Two problems on fluid dynamics:
Problem 2: Water in a tank being siphoned off. To determine velocity, pressure, flow rate etc..
See the attached file.
Please refer to the attachment.
1. a) Pressure at section 1 = Atmospheric pressure + Pressure due to the 3 ft water column
Pressure due to 3 ft water column (in psi) = Height of water column in inches x Density of water in pounds/cu.inch = 36 x 62.5/123 = 1.3 psi (Density of water = 62.5 lb/ft3)
As atmospheric pressure = 14.7 psi, p1 = 14.7 + 1.3 = 16 psi
b) Pressure at section 2 is atmospheric pressure as it is open to atmosphere.
Hence, p2 = 14.7 psi.
c) From continuity equation we have : a1v1 = a2v2 where a1 and a2 are the cross section area and v1 and v2 are the flow velocities at the two sections.
Or v2/v1 = a1/a2 = (Πx0.0752)/(Πx0.052) = 2.25 .......(1)
From Bernoulli's theorem we have : p1 + ½ d x v12 = p2 + ½ d x v22 (d = density of water)
(Note : As the pipe is horizontal, dgh terms on the two sides get cancelled out)
Or p1 - p2 ...
Step-by-step solutions to the problems regarding fluid dynamics are provided.