Question

Water at 298 K discharges from a nozzle and travels horizontally hitting a flat, vertical wall. The nozzle diameter is 12 mm and the water exits the nozzle with a flat velocity profile at a velocity of 6.0 m/s. Use macroscopic balance for momentum to compute the force [N] on the wall neglecting gravitational and frictional effects

Answer 1

Answer:

0.00407 N

Explanation:

Let's assume that the water from the nozzle travels in the form of a cylinder and hits the wall. Therefore it has a momentum in the direction of flow, which we take as the x direction. After hitting the wall, the water splashes randomly. We assume that none of the water particles travels backward and that the momentum of water in the x direction after hitting the wall becomes 0. Therefore, the force experienced on the stream of water is rate of change of its momentum. The momentum(p) of a particle is given by, p =mv, where m is the mass and v the velocity.

Take a time interval of 1 second. In this one second, a stream of water with volume, V = Av (where 'A' is the area of the nozzle and 'v' is the exit velocity) is coming out of the nozzle. The mass(m) of water thus coming out of the nozzle in a single second is given by, m = Avρ, where 'ρ' is the density of water ( which is ≈ 1kg/m³).  

This volume of water has a momentum(p) given by: p = mv = Avρv = Aρv².

Thus in 1 second, the change in momentum of the stream of water is given by Aρv². Thus the rate of change of momentum is Aρv²/ 1 second = Aρv².

Here, A = πr², where r = 6 mm.

         ρ = 1 kg/m³.

         v = 6 m/s

Therefore, Force(F) = rate of change of momentum =  Aρv² .

Thus, F = 0.00407N.