PLAIN AND CIVIL: April 2019

Monday 22 April 2019

6.6. MINOR LOSSES

Considering friction loss as the major loss of head in pipes, there are also some minor losses present. These minor losses affect the velocity of the flow, thus they become deviations of the velocity head. They come from various conditions:

Contraction
Enlargement
Gate
Bend
Change in direction
Obstructions

6.6.1. LOSS OF HEAD DUE TO CONTRACTION

Sudden contraction of the pipe size can cause head loss, in which it is equated as below:

   $h_{c}=K_{c}\frac{V^{2}}{2g}$

where:   $h_{c}$ = head loss due to contraction
   $K_{c}$ = empirical coefficient of contraction
V = velocity in the smaller pipe

Values of the coefficient Kc for sudden contraction (English units) are tabulated as follows. Interpolation is required for exact values.

VELOCITY IN SMALLER PIPE, V			RATIO OF	SMALLER	TO	LARGER	DIAMETER
	0.0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9
2	0.49	0.49	0.48	0.45	0.42	0.38	0.28	0.18	0.07	0.03
5	0.48	0.48	0.47	0.44	0.41	0.37	0.28	0.18	0.09	0.04
10	0.47	0.46	0.45	0.43	0.40	0.36	0.28	0.18	0.10	0.04
20	0.44	0.43	0.42	0.40	0.37	0.33	0.27	0.19	0.11	0.05
40	0.38	0.36	0.35	0.33	0.31	0.29	0.25	0.20	0.13	0.06

*Values of the coefficient for sudden contractions will be much smaller.

A different case for sudden contraction is used for pipe entrances. The formula for the loss of head at pipe entrance is as follows:

$h_{c}=\left ( \frac{1}{C_{v}^{2}}-1 \right )\frac{V^{2}}{2g}$

where: Cv = coefficient of velocity

ENTRANCE TO PIPE	Cv

Inward projecting	0.75
Square cornered	0.82
Slightly rounded	0.90
Bell mouth	0.95

EXAMPLE 6.6.1.1. EXAMPLE OF HEAD LOSS DUE TO SUDDEN CONTRACTION

EXAMPLE 6.6.1.2. EXAMPLE OF HEAD LOSS DUE TO SUDDEN CONTRACTION

6.6.2. LOSS OF HEAD DUE TO ENLARGEMENT

The enlargement of the pipe's cross section would cause the reduction in velocity. The loss of head due to sudden enlargement is given in the formula:

$h_{e}=K_{e}\frac{V^{2}}{2g}$

where: velocity head is taken from the smaller pipe;
Ke is identified by Archer as:

$K_{e}=\frac{1.10}{V^{0.081}}\left [ 1-\left ( \frac{D_{S}}{D_{L}} \right )^{2} \right ]^{1.92}$

The case of sudden enlargement from pipe to a reservoir is a special case in which the ratio of diameters is taken as zero.

EXAMPLE 6.6.2.1. EXAMPLE OF HEAD LOSS DUE TO SUDDEN ENLARGEMENT
EXAMPLE 6.6.2.2. EXAMPLE OF HEAD LOSS DUE TO SUDDEN ENLARGEMENT

6.6.3. LOSS OF HEAD DUE TO OBSTRUCTIONS

6.6.3.1. GATES OR VALVES PARTIALLY CLOSED

Loss of head can also happen due to partially closed gates or valves. Head loss due to obstruction follows similar formula with the velocity head as the others:

$h_{g}=K_{g}\frac{V^{2}}{2g}$

where: V = mean velocity in the pipe

Diameter of valve, inches	Ratio of	opening	to	diameter	, d/D
	1/8	1/4	3/8	1/2	3/4	1

1/2	450	60	22	11	2.2	1
3/4	310	40	12	5.4	1.1	0.29
1	230	32	9	4.1	0.9	0.23
1-1/2	170	23	7.2	3.3	0.75	0.18
2	140	20	6.5	3	0.68	0.16
4	92	16	5.5	2.6	0.55	0.14
6	73	14	5.3	2.4	0.49	0.12
8	66	13	5.2	2.3	0.46	0.1
12	56	12	5.1	2.2	0.42	0.07

6.6.3.2. OBSTRUCTION IN A PIPE

For an obstruction inside the pipe, the head loss will be as follows:

$h_{g}=\left [ \frac{A}{C_{c\left ( A-a \right )}}-1 \right ]\frac{V^{2}}{2g}$

where: A = area of pipe
a = largest area of obstruction
Cc = coefficient of contraction
V = velocity of flow in pipe

6.6.4. LOSS OF HEAD DUE TO BENDS

6.5. OTHER PIPE FORMULAS

Other formulas apart from Darcy Weisbach's have been used in computation of head loss in flowing fluids. These formulas, based from experiments, are intended for water only at a specific temperature range - less than about $75^{o}F$ .

With ordinary engineering hydraulic problems involving water as fluid, the choice of formula is a matter of personal preference which would be based on simplicity and convenience.

6.5.1. MANNING'S FORMULA

This formula is more used on open channels but still quite common in pipes.

$V=\frac{1.486}{n}R^{\frac{2}{3}}S^{\frac{1}{2}}$

where: V = velocity of fluid flow
n = roughness coefficient
R = hydraulic radius
$R=\frac{A}{P}$
S = slope of the energy gradient
$S=\frac{h_{f}}{L}$

Note: The roughness coefficient increases with the conduit's degree of roughness.

6.5.2. ROUGHNESS COEFFICIENT OF MANNING'S FORMULA

KIND OF PIPE	VARIATION		USE IN DESIGNING
	FROM	TO	FROM	TO
Brass and glass pipe	0.009	0.013	0.009	0.011
Asbestos-cement pipe	--	--	0.010	0.012
Wrought-iron and welded-steel pipe	0.010	0.014	0.011	0.013
Wood-stave pipe	0.010	0.014	0.011	0.013
Clean cast-iron pipe	0.010	0.015	0.011	0.013
Concrete pipe	0.010	0.017	--	--
very smooth	--	--	0.011	0.012
"wet mix," steel forms	--	--	0.012	0.014
"dry mix'" rough forms	--	--	0.015	0.016
with rough joints	--	--	0.016	0.017
Common-clay drainage tile	0.011	0.017	0.012	0.014
Vitrified sewer pipe	0.010	0.017	0.013	0.015
Riveted-steel pipe	0.013	0.017	0.015	0.017
Dirty or tuberculated cast-iron pipe	0.015	0.035	--	--
Corrugated-iron pipe	--	--	0.020	0.022

6.5.3. THE HAZEN-WILLIAMS FORMULA

This formula is the one adopted in United States for the design of water supply systems. Although it can also be used for pipes and open channels, this formula is more commonly adopted for pipes.

$V=1.318C_{1}R^{0.63}S^{0.54}$

where: V = velocity of fluid flow
C1 = roughness coefficient
R = hydraulic radius
$R=\frac{A}{P}$
S = slope of the energy gradient
$S=\frac{h_{f}}{L}$

6.5.4. VALUES FOR $\mathbf{C_{1}}$

DESCRIPTION OF PIPE	C1

Extremely smooth and straight	140
Very smooth	130
Smooth wooden or wood stave	120
New riveted steel	110
Vitrified	110

After a series of years in use of the pipe, the controlling factor somehow varies:

PIPES AFTER YEARS OF USE	C1

Cast-iron pipe	100
Riveted Steel	95
Old iron pipes in bad condition	80 to 60
Small pipes badly tuberculated	40
Asbestos-cement pipe	140

6.5.5. PIPE DIAGRAMS

Pipe diagrams are also used to solve problems on flow. For both the diagrams below, the procedure is to plot the given values and check on the intersection to find another parameter. Additionally, both graphs below are just representations of a specific value of the roughness coefficient. For example, the manning diagram is only for n=0.013. There are also different diagrams for different roughness coefficients. Similarly, Hazen-William's nomograph is for C1=130 only. Another set of nomograph is used for another C1.

A. THE MANNING DIAGRAM

B. HAZEN-WILLIAM'S NOMOGRAPH

EXAMPLE 6.4.4. HYDRAULIC RADIUS FOR PIPES FLOWING MORE THAN HALF FULL

Calculate the hydraulic radius for water flowing 3.4 ft deep in a 48- inch diameter storm sewer.

   $\theta =2cos^{-1}\left ( \frac{r-h}{r} \right )$

The angle formula remains the same, but this time, h is the available height of air, not water. The value for h is:

$h=4ft-3.4ft =0.60ft$

So the angle can be computed as [calculator in radians mode]:

   $\theta =2cos^{-1}\left ( \frac{2-0.60}{2} \right )=1.5908$

So the area and wetted perimeter can be computed as follows:

$A=\pi r^{2}-\frac{r^{2}\left ( \theta -sin\theta \right )}{2}$

   $A=\pi 2^{2}-\frac{2^{2}\left ( 1.5908 -sin\left (1.5908 \right ) \right )}{2}=11.3844ft^{2}$

$P=2\pi r-r\theta$

$P=2\pi \left ( 2 \right )-2\left ( 1.5908 \right )=9.3848ft$

The hydraulic radius is then computed as:

   $R=\frac{A}{P}=\frac{11.3844ft^{2}}{9.3848ft}$

${\color{Red} \mathbf{R=1.2131ft}}$

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

EXAMPLE 6.4.3. HYDRAULIC RADIUS FOR PIPES FLOWING LESS THAN HALF FULL

Calculate the hydraulic radius (m) for water flowing 20 mm deep in a pipe of 100 mm diameter.

Solve for the angle $\theta$ : [setting your calculator at Radians mode]

$\theta =2cos^{-1}\left ( \frac{r-h}{r} \right )$

$\theta =2cos^{-1}\left ( \frac{50mm-20mm}{50mm} \right )=1.8546$

Then proceed with solving the area and wetted perimeter:

$A=\frac{r^{2}\left ( \theta -sin\theta \right )}{2}$

$A=\frac{0.050^{2}\left ( 1.8546 -sin1.8546 \right )}{2}=1.1183x10^{-3}m^{2}$

And,

$P=r\theta =0.05\left ( 1.8546 \right )=0.09273m$

So the hydraulic radius becomes:

$R=\frac{A}{P}=\frac{1.1183x10^{-3}}{0.09273}$

${\color{Red} \mathbf{R=0.01206m}}$

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

EXAMPLE 6.4.2. HYDRAULIC RADIUS FOR PIPES FLOWING LESS THAN HALF FULL

Calculate the hydraulic radius (ft) for water flowing 6 inches deep in a 48-inch diameter storm sewer.

The diameter is 48 inches, so radius = 24 inches or 2 ft.

h= 6 inches or 0.5ft.

With only 6inches depth of water flowing in the 48-inch diameter pipe, consider the pipe less than half full. Thus hydraulic radius formula becomes:

The formulas for area and perimeter are in function of theta which is in radians.

$\theta =2\left [cos^{-1}\left ( \frac{2-0.5}{2} \right ) \right ]\left ( \frac{\pi rad}{180degrees} \right )=1.4455rad$

Then the area and wetted perimeter will be solved as functions of Radians [calculator should be in Radians mode]

$A=\frac{r^{2}\left ( \theta -sin\theta \right )}{2}$

$A=\frac{2^{2}\left ( 1.4455 -sin1.4455 \right )}{2}=0.9146ft^{2}$

$P=r\theta$

$P=2ft\left ( 1.4455 \right )=2.891ft$

Thus, the hydraulic radius becomes:

$R=\frac{A}{P}=\frac{0.9146ft^{2}}{2.891ft}$

${\color{Red} \mathbf{R=0.3164ft}}$

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

Sunday 21 April 2019

EXAMPLE 6.4.1. HYDRAULIC RADIUS FOR PIPES FLOWING IN FULL AND HALF FULL

a. Calculate the hydraulic radius (ft) for water flowing full in a 48-inch storm sewer.

b. What is the hydraulic radius (ft) if water flows only half the storm sewer?

From diameter = 48 inches, radius = 24 inches or 2 ft.

A. WATER FLOWING FULL

$A=\pi r^{2}=\pi \left ( 2ft \right )^{2}=12.5664ft^{2}$

$P=2\pi r=2\pi \left ( 2ft \right )=12.5664ft$

Thus, for hydraulic radius:

$R=\frac{A}{P}=\frac{12.5664ft^{2}}{12.5664ft}$

${\color{Red} \mathbf{R=1ft}}$

Alternately, from the formula:

$R=\frac{D}{A}=\frac{4ft}{4}=1ft$

B. PIPE IS HALF-FULL

$A=\frac{\pi R^{2}}{2}=\frac{\pi \left ( 2ft \right )^{2}}{2}=2\pi ft^{2}$

$P=\frac{2\pi r}{2}=2\pi ft$

So, the hydraulic radius will become:

$R=\frac{A}{P}=\frac{2\pi ft^{2}}{2\pi ft }$

${\color{Red} \mathbf{R=1ft}}$

Same formula can be applied for pipes half full - $R=\frac{D}{4}$ .

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

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