The figure shows a siphon, which is a device for removing liquid from a containe

Question

The figure shows a siphon, which is a device for removing liquid from a container. Tube ABC must initially be filled, but once this has been done, liquid will flow through the tube until the liquid surface in the container is level with the tube opening at A. The liquid has density 1000 kg/m^3 and negligible viscosity. The distances shown are h_1 = 27.0 cm, d = 14.0 cm, and h_2 = 46.0 cm. (a) With what speed does the liquid emerge from the tube at C? (b) If the atmospheric pressure is 1.00 times 10^5 Pa, what is the pressure in the liquid at the topmost point B? (c) Theoretically, what is the greatest possible height that a siphon can lift water? Number Units Number Units Number Units

Explanation / Answer

given data

density =1000 kg/m^3

h1 = 27cm
d=14cm
h2 = 46cm
Patm =1.00*10^5 Pa

Patm +1/2 V^2 =PA + 1/2 v^ -gd [S<---->A]

but V = va/A

Patm = PA + 1/2 v^2 (1-a^2/A^2) - gd

as a <<<A

so,

Patm = PA + 1/2 v^2 - gd

PA +1/2 v^2 - gd =Patm +1/2 v^2 - g (d+h2) [A<--->C]

1/2 v^2 =  g (d+h2)

v = sqrt (2*g*(d+h2))

v = sqrt (2*9.8*(0.14+0.46))

v = 3.42 m/s

PB + 1/2 v^2 +gh1 = Patm +1/2 v^2 -  g (d+h2) [B<----->A]

h1 = ((Patm - PB )/g) - (d+h2)

h1(max) = Patm /g -(d+h2)

=(1.0*10^5/(1000*9.8)) - (0.14+0.46)

=9.60 m

Part B)

h1 = ((Patm - PB )/g) - (d+h2)

0.27 = ((1.0*10^5-PB)/(1000*9.8)) - (0.14+0.46)

PB = 9.1*10^5 Pa

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