Recent content by previah

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...