Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.4CP
To determine

The feasibility of installing such a weir which will be accurate and yet not cause the water to overflow to the side of the channel.

Expert Solution & Answer
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Answer to Problem 10.4CP

The first two of these are plausible although H/Y is large than 2 . The third result is not recommended because Y is too far below 9 cm. We can say that reasonable designs are possible.

Explanation of Solution

Figure:

Fluid Mechanics, Chapter 10, Problem 10.4CP

Given:

For asphalt take n=0.016, b=1.5m, g=9.81m/s2

So=0.0036, Depth y=0.7m

Concept Used:

Annuity problem requires the use of the formula as follows:

Q=1nARh2/3So1/2

Here

Area = A, Slope = So, Hydraulic radius =Rh

Annuity problem requires the use of the formula as follows:

Rh=byb+2y

Here,

Hydraulic radius =Rh, Depth =y, width =b

Annuity problem requires the use of the formula as follows:

A=by

Here,

Area= A, Depth =y, width =b

Annuity problem requires the use of the Weir formula as follows:

Qweir=0.581(b0.1H)g1/2H3/2

Here, width =b, acceleration gravity =g

Calculation:

As per the given problem

b=1.5m, y=0.7m

Annuity problem requires the use of this formula:

A=by

Substitute these values in the formula

A=1.5×0.7=1.05m2

A=1.05m2

Annuity problem requires the use of this formula:

b=1.5m, y=0.7m

Rh=byb+2y

Substitute these values in the formula:

Rh=1.5×0.71.5+2×0.7=0.362m

Rh=0.362m

Annuity problem requires the use of this formula

Rh=0.362m, A=1.05m2, So=0.0036, n=0.016

Q=1nARh2/3So1/2

Substitute these values in the formula

Q=10.016(1.05)(0.362)2/3(0.0036)1/2=2.0m3/s

Q=2.0m3/s

The weir discharge must equal this flow rate. Let us being by making b equal to the full available width of 1.5 m and holding Y to the minimum height of 9 cm.

Annuity problem requires the use of the Weir formula as follows

b=1.5m, g=9.81m/s2, Q=2.0m3/s

Qweir=0.581(b0.1H)g1/2H3/2

Substitute these values in the formula

2.0m3/s=0.581(1.5m0.1H)(9.81m/s)1/2H3/2

H=0.845m

H=0.845m

H is independent of Y ,

To find out the ratio H/Y

HY=0.8450.09=9.40

Which far exceeds Sturm’s recommendation H/Y 2.0. If we rise Y to H/2=0.423 m,

The total upstream water depth

H+Y=1.27m which overflow the channel walls?

If we back down to Y =0.35m,

The upstream depth

Y =0.35m, H=0.845m

H+Y=?

0.845+0.35=1.195m

So, a wide weir design is possible with HY=2.4 not bad

Similarly,

To let shorter values of b

* Upstream depth will exceed 1.2 m

* The ratio HY will exceed 2.0

Annuity problem requires the use of this formula:

b=1.4m, y=0.7m

A=by

Substitute these values in the formula:

A=1.4×0.7=0.98m2

A=0.98m2

Annuity problem requires the use of this formula:

b=1.4m, y=0.7m

Rh=byb+2y

Substitute these values in the formula:

Rh=1.4×0.71.4+2×0.7=0.35m

Rh=0.35m

Annuity problem requires the use of this formula:

Rh=0.35m, A=0.98m2, So=0.0036, n=0.016

Q=1nARh2/3So1/2

Substitute these values in the formula:

Q=10.016(0.98)(0.35)2/3(0.0036)1/2=1.837m3/s

Q=1.837m3/s

Annuity problem requires the use of the Weir formula as follows:

b=1.4m, g=9.81m/s2, Q=1.837m3/s

Qweir=0.581(b0.1H)g1/2H3/2

Substitute these values in the formula

1.8375m3/s=0.581(1.4m0.1H)(9.81m/s)1/2H3/2

H=0.89m

H=0.89m

H+Y1.2m if Y =0.31m

To find out the ratio H/Y

HY=0.890.31=2.9

Annuity problem requires the use of this formula:

b=1.25m, y=0.7m

A=by

Substitute these values in the formula:

A=1.25×0.7=0.875m2

A=0.875m2

Annuity problem requires the use of this formula:

b=1.25m, y=0.7m

Rh=byb+2y

Substitute these values in the formula:

Rh=1.25×0.71.25+2×0.7=0.33m

Rh=0.33m

Annuity problem requires the use of this formula:

Rh=0.33m, A=0.875m2, So=0.0036, n=0.016

Q=1nARh2/3So1/2

Substitute these values in the formula:

Q=10.016(0.875)(0.33)2/3(0.0036)1/2=1.575m3/s

Q=1.575m3/s

Annuity problem requires the use of the Weir formula as follows:

b=1.25m, g=9.81m/s2, Q=1.575m3/s

Qweir=0.581(b0.1H)g1/2H3/2

Substitute these values in the formula

1.575m3/s=0.581(1.4m0.1H)(9.81m/s)1/2H3/2

H=0.97m

H=0.97m

H+Y1.2m if Y =0.23m

To find out the ratio H/Y:

HY=0.970.23=4.2

Conclusion:

The first two of these are plausible although H/Y is large than 2.0.The third result is not recommended because Y is too far below 9 cm .We conclude that reasonable designs are possible.

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Chapter 10 Solutions

Fluid Mechanics

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