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Chapter 14, Problem 24P

(a)

To determine

The volume flow rate as a function of ΔP .

(a)

Expert Solution
Check Mark

Answer to Problem 24P

The volume flow rate as a function of ΔP is IV=3.933×106ΔPm3/s .

Explanation of Solution

Given information:

Density of liquid is 850kg/m3 , radius of the first horizontal tube is 1.0cm and the radius of second horizontal tube is 0.50cm .

The cross section area of first pipe is,

A1=πr12

A1 is the cross section area of first pipe.

r1 is the radius of the first pipe.

The cross section area of the second pipe is,

A2=πr22

A2 is the cross section area of second pipe.

r2 is the radius of the second pipe.

From the equation of continuity,

A1v1=A2v2

v1 is the speed of liquid in first pipe.

v2 is the speed of liquid in second pipe.

Substitute πr12 for A1 and πr22 for A2 in the above equation.

πr12v1=πr22v2v2=v1(r1r2)2

Substitute 1.0cm for r1 and 0.50cm for r2 in the above equation.

v2=v1(1.0cm0.50cm)2=4v1 (I)

Apply Bernoulli’s equation for two pipes,

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2

Since both the pipes are at same elevation then,

h1=h2=h

Therefore,

P1+12ρv12+ρgh=P2+12ρv22+ρghP1+12ρv12=P2+12ρv22P1P2=12ρ(v22v12)ΔP=12ρ(v22v12)

Substitute 4v1 for v2 in the above equation.

ΔP=12ρ((4v1)2v12)=152ρv12v1=(2ΔP15ρ)12 (II)

Formula to calculate the volume flow rate is,

IV=A1v1

IV is the volume flow rate.

Substitute πr12 for A1 and (2ΔP15ρ)12 for v1 in the above equation.

IV=πr12(2ΔP15ρ)12

Substitute 1.0cm for r1 and 850kg/m3 for ρ in the above equation.

IV=3.14×(1.0cm×102m1cm)2(2ΔP15×850kg/m3)12=3.933×106ΔPm3/s (III)

Conclusion:

Therefore, the volume flow rate as a function of ΔP is IV=3.933×106ΔPm3/s .

(b)

To determine

The volume flow rate for ΔP=6.0kPa .

(b)

Expert Solution
Check Mark

Answer to Problem 24P

The volume flow rate for ΔP=6.0kPa is 3.046×104m3/s .

Explanation of Solution

Given information:

Density of liquid is 850kg/m3 , radius of the first horizontal tube is 1.0cm and the radius of second horizontal tube is 0.50cm .

Substitute 6.0kPa for ΔP in equation (III).

IV=3.933×106(6.0kPa×103Pa1kPa)m3/s=3.046×104m3/s

Conclusion:

Therefore, the volume flow rate for ΔP=6.0kPa is 3.046×104m3/s .

(c)

To determine

The volume flow rate for ΔP=12.0kPa .

(c)

Expert Solution
Check Mark

Answer to Problem 24P

The volume flow rate for ΔP=12.0kPa is 4.31×104m3/s .

Explanation of Solution

Given information:

Density of liquid is 850kg/m3 , radius of the first horizontal tube is 1.0cm and the radius of second horizontal tube is 0.50cm .

Substitute 12.0kPa for ΔP in equation (III).

IV=3.933×106(12.0kPa×103Pa1kPa)m3/s=4.31×104m3/s

Conclusion:

Therefore, the volume flow rate for ΔP=12.0kPa is 4.31×104m3/s .

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Chapter 14 Solutions

Bundle: Physics For Scientists And Engineers With Modern Physics, Loose-leaf Version, 10th + Webassign Printed Access Card For Serway/jewett's Physics For Scientists And Engineers, 10th, Single-term

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