Question

Water flows steadily from a large tank as shown in the figure. The height of water in the tank is H = 4 m represents the height from the centerline of the horizontal pipe to the surface of the water in the tank. The water flows out of the tank through a pipe with diameter of 4 cm. The stream of water flows through a bend (= 30 degrees, L = 2.88 m) and exits as a free jet through a filter. The filter is designed as holes of 3 mm diameter in a showerhead. A manometer containing mercury (SG = 13.6) is used to control the pressure in the flow. The reading on the manometer is h = 10 cm and hl= 30 cm. Calculate the force on the bend. Type your response * Submit

Answer 1

solution:

By using Bernoulli equation, continuity equation and impulse momentum equation, we will solve the question.

Solution 13-611 lm Han Givent H= 4m = Height of tank | h= 10 cm; h=30cm Toom, h = 30cm O_ 309 Se = 103 kg Im; I; Im = 13.6 xlor For Manometen; 300- 1 lo- 30 cole y =2.8&m Pop Y= (2-4) Chy-h) sy pes (136+1)x (0.3-0.1) = 2.52m Pez 24.7212 kPa] = (l) gauge Applying Bernouli at point A and C; oy 0+ 0 + 45 2.52 + ve to - Vez 5.38 m/s] =v, How; p= A pipex Ve= IX dpipe XV c = I x (0.04) x 5:38 >[dz 6.47 X10'mols 7 Mow; considering shower head; the volume of liquid which entory in the bend portion of pipe in Isecond will be the liquid which exit from the shower head in one second

het, N be the number of holes in shower head- . NX A dhale x L = 9 > MX Y (0.003)*x 2.98 6.492103Hz. 332.6 My 333] { to N A Prteger) How, het; Vz be the velocity of liquid exiting from shower 37 NYR 60.083? x Vs 0 [ continuity equation] 3 TV2= 2.88 mis wed How; Applying momentum equation at pipe bend; In x direction; (Efex) ev- PIA + fex - P, A, Cegjo= m (=v₂ Cos3o+V) m = fx d = 8.77 Kg/ Az IX (0.04)² = 1.25X103 m2 A2 I 2003 x 333= 2.35 x10 m2 Piza24.7212 +101.325 = 126.045 K Pa. P2= 101.325 kPa. letting values in above equation we get for -0.04865= 0.019537 / FR x = 68.187 M) fer- ► In y direction (EFR) = 0 + fry+ PAL sin 30 = m EV2 sin 30-0) => fry + 0.11911 - -971488x1053 - Fey-128.8 MB

force on Bend Poi In x direction: Fox = 68.1874 In y direction : fay=-128.0M - it means; fry is in dousnewound cliration . Resultant force on Bendi- (Fret) = J (FRx)² +C PRy)? [ (finer) Gand = 145.74 N - 145.74 N & Fry=68-187M tanica= Fay = 128.8 FRx 68.187 0:62.103 -40=62.103 in 4th qualsant. I nel 14574 M Frye 128.0N.