Chapter 14b Dynamic Fl... x Water moves through a c... X C Water Moves Through A

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Chapter 14b Dynamic Fl... x Water moves through a c... X C Water Moves Through A X iD b/Si Responses/last?dep- 13128388 OG Google f book M Gmail Trolino Funny pict Twitter Y azoomer2 StallTactic... B Blackboard Learn wA WebAssig Spotify Web Player Ebay YouTube Broadcast D Pinterest NHL. e TextN Assignment Submission For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Assignment Scoring Your last submission is used for your score. +0 -3.33 point SerPSE9 14. P.042. My Notes Ask Your Teacher Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P1 3 1.90 x 104 Pa, and the pipe diameter is 5.0 cm. At another point y 0.40 m high the pressure is P2 3 1.25 x 104 Pa and the pipe diameter is 2.50 cm. (a) Find the speed of flow in the lower section. (b) Find the speed of flow in the upper section. (c) Find the volume flow rate through the pipe. Submit Answer Save Progress 2. -3.33 point SerPSE9 14. P.044. My Notes Ask Your Teacher A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.40 cm. The hose ends with a nozzle of diameter 2.10 cm. A rubber stopper is inserted into the nozzle. The water level in the tank is kept 7.50 m above the nozzle. (a) Calculate the friction force exerted by the nozzle on the stopper. (b) The stopper is removed. What mass of water flows from the nozzle in 2.00 h

Explanation / Answer

let,

pressure, P1=1.9*10^4 Pa, diameter d1=5cm

pressure, P2=1.25*10^4 Pa, diameter d2=2.5cm

hieght, y=0.4 m

===>

a)

from the Bernoulli's equation,

P+(1/2*rho*v^2)+(rho*g*h)=constant

=====>

P1+(1/2*rho*v1^2)+rho*g*h1=P2+(1/2*rho*v2^2)+rho*g*h2

P1-P2=1/2*rho*(v2^2-v1^2)+rho*g*(h2-h1)

(1.9*10^4) - (1.25*10^4) =1/2*1000*(v2^2-v1^2)+1000*9.8*(0.4) ---(1)

and

A1*v1=A2*v2

===> d1^2*v1=d2^2*v2

===> (v2/v1)=(d1/d2)^2

(v2/v1)=(5/2.5)^2

v2=4*v1 ----(2)

from (1) and (2),

(1.9*10^4) - (1.25*10^4) =1/2*1000*((4*v1)^2-v1^2)+1000*9.8*(0.4)

===> v1=0.586 m/sec

speed of flow in the lower section, v1=0.586 m/sec

b)

now,

v2=4*v1

v2=4*(0.586)

v2=2.344 m/sec

speed of flow in the lower section, v2=2.344 m/sec

c)

volume flow rate =A1*v1

=(pi*r1^2)*v1

=(pi*d1^2/4)*v1

=(pi*0.05^2/4)*0.586

=1.15*10^-3 m^3/sec

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