Q 025 galmin and 3 rpmSolutionThe torque from the water f

Solution

The torque from the water flow through 1/4 inch diameter holes is a superimposted condition.

The total resistive torque is given by,

T tot = F ( x1 + x2 + x3 + x4)

The force due to flow rate of water is rho x Av^2

Considering flow rate pre unit volume to be Q = 0.25 gal / min, and

the velocity v = Q / A = 0.25 x 0.16 / (3.14 x (1/8)^2) = 371 inch / min or 0.51 ft / s or 0.15 m / s

Torque due to water = rho x A x v^2 where rho = 983 kg / m^3

F = 983 x 3.14 x (1/8 x 0.083 x 0.3)^2 x 0.15 ^2 = 0.23 kg or 2.26 N

Total resistive torque is F ( x1 + x2 + x3 + x4) = 2.26 ( 3+5+7+9) x 0.083 x 0.3 = 1.35 N-m