Question

i need clear writing please

1) Water from a stationary horizontal nozzle strikes a lat plate which is aligned at 30 from the vertical. 70% of the flow exits past the upper side of the plate and 30% the lower. Water leaves the nozzle at 15 m/s, and the area of the nozzle is 0.1 m2. The flow of the water is along the plate. It was determined that the area of the flow leaving the upper side of the plate is 0.07 m2 and leaving the lower side is 0.03 m2. Assume that the density of the water is 999 kg/m Determine the horizontal and vertical components of the force required to keep the plate in position. 30 777

Answer 1

We know that F= dp = Force S (change in momentum) on of change of momentum. o time. rate a momentum = mass x velocity Let's do calculate change in momentum in 1 sec (that will the Force) Initial momentum = mona massx velocity mass in 1sec = P A ~ volume in isec. horizontal direction Initial momentum in = PAUxu = 999 X0.1 x 15 X15 = 22477.5 kg-m/s 22477.5 Î

mass going to upper side (1 sec) = 0.7X (PAV) = 0.7 x999x 0.1 x 15 = 1048.95 kg/sec. mass leaving lower side = 0.3% (999x0:1X15) = 449.55 k91. final momentum, 1018.95 oony 300 < 449.55 If mini-manz Pfaz mix, - me Vxz (in x direction) €4={(1098.95x15) Sin 30-449.55% 15% Singo I 9: = (4495.5) . Pfy = (1048.95x15) (0530 -(449-55X15 )(0330 = 7786.435 Change in momentum in horizontal direction = Per-Pi = 4495-5 î – 22477.5 ì. = 17982 Mence force in horizontal directions 17982N change in momentum in vertical direction = Best 7786.439 -0 = 77864 3 2 Horizontal Forso = 17982 & Vertical Force =7786.43 H