Question

Water sprays out of a sprinkler as shown in the picture.

At point 1, the diameter is D1=4in, Q=245 gpm, P1=34 psig.

At point 2 , the diameter of the nozzle is D2=1 in, Flow rate is still Q=245 gpm.

Ignoring the weight of water, calculate:

a) V1 N=AND V2

b) force on the flange @section 1

c) total head loss of elbow and nozzle, hL

1 40° (2)

Answer 1

Given D = 4 in Q,= 2459pm = 245X231 inp/min 245X231 943.25 ing in3 60 sec 409 [1'1gpm= 231 ins/min] P, = 34 psig (Gauge press]: Dis '1'in 0,- 2459 pm = 943.25 in/s discharge(Q) Area (A) velocityx 90s 943.25 75.06 în/sec @s 'A & X 42 1943:25 $ 1200.98. indsec. V II 12

X @ Force on the Hange @ section 1. Rate of change of linear momentum Forte on x dirhRate of change of linear momentum in medizin F = mx invx where mia mass flow rate: 8x8 8- density of water reserved Specific ws of water (= 29: 62.4 W/?. g= 32.17 44/32 ~ 32+1/32 62.416. s2 62.4 Los? ft3 x 32 ft 32x124 in V or 328124 is ex Now, Ty mly Vzaje VE V = 75.06 in's Vex tv, tos 40) 62.4 1682 94.25 in (75-06-4 2004 set COS 40) 1 - Fun88-26 tot. Now, Ty in Cuby key] { y = 0]. 62.4 *943.25(041200-98547407 32x124 Fy = 68.44 164 Х

.. - t + 온 국민물 Š Net Force (IR)= VE 2472 111.7 16f Ang @ Total bead loss of elbow and nozzle, be Apply Bernaulle's Foun:- 22 + z = +8the Ya 2g Vo zg R- Atmospheris press Given P,- R = 34 psig (Gäuge press). OR Pi-P2 1, 2v, ². the ti loch). 8 2g (75:06)2-(1200.9892 OR 62.4 2x 32.17X12 (1203 919.31 inch ans + +