Water ,at a constant flow rate,F,goes into a long fire hose of cross sectional a

Question

Water ,at a constant flow rate,F,goes into a long fire hose of cross sectional area,A1=7.50*10^3m^3 and exits through a nozzle of cross sectional area A2=1.96*10^3 m^3.The hose and nozzle are both at ground level(y1=y2=0) a)if it takes 12.0s to fill a 360L pail with water,What is the flow rate,F? b)What is the speed ,v,of the water exiting the nozzle? c)Calculate the change in pressure, dp=p2-p1, in the water,from when it enters the hose to when it exits the nozzle d)Suppose we take a cubic object (10.0cm on the side) and placed "unknown" fluid.At one instant in time,the pressure at the top of the candle is P top =5.25*10^4 pa and at the bottom of the candle is P bot =5.34*10^4 pa. What is the "density of the unknown fluid?

Explanation / Answer

a)
volume flow rate = volume filled/time

volume flow rate = 360 * 10^-3/12 m^3/s

volume flow rate = 0.03 m^3/s

b)

A1 = 7.5 *10^3 m^2

let the speed of flow is v1

v1 * A1 = volume flow rate

v1 * 7.5 *10^3 = 360 * 10^-3/12

v1 = 4 *10^-6 m/s

the speed of flow is 4 *10^-6 m/s

speed of water at the exit is v2

v1 * A1 = v2 * A2

4 *10^-6 * 7.5 *10^3 = 1.96 *10^3 * v2

v2 = 1.53 *10^-5 m/s

the water of exit velocity is 1.53 *10^-5 m/s

c)

change in pressure = -0.5 * p * (v1^2 - v2^2)

change in pressure = -0.5 * 1000 * ((4 *10^-6)^2 - (1.53 *10^-5)^2)

change in pressure = 1.09 *10^-7 Pa

d)

Ptop = 5.25 *10^4 Pa

Pbot = 5.34 *10^4 Pa

let the density of the object is p

p * 0.10^3 * g = (Pbot - Ptop) * 0.10^2

p * 0.10 * 9.8 = (-5.25 *10^4 + 5.34 *10^4)

p = 918.4 Kg/m^3

the density of object is 918.4 Kg/m^3

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