Question

 

(Problem 4.3-3) Water at 298 K discharges from a nozzle and travels horizontally, hitting a flat, vertical wall. The nozzle has a diameter of 12 mm and the water leaves the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Neglecting frictional resistance of the air on the jet, calculate the force in newtons on the wall.

Ans. –Rx = 4.059 N

Repeat Problem 4.3-3 for the same conditions except that the wall is inclined 45° to the vertical. The flow is frictionless. Assume no loss in energy. The amount of fluid splitting in each direction along the plate can be determined by using the continuity equation and a momentum balance. Calculate this flow division and the force on the wall.

Ans. m2 = 0.5774 kg/s, m3 = 0.09907 kg/s, –Rx = 2.030 N, –Ry = –2.030 N (force on wall).

Answer 1

Inlet velocity V1 = 6 m/s

Diameter D1 = 12 mm x 1m/1000mm = 1.2 x 10^-2 m

Area A1 = (3.14/4) x D1² = (3.14/4) x (1.2 x 10^-2 m)²

A1 = 1.13 x 10^-4 m2

Temperature T = 298 K

Density of water = 996.9 kg/m3

Mass flow rate at inlet

m1 = density x Area x velocity

= 996.9 kg/m3 x 1.13 x 10^-4 m2 x 6 m/s

m1 = 0.6765 kg/s

Mass flow rate at point 2

m2 = m1/2 + m1/2 * Cos 45°

m2 = 0.6765/2 + 0.6765/2 * 0.7071

m2 = 0.5774 kg/s

Mass flow rate at point 3

m3 = m1/2 - m1/2 * Cos 45°

m3 = 0.6765/2 - 0.6765/2 * 0.7071

m3 = 0.09907 kg/s

Now reaction in X-direction and do the force balance

Rx = m2*V2*Cos45° - m1*V1 + m3*V3 * (-Cos 45°)

Rx = 0.5774*6*0.7071 - 0.6765*6 + 0.09907*6 * (-0.7071)

Rx = - 2.03 N

Now reaction in Y-direction and do the force balance

Ry = m2*V2*Sin45° - m1*V1*Sin0° + m3*V3 * (-Sin 45°)

Ry = 0.5774*6*0.7071 - 0.6765*6*0 + 0.09907*6 * (-0.7071)

Ry = 2.03 N

force on wall

-Rx = 2.03 N

-Ry = - 2.03 N