Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
Question
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Chapter 10, Problem 10.93P
To determine

i.

The velocity V1 for the given flow.

Expert Solution
Check Mark

Answer to Problem 10.93P

The velocity V1for the given flow is 1.5 m/s 

Explanation of Solution

Given:

y1 = 1 m

y3 = 0.4 m

Concept Used:

E1=y1+V122gForuniformflow,q=y1V1=y2V2

Calculation:

E1=E3y1+V122g=y3+V322g1+V122*9.81=0.4+V322*9.81........(1)Foruniformflow,q=y1V1=y3V31*V1=0.4*V3.....(2)Onsolving(1)and(2),wegetV1=1.5m/sV3=3.74m/s

Conclusion:

The velocity V1 for the given flow is 1.5 m/s.

To determine

ii.

The velocity V3 for the given flow.

Expert Solution
Check Mark

Answer to Problem 10.93P

The velocity V1for the given flow is 3.74 m/s

Explanation of Solution

Given:

y1 = 1 m

y3 = 0.4 m

Concept Used:

E1=y1+V122gForuniformflow,q=y1V1=y2V2

Calculation:

E1=E3y1+V122g=y3+V322g1+V122*9.81=0.4+V322*9.81........(1)Foruniformflow,q=y1V1=y3V31*V1=0.4*V3.....(2)Onsolving(1)and(2),wegetV1=1.5m/sV3=3.74m/s

Conclusion:

The velocity V1 for the given flow is 3.74 m/s.

To determine

iii.

The depth y4 for the given flow.

Expert Solution
Check Mark

Answer to Problem 10.93P

The depth y4for the given flow is 0.89 m

Explanation of Solution

Given:

y1 = 1 m

y3 = 0.4 m

Concept Used:

Fr=VVc 2 y 4 y 3 =1+ (1+8F r 3 2 ) 1/2

Calculation:

Fr3=V3V3cFr2=3.74g* y 23.749.81*0.4Fr2=1.89Theflowissupercritical 2 y 4 y 3 =1+ (1+8F r 3 2 ) 1/2 2 y 4 0.4 =1+ (1+8* 1.89 2 ) 1/2 y4=0.89m.

Conclusion:

The depth y4 for the given flow is 0.89 m.

To determine

iv.

The bump height h for the given flow.

Expert Solution
Check Mark

Answer to Problem 10.93P

The height of the bump is 0.2m 

Explanation of Solution

Given:

y1 = 1 m

y3 = 0.4 m

Concept Used:

Fr=VVcE1=y1+V122gForuniformflow,q=y1V1=y2V2

Calculation:

Fr2=V2V2cForthebumpflow,theflowmustbecrititcalFr2=1V2=V2c*Fr2V2=gy2*1V2=9.81*y2.....(3)Foruniformflow,q=y1V1=y2V21*1.5m/s=y2*V2Substituteequation(3)9.81*y23/2=1.5y2=0.612mV2=9.81*y2V2=9.81*0.612V2=2.45m/sE1=E2y1+V122g=y2+V222g+h1+1.522*9.81=0.612+2.4522*9.81+hh=0.2m

Conclusion:

The height of the bump is 0.2m .

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Chapter 10 Solutions

Fluid Mechanics

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