Question

Required information Water enters vertically and steadily at a rate of 31 L/s into the sprinkler with unequal arms and unequal discharge areas. The smaller jet has a discharge area of 3 cm2 and a normal distance of 50 cm from the axis of rotation. The larger jet has a discharge area of 5 cm2 and a normal distance of 35 cm from the axis of rotation. Take the density of water to be 1000 kg/m3 = 1 kg/L. SO cm Disregarding any frictional effects, determine the rotational speed of the sprinkler in rpm. The rotational speed of the sprinkler is rpm.

Answer 1

Assuming equal flow rate through both the arms. First we find absolute velocity of flow at each outlet. Since no external torque is acting on the body, equating net torque to zero we get the rotational speed of the sprinkler. Detailed calculation is given below.

hi - 50cm 4 35 con А, ах 3х б“ Essuming equal flow sali Relative relocity of watu with spocitler, Vi = = 21x102 = 51.67 m/s ming. Vo = - 346 = 31 ms Absoleli velocity at both ende is geven by, Viabs = V, - 40W - 51.67 - 0-5XW = 31+8.350 Since external torque to sprinkler is zero, we have ET=0 mi, Viabstep - m Vzabeth 50 | JxQ [ (51-67-0.54)*9*814 0.35w)x 0-36) = 0 Vrabs = Vz + bzw 14.985 - 0: 37250 = 0 w = 40.228 sad's hotational speed, N- wx60 = 40-228860 = 384.15 spm