Question

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A nozzle is often added to the end of a garden hose to produce a larger exit velocity for the water. Consider a nozzle attached to the end of a 3/8-in. internal diameter hose flowing 0.2 gallons per minute. The exit of the nozzle has an exit diameter 1/5W of its entrance diameter. 1. Draw a schematic of the problem labeling all pertinent information. 2. Use Bernoulli's equation to predict the pressure difference in units of psi, between the entrance and exit of the nozzle. 3. If only 81% of the pressure difference found in part b is available across the nozzle, express the new flow rate as a percentage of the flow rate of part b.

Answer 1

Applying Bernoulli's equation and Continuity equation between entrance and exit of nozzle we can solve this question as below-
sol :- The data provided in the question are as below- - diameter of hose, di Internal 33 in ins 3 Nozzle exit diameter ,dze Ś d IS ) in. 8 3 40 in. flowrate of wate 0,2 gallons / min. we have to find out schematic of the information problem labeling all pertinent 2. in Pressure difference psi b/w the entrance and exit of the nozzle. 3 as a percentage of the old New flow rate flow rate 1. schematic diagram :- Nozzle Garden Hose water flow P2 Po let, section 0 - 0 :- Entry to the nozzle section Exit of the 2 nozzle. - 2. Assumptions : Steady Incompressible , Irrotational fluid flow. Flow along streamline Ideal ie fluid flow are nozzle losses neglected. a

Now, applying Ideal 4 (2 can write equation b/w section we Pi P2 + v,2 29 + Z, 8.3 А. + Z2 Swg 29 where, V = Velocity at section ① - 0 = 4 4 di? 0,77 TI 2 “x) 0.2 x 231 = Q = 0.2 gallons/min. = (0.2*231 in 3/min) = 6.972. in./ sec. in 3/sec 0.77 in?/sec V2 = Velocity at section = 60 (2) 2 0.77 174.292 in. /sec 7 x ( 2 ) hose As, the Pipe and nozzle are in horizontal So, Z, = Zz

evom) in). on 1 Density of wate Sw 0.036 127 ebm in 3, = 2.68 insec?! Putting an Values in ean (A. 6.972² P2 0.036127 x 2.68 2X2.68 0.036127 X 2-68 we can get pl 174-292 + 2 x 2-68 or Pi-P2 547 85 ebm in-sec? (547.85 x12) ebm-st {arin ain=1 / tt ft² sec? 547.85 X 12 lbf 32.174 I ebf= 32.174 lbm-st [ oft2 Sec? 547.858 12 Psi {"I psi = 144 ewolf) 32-174 x 144 1.419 Psi Ans (2) so, the pressure difference b/w the entrance and exit of the nozzle 1.419 psi (3) when the Pressure difference available across the nozzle is 81% then the new Pressure difference Pi-P / - (54 147 85 x 0.8$ eum in-sec 44 3.7585 = ebm in-sec?

nozzle - Now, applying Bernoulli's eah b/w entrance and exit of the P vil Pa' Swa 2g wa 2 g P-PM V. ²_vi? Sw + + kerana or, s 2 443.7585 [ V₂' Vi Or, dz = { x V (251) 0.036127 )=(5? >> Vá= 25ví or, v' = 6.275 in/sec. so, New flow rate, al di? xvi! x ()** 6.275 П 4 = = 0.693 insec

Therefore, new flow rate as a % of old flow rate 0693 - X100 X100 0 0.77 90. ANS (3).