for large rectangles/slots say 300x1000mm and above, the maths is miles out. It is equations 1a & 2a I’m particularly stuck with as I don’t know how to calculate the value of β Beta?

Appreciate any help.xxxx
lhdwce2018 is offline  
July 16th, 2018, 10:13 AM   #2
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You can take the logarithm of each side of equation 1 (or 1a).
What exactly were you unable to do with equation 2?
Equation 3a should be a = √)100Vh-93Vc)/(√858*√Vc).
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