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 g = 9.8 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? 3D 2.6 ст d Н-? v2= 0.42 m Figure 2: Schematic graph for Problem 2.

Answer 1

(a ) From Bernoulli's principle pot sv, 't S 9H = Po+ Sve V, V +29H i thow must be downward (6) R = Td yu 127x166 = x 6-76X164 X V Vi = 0.24 mis " =v²+ 2gH (1.2) =66.24)+2x918xH rH=7o1cm (c) V2 = 2 vi Pot & guir +39 H = Pot 13401 Igt=2 / suit 29,8X0:07 - 3 Vi = 0.68 m/s Rnew = Td 80.68 = A 6.76 x 109 x0.68 Rnew = 361 cm ls