Question

graph lor Problem 1. Problem 2 (3pts.) 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 cm /s, at what distance H below the tap its speed will be ty = 1.2 m/s? (c) For which flow rate Rrew 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-? 02 V2 = 0.42 Figure 2: Schematic graph for Problem 2.

Answer 1

(a) Inside tube , water flow is under gauge pressure P_g . After exit of faucet, water is under atmospheric pressure only.

By bernouli's eqn :- P_g + (1/2)$\rho$u_o² = P_a +(1/2)$\rho$u²

Left hand side is inside pipe line , u_o is velocity of water flow inside pipe line, P_a is atmospheric pressure and u is exit velocity of water flow.

Hence after exit, we assume water is flowing down with free-fall acceleration g = 9.8 m/s²

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(b) exit velocity = flowrate / cross section area = ( 127 $\times$ 10^-6 ) / ( 0.25 $\times$ $\pi$$\times$ 2.6 $\times$ 2.6 $\times$ 10^-4 ) = 0.24 m/s

If H is the distance at which we get speed 1.2 m/s , then we use the formula " v² = u² + 2gH " to get the distance.

v is final speed 1.2 m/s and u is initial speed = 0.24 m/s

H = ( 1.2 $\times$ 1.2 - 0.24 $\times$ 0.24 ) / ( 2 $\times$ 9.8 ) = 0.07 m = 7 cm

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(c) if we need speed of water flow 2u m/s at H = 7 cm, where u m/s is the speed at faucet exit,

then we have , H = 0.07 = ( 4 u² - u² ) / (2g) = 3u² / (2g)

Hence, we get, u = [ (2 $\times$ 9.8 $\times$ 0.07 ) / 3 ]^1/2 = 0.676 m/s

flow rate = speed $\times$ cross-section area = 0.676 $\times$ (0.25 $\times$ $\pi$$\times$ 2.6 $\times$ 2.6 $\times$ 10^-4 ) = 3.59 $\times$ 10^-4 m³ /s

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