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Surge Tanks

An Analysis of Surge Tanks ( Frictionless and Flow with allowance made for Friction)

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Contents

Introduction
Instantaneous Closure Neglecting Friction
A Sudden Complete Valve Closure Allowing For Friction
Gradual Valve Closure
Sudden Or Gradual Partial Closure.
Sudden Valve Opening On Increased Load.
Refinements To Simple Surge Tanks
Page Comments

Introduction

When the rate of flow of a fluid passing down a pipe-line changes, there is a change in pressure. The severity of this effect depends upon the rate of change in the flow rate, the length of the pipe and its diameter. In small bore pipes there is no real problem other than maybe an annoying hammering sound. In large water mains the rate of change in flow is carefully controlled to avoid damage to pipes and valves. Unfortunately this solution will not work with Turbines where a sudden change in load requires a rapid change in water demand.

Here Surge Tanks or Stand Pipes are used to reduce the pressure surges. When the flow to the turbine is reduced, water flows into the surge tank and conversely for increased load , the initial extra water required is from the surge tank.

The size of the tank should be such that water will not overflow when the turbine is suddenly shut down, nor allow air to be drawn into the system following a sudden increase in demand. In addition it must be sited as close to the turbines as possible to avoid surges in the length of pipe between the surge tank and the turbine.

A surge tank (or surge drum) is a standpipe or storage reservoir at the downstream end of a closed aqueduct or feeder pipe to absorb sudden rises of pressure as well as to quickly provide extra water during a brief drop in pressure.

A valve is a device that regulates, directs or controls the flow of a fluid by opening, closing, or partially obstructing various passageways.

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Instantaneous Closure Neglecting Friction

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Let:

$\inline R=$ Surge tank area $\inline A$ / Pipe line area $\inline a$
$\inline v_0$ = the initial velocity of water in the pipe.
$\inline v$ = the instantaneous velocity in the pipe line t secs. after the change in load.
$\inline y$ = The height of the surge level above the surface of the reservoir t secs. after the change in load.
$\inline w$ = density of water.
$\inline T$ be the periodic time of the oscillation in the Surge Tank.

$T=2\pi \sqrt{\frac{LR}{g}}$

$y = v_0 \sqrt{\frac{L}{gR}}\ cos \sqrt{\frac{g}{LR}}$

$v = v_0\;cos\sqrt{\frac{g}{LR}}\times t$

At time $\inline t$ the pressure at $\inline P$ above normal flow conditions $\inline = w y$

By Newtons second Law The pressure force on the water column in the pipe line = Mass multiplied to Acceleration

$w\,y\,a=\frac{w\,a\,L}{g}\times - \frac{dv}{dt}$

$\therefore\;\;\;\;y = -\frac{L}{g}\;\frac{dv}{dt}$

(2)

( $\inline y$ is called the Acceleration or Inertia head )
By Mass Continuity Flow in the surge tank = Flow in the pipe, i.e.

$A \frac{dy}{dt} = a\,v$

(3)

Or this can be written as:

$v = \frac{A}{a}\,\frac{dy}{dt} = R\;\frac{dy}{dt}$

Differentiate:

$\frac{dv}{dt} = R\,\frac{d^2y}{dt^2}$

(4)

Substituting equation (4) in equation (3)

$y = - \frac{L}{g}\times R\,\frac{d^2y}{dt^2}$

(5)

$\therefore\;\;\;\;\;\frac{d^2y}{dt^2} + \frac{g}{LR}y=0$

But this is the equation for simple harmonic motion whose solution is:

$y = C \cos\sqrt{\frac{g}{LR}t^2} + D \sin\sqrt{\frac{g}{LR}}t$

(6)

When $\inline t = 0$ , $\inline y = 0$ and by inspection $\inline C = 0$ .

Hence:

$y=D\sin \sqrt{\frac{g}{L\,R}.}\times t$

Differentiate the above equation:

$\frac{dy}{dt} = D\sqrt{\frac{g}{LR}}\times \cos\sqrt{\frac{g}{LR}}\times t$

(7)

But from equation (3) when $\inline t=0$ :

$\frac{dy}{dt} = \frac{v_0}{R}$

And also:

$\cos \sqrt{\frac{g}{L\,R}}=1$

$\therefore\;\;\;\;\frac{v_0}{R} = D\sqrt{\frac{g}{LR}}$

$\therefore\;\;\;\;D = v_0\sqrt{\frac{L}{gR}}$

Substitute in equation (7)

$y = v_0 \sqrt{\frac{L}{gR}}\times \sin \sqrt{\frac{g}{LR}}\times t$

If $\inline T$ is the period of a complete oscillation:

When:

$t=\frac{T}{2}\;\;\;\;\;\;\;\;y=0$

From equation (5)
And since $\inline D\neq 0$

$\ sin\;\sqrt{\frac{g}{LR}}\times \frac{T}{2}=0$

$\therefore\;\;\;\;\;\sqrt{\frac{g}{LR}}\times \frac{T}{2}=\pi$

$\therefore\;\;\;\;\;T=2\pi \sqrt{\frac{LR}{g}}$

From equation (6)

$v=R \frac{dy}{dt}=R v_0 sqrt{\frac{L}{gR}}\times \sqrt{\frac{g}{LR}} \cos\sqrt{\frac{g}{LR}}\times t$

$\therefore\;\;\;\;\;v = v_0\;cos\sqrt{\frac{g}{LR}}\times t$

The following two graphs show the variation of both $\inline y$ and $\inline v$ with time.

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Notes:

The maximum surge height occurs at time $\inline t$ , i.e. $\inline \left(\displaystyle\frac{T}{4} \right)=\displaystyle\frac{\pi }{2}\sqrt{\displaystyle\frac{LR}{g}}$

$y_{max}=v_0 \sqrt{\frac{L}{gR}}$

A large value of $\inline R$ i.e. a large surge tank area, means a small $\inline y_{max}$ but a longer periodic time $\inline T$ .
Changes in reservoir level and the inertia of the water column in the surge tank have been neglected.

A Sudden Complete Valve Closure Allowing For Friction

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The height of the first maximum surge can then be found by putting $\inline v = 0$ and $\inline y = y_(max.)$ in the following equation.

$v^2 = -\frac{L}{2gRC^2} e^{\frac{- 2gRC}{L}(y\,+\,CV_0^2)}$

This equation can only be solved by trial and error but a first approximation neglecting friction and using $\inline y_{max}=v_0\sqrt{\displaystyle\frac{L}{gR}}$ will save work!

And for the first minimum surge

$v^2-\left(\frac{y_{max}}{C}+\frac{L}{2gRC^2}\right ) e^{\frac{2gRC}{L}(y_{max}\,-\,y)} + \frac{y}{C} + \frac{L}{2gRC^2}$

Note that in the theory $\inline y$ is measured positively upwards from the reservoir level. $\inline \displaystyle\frac{dy}{dt}$ is positive or negative depending upon whether the surge level is rising or falling. $\inline v$ is positive or negative depending whether the flow is towards the surge tank or away from it. $\inline Cv^2$ is added when the flow is towards the surge tanks and is subtracted if the flow is towards the reservoir.

At the initial steady flow state the level in the surge tank will be below the reservoir level by an amount equal to the friction head lost in the pipe.
From the Darcy equation the friction head lost is:

$\frac{4fl\,{v_{0}}^{2}}{2dg}$

(8)

This is usually written as $\inline C{v_{0}}^{2}$ where $\inline C$ is a constant for the pipe line.

At a time $\inline t$ secs, after closure let the surge level be at a height $\inline y$ above the reservoir level and let the pipe velocity be $\inline v$ .

For steady flow at velocity $\inline v$ the level in the surge tank would be $\inline C v^2$ below the reservoir level.

Thus the excess pressure head at $\inline P$ causing the deceleration of the water column is $\inline (y+C\,v^2)$

from Newton's second Law:

Pressure force = Mass multiplied to Acceleration

$w(y + Cv^2)a=\frac{wal}{g}\times - \frac{dv}{dt}$

$\therefore\;\;\;\;\;y + Cv^2=- \frac{l}{g}\;\;\frac{dv}{dt}$

(9)

But by continuity:

$av=A\;\frac{dy}{dt}\;\;\;or\;\;\;v=R\frac{dy}{dt}$

(10)

Hence:

$\frac{dv}{dt}=R\;\frac{d^2v}{dt^2}$

(11)

Substituting from equations (10) and (11) in (9)

$\therefore\;\;\;\; y+C\left(R\,\frac{dy}{dt} \right)^2=- \frac{L}{g}R\frac{d^2y}{dt^2}$

i.e.

$\frac{d^2y}{dt^2}+\frac{CRg}{L}\left(\frac{dy}{dt} \right)^2+\frac{g}{LR}=0$

(12)

This is insoluble as it stands since we can not deal with the friction term.

To eliminate $\inline t$ :

$\frac{dv}{dt}=\frac{dv}{dy}\times \frac{dy}{dt} = \frac{v}{R} \frac{dv}{dy}$

But:

$v\,\frac{dv}{dy} = \frac{1}{2} \frac{d(v^2)}{dy}$

$\therefore\;\;\;\;\frac{dv}{dt} = \frac{1}{2R}\times \frac{d(v^2)}{dy}$

(13)

Substitute in equation (8)

$y+Cv^2=-\frac{L}{2gR}\times \frac{d(v^2)}{dy}$

Or:

$\frac{d(v^2)}{dy} + \frac{2gRC}{L}\times v^2 + \frac{2gR}{L}y=0$

The solution of this equation is:

$v^2=K\;e^\frac{-2gRCy}{L} - \frac{y}{C} + \frac{L}{2gRC^2}$

It is now necessary to evaluate the constant $\inline K$ :

When: $\inline t = 0$ , $\inline v=v_0$ and $\inline y=-Cv_0^2$

$\therefore\;\;\;\;\;K = -\frac{L}{2gRC^2} e^\frac{- 2gRC^2\,v_0^2}{L}$

$\therefore\;\;\;\;\;v^2 = -\frac{L}{2gRC^2} e^{\frac{- 2gRC}{L}(y\,+\,CV_0^2)}$

The height of the first maximum surge can then be found by putting $\inline v = 0$ and $\inline y = y_(max.)$ in the above equation. The equation can then only be solved by trial and error but a first approximation neglecting friction and using $\inline y_{max}=v_0\sqrt{\displaystyle\frac{L}{gR}}$ will save work!

First Minimum Surge.

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Head at $\inline p$ accelerating the column towards the reservoir is $\inline y - Cv^2.$

Equation (8) now becomes:

$y-Cv^2 = -\frac{L}{g} \frac{dv}{dt}$

And by continuity:

$-av = A\times - \frac{dv}{dt}$

Or:

$v = R\;\frac{dy}{dt}$

NOTE: the net effect of the flow reversal on the above equations is to change the sign of the $\inline C\;v^2$ term, so the equations generated, during the consideration for sudden complete closure are modified as follows.

Eliminating $\inline v$ equation (13) becomes:

$\frac{d^2y}{dt^2} - \frac{CRg}{L}\left(\frac{dy}{dt} \right)^2 + \frac{g}{LR}y = 0$

Eliminating $\inline t$ :

$\frac{d(v^2)}{dy} - \frac{2gRC}{L}v^2 + \frac{2gR}{L}y=0$

Putting $\inline v=0$ when $\inline y=y_{max}$ :

$v^2=-\left(\frac{y_{max}}{C}+\frac{L}{2gRC^2}\right ) e^{\frac{2gRC}{L}(y_{max}\,-\,y)} + \frac{y}{C} + \frac{L}{2gRC^2}$

The following graphs show the variations of $\inline y$ and $\inline v$ with time.

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In all the above theory $\inline y$ is measured positively upwards from the reservoir level. $\inline \displaystyle\frac{dy}{dt}$ is positive or negative depending upon whether the surge level is rising or falling.
$\inline v$ is positive or negative depending whether the flow is towards the surge tank or away from it.
$\inline Cv^2$ is added when the flow is towards the surge tanks and is subtracted if the flow is towards the reservoir.

Gradual Valve Closure

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Consider the instantaneous conditions at a time $\inline t$ as shown. The head at $\inline P$ decelerating the column is $\inline y + Cv^2$ .

Thus:

$y+Cv^2 = - \frac{L}{g} \frac{dv}{dt}$

And from the Continuity equation:

$a(v-\omega ) = A \frac{dy}{dt}$

Or:

$v-\omega = R \frac{dy}{dt}$

After the valve has closed in a time $\inline t_c, \omega$ remains zero and the equation becomes: $\inline v = R \displaystyle\frac{dy}{dt}$ as before

This can only be dealt with by numerical integration and even then the variation of $\inline \omega$ with time must be known. It is usual to assume that $\inline \omega$ decreases uniformly from $\inline v_0$ to zero in a time $\inline t_c.$ i.e. at a time $\inline t$ .

$\omega = v_0 \left(1-\frac{t}{t_c} \right)$

Sudden Or Gradual Partial Closure.

Equations (12) and (5) still apply but $\inline \omega$ does not now fall to zero.

For a sudden partial closure $\inline \omega$ is assumed to fall instantaneously to the new steady value. For a gradual partial closure $\inline \omega$ is assumed to fall linearly with time to the new constant value of $\inline \omega_c\; at\; time\; t.$

$\omega =\omega _c + (v_0 - \omega _c)\left(1 - \frac{t}{t_c} \right)$

Sudden Valve Opening On Increased Load.

Note : Assume that the velocity at the valve increases instantaneously to the final steady velocity $\inline \omega_c$

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Consider the position shown. Take $\inline y$ as positive downwards.

$\left(y-Cv^2 \right)=\frac{L}{g} \frac{dv}{dt}$

$\left(\omega _c - v \right)=A \frac{dy}{dt}$

$\left(\omega _c - v \right) = R \frac{dy}{dt}$

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Notes: The interaction of the turbine governing mechanism and the surge tank frequency must be studied so that surges are damped out by friction and not perpetuated and amplified by the action of the governor. The following equation gives the critical area ratio for stability.

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$R_{critical} = \frac{L}{H_tCv_0^2}\times \frac{v_0^2}{2g}$

Where $\inline H_t$ is the initial steady flow level in the surge tank above the turbine gate.

Example:

[imperial]

Example - Example 1

Problem

Describe the operation of a simple Surge Tank communicating with the pipe-line supplying the turbine in a hydro-electric plant.

Show that if the friction head is proportional to velocity squared , the oscillatory motion of the level in the Surge tank following a sudden complete shut down of then turbines is given by the equation taking the form:

$\frac{d^2H}{dt^2}+F\left ( \frac{dH}{dt} \right )^2+GH=0$

in which $H$ is the height at any instant of the Surge Tank level with reference to the reservoir level, $F$ and $G$ are constants, the former having positive values when the flow along the pipe-line is towards the Surge Tank and negative when reversed.

Find $A$ and $B$ if the Surge Tank diameter is 100 ft. pipe line diameter 15 ft. and the length of the pipe-line from reservoir to Surge Tank 2,400 ft. At the instant when the turbines are completely shut down, the flow along the pipe-line from reservoir to surge tank is $1,500\;ft.^3/sec.$ and the level in the Surge Tank is stationary, 3 ft. below the level in the reservoir.

Workings

The head producing the deceleration of the water in the pipe is, using equation (7)

$H+Cv^2=-\frac{L}{g}\times \frac{dv}{dt}$

(1)

And from the continuity equation for an instant valve closure . Equation (1)

$av=A\frac{dH}{dt}$

(2)

Differentiating:

$\frac{dv}{dt}=\frac{A}{a}\times \frac{d^2H}{dt^2}$

(3)

Substituting equations (3) and (2) in equation (1)

$H+C\left ( \frac{A}{a} \right )^2\times \left ( \frac{dH}{dt} \right )^2+\frac{L}{g}\times \frac{A}{a}\times \left ( \frac{d^2H}{dt^2} \right )=0$

$\therefore \;\;\;\;\frac{d^2H}{dt^2}+F\left ( \frac{dH}{dt} \right )^2+GH=0$

Where:

$F=\frac{cAg}{aL}$

And:

$G=\frac{ga}{LA}$

When $v$ is towards the reservoir, $Cv^2$ will be above $XX$ and the equations corresponding to (1) and (2) will now by:

$H-Cv^2=\frac{L}{g}\times \frac{dv}{dt}$

And:

$av=-A\frac{dH}{dt}$

Which together will make $F$ negative.

$Q=1,500=av$

$\therefore \;\;\;\;v=\frac{1500}{\displaystyle\frac{1}{4}\pi\times 15^2}=8.5\;ft./sec.$

The head drop due to friction is 3 ft.

$\therefore \;\;\;\;C=\frac{3}{8.5^2}=0.0415$

And:

$\frac{A}{a}=\frac{10^4}{15^2}$

$\therefore \;\;\;\;F=\frac{0.0415\times 10^4\times 32.2}{225\times 2400}=0.0246$

And:

$G=\frac{32.2\times 225}{2400\times 10^4}=3.02\times 10^{-4}$

Refinements To Simple Surge Tanks

Here are some refinements to simple surge tanks.

Variable Area Chamber

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The object is to limit the extremes of the surges. This arrangement provides a quick initial response followed by a slower change in levels and almost constant pressure in the larger chamber. Note that:

Chamber $\inline A$ caters for sudden valve closure.
Gallery $\inline B$ assists demand for more water caused by an increased load on the turbine.

Throttled Surge Tank

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The orifice will limit the total height of the surge by increasing the friction and velocity head losses at entry to the chamber.

Johnson Differential Surge Tank

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On a change of load the Surge level will change rapidly in the riser thus causing a rapid deceleration or acceleration in the pipe.The main chamber level lags behind. The net flow into or out of the Surge Tank is thus less than in the simple design.

Note: There are various other designs of Surge Tank with Air tight Chambers, Conical shapes and multiple chambers. In some designs where the height of the Tank is limited , it is common to have chambers with an overflow spill way.