Fluid Mechanics: Bernoullis Theorem Demonstration?

[Andrew871]

Having trouble using the bernoulli equation to work out the dynamic head for this experiment.

the equation I am using is as follows:

P1/ρg+(v1^2)/2g+z1=P2/ρg+(v2^2)/2g+z2

Where:

P= static pressure
v = fluid velocity
z = vertical elevation of fluid so z1 = z2 for horizontal tube

so equation is now:

P1/ρg+(v1^2)/2g=P2/ρg+(v2^2)/2g

now the measurements taken for static presure head (h), in meter which is related to P using this realtionship:

h = P/ρg

which allows the bernoulli equation to be rewritten as:

h1+(v1^2)/2g=h2+(v2^2)/2g

now i am having trouble rearanging this part of the equation to work out the the dynamic head for the experiment

 

[Navin A S]

.The dynamic head can be calculated using the following equationd = h2 - h1where Hd is the dynamic head, h1 is the static pressure head at the beginning of the tube and h2 is the static pressure head at the end of the tube.Therefore, the dynamic head can be calculated asd = h2 - h1 = (P2/ρg)-(P1/ρg) = (P2-P1)/ρg