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- - **HVAC finding the value of BETA help**
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HVAC finding the value of BETA helpGood 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) Key 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 |

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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