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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.

## Homework Statement

## Homework Equations

Hydrostatics/ideal fluid flow stuff.

## 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):

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 s

_{1}= s

_{2}= s then P

_{L}(s

_{1}) = P

_{R}(s

_{2}) = 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) = P

_{1}; P(c) = P(d) = P

_{2}

Therefore W

_{d}= P

_{1}

*integral a to b*A(s)ds - P

_{2}

*integral c to d*A(s) ds

= P

_{1}V

_{1}- P

_{2}V

_{2}

Since the fluid is incompressible, V

_{1}= V

_{2}= V.

So W

_{d}= (P

_{1}-P

_{2})V

Density of fluid = D = mass of fluid element / volume of fluid element = m / V; V = m/D

W

_{d}= (P

_{1}-P

_{2})m/D

= [tex]\Delta[/tex]P

_{e}+ [tex]\Delta[/tex]K

_{e}

= mg(y2-y1) + 1/2 m(v

_{2}

^{2}- v

_{1}

^{2})

(P

_{1}-P

_{2}) = Dg(y

_{2}-y

_{1}) + 1/2 D(v

_{2}

^{2}- v

_{1}

^{2})

P

_{1}+ Dgy

_{1}+ 1/2Dv

_{1}

^{2}= P

_{2}+ Dgy

_{2}+ 1/2Dv

_{2}

^{2}

So, having Bernoulli's equation, rearranging for v

_{2}

^{2}:

P

_{1}-P

_{2}+ Dg(y

_{1}-y

_{2}) + 1/2Dv

_{1}

^{2}= 1/2Dv

_{2}

^{2}

The pressures are the gauge pressures, so P

_{1}= 0Bar.

y1-y2 = h, and v

_{1}~= 0ms

^{-1}

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

v

_{2}

^{2}= -2P

^{2}/D + 2g(h)

*The next part of the question is a numeric calculation which gives a value for h and gauge pressure P*

_{2}- along with the front matter's values for g and density of water. I reckon I've got the +/- signs alright because v_{2}increases as h increases, which is expected, and decreases as the back-pressure, P_{2}, increases, also expected. Any feedback would be most appreciated.*In the numeric part of the question, the gauge pressure P*

_{2}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)?