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

## 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 called The Hydraulic Gradient

A pipe is a tubular section or hollow cylinder, usually but not necessarily of circular cross-section, used mainly to convey substances which can flow-liquids and gases (fluids), slurries, powders, masses of small solids.

The hydraulic gradient is a vector gradient between two or more hydraulic head measurements over the length of the flow path.

## 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.

Example:
[imperial]
##### Example - Example 1
Problem
Two tanks $\inline&space;A$ and $\inline&space;B$ 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. $\inline&space;f$ = 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 $\inline&space;A$ in terms of the velocity $\inline&space;v_2$ and hence find the flow in gallons/min. Draw in Hydraulic Gradient diagram.

Workings
From the diagram:

$a_1\;v_1=a_2\;v_2$

Substituting given values:

$\frac{\pi}{4}\times&space;3^2=\frac{\pi}{4}\times&space;2^2$
$\therefore&space;\;\;\;\;\;v_1=\frac{4}{9}\;v_2$
And:
${v_{1}}^{2}=0.197\;{v_{2}}^{2}$

• The loss at the pipe entry $\inline&space;A$
$=\left&space;(&space;\frac{1}{C_C}-1&space;\right&space;)^2\times&space;\frac{{v_{1}}^{2}}{2g}=\left&space;(&space;\frac{1}{0.58}-1&space;\right&space;)^2\times&space;\frac{{v_{1}}^{2}}{2g}\;ft.$

$=\frac{0.52\;{v_{1}}^{2}}{2g}=0.52\times&space;0.197\;\frac{{v_{2}}^{2}}{2g}=0.103\;\frac{{v_{2}}^{2}}{2g}$

• The frictional loss along $\inline&space;AC$ (The Darcy equation):
$=\frac{4\times&space;0.005\times&space;70\times&space;{v_{1}}^{1}}{2\times&space;\tfrac{3}{12}\times&space;g}=\frac{1.4\times&space;0.197}{0.25}\times&space;\frac{{v_{2}}^{2}}{2g}=1.105\frac{{v_{2}}^{2}}{2g}\;ft.$

• The loss caused by the sudden contraction at $\inline&space;C$:
$=\frac{0.52\;{v_{1}}^{2}}{2g}=0.52\times&space;0.197\;\frac{{v_{2}}^{2}}{2g}=0.103\;\frac{{v_{2}}^{2}}{2g}\;ft.$

• The frictional loss along $\inline&space;CB$:
$=\frac{4\times&space;0.005\times&space;30}{2\times&space;\tfrac{2}{12}\times&space;g}=3.6\times&space;\frac{{v_{2}}^{2}}{2g}\;ft.$

• The velocity loss at the exit into $\inline&space;B$:
$=&space;\frac{{v_{2}}^{2}}{2g}\;ft.$

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

$8=(0.103+1.105+0.53+3.6+1)\;\frac{{v_{2}}^{2}}{2g}$
Hence:
${v_{2}}^{2}=81.3$
And:
$v_2=9.02\;ft./sec.$

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

The flow is:
$\frac{\pi}{4}\times&space;\left&space;(&space;\frac{2}{12}&space;\right&space;)^2\times&space;9.02=0.197\;ft.^3/sec.$
$=0.197\times&space;60\times&space;\frac{62.4}{10}=74\;gal./sec.$

Note that 1 $\inline&space;ft.^3$ of water weighs 62.4 lb. and 1 English gallon weighs 10 lbs.

${v_{1}}^{2}=0.197\times&space;{v_{2}}^{2}=0.197\times&space;81.3=16.02$

• The velocity head in $\inline&space;AC$:
$=\frac{{v_{1}}^{2}}{2g}=\frac{16.02}{2g}=0.249$

• The velocity head in $\inline&space;CB$:
$=\frac{{v_{2}}^{2}}{2g}=\frac{81.3}{2g}=1.263$

The various head losses have already been written down in terms of $\inline&space;\displaystyle&space;\frac{{v_{2}}^{2}}{2g}$

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

The Hydraulic Gradient is the line shown in Blue

Solution
The losses are:

• Entry at $\inline&space;A$: 0.130 ft.
• Pipe friction: 1.395 ft.
• Loss at $\inline&space;C$: 0.657 ft.
• Pipe Friction: 4.550 ft.
• Exit loss: 1.263 ft.