Question

1 (a) A steady stream of water jet discharges from the nozzle with a uniform velocity of 10 m/s at angle 60° as shown in Figure 1. The jet strikes a horizontal plate 2 m above the nozzle exit. Upon impact, part of the flow is guided towards the left while the remainder towards the right, and the plate is suspended in the air. The weight of the plate is 25 N. Assume frictionless flow along the plate and weight of water to be neglected. (i) Write down the momentum equation for x and y direction. (ii) Determine the diameter of the nozzle (at the exit). (iii) Determine the mass flow rate A and B. (12 marks) Plate is stationary ON Left Right Mass flowrate A Mass flowrate B V 2 m Flange 60° Nozzle exit diameter (?). pipe nozzle Figure 1

Answer 1

i) Momentum equation for x and y direction (Initial momentum) = P = mxv. costo final momentum) = Par my, Lesbo Ap=o na (Initial Momentur), - Pyi mty (final momentum), - P/f = PAX O. App - (my sin6o - o) (APy = mv, sinbo] ii) A Santo, cosso) nie = Sax -Cui caso) roma 1000x 2.8 x107ma ☺= 1.4 m/s mig = 1000x2.8 x 101m-(5)- -|4ms B. La f A. U, Cosão- Vitrino 060

given data: 19 = 100/ 0 60 h = am. Wt plate - 250 d = diameterd n pet at exlt w -25 4 case a areag nogle vi cose su costo u, simbo ) By Impulse Momentum principle : ty= par (usino -o). (=par (Initial momentum - final . momentum) (Fy = parsino = 1000 x π d²x (10) 2 bin 60 Fy = 100000 * Id ? x 1in600 4 ZFy=0 since the plate is in equillibrium. W = FY 25 = 100x I xd>x fin6o 4 al2 = 25x10 x4 π x sinbo d = 0.019 m = 1.91cm //