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

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 (W_d) to be done.

If the force on the left is F_L(s) = P_L(s)A_L(s) and the right is F_R(s) = P_R(s)A_R(s), then

W_d = integral a to c [P_L(s)A_L(s)] ds - integral b to d [P_R(s)A_R(s)] ds

= integral a to b [P_L(s)A_L(s)] ds + integral b to c [P_L(s)A_L(s)] ds - integral b to c [P_R(s)A_R(s)] ds - integral c to d [P_R(s)A_R(s)] ds

If s₁ = s₂ = s then P_L(s₁) = P_R(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 W_d = 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 W_d = (P₁-P₂)V
Density of fluid = D = mass of fluid element / volume of fluid element = m / V; V = m/D
W_d = (P₁-P₂)m/D
= [tex]\Delta[/tex]P_e + [tex]\Delta[/tex]K_e
= 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)?