|Aug5-10, 02:57 PM||#1|
Deriving equation for ideal fluid flow problem (~bernoulli equation)
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.
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
The fluid element is small enough such that the pressure difference across it is ~= 0Pa.
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
= [tex]\Delta[/tex]Pe + [tex]\Delta[/tex]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)?
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