# Deriving equation for ideal fluid flow problem (~bernoulli equation)

1. Aug 5, 2010

### Villhelm

Could anyone offer some feedback?
I'm reasonably sure I got the gist of it, but I'm not sure how neat a solution it is (i.e. nothing relevant missing, or irrelevancies included). It's a past paper question so I haven't tried to tidy it up.

1. The problem statement, all variables and given/known data

2. Relevant equations
Hydrostatics/ideal fluid flow stuff.

3. The attempt at a solution
Derived Bernoulli's equation for arbitrary tube of flow then subbed in appropriate variables.

The simplified diagram (I drew an equivalent sketch by hand):

What I wrote:

This is a/the flow tube (not representative of cross section or elevation, includes the reservoir itself) representing the above system.

Assumptions:
Flow is steady streamline (large reservoir, so little change in pressure over short time);
Fluid is inviscid and incompressible.

Other premises (?):
From the assumption of steady flow -> Pressure in the flow tube is constant for any particular position therein (i.e. Pressure = P(s) for some position s in the flow tube).

The cross section of the flow tube is similarly only a function of position (it doesn't change as a result of pressure etc), so cross sectional area = A(s).
P(s) >= 0 and A(s) >= 0 for all s.

y1 is the height of the reservoir water;
y2 is the turbine height (=yt);
so y1-y2 = h.
v1,v2 are the velocity of the fluid elements at positions 1 and 2 respectively.

The fluid element moves from position 1 to position 2. The arrows inside the tube represent the internal pressure acting on either "side" of the fluid element producing a force on either side, which will cause some net work (Wd) to be done.

If the force on the left is FL(s) = PL(s)AL(s) and the right is FR(s) = PR(s)AR(s), then

Wd = integral a to c [PL(s)AL(s)] ds - integral b to d [PR(s)AR(s)] ds

= integral a to b [PL(s)AL(s)] ds + integral b to c [PL(s)AL(s)] ds - integral b to c [PR(s)AR(s)] ds - integral c to d [PR(s)AR(s)] ds

If s1 = s2 = s then PL(s1) = PR(s2) = P(s) [is something like this too trivial to mention, or better safe than sorry esp. for an exam question that will be ~ similar?]

=integral a to b [P(s)A(s)] ds - integral c to d [P(s)A(s)] ds

More assumptions:
The fluid element is small enough such that the pressure difference across it is ~= 0Pa.

So approximately:
P(a) = P(b) = P1; P(c) = P(d) = P2

Therefore Wd = P1 integral a to b A(s)ds - P2 integral c to d A(s) ds
= P1V1 - P2V2

Since the fluid is incompressible, V1 = V2 = V.

So Wd = (P1-P2)V
Density of fluid = D = mass of fluid element / volume of fluid element = m / V; V = m/D
Wd = (P1-P2)m/D
= $$\Delta$$Pe + $$\Delta$$Ke
= mg(y2-y1) + 1/2 m(v22 - v12)

(P1-P2) = Dg(y2-y1) + 1/2 D(v22 - v12)

P1 + Dgy1 + 1/2Dv12 = P2 + Dgy2 + 1/2Dv22

So, having Bernoulli's equation, rearranging for v22 :

P1-P2 + Dg(y1-y2) + 1/2Dv12 = 1/2Dv22

The pressures are the gauge pressures, so P1 = 0Bar.
y1-y2 = h, and v1 ~= 0ms-1

Therefore, an expression for the fluid flow velocity at the entrance to the turbine is:

v22 = -2P2/D + 2g(h)

The next part of the question is a numeric calculation which gives a value for h and gauge pressure P2 - along with the front matter's values for g and density of water. I reckon I've got the +/- signs alright because v2 increases as h increases, which is expected, and decreases as the back-pressure, P2, increases, also expected. Any feedback would be most appreciated.

In the numeric part of the question, the gauge pressure P2 at the turbines is given as 0.5bar - in a physical system would this pressure above atmospheric be associated with (or at least contributed to by) the resistance of the turbine and connected equipment? If so, would there be a situation where there is maximum energy generated per unit of fluid passing (or transferred from the fluid to the turbine and/or generating equipment ~efficiency?) and also a peak power output (not necessarily under the same conditions as peak energy efficiency)?