Lubricating oil, with a relative density of 0.79, flows around a 90º bend. The pipe diameter is 0.45m, and the oil has a pressure head of 7m and the flow is 1.7m3/s. Find the force exerted by the oil on the bend.

Answer 1

density = .79 * 10^3 kg/m^3
velocity of oil = v
v*pi r^2 = 1.7
v =1.7 /(pi*.45^2) = 2.67 m/s
x momentum of oil coming down the pipe = m v = 1*2.67 = m 2.67 m/s
Since the oil turns 90 degrees, the x momentum out is zero
so
what is the change of momentum per second (the force)?
mass per second = .79*10^3 kg/m^3 * 1.7 m^3/s
= 1343 kg/s
so force due to change of momentum/second = 1343 kg/s*2.67 m/s
= 3586 Newtons
Now add to that the pressure times the area of the pipe for the x component of pressure force.
Note there is a similar y force.

Answer 2

Thank you, makes so much sence now! lol :-)