a review of some dynamic problems in hydropower plants - IHS

IAHR WG1 Stuttgart Oct.8-10, 2003. (Corrected version).
A REVIEW OF SOME DYNAMIC PROBLEMS IN HYDROPOWER PLANTS
Hermod Brekke Professor Emeritus
NORCONSULT A.S.
SUMMARY
The paper deals briefly with the draft tube problems and high frequency problems caused by the blade passing
frequency of Francis turbines. The theory will not be presented in the paper, but will be presented in the
audience.
References will be given in the paper together with illustrations of the developed geometry of Francis runners.
The main body of the paper will describe an unpublished problem caused by water column separation in the
penstock of Muna pumping station in Bogota Colombia and a demonstration of how a stability could be obtained
by a pressure feed back system in an unstable power plant with a very long penstock.
LOW CYCLE DRAFT TUBE PROBLEMS IN FRANCIS TURBINES.
Part load surge problems in the Francis turbines are well known and this problem has its root in the shape of the
runner. No draft tube can save a runner with very unstable behaviour.
Normally the surge problem starts at part load not far from the best efficiency point with the so called inter-blade
separation.
The inter-blade separation is also, besides the runner blade geometry, affected by the cone extended from the
runner hub and turbines with the shaft going through the draft tube bend seems have less problems with
instability.
At lower load, the well known cork screw swirl is building up in all Francis turbines without shaft through the
draft tube, caused by instability when the tangential velocity component have established a potential swirl with
velocity that increases with decreasing radius towards the centre.
The surge problems caused by the cork-screw swirl is strongly influenced by the runner geometry.
Recently the runners of Guri power plant in Venezuela was substituted by new runners of the so called X-blade
type reducing the dynamic part load problems dramatically.
The new so called X-blade runners have a relatively strong negative blade lean angle towards the band i.e. the
blade joint at the band is running in front of the joint at the hub, but the blade is twisted back at the outlet in
order to obtain a stabilized balanced pressure between the band and the hub and thus reduce the cross flow on the
pressure side of the blades. The outlet edges of the blades are also skewed i.e. the joint at the band is running in
front of the joint at the hub. When the X- blade runners was presented first time they were described as
“PRESSURE BALANCED RUNNERS WITH SKEWED OUTLETS “, but the name was improved to X-blade
runners by the Chinese.
In fig 1 a X-blade runner is compared with a traditional runner.
Fig. 1 A traditional runner (Left) compared with a X- blade runner (Right)

The main reason for the improved behaviour of the X-blade runners is the reduced cross flow which is almost
zero at best efficiency, and thus it is also reduced at part load compared with a traditional runner.
Also the skewed outlet edge has a stabilizing effect on the flow at part load.
It can be proven theoretically, that the pressure gradient from hub to band is a function of the blade lean, i.e. the
slope of the blade normal to the meridian velocity vector component, and in addition the geometry of the blade
hub.
dh/dm= f( geometry, Cm and ) where is the blade lean angle. (Ref.1.)
In fig 2 is shown the recorded draft tube pressure surges at part load of a runner with skewed outlet edges
compared with a runner with radial outlet edges.
Fig. 2. Recorded draft tube surge at part load of a runner with skewed outlet edges (Left) compared with a
runner with radial outlet edges (Right). The two turbines are installed in the same power house with the same
operating conditions, but are delivered from different manufacturers.
Full load surges.
At full load a vapour –air filled void is normally built up in the centre of the runner outlet rotating in opposite
direction of the runner. Such void photo taken from a model test is showed in fig 3.
At the power plant Kvilldal in Western Norway, power oscillations of 50-60 MW occurred at full load of two
turbines each at 315 MW.
Fig 3 The natural frequency of penstock fr=1/Tr=1/(4Lp /a), draft tube frequency fdr=(gA H0/(L0 V0)) 0.5 /(2 ) ,
and the generator rotor that was approximately 1.4 Hz and very close to fr and fdr .
The reason for this problem was proven to be three natural frequencies very close to each other.

The first frequency was the resonance frequency of the penstock.
The other was the natural frequency of the generator rotor in the magnetic field of the stator, depending on the
load and setting of the voltage regulator and the external electric load situation.
Finally, the natural frequency of the draft tube mass oscillation cushioned by the air-vapour volume under the
runner.
In fig 3 the physical problem is illustrated.
High frequency pressure pulsations
The well known blade passing pressure pulsations will not be discussed in this paper.
However, it should be noted that the majority of blade cracking are caused by pressure pulsations caused by the
impulses from the blades passing the wakes from the guide vanes.
Examples are given in figure 4 and figure 5.
Figure 4 illustrates pressure pulsations measured between the band and bottom cover of the high head turbine at
Hemsil I with number of guide vanes and runner blades (including splitter blades) were 28 and 30 respectively.
The problem was solved by installing a new runner with the same blade geometry as the existing runner, but
with 32 blades instead of 30 blades. The sound level dB A was reduced from 120 to 85. The problem was not
blade cracking but sound in this case.
In fig 5 is shown measured stress amplitudes at the outlet of a high head Francis turbine blade, 25 MPa.
Fig 4 Sectional view of the turbine at Hemsil I with recorded pressure pulsations between runner and lower
cower. H=510 m, P= 48 500 HP, N=750 rpm.
Fig 5 Stress amplitudes measured at the blade outlet edge of the runner for Tonstad power plant. P=160 MW.

PRESSURE SURGES AT MUNA PUMPING PLANT
The Muna pumping station is located near Bogota city in Colombia where three pumps have been installed.
The main problem for this pumping plant was an unacceptable reverse runaway speed if two or three pumps in
operation tripped simultaneously caused by defects in the electricity supply.
In such case the water hammer pressure rise was not the problem, because the increased pressure causing the
high backwards runaway speed could only be explained by water column separation and air inlet through the air
inlet valve followed by an air compression caused by the returning water column from the upper 280 long
concrete pipe. The system is illustrated in figure 6.
Fig 6. The lay out of Muna pumping plant. Head from air inlet valve (critical point) to tail race level =18 m, and
from air inlet valve to head race level = 9 m at minimum level in the reservoir. Length of upper horizontal and
slight inclining concrete pipe =280 m .(NB! Length of pipes mot in scale.)
The technical data for the three pumps are: P= 7.07 MW, Q= 18.0 m 3 /s, H= 37.5 m, (For increased head and
max head loss), n= 223,7 rpm, Inertia mass for pump and motor: GD 2 = 169.2 tm 2 , inertia time constant:
Ta=3.28 sec. (It should be noted that the atmospheric pressure in Bogota is approximately 7.5 m WC (Altitude
2557. m above sea level.)
By regarding the pump characteristic diagram enclosed at the end of this paper, we find that the flow will be zero
if the speed is decreased to 88% of rated speed and the retarding torque drops from 100% to 52% if the head is
constant at 37.5 m, which is not the case during tripping of the pumps.
If the speed drops further the retarding torque will increase to 124% at zero speed and the flow will be reversed
at -91% of rated flow at rated head.
Because this problem had to be solved by means of a pocket calculator at night during my stay in China a
simplified solution had to be made by a rough calculation with advices to solve the problem.
Following rough calculations was made based on a study of the pump characteristic diagram and the conduit
system:
Time from rated speed to stop with assumed breaking torque -91%, Tstop 3.28/0.91= 3.5 sec. The reversed
runaway speed of –143% of nominal speed, will be reached within 4.7 sec from zero speed at a constant head of
37.5 m .
Because zero flow occurs at 88% speed with an average torque of 76% according to the pump characteristic
diagram, the time to zero flow will be Tstop 3.28(1-0.88)/0.76=0.52 sec.
However, the low pressure will reach the critical point at the air inlet valve approximately 1.0 sec later due to the
water hammer reflection time i.e. 1.5 sec after the tripping of the pumps.(Lengths for control of these data is not
included.)
Prevention of separation at the critical point will not be possible in this case because of the small ratio between
the inertia time constant of the pump unit and the inertia time constant of the penstock: i.e. Ta/Tw 3.0.
It should be noted that the time before the pumps reaches reversed runaway speed, will be increased because the
pressure will be reduced to 18 m WC after separation because the torque will be reduced to approximately 34%
of the torque at rated head and with a reduction of the maximum reversed runaway speed by 70% i.e. –100%
instead of –143%.
However, in this case the high reversed speed is not caused by water hammer, but by compression of the trapped
air volume when the water upstream of the trapped air starts to accelerate downwards driven by gravitation.
The highest pressure will be reached when the upstream water column stops after the compression of the air
between the sinking water level up-streams of the pumps and the upper water column. This high pressure
increases the speed of the pumps that are rotating in reversed direction.

Because of the decreased torque of the pumps before the compression of the trapped air, the maximum reversed
runaway speed would be reached after a time Trunaway (3.5+4.7)/0.34=24 sec.
However, the pressure rise from the air compression occurs after approximately 20 sec for two pumps and the
dangerous reversed runaway can be avoided by closing the butterfly valve in 10 sec or in 15 sec as a
maximum time.
Calculations:
The pressure drop at the critical point where the air inlet valve are located, will be:
h=a C/g m where: C=-(Q/A)(2L/a)/Tsh
Tsh 1.5/0.34= 4.4 sec = the time from the pumps loose the electricity supply until the flow is zero. (1.5 sec is
stopping time + water hammer propagation time)
The following presented analysis of the compression of the trapped air is based upon an assumption that the
outlet of the concrete pipe into the reservoir is furnished with a threshold so no water will enter the pipe during
reversed flow. If that is not the case the situation will be worse.
It is assumed that the velocity at the time of separation will be C0 (m/s), the cross section of the pipe to be
constant =A (m 2 ), the length of the pipe upstream of the critical point to be L (m) and the length of the cavity
filled with air =X (m) (The volume of air =XA (m 3 )) See fig 6.
The following equations may then be established when it is assumed that the concrete pipe is filled with water to
the inlet from the reservoir:
Equilibrium of forces: F=m dC/dt
The energy equation (Friction neglected): Fdx=m (dx/dt) dC
In this case the mass m is not constant and x is the length of water volume leaving the pipe at the pipe outlet at
the reservoir Muna (See fig 6). The elasticity of the pipe is neglected in the following calculation.
When taking into account that the force from gravity is negative and that the mass is reduced by Ax from the
point of time when the flow separates at the critical point, the following equation yields:
gHA dx= A(L-x) c dc
Rearranging and integrating to the maximum value of x=X where the water flow upstream of the separation
point stops:
X was then found by integration:
− gH
X
0
dx
=
( L − x)
0
C0
cdc
( C /( 2gH
)
( ) )
2
− 0
X = L 1−
e
(1)
When the maximum length of the air volume was found by equation 1 the air compression caused by the
reversed flow upstream of the separation point was be studied in order to find the maximum pressure that gives a
high runaway speed of the pumps in reversed direction
.

Following assumptions was made:
1. The cavity is located in the horizontal part of the pipe.
2. The upper horizontal part of the pipe is filled with water and will not be drained of water. (See fig. 6)
3. A threshold will prevent additional water to enter the pipe from the reservoir. (Without a threshold the
situation will be worce.)
4. The air inlet valve closes immediately when the water column upstream stops and the inner pressure is equal
to outer pressure.
5. Frictional damping is neglected.
6. the value X found by equation 1 is used as a boundary condition input value for the calculation of the
compression.
During reversed flow following equation for the compression yields:
C0 X0
A(L-X) c dc= gA(P0-p) dx (2)
0 x
Here =density of water, A= cross section, P0= atmospheric pressure, p= absolute pressure during compression
And X= the maximum length of the air volume at the lowest pressure (ref eq.1),X0= the length of the air volume
when the pressure of the compressed air is equal to the driving back pressure i.e. =P0-p and finally C0=maximum
velocity during the reversed flow period when air and head is equalized i.e. p=P0 and x=X0)
During the acceleration period the kinetic energy that is accumulated in the reversed flowing water column, will
reach its maximum value as shown in equation 2 at a length=X0 and a volume =X0A at a pressure =P0 as
illustrated in figure 7.
From the length X0 a compression of the air with closed air inlet valve, will decelerate the flow and at volume
AX2 and pressure P2 the velocity of the water column will be zero.
The compression is assumed to be adiabatic with a compression exponent =1.4 giving a maximum pressure of
40.08 m WC. However, a calculation with a polytropic exponent of 1.3 was also tested, giving a higher
maximum pressure of 41.3 m WC.
The equation of adiabatic compression yields:
P(xA) = P(XA = P0(X0A)
Or p=(X/x) P (3)
When substituting for p in the left hand side in equation 2 following equations can be established:
X0 X2
(P0-PX /x )dx= (PX /x – P0)dx (4)
X X 0
After integration and rearrangement following equation can be established in order to find X2=f(X,P) and then
the maximum pressure could be found by a simple iteration on a pocket calculator.
The final equation yield:
X2 - ((1+P/(P0( -1)))X)X2 ( -1) + PX /( -1) = 0 (5)
And P2 = P(X/X2) (6)
The result of the calculations is illustrated in figure 7.

The maximum pressure reached 40.08 m WC caused by compression of the air and this was the main reason for
the non-acceptable reversed runaway speed. It should be noted that by using a polytropic exponent of 1.3 instead
of the adiabatic exponent 1.4, the maximum pressure increased to 41.28 m WC. It was also surprising that the
maximum pressure with tripping of 3 pumps was equal to the tripping of 2 pumps.
Physically this can be explained by longer distance with air filled pipe and less water in reversed flow because it
was assumed that no water from the reservoir entered the pipe during the reversed flow period.
RESULT OF CALCULATIONS.
2 pumps tripping: Flow 36=m 3 /s 2 pumps tripping Flow=36m 3 /s 3 pumps tripping: Flow=53 m 3 /s
= 1.4 =1.3 =1.4
X2=6.449 X2=5.526 X2=13.823
X0=12.156 X0=11.189 X0=26.055
X=21.348 X=20.52 X=45.760
P2=40.08 m WC P2=41.28 m WC P2=40.08 m WC
H2=40.08-7.5=32.58 m H2=41.28-7.5=33.78 m H2=40.08-7.5=32.5
Fig 7 Illustration of the compression history caused by reversed flow after water column separation.
IMPROVING GOVERNING STABILITY BY PRESSURE FEED BACK SYSTEMS.
In figure 8 is shown the general block diagram for a pressure feed back system where the algorithm indicates
the algorithm that transforms the pressure signal in the feed back loop to the governor.
Fig. 8 Block diagram for the governing system of a hydropower plant equipped with a Francis turbine. For a
plant with a Pelton turbine the value of Qn = 0.

It is well known that the inertia mass and elasticity of the penstock are the main reason for the stability problems
of hydropower plants.
The first known introduction of a pressure feed back system was made by Dr. Lein (Ref 2).
Lein introduced a so called DT1 that is a derivative element with one time constant, element in the compensator
element shown in figure 8.
The author of this paper worked on the power plant SKJAAK in Norway that had a very difficult stability
problem and an improvement of compensator element was needed.
The element introduced by Dr. Lein yields:
The function of the new element yields:
= kTds/(1+THs) (7)
=kT1s/((1+T1s)(1+Tds)(1+0.5s)) (8)
However, it is of interest to study and simplify the block diagram in figure 8 by studying a Pelton turbine where
the influence from the speed on the flow disappears i.e. Qn=0.
Note that in the block diagram in fig.8, =2hw tanh(Ls/a), where s=j
n
n
ref
( 1+
Tds)(
1−
Φ)
= (9)
btT s(
1+
0.
5Φ(
1−
2((
1+
T s)
/( T s)))
Θ)
) T s
d
Introducing: b = 0.5(1-2((1+Tds)/Tds) ) (10)
Then we obtain: n/nref = ((1+Tds)/btTdsTas) (1- )/(1+b ) (11)
Here the function (1- )/(1+b )=P/y i.e. the transfer function of turbine power as function of needle
opening.
We can now re write the transfer function P/y(s) as follows:
P ( 1−
Φ)
( b −1)
( 1+
bΦ)
( b + 1)
( 1−
bΦ)
( b −1)
( b + 1)
( 1−
bΦ)
= =
+
= +
(12)
y ( 1+
bΦ)
2b
( 1+
bΦ)
2b
( 1+
bΦ)
2b
2b
( 1+
bΦ)
We find now that the power response forms a circle in the complex plane plot with the centre in (b-1)/(2b) on
the real axis.
As a demonstration of the pressure feedback system let us use the DT1 element proposed by Dr Lein and let
TH=Td in equation 7. Then let us look at the result by making variation of the constant k as illustrated in
figure 9.
As expected we find that for k=0, i.e. no pressure compensating system, we get a circle through +1.0 and –2.0 on
the real axis and the circle has a radius =1.5.
When we decreases the value of k, the radius of the circle decreases, but the circlr will always go through the
point 1.0, 0.0. It is also interesting to see that for negative values of k the radius increases and the system will be
more unstable than without a pressure feed back system as expected because we now get a positive feed back
signal.
d
d
a

Fig 9. The transfer function of P/y(s) with the DT1 element (equation 7) where TH=Td for k=3,1, 0 and-0.5.
In figure 10 is illustrated the Nyquist diagram for the stability on isolated load with variations of the constant k
of a DT1 element with Td=TH. We find here that a decreased circle as shown in fig 9 may decrease the stability.
However, an improved setting might have been obtained by adjusting Td TH , but our work was aimed at a
more complex function of the pressure feed back function.
Fig.10 Nyquist diagram of the transfer function n/nref with a DT1 element (equation 7)in the pressure feed back
system with variation of k=0.01,0.5 and 1.0.
Finally the more complex pressure feed back function was analysed. The function is shown in equation 8.
By fine-tuning the time constants T1 and letting Td be equal to the integral time constant in the turbine governor
we find that the stability margin is very good. (See distance to the point (–1,0) on the real axis.
The result of the computation is illustrated in figure 11 showing the influence of variation of the constant k.

Fig 11. Nyquist diagram of n/nref with the authors approach with different values of the amplification
k =0.5,1.0 and 2.0.
ACKNOWLEDGEMENTS.
The author wants to thank his colleagues at Kvaerner Energy and at NTNU for collaboration during the work
presented in this paper.
A special thank to Dr ing Li Xin Xin for his collaboration with regulating stability programming.
REFERENCES
1. Brekke H:“Hydraaulic design strategy of Francis turbines”. Hydropower 1996. Lausanne July 1996.
2. Lein G. and Maurer W.: “Advances in control of hydropower plants” Report University of Stuttgart.