Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.51P
<
To determine

(a)

If the flow of the unfinished concrete duct is critical.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

The flow is super critical.

Explanation of Solution

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Froude number is given as:

Fr=Vgy

(1)

Where,V=flow velocityg=specific densityy=hydraulic depth

Velocity of the flow

V=QA

(2)

A=areaQ=discharge

Calculation:

From equation (2)

V=QA

=QA=Qπ4(1.5)283.144(1.5)2=83.144(2.25)=4.527m/s Angle is calculated for the circular channel which is partially filled:

θd=900250650

θr=65*π1801.13rad

[units to radians]

Area of the flow is given as:

A=(D2)2(θrsinθdcosθd)

Substituting the values:

=(1.52)2(1.13sin65cos65)=(2.252)2(1.13sin65cos65)=1.2656(1.13sin65cos65)=0.420m2

Hydraulic depth is given as:

y=Awidth at top

=A2(D2)sinθ0.4202(1.52)sin65=0.3m

Substituting all equations (1):

Fr=Vgy

=Vgy=4.5279.81*0.32.685

The flow is determined to be super critical as the Froude’s number is more than 1.

Conclusion:

Thus, the flow of the unfinished concrete duct is determined.

<
To determine

(b)

The critical flow rates.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Critical flow rate is Qc=3.07m3/sec

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Concept Used:

Qc=αARb2/3*Sc1/2n

(1)

where,A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

Substituting we have,

8=1.76×1.486×0.25×1.5×Sc1/20.014Sc1/2=8×0.0141.76×1.486×0.25×1.5Sc1/2=8×0.0140.980760.1120.98076

Sc=0.001

From equation (1) we have,

=1.486×1.76×(0.25×1.5)2/3×0.0011/20.014=2.61536×(0.375)2/3×0.0011/20.014=3.07m3/sec

Conclusion:

Thus, the critical flow rate is 3.07m3/sec.

<
To determine

(c)

The critical slope.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Critical slope is Sc=0.0019

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s

Concept Used:

Qc=αARb2/3*Sc1/2n

(1)

where,A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

Substituting we have:

nQcαARb2/3=Sc1/2Sc1/2=3.07×0.0140.98076Sc=0.0019

Conclusion:

Thus, the critical slope is determined.

<
To determine

(d)

The Froude’s number.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Froude’s number is Fr=1.027

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s.

Concept Used:

Froude number is given as:

Fr=Vgy

(1)

Where,V=flow velocityg=specific densityy=hydraulic depth

Velocity of the flow

V=QA

(2)

A=areaQ=discharge

Calculation:

Substituting all equation (1)

Fr=Vgy

=VgyQAgy=3.071.767×19.81×0.31.0127

The flow is determined to be critical.

Conclusion:

Thus, the Froude number is determined.

<
To determine

(e)

The slope of the duct when the flow is uniform.

<
Expert Solution
Check Mark

Answer to Problem 10.51P

Sc=0.2698

Explanation of Solution

Given Information:

The concrete duct which is unfinished has a diameter of 1.5m that flows at 8.0m3/s .

Concept Used:

nQcαARb2/3=Sc1/2where:A=areaSc=Critical slopeR=hydraulic channelα=conversion factornn=mannings factor 

Calculation:

The slope is equal in the duct with the uniform flow. So the duct slope with the critical flow is given as:

Sc=0.2698

Conclusion:

Thus, the slope of the duct when the flow is uniform is determined.

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Chapter 10 Solutions

Fluid Mechanics

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