For a bent galvanized iron pipe what minimum pressure P is n

For a bent galvanized iron pipe, what minimum pressure P is necessary to levitate the smooth plastic ball of density p_out =1665 kg/m^3 with an air stream at 20C? (rho_air = 1.2 kg/m^3, mu_air = 1.82 Times 10^-5 kg/m.s, D_ball = D = 4 cm, D_pipe = 8 cm, L = 3m, and the bend radius to diameter ratio is 1:1. Assume fully developed flow and ignore entrance and exit losses. Note sketch is not drawn to scale. Provide two reasons to suspect a degree of error in your prediction.

Solution

Given, D _ball = 4 cm = 0.04 m

D _pipe = 8 cm = 0.08 m

_ball = 1665 kg/m³

_air = 1.2 kg/m³

Pressure exerted by the air = P N/m²

Bend radius of the pipe = 0.08 m

Radius of the ball, R _ball = 0.04/2 = 0.02 m

Volume of the ball, V _ball = (4/3)* *R³_ball = (4/3)* *(0.02)³ = 0.0000335 m³

Weight of the ball = V _ball* _ball*g = 0.0000335*1665*9.81 [g = acceleration due to gravity = 9.81m/s²]

So, the air-pressure through the pipe has to encounter exactly 0.547 N of downward force exerted by the plastic ball.

Now, area of the pipe section, A _pipe = (/4)* D²_pipe = (/4)*(0.08)² = 0.005 m²

The total pressure that is need to be created at the exit of the pipe = 0.547/0.005 = 109.4 N/m²

Total length of the pipe = 2L + D _pipe = 2*3 + 0.08 = 6.08 m

Required pressure at the entry of the pipe = pressure drop through the length 6.08 m + 109.4 N/m²

Let, the co-efficient of drag of the ball, C_D = 0.47 and the velocity of air flow = V m/s

We know that the force required to hold the ball in air = (C_D*A _ball* _air*V²)/2

                                                                                        =( 0.47*.00126*1.2* V²)/2

Now, ( 0.47*0.00126*1.2* V²)/2 = 0.547

[projected area of the ball = A _ball =(/4)*(0.04)² = 0.00126 m²]

Reynolds number, R_e = ( _air*V*D _pipe)/ = (1.2*39.24*0.08)/(1.82*10^-5)

= 206980

Since R_e > 4000, the flow is turbulent.

Co-efficient of friction, f = 0.0791/( R_e)^1/4 = 0.0791/(206980)^1/4 = 0.0037

Pressure loss due to friction = (4 _air *f*L*V²)/2D = 1048.7 N/ m²

Total pressure = 1048.7 + 109.4 = 1158 N/ m²