(a) Using reasoning similar to that employed inobtaining Torricelli\'s theorem,
ID: 1726655 • Letter: #
Question
(a) Using reasoning similar to that employed inobtaining Torricelli's theorem, derive an expression for the speedv of the fluid emerging from the tube. This expressionshould give v in terms of the vertical height yand the acceleration due to gravity g. (Note that thisspeed does not depend on the depth d of the tube below thesurface of the liquid.)v =
(b) At what value of the vertical distance y will thesiphon stop working?
y =
(c) Derive an expression for the absolute pressure at the highestpoint in the siphon (point A) in terms of the atmospheric pressureP0, the fluid density ,g, and the heights h and y. (Note thatthe fluid speed at point A is the same as the speed of the fluidemerging from the tube, because the cross-sectional area of thetube is the same everywhere.) (Use the following variables asnecessary: g, h, y, P_0 for P0, and rho for.)
PA = A siphon tube is useful for removing liquid from a tank. The siphontube is first filled with liquid, and then one end is inserted intothe tank. Liquid then drains out the other end, as the drawingillustrates. (a) Using reasoning similar to that employed inobtaining Torricelli's theorem, derive an expression for the speedv of the fluid emerging from the tube. This expressionshould give v in terms of the vertical height yand the acceleration due to gravity g. (Note that thisspeed does not depend on the depth d of the tube below thesurface of the liquid.) v = (b) At what value of the vertical distance y will thesiphon stop working? y = (c) Derive an expression for the absolute pressure at the highestpoint in the siphon (point A) in terms of the atmospheric pressureP0, the fluid density p,g, and the heights h and y. (Note thatthe fluid speed at point A is the same as the speed of the fluidemerging from the tube, because the cross-sectional area of thetube is the same everywhere.) (Use the following variables asnecessary: g, h, y, P_0 for P0, and rho for p.) PA =
Explanation / Answer
(a)
from the given problem let the surface of thewater will be the position 1 and the exit of the hose
be the position 2 then we can see that thepressures are equal so we get
P1 = P2
from the bernoulis equation we get
(1 / 2) v12 + g y = (1 / 2) v22
as at the surface of the water the speed is zero wecan write v1 = 0 so that
v2 = (2 g y)
(b)
the siphon will stop working when v2 =0 m / s or y = 0
(c)
at point A we get
PA + g (h + y) + (1 / 2) vA2 = Po + (1 / 2) v22
as vA = v2
PA = Po - g (y +h)
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