Question

Water flows through a pipe of 100 mm at the rate of 0.9 m3 per minute at section A. It tapers to 50mm diameter at B, A being 1.5 m above B. The pressure in the pipe at A is 70 kPa above atmospheric. Assuming no loss of energy between A and B, determine, a)- the velocity of A and B b)- Find pressure at B

Answer 1

Answer:

The velocities in points A and B are 1.9 and 7.63 m/s respectively. The Pressure at point B is 28 Kpa.

Explanation:

Assuming the fluid to be incompressible we can apply for the continuity equation for fluids:

$Aa.Va=Ab.Vb=Q$

Where A, V and Q are the areas, velocities and volume rate respectively. For section A and B the areas are:

$Aa=\frac{pi.Da^2}{4}= \frac{\pi.(0.1m)^2}{4}=7.85*10^{-3}\ m^3$

$Ab=\frac{pi.Db^2}{4}= \frac{\pi.(0.05m)^2}{4}=1.95*10^{-3}\ m^3$

Using the volume rate:

$Va=\frac{Q}{Aa}=\frac{0.9m^3}{7.85*10^{-3}\ m^3} = 1.9\ m/s$

$Vb = \frac{Q}{Ab}= \frac{0.9m^3}{1.96*10^{-3}\ m^3} = 7.63\ m/s$

Assuming no losses, the energy equation for fluids can be written as:

$Pa+\frac{1}{2}pa.Va^2+pa.g.za=Pb+\frac{1}{2}pb.Vb^2+pb.g.zb$

Here P, V, p, z and g represent the pressure, velocities, height and gravity acceleration. Considering the zero height level at point A and solving for Pb:

$Pb=Pa+\frac{1}{2}pa(Va^2-Vb^2)-pa.g.za$

Knowing the manometric pressure in point A of 70kPa, the height at point B of 1.5 meters, the density of water of 1000 kg/m^3 and the velocities calculated, the pressure at B results:

$Pb = 70000Pa+ \frac{1}{2}*1000\ \frac{kg}{m^3}*((1.9m/s)^2 - (7.63m/s)^2) - 1000\frac{kg}{m^3}*9,81\frac{m}{s^2}*1.5m$

$Pb = 70000\ Pa-27303\ Pa - 14715\ Pa$

$Pb = 27,996\ Pa = 28\ kPa$