Answer to Question #87686 in Mechanics | Relativity for Sjsjjx

Question

Sections 1 and 2 are at the beginning and end of the bend of the 200 mm diameter pipe in which the quantity of flow is 0.28 m3/s. The angle of deflection of the water is 40 degrees.Calculate the force that the liquid exerts on the bend if the pressure in the pipe is 50 kPa. Assume no loss of pressure round the bend.

Please help me

Answer 1

$$D=200mm=0.2m, A= \pi D^2 / 4, Q=0.28 m^3/s, V=Q/A$$ $$p=50kPa=50000 N/m^2, \rho = 1000 kg/m^3, \theta=40^o=2 \pi /9$$

From the equation of momentum in the x direction:

$$F_x= \rho Q V (1-cos \theta ) + p A (1-cos \theta )=(\rho Q^2 / A + pA)(1-cos \theta ) \approxeq 951.344 N$$

From the equation of momentum in the y direction:

$$F_y= \rho Q V (-sin \theta ) + p A (-sin \theta )=(\rho Q^2 / A + pA)(-sin \theta ) \approxeq -2613.797 N$$

The resultant force:

$$F= \sqrt{F_x^2 + F_y^2}= \sqrt2 (\rho Q^2 / A + pA) \sqrt{1-cos \theta } \approxeq 2781.544 N$$