Could anyone offer some feedback?(adsbygoogle = window.adsbygoogle || []).push({});

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 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 bA(s)ds - P_{2}integral c to dA(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)?

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

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