Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.111P
To determine

(a)

The water depth one mile upstream for initial depth of 7.7ft.

Expert Solution
Check Mark

Answer to Problem 10.111P

The water depth one mile upstream of the dam is 7.325ft.

Explanation of Solution

Given information:

The average flow rate is 900ft3/s, upstream width is 150ft, slope is 10ft/mile and water depth just upstream of the dam is 7.7ft.

Write the expression for the flow:

Fr=Vgy     …… (I)

Here, Froude number is Fr, velocity of flow is V, acceleration due to gravity is g and water depth is y.

Write the expression for flow rate in open channel.

q˙=w×y×V     …… (II)

Here, flow rate in open channel is q˙, width of channel is w and velocity of flow in open channel is V.

Write the expression for width upstream.

wm=w+m×s     …… (III)

Here, width smile upstream is wm, width at flash board is w, slope of width is m and distance from flashboard is s.

Calculation:

Substitute 900ft3/s for q˙, 150ft for w, and 7.7ft for y .in Equation (II)

900ft3/s=150ft×7.7ft×V900ft3/s=1155ft2×V

V=900 ft3/s1155ft2=0.78ft/s

Substitute 0.78ft/s for V, 32.2ft/s2 for g and 7.7ft for y in Equation (I).

Fr=0.78ft/s( 32.2 ft/ s 2 )×( 7.7ft)=0.78ft/s247.94 ft 2 / s 2 =0.05

Substitute 150ft for w, 10ft/mile for m and 1mile for s.

wm=150ft+(10ft/mile)×(1mile)=150ft+10ft=160ft

Substitute 900ft3/s for q˙, 160ft for w, V for velocity and y for depth in Equation (II).

900ft3/s=160ft×y×VV=900 ft3/s160ft×y

Substitute 900ft3/s160ft×y for V, 32.2ft/s2 for g, 0.05 for Fr and y for depth in Equation (I).

0.05=900 ft 3 /s160ft×y32.2 ft/ s 2 ×y=0.9912ft3/2y3/2

y3/2=0.9912ft3/20.05

y=( 0.9912 ft 3/2 0.05)2/3=(19.824 ft 3/2 )2/3=7.325ft

Conclusion:

The water depth one mile upstream of the dam is calculated by the flow expression: Fr=Vgy.

To determine

(b)

The water depth one mile upstream for initial depth of 10.7ft.

Expert Solution
Check Mark

Answer to Problem 10.111P

The water depth one mile upstream of the dam is 10.3ft.

Explanation of Solution

Given information:

The average flow rate is 900ft3/s, upstream width is 150ft, slope is 10ft/mile and water depth just upstream of the dam is 10.7ft.

Write the expression for the

Fr=Vgy     …… (I)

Here, Froude number is Fr, velocity of flow is V, acceleration due to gravity is g and water depth is y.

Write the expression for flow rate in open channel.

q˙=w×y×V     …… (II)

Here, flow rate in open channel is q˙, width of channel is w and velocity of flow in open channel is V.

Write the expression for width upstream.

wm=w+m×s     …… (III)

Here, width smile upstream is wm, width at flash board is w, slope of width is m and distance from flashboard is s.

Calculation:

Substitute 900ft3/s for q˙, 150ft for w, and 10.7ft for y .in Equation (II)

900ft3/s=150ft×10.7ft×V900ft3/s=1605ft2×V

V=900 ft3/s1605ft2=0.56ft/s

Substitute 0.78ft/s for V, 32.2ft/s2 for g and 7.7ft for y in Equation (I).

Fr=0.56ft/s( 32.2 ft/ s 2 )×( 10.7ft)=0.56ft/s344.54 ft 2 / s 2 =0.03

Substitute 150ft for w, 10ft/mile for m and 1mile for s.

wm=150ft+(10ft/mile)×(1mile)=150ft+10ft=160ft

Substitute 900ft3/s for q˙, 160ft for w, V for velocity and y for depth in Equation (II).

900ft3/s=160ft×y×VV=900 ft3/s160ft×y

Substitute 900ft3/s160ft×y for V, 32.2ft/s2 for g, 0.05 for Fr and y for depth in Equation (I).

0.03=900 ft 3 /s160ft×y32.2 ft/ s 2 ×y=0.9912ft3/2y3/2

y3/2=0.9912ft3/20.03

y=( 0.9912 ft 3/2 0.03)2/3=(33.04 ft 3/2 )2/3=10.3ft

Conclusion:

The water depth one mile upstream of the dam is calculated by flow expression in open channel : q˙=w×y×V.

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Chapter 10 Solutions

Fluid Mechanics

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