52860284 the hydraulic gradient

8/6/2019 52860284 the Hydraulic Gradient

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The Hydraulic Gradient

Describes how the friction and other losses effect the Hydraulic Gradient in a pipe subjected

to a constant head

Contents

1. Introduction2. The Significance Of The Hydraulic Gradient3. The Hydraulic Gradient4. Worked Examples5. Page Comments

Introduction

In the pages on the flow of a fluid through pipes, it is seen that there is a loss of head. Whilst some of

this is due to the effect of sudden contraction or expansions in the pipe diameter, pipe fittings such as

bends and valves and entry and exit losses,a loss of potential head (i.e. The input of the pipe is higher

than the outflow) a significant portion is due to the friction in the pipe (The Darcy Equation). However

in a pipe of uniform cross section, there will be no loss of velocity head and so the loss of Total

Energy will be the result of a loss in Pressure Head. The following diagram shows a uniform pipe

and the value of pressure head at three points down its length. The line joining these points is

calledThe Hydraulic Gradient
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The Significance Of The Hydraulic Gradient

Normally when a pipe is laid, attempts are made to keep the pipe at or below the hydraulic Gradient

However in some cases this may not be possible but provided that the pipe does not rise by more than

about 26 ft. (8 mtr.) water will still flow. Above this height air comes out of solution and an airlock is

formed.


The above diagram is of an extremely simple system. On the next diagram the pipe has both a sudden

contraction and an enlargement. The various losses in energy are shown and this is used to construct

the Total Energy Line which is shown in red on the diagram.

The velocity head at the salient points in the pipe are also calculated and these are subtracted from

the Energy Line to give the Hydraulic Gradient which is shown in blue.

Worked Examples

To see the workings please click on the red buttons.
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Example 1:

Two tanks A and B are connected by a pipe 100 ft. long.

The first 70 ft. has a diameter of 3 in. and then the pipe

is suddenly reduced to 2 in. for the remaining 30 ft. The

difference of levels between the tanks is constant at 30

ft. f = 0.005 and the coefficient of contraction at all

sudden changes of area is 0.58

Find all the head losses including that at the sharp

edged pipe entry at A in terms of the velocity and

hence find the flow in gallons/min. Draw in Hydraulic

Gradient diagram.

To see the workings please click on the red button

From the diagram:-

(1)

Substituting given values

(2)

(3)

The loss at the pipe entry A
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(4)

(5)

The frictional loss along AC (The Darcy equation)

(6)

The loss caused by the sudden contraction at C

(7)

The frictional loss along CB

(8)

The velocity loss at the exit into B

=

(9)

It can be seen from the diagram that the total head lost is 8 ft. and therefore:-

(10)

(11)

The flow through the pipes is the product of the cross sectional area of the pipe and the velocity. i.e.


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(12)

(13)

Note that 1 of water wqeighs 62.4 lb. and 1 English gallon weighs 10 lbs


(14)

The velocity head in AC

(15)

The velocity head in CB

(16)

The various head losses have already been written down in terms of

It is thus now possible to tabulate actual values and enter them on a diagram of the two reservoirs

and the connecting pipes.


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The losses are:-

Entry at A 0.130 ft.

Pipe friction 1.395 ft.

Loss at C 0.657 ft.

Pipe Friction 4.550 ft.

Exit loss 1.263 ft.

The Hydraulic Gradient is the line shown in Blue

Example 2:

Two tanks, A and B are 100 ft. apart and are connected

by two pipes of 2 in. and 4 in. diameter. The height of

water in tank A is maintained at 25 ft. above that in

tank B. f = 0.008 for both pipes. The coefficient of

contraction at pipe entry is 0.585

Calculate the rate of flow in each pipe and draw the

Hydraulic Gradient. If the pipes are to be replaced by

one larger pipe, calculate the diameter of the pipe

required for the total rate of flow from A to B, to remain

unchanged.

Drawing of twin pipes from notes

To see the workings please click on the red button
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The head lost at pipe entry.

(17)Applying Bernoulli's and Darcy's equations.

(18)

For the 2 in. pipe

(19)

(20)

(21)

(22)

For the 4 in. pipe

The arithmetic is identical to that shown above except

the diameter of the pipe is now

(23)

(24)

(25)

(26)

A single pipe


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The total quantity of water flowing is

Applying Bernoulli and Darcy to the proposed pipe

(27)

(28)

(29)

But the required volume of water delivered is

(30)

(31)

Substituting in Equation (29) and solving for D

(32)

The Hydraulic Gradient for the two pipes is shown in

blue

Drawing of twin pipes modified.

Example 3:

An 8 in. diameter pipeline which connects two reservoirs

whose difference of level is 37 ft., consists of a 1000 ft.

length AB and a straight 1500 ft. length Bc having a

valve 500 ft. from the lower end C, as shown. The pipesAB and BC have each constant coefficients but BC is

slightly rougher than AB. When the flow is 600 gal./min.

a gauge at B shows that the pressure head of water is

1.7 ft. and the loss of head across the valve is 15 ft.


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Determine the maximum rate of flow if the loss of head

across the valve when fully open is 4 ft. Also draw the

Hydraulic Gradient for this quantity.

Allow for the effects of velocity head in the pipe at exit

but neglect the loss of head at pipe entry. Assume that

the horizontal projection of BC equals its length. (B.Sc.

Part1)

To see the workings please click the red button

When the flow is 600 gal./min. let the velocity of flow be and let and the the pipe friction

coefficients fo AB and BC respectively.

Thus.

(33)

Applying Bernoulli and Darcy to AB

(34)

(35)

Similarly for BC

(36)

(37)

When the flow is maximum let the velocity by

Then applying the Bernoulli and Darcy equations to the whole pipe system.
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(38)

(39)

(40)


(41)

(42)

(43)

As DC is half the length of BD

(44)

The Hydraulic gradient is shown in blue on the following diagram. Note that it has not been drawn to

scale. The difference between 37 ft. and 36.562 is due to rounding errors.


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52860284 the hydraulic gradient

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