A siphon is a device for removing liquid from a container that is not to be tipped. It operates as shown in below figure. The tube must initially be filled, but once this has been done the liquid will flow until its level drops below the tube opening at A. The liquid has density ρ and negligible viscosity. (a) With what speed does the liquid emerge from the tube at C? (b) What is the pressure in the liquid at the topmost point B? (c) What is the greatest possible height h that a siphon may lift water?
Answers
Answer:
(a) Water will flow in the pipe and emerges from the point C. It is considered that water flows from depth equal to the d+h2.
Here, the equivalent depth of the water is
h = d+ h2
From equation v =√2gh, the speed of the water is
v =√2gh
=√2g(d+h2)
Therefore, the speed of the water emerges from the point C is √2g(d+h2).
(b) The pressure at the point C is
pC = ρg (h1+d+h2)
Now, the pressure at the point B is
pB = p0 – pC
Insert pC = ρg (h1+d+h2) in the above equation gives
pB = p0 – pC
= p0 - ρg (h1+d+h2)
Therefore, the pressure at the point B is p0 - ρg (h1+d+h2).
(c) Now, the maximum possible height is obtained as
p0 = ρgh
h = p0/ρg
Substitute 1.01×105 kg/m.s2 for p0, 1000 kg/m3 for ρ and 9.8 m/s2 for g in the above equation gives
h = p0/ρg
= (1.01×105 kg/m.s2)/( 1000 kg/m3) (9.8 m/s2)
= 10.306 m
Rounding off to three significant figures, the greatest possible height that the siphon may lift water is 10.3 m.
Hope it will help you
Answer:
Now, the lines represented here are perpendicular to each other and hence,
The coefficient of x2 + the coefficient of y2 = 0
This gives, 3c + 2m + 4 – c = 0
Hence, c+m+2 = 0
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