Physics
posted by Henri .
Lubricating oil, with a relative density of 0.79, flows around a 90¨¬ 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.
Oil density = 0.79 X 103
Volumetric mass flow rate = 1.7 m3/s
Pipe diameter = 0.45 m
Pressure head (¥÷) = 7m
Oil mass flow=density*Vol/sec
Mass flow=0.79X10^3*1.7
Mass flow=1343kg/sec
Oil velocity in the pipe:
Vel.*pir^2=1.7m^3/sec
vel.=1.7/pi*(0.45/2)^2
vel of oil=10.69m/s
Force(F1) on the bend due to oil flow:
F1=Oil mass flow*Oil velocity
F1=1343*10.69
F1=14356kgm/s^2 or 14.356kN
Force(F2) due to oil pressure:
Pressure=press. head*(density*gravity)
p=7m*0.79X10^3kg/m^3*9.81m/s^2
p=54249kg/ms^2
F2=pA
F2=54249kg/ms^2*(0.7854X0.45^2)
F2=8627kgm/s^2 or 8.627kN
Total Force(Ft) acting on elbow:
Ft=F1+F2
Ft=14.3456kN+8.627kN
Ft=22.983kN (ans.)
Thanks for taking a look.

It was not clear to me if the "pressure head" of the fluid was in meters of mercury, water or the oil. Your assumption that it the fluid flowing (oil) makes sense.
I agree with your mass flow and velocity numbers. There are two components to the reaction force on the 90 degree bend. To change the direction of the momentum flow by 90 degrees, you need a force equal to the magnitude of the momentum change vector, which is sqrt2 times the flow scalar momentum. The force acts at 45 degrees to the two flow directions.
As you have noted, there may be an additional force due to the loss of pressure at the turn. The loss will be approximately equal to the pressure head for a 90 degree turn. I do not believe this is an exact rule. It seems to me that this should be treated as a vector also, since the two pressure forces before and after the bend act in perpendicular directions.
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