Question

-2 2 Pipe flow [6 points) A liquid with SG 0.96, 1 = 9.2 x 10-4 Nsm-2, and vapor pressure pu = 8000 Nm (absolute) is drawn into the syringe as is indicated in figure 3. What is the maximum flowrate if cavitation is not to occur in the syringe? Assume the flow is laminar. Neglect minor losses. Assume that the velocity of the piston is negligible compared to the velocity in the needle. Atmospheric pressure is Patm 101 325 Nm-2.

Answer 1

Dear student,

Inorder to find maximum flow rate, we need to apply bernoulis equation between two points.

Here only major frictional headless is considered.

Numerical solved below,

Soen: 31-07-2020 H Given! - S.G=0.96 M= 9.2x104 N8/m2 Pu= 8000 N/m² Parm= 101825 N/m2 Plomm. 0.12m. 0.25 mm. o 1 0.1m. 0. Apply Bernoulis equation b/w 0 -0 & ③0. -Maior head Loss. Pi v/ P2 + C V22 + Z2th 29 29 Pi = Patm = 101325 N/m2. | P2 =Pv=8000 N/m2 V=0. (In order to get Max. Z=0 flow rate; we need Max. pressure diffrence) 4 And this pressure diftente limited to trv. to Length of needle avoid Canitation Here; hu= JL VE Z = 0.12 29 D. the 0. 8000 + 0.12t 101325 0.25X10) (0.96 19.88 103) (0.96X9.87X103)

101325 8000 0.12 + (2 v² 12x9.81 (0.46 x 9-81X103) (0.25x103 9.9096 = 0.12 + (400f + 1) v² 2x9.81 192.0719 = (400f +1) 8² Assume; because of very Smay dammeror 0.25mm; flow is laminar, f= 64 64 H Re. SXV2XD. 0.2453 f= 64 (9.20104) 0.96 xlooo x V2 x (0.25X103) V2. 192.0719 = +1) V₂ 2 192.0719 V₂ (400 x 0.2453+1 (93:12+1) :) x v² 142.07198 (98-12 + V2) VX 192.0719 98.12 + V2 Vz + 98.12 V₂ - 192.0719 = 0. V2 = -98.12 +V98-12 +4 (192.0719) (V2 = 1.9199 m/s

HeneQ= AV = (02SxTow) x (19199) Q = 9.4242*108 m3/s. Check the flow is laminar (or not. Re= PVD 0.96 x 1000 x 1.9199 x 0.25X103) (9.2x10-4 ) Re= 500.8.4 < 2100: - Our assumption is correct, lamsrom flow-

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