Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 3, Problem 3.127P
To determine

(a)

The mass flow rate.

Expert Solution
Check Mark

Answer to Problem 3.127P

The mass flow rate is 5.26kg/s.

Explanation of Solution

Given information:

The diameter at section (1) is 12cm . The diameter of the jet is 4cm . The pressure at the centerline of section (1) is 110kPa . The pressure at the exit of the jet is 101.325kPa.

The Figure-(1) shows the section (1) and the section (2) for the given system.

Fluid Mechanics, Chapter 3, Problem 3.127P

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

p1ρg+V122g+z1=p2ρg+V222g+z2( p 1 p 2ρg)+( V 1 22g V 2 22g)+(z1z2)=0 …… (I)

Here, the pressure at inlet is p1, the velocity at inlet is V1, the height of the inlet from the datum is z1, the pressure at outlet is p2, the velocity at outlet is V2, the height of the outlet from the datum is z2, the density of water is ρ, and the acceleration due to gravity is g.

Since the section (1) and the section (2) are at the same level, so the difference between the datum at the two sections is zero.

z1z2=0

Substitute 0 for z1z2 in Equation (I).

( p 1 p 2ρg)+( V 1 22g V 2 22g)+0=0( p 1 p 2ρg)+( V 1 22g V 2 22g)=0 …… (II)

Write the expression for the flow rate at section (1).

Q=A1V1 …… (III)

Here, the flow rate is Q, the area at section (1) is A1 and the velocity at section (1) is V1.

Write the expression for the area at section (1).

A1=π4d12

Here, the diameter at section (1) is d1.

Substitute π4d12 for A1 in Equation (III).

Q=π4d12V1

Write the expression for the flow rate at section (2).

Q=A2V2 …… (IV)

Here, the flow rate is Q, the jet area at section (2) is A2 and the velocity at section (2) is V2.

Write the expression for the area at section (2).

A2=π4d22

Here, the diameter at section (2) is d2.

Substitute π4d22 for A2 in Equation (IV).

Q=π4d22V2 …… (V)

Since the flow rate at both the sections is equal, it satisfies the continuity equation.

Substitute π4d12V1 for Q in Equation (V).

π4d12V1=π4d22V2d12V1=d22V2V1=d22d12V2

Substitute d22d12V2 for V1 in Equation (II).

( p 1 p 2ρg)+( ( d 2 2 d 1 2V2 ) 22g V 2 22g)=0( p 1 p 2ρg)+( d 2 4 d 1 4 V 2 22g V 2 22g)=0( p 1 p 2ρg)+V222g( d 2 4 d 1 41)=0 …… (VI)

Write the expression for mass flow rate.

m˙=ρQ …… (VII)

Calculation:

Substitute 110kPa for p1, 101.325kPa for p2, 12cm for d1, 4cm for d2, 9.81m/s2 for g and 998kg/m3 for ρ in Equation (VI).

{( 110kPa)( 101.325kPa)( 998 kg/ m3 )( 9.81m/ s2 )}+V222(9.81m/ s 2 )( (4cm ) 4 (12cm ) 41)=0{( 8.675kPa)( 1000Pa 1kPa )( 998 kg/ m3 )( 9.81m/ s2 )}+V22(0.05034s2/m)=0(0.88607388Nm/kg)=(0.05034s2/m)V22V2=( 0.88607388Nm/kg )( 0.05034s2/m )

Further solve the above expression.

V2=( 0.88607388Nm/kg )( 0.05034s2/m )=17.60m2/s2=4.195m/s

Substitute 4.195m/s for V2 and 4cm for d2 in Equation (V).

Q=π4(4cm)2(4.195m/s)=π4(16cm2)(1 m 210000 cm 2)(4.195m/s)=0.00527m3/s

Substitute 998kg/m3 for ρ and 0.00527m3/s for Q in Equation (VII).

m˙=(998kg/m3)(0.00527m3/s)=5.26kg/s

Conclusion:

The mass flow rate is 5.26kg/s.

To determine

(b)

The height of the fluid in stagnation tube.

Expert Solution
Check Mark

Answer to Problem 3.127P

The height of the fluid in stagnation tube is 0.917m.

Explanation of Solution

Given information:

The diameter at section (1) is 12cm . The diameter of the jet is 4cm . The pressure at the centerline of section (1) is 110kPa . The pressure at the exit of the jet is 101.325kPa.

Write the expression for the velocity of the jet.

V2=2g(H d 22) …… (VIII)

Here, the height of the water in stagnation tube is H.

Calculation:

Substitute 4cm for d2, 4.195m/s for V2 and 9.81m/s2 for g in Equation (VIII).

(4.195m/s)=2(9.81 m/s 2){H ( 4cm )2}(4.195m/s)=(19.62 m/s 2){H( 2cm)}

Square on both the sides in above expression.

(4.195m/s)2=[( 19.62m/s2 ){ H( 2cm )}]2(17.59m2/s2)=(19.62m/s2){H(2cm)( 1m 100cm)}H(0.02m)=0.8969mH=0.917m

Conclusion:

The height of the fluid in stagnation tube is 0.917m.

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Chapter 3 Solutions

Fluid Mechanics

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