Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.114P
To determine

(a)

The velocity at section (1).

The velocity at section (2).

Expert Solution
Check Mark

Answer to Problem 3.114P

The velocity at section (1) is 0.896m/s.

The velocity at section (2) is 9.96m/s.

Explanation of Solution

Given information:

The force required to hold the plate steady is 70N, diameter of the nozzle at section (1) is 10cm and diameter of nozzle at section (2) is 3cm.

Concept Used:

The below figure shows the two sections of the nozzle as section (1) and section (2).

Fluid Mechanics, Chapter 3, Problem 3.114P

Figure (1)

Write the expression for the force with which the jet strikes a plate.

F=ρWA2V22V22=FρWA2V2=F ρ W A 2 .... (I)

Here, the jet force on the plate is F, density of water is ρW, cross section area at point (2) is A2, and the velocity of jet at section (2) is V2.

Write the expression for cross section area in terms of nozzle diameter for section (1).

A1=π4d12 .... (II)

Write the expression for cross section area in terms of nozzle diameter for section (2).

A2=π4d22 .... (III)

Here, the diameter of the nozzle at section (1) is d1, area of the nozzle at section (1) is A1, and the diameter of the nozzle at section (2) is d2.

Write the expression for momentum equation at section (1) and section (2).

A1V1=A2V2V1=A2V2A1

.... (IV)

Here, the cross-section area of the nozzle at section (1) is A1, and velocity of water at section (1) is V1.

Calculation:

Substitute 10cm for d1 in Equation (II).

A1=π4×(10cm)2=( 3.1415×100 cm 2 4)( 1 m 2 10000 cm 2 )=314.15m240000=7.853×103m2

Substitute 3cm for d2 in Equation (III).

A2=π4×(3cm)2=( 3.1415×9 cm 2 4)( 1 m 2 10000 cm 2 )=28.2735m240000=7.068×104m2

Substitute 7.068×104m2 for A2, 70N for F and 998kg/m3 for ρW in Equation (I).

V2= 70N ( 998 kg/ m 3 )( 7.068× 10 4 m 2 )=( 70N 0.7053 kg/m )( 1 kgm/ s 2 1N )=99.25 m 2/ s 29.96m/s

Thus, the velocity at section (2) is 9.96m/s.

Substitute 7.068×104m2 for A2, 7.853×103m2 for A1 and 9.96m/s for V2 Equation (III).

V1=( 7.068× 10 4 m 2 )( 9.96m/s )7.853× 10 3m2=70.4 m 3/s78.53m2=0.896m/s

Conclusion:

Thus, the velocity at section (1) is 0.896m/s.

Thus, the velocity at section (2) is 9.96m/s.

To determine

(b)

The mercury manometer reading.

Expert Solution
Check Mark

Answer to Problem 3.114P

The mercury manometer reading is 0.398m.

Explanation of Solution

Concept Used:

Write the expression for Bernoulli equation at section (1) and section (2).

P1+ρWV122=P2+ρWV222P1P2=ρWV222ρWV122P1P2=ρW2(V22V12) .... (IV)

Here, pressure at section (1) is P1, and the pressure at section (2) is P2.

Write the expression for continuity equation.

P1+ρWgh1=P2+ρWgh2+ρHgg(h1h2)(P1P2)+ρWg(h1h2)=ρHgg(h1h2)(P1P2)=g(h1h2)(ρ HgρW)(h1h2)=( P 1 P 2 )( ρ Hg ρ W )g .... (V)

Here, height of lower mercury level from the axis of the nozzle is h1, the height of the upper mercury level from the axis of the nozzle is h2, density of water is ρW and density of mercury is ρHg.

Calculation:

Substitute 998kg/m3 for ρW, 9.96m/s for V2 and 0.896m/s for V1 in Equation (IV).

P1P2=998kg/ m 32×[( 9.96m/s )2( 0.896m/s )2]=499kg/m3×98.398m2/s249101Pa

Substitute 49101Pa for P1P2, 998kg/m3 for ρW and 998kg/m3 for ρHg in Equation(V).

h1h2=49101Pa( 13550 kg/ m 3 998 kg/ m 3 )×9.81m/ s 2=( 49101Pa 123135 kg/ m 2 s 2 )( 1N/ m 2 1Pa)( 1 kgm/ s 2 1N)=0.398m

Conclusion:

Thus, the mercury manometer reading is 0.398m.

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Chapter 3 Solutions

Fluid Mechanics

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