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(a) What is the pressure drop (in N/m2) due to the Bernoulli effect as water goes...

(a) What is the pressure drop (in N/m2) due to the Bernoulli effect as water goes into a 3.20 cm diameter nozzle from a 9.10 cm diameter fire hose while carrying a flow of 42.0 L/s?

(b) To what maximum height (in m) above the nozzle can this water rise? (The actual height will be significantly smaller due to air resistance.)

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Answer #1

Ans: We know the Bernoulli equation

P1 + 1/2( V12) + gh1 = P2 + 1/2( V22) + gh2 -------------------------------- (1)

Where,

P1, P2 be the pressure at the fire hose and nozzle

is the density of fluid

V1 , V2 be the velocities of the fluid at fire hose and nozzle

when two points are at the same height (h1 = h2 )

P1 + 1/2( V12) = P2 + 1/2( V22) -------------------------------- (2)

Given:

Acceleration due to gravity (g) = 9.8 m/s2

Diameter of the fire hose (d1) = 9.10 cm = 0.0910 m.

Radius of the fire hose (r1) = 0.0455 m.

Diameter of the nozzle (d2) = 3.20 cm = 0.0320 m.

Radius of the fire nozzle (r2 ) = 0.016 m.

Density of fluid () = 1000 kg/m3.

Flow (Q) = 40 L/s

Q=A1V1

V1 = Q/ A1 = 40 x 10-3/ r12

= 40 x 10-3 / 3.14 x (0.0455)2

= 6.153 m/s

  Q=A2V2

V2 = Q/ A2 = 40 x 10-3/ r12

= 40 x 10-3 / 3.14 x (0.016)2

= 57.6 m/s

Therefor eq. (2) becomes

P1 + 1/2( V12) = P2 + 1/2( V22)

P2 -   P1 = 1/2( V22) - 1/2( V12)

  P = (1/ 2) (V22 - V12)

= (1/2) x 1000 ( 57.62 - 6.152)

= 1.639 x 106 N/m2

therefor the pressure drop is 1.639 x 106 N/m2

We have to calculate the max height of the fluid therefor eq. becomes

Patm + 1/2( V12) + gh1 = Patm + 1/2( V22) + gh2 ----------- (3)

Here, h1 is the height of the fluid that leaving

and h2 is the max height that fluid reaching

Eq (3) becomes

  

Patm + 1/2( V12) + 0 = Patm + 0 + gh2

h2 = {(1/2)V12 } / g

h2 = {0.5 x (57.6)2 }/ 9.8

h2 = 169.2 m

Therefor the max height that fluid raise to the above nozzle is 169.2 m

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