Hi all,
The drag of a stationary obstacle due to incoming current versus the drag of the same obstacle that moves in a stationary (but the same) medium. Will these two result in a same drag?
best regards,
-Arman-
Bernoulli equation with pressure loss:
p_{1}+0.5\rho v_{1}^{2}=p_{2}+0.5\rho v_{2}^{2}+p_{L}
With p_{L} the pressure loss. The head loss, h_{L} can be computed by dividing the equation with \rho g , and after re-arranging:
h_{L} =\frac{p_{1}-p_{2}}{\rho g...
Well, the g should be there, otherwise the dimensions of your equation does not match:
m (n.e.) m^2/s^2
(how do you post an equation in a thread by the way?)
About the pressure: yes the (static) pressure is not the same between a and b. What I mean is the (static) pressure at border...
Hi all,
I'm sure a lot of you know about the head loss due to sudden expansion:
Hl = (1/2g)*(v1-v2)^2
This equation can be derived from Bernoulli, continuity and momentum balans equations. The underlying assumption in deriving this equation is that the pressure just after the expansion...