Water, with a density of 1000 kg/m^3, flows out of a spigot, through a hose, and

Question

Water, with a density of 1000 kg/m^3, flows out of a spigot, through a hose, and out a nozzle into the air. The hose has an inner diameter of 2.25 cm. The opening in the nozzle that the water comes out of has a diameter of 2.00 mm. The water coming out of the nozzle, which is held at a height of 7.25 meters above the height of the spigot, has a velocity of 11.2 m/s. Neglecting viscosity and assuming that the water flow is laminar (not necessarily good assumptions, but let's not make this any harder than it already is), what is the pressure of the water in the hose right after it comes out of the spigot where the water enters the hose (to three significant digits)? Assume that g=9.80m/s^2 and that the surrounding air is at a pressure of 1.013×10^5N/m^2.

Answer Choices

A. 1.33×10^5 N/m^2

B. 1.01×10^5 N/m^2

C. 1.72×10^5 N/m^2

D. 2.29×10^5 N/m^2

E. 2.35×10^5 N/m^2

F. 1.78×10^5 N/m^2

G. None of the above

Explanation / Answer

Given

data

P1 =?, P2 = 1.013*10^5 Pa

rho = 1000 kg/m3

h1=0 m, h2 = 7.25 m

v1= ?

v2 = 11.2 m/s

d1 = 2.25 cm ==>r1 =1.125*10^-2 m

d2 = 2 mm ===> r2 = 1*10^-3 m

area A1 = pi*r1^2

A2 = pi*r2^2

From continuity equation

A1v1 = A2v2

v1 = A2v2/A1

= pi*(1*10^-3)^2*11.2 /(pi*1.125*10^-2)^2

= 0.02816846 m/s

From Bernouli's equation

P1 + 0.5*rho*v1^2+rho*gh1 = P2 + 0.5*rho*v2^2+rho*gh2

substituting the values

P1 + 0.5*1000*0.02816846 ^2+1000*9.8*0 = 1.013*10^5 + 0.5*1000*11.2^2+1000*9.8*7.25

P1 = 235069.60 N/m2 = 2.35*10^5 N/m2 <<<<<<<----------Answer

option E

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