Hello :) The question seems simple, but I'm not sure, so I'm curious about the answer ... it's about converting the flow (air stream) m3 / h to m / s ... This is needed to select a windmill :) - that's a hint ...

It is impossible to calculate, how do you want to change the volume to length? One would have to know the cross-sectional area of the tunnel in which this air flows

Hello m3 / h is the t unit of flow rate (Qv), and Qv = S * V, where S conduit section [m2], V - air velocity [m / s] after transformations: V = Qv / S [m / s] so much that Qv must be in m3 / s m3 / h to m3 / s, you probably already know how to convert? :) Regards

Exactly ... I agree ... so for this information I need the area of the windmill (square meters), then I will have the air flow speed (because m3 / h is the windmill efficiency - that's how I understand it, am / s air flow speed forced by a windmill with a given area) ...

How to convert cubic meters into hours into meters per second

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#1 641055 14 May 2004 16:06

mito mito

Level 11

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Post #1
641055 14 May 2004 16:06

Hello The question seems simple, but I'm not sure, so I'm curious about the answer ... it's about converting the flow (air stream) m3 / h to m / s ... This is needed to select a windmill - that's a hint ...
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#2 641060 14 May 2004 16:17

Gorgoroth Gorgoroth

VIP Meritorious for electroda.pl

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Post #2
641060 14 May 2004 16:17

It is impossible to calculate, how do you want to change the volume to length? One would have to know the cross-sectional area of the tunnel in which this air flows
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#3 641072 14 May 2004 16:27

sokmarcin sokmarcin

Level 20

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Post #3
641072 14 May 2004 16:27

Hello
m3 / h is the t unit of flow rate (Qv), and Qv = S * V, where S conduit section [m2], V - air velocity [m / s]
after transformations: V = Qv / S [m / s] so much that Qv must be in m3 / s
m3 / h to m3 / s, you probably already know how to convert? :)

Regards
#4 641075 14 May 2004 16:28

mito mito

Level 11

» | Topic author Positive voted Helpful post? (+7)

Post #4
641075 14 May 2004 16:28

Exactly ... I agree ... so for this information I need the area of the windmill (square meters), then I will have the air flow speed (because m3 / h is the windmill efficiency - that's how I understand it, am / s air flow speed forced by a windmill with a given area) ...
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