Question

5 points Save Answer Water leaves the shown large open tank to the atmosphere through a short length section that consists of a sharp inlet and two sudden contraction fittings from d, to d2 and then from da to dz. Neglecting major losses determine the height h (m) needed to maintain a steady flow rate Q, when the pipe diameters are d = 5 cm, d2= 2 cm and d3= 1cm and the flow rate is Q=3 Us. h d, d, d,

Answer 1

Here, we have to neglect the major losses.

So, only the diameter of the outermost hole will matter.

The flow rate is

Q = 3 L/s = 0.003 m^3/s

So, the flow velocity is given by

v = Q/A

A is the area of the opening

We have

d3 = 1 cm = 0.01 m

So,

A = 3.14*0.01^2/4 = 7.85*10^-5 m^2

So,

v = Q/A = 0.003/(7.85*10^-5) = 38.22 m/s

USing bernolli's principle,

$P + \rho hg + \rho v^2/2 = constant$

Considering this equation for the top of the tank and the opening and that the area of the tank is very large compared to the opening,

$\rho hg = \rho v^2/2$

h = w²/29

Using the values,