Question

Problem 2 (3 pts.) Water is emerging from a circular faucet with a diameter 2.6 cm, as shown in Fig. 2. (a) In order to solve this problem, you may assume that the water is flowing down with the free-fall acceleration g9.8 m/s2. Please prove this assumption using Bernoulli equation. (b) If the flow rate (volume of the water which exits the faucet per unit time) is R 127 cm3/s, at what distance H below the tap its speed will be v2 1.2 m/s? (c) For which flow rate Rnew the speed of the water at distance H below the tap would be exactly twice compared to its initial speed right next to the tap? d 2.6 cm H-? 2 20.42m Figure 2: Schematic graph for Problem 2.

Answer 1

Alo to Bernoull's equation between 102 0 P + z = P + + Z₂ - 2.6cm Whene, P= Pressure V = Velocity Z datum height: on a datum point- & Pi = P2: Patm tv - (Palin- atmospheric + Z, Palm + V₂ to logo) taking , 2 Talin & pressure) eg mig to l'ó 2-0) va? = vil +2924-ti This is the equation of free fall where 2 is final Velocity W is initial velocity, Z, is displacement and acceleration 7 (B) Volume flow rate (R) = 127 canl - A l l i velocity at point (4) = 2 we where A, = Cross-section area att Vi = 127 A. Ad 2 = { x (26)? Vi = 23.93 cm/s [A, = 5:31 am = 24 m/s Applying bernaulle's equation or equalion No. 1 gh ( 2 = H) 1TM= 0.07053 m 2 2 V + (042) ² = (01 244 2x9.8 xH (1.2) ² = (0.24) "+ 2x 9.8 xH H = 7.053 cm

RNew - A, XY, NOW A - areal at I li New RNow New Velocily att given, Van New = 27, New According to equation-2. (V2, New) = (Vignew) ² + 29 H 4 (VisNew) ² - (\ , New ) ² = 2gH 3 VI, New = 2gH U RNew? = 29H A? Rne w 2 x 9.8 x 0. 0705 3 X 5.31) X10 Rnew - 3.6045 x10 m/s or Rnew = 360.45