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Question

Water flows out of a big tank along a tube bent at right angles; the inside radius of the tube is equal to r=0.50cm (figure shown above). The length of the horizontal section of the tube is equal to l=22cm. The water flow rate is Q=0.50 litres per second. Find the moment of reaction forces of flowing water, acting on the tube's walls, relative to the point O.
161402_dce311be319b445ab9d1d86eeade01df.png

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Solution

Let the velocity of water flowing through the tube at a certain instant
of time be u, then u=Qπr2, where
Q is the rate of flow of water and πr2 is the cross section
area of the tube.
From impulse momentum theorem, for the stream of water striking the tube corner, in x-direction in the time interval dt,
Fxdt=ρQudt or Fx=ρQu
and similarly, Fy=ρQu
Therefore, the force exerted on the water stream by the tube,
F=ρQui+ρQuj
According to third law, the reaction force on the tube's wall by the stream equals (F)
=ρQuiρQuj.
Hence, the sought moment of force about 0 becomes
N=l(i)×(ρQuiρQuj)=ρQulk=ρQ2πr2lk
and |N|=ρQ2lπr2=0.70Nm=70 Ncm
274744_161402_ans_fccc1dc9dd1945a68d060fcbf7378a2e.png

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