I think it is fairly simple but I don't understand how the teacher got to this answer:
L= 0,5m
D = 0,2m
Phi,out = Integral v(z) * pi *D*dz = pi*D integral (v/2L) * z *dz
= (pi*D*v)/(2L) [1/2 z²] = ((pi*D*v)/(4))*L
the question is about a tube which is porous, which has lengt L and diameter D
The entrance speed is 6m/s
The velocity of the liquid increases while going up the tube lineairy.
At the bottom speed is 0 and at the top the speed is V/2
So when calculating the v(z) then z = V/2L
As y = ax+b one can calculate the b = 0 and then fill in V/2 = aL+b
Then a = V/2L but I just don't get how one goes from here to here:
pi*D integral (v/2L) * z *dz
= (pi*D*v)/(2L) [1/2 z²] = ((pi*D*v)/(4))*L
PS the integral goes from 0 to L
L= 0,5m
D = 0,2m
Phi,out = Integral v(z) * pi *D*dz = pi*D integral (v/2L) * z *dz
= (pi*D*v)/(2L) [1/2 z²] = ((pi*D*v)/(4))*L
the question is about a tube which is porous, which has lengt L and diameter D
The entrance speed is 6m/s
The velocity of the liquid increases while going up the tube lineairy.
At the bottom speed is 0 and at the top the speed is V/2
So when calculating the v(z) then z = V/2L
As y = ax+b one can calculate the b = 0 and then fill in V/2 = aL+b
Then a = V/2L but I just don't get how one goes from here to here:
pi*D integral (v/2L) * z *dz
= (pi*D*v)/(2L) [1/2 z²] = ((pi*D*v)/(4))*L
PS the integral goes from 0 to L