Referensi Waterhammer 1 Manual on Pumps Used as Turbines

4.4 Waterhammer / Pressure Surges4.4.1 GeneralIn hydropower installations, pressure surges occur as a result of changes in velocity of flow in the penstock. Caused by either rapid closure or opening of the control valve or by a sudden load rejection of the turbine / PAT (failure of coupling, transmission belt, grid failure in electricity generation), these pressure surges can reach several times the static pressure of the installation. Precautionary measures must be taken to avoid damage to pipelines and machines.Waterhammer and pressure surges have been introduced in Appendix A for the case of sudden and gradual valve closure. This chapter deals mainly with load rejection of the PAT and the subsequent pressure transients induced in the penstock.4.4.2 Gradual Closure/Opening for Nonlinear Valve CharacteristicsFigure 4.8 above has shown that valve characteristics are seldom linear over the whole range of valve stroke. For many valve/penstock configurations the rate of change of flow is low at the beginning of valve closure and will increase rapidly towards complete closure. The magnitude of possible pressure surges (waterhammer) induced by valve closure will therefore depend on the maximum rate of chance of flow rather than on the mean value. For practical purposes an equivalent linear valve characteristic is defined as shown in Figure4.10 below. The corresponding closure time should be used for the waterhammer computations as shown in Appendix A for gradual valve closure. Note that the valve opening(angle Delta ) might not always correspond to the valve stroke (revolutions of wheel) because of the gate mechanism; a diagram of flow versus stroke rather than versus opening should be employed to define the equivalent valve closure time.Generally, a lever-actuated valve as sometimes offered for small butterfly or ball valves should not be used on PAT installations since very short closure times and subsequently large pressure surges can be produced with this mechanism.

FIGURE 4.10:Equivalent gate closure time for waterhammer computations4.4.3 Pressure Transients due to Load Rejection of the PATIn section 3.6.1, the problem of the runaway speed of a turbine or PAT has been introduced.It was shown that the PAT is accelerated to runaway speed when all external loads are removed. The values given for runaway speed of different machines corresponded to the final steady state conditions of the PAT. It was also shown that a change of speed is in most cases accompanied by a change of flow (except for turbines/PATs of nq ~ 80) which, as stated above, basically bears the danger of inducing pressure surges into the pipeline if the rate of change (DeltaQ/Delta t) is high. These pressure surges or pressure transients might drive the PAT temporarily to a higher speed than according to the steady state runaway speed. Figure 4.11 below presents this transient behaviour of a radial flow PAT and its interrelation with the penstock system in a head-flow (H/Q) diagram (note the difference toFigure 3.31, where the change of flow was assumed to take place very slowly (along the system resistance curve) without inducing pressure transients).

FIGURE 4.11:Transient pressure development and runaway speed after total load rejection of a radial flow PATAs can be imagined, the determination of the maximum PAT speed and pressure head after full load rejection including waterhammer is rather involved. Somewhat more analysis is required than for the computation of the waterhammer due to valve closure/opening where a simple formula could be given for a rough estimate (see Appendix A). While the gate movement was determined externally by an operator, PAT speed/flow variations after load rejection is itself affected by the development of the waterhammer; this requires an iterative solution of the problem.In the following, we will present a simplified graphical method to estimate maximum waterhammer due to load rejection of a PAT. Further development of this method would lead to the general solution of transient problems in hydraulics according to Schnyder / Bergeron.4.4.4 Machine ParametersWhen the load on the PAT shaft is suddenly removed, the torque produced by the water flowing through the impeller is no longer in equilibrium with load torque. This torque difference therefore accelerates the PAT until a new equilibrium between the driving hydraulic and the resistive forces (friction) is reached. Figure 4.12 b) shows the development of the torque from nominal to runaway speed under constant head. As speed changes so does the flow through the PAT. Figure 4.12 c) shows the speed versus flow relationship for a radial flow PAT. Note that the flow through an axial flow machine might be increasing with rising speed. Both curves T/n and Q/n are developed from the PAT performance curve calculated in the course of the PAT selection procedure discussed in Chapter 3. (Figure 4.12 a) To simplify calculations, a linear relationship between speed and torque, and flow respectively has been assumed.

FIGURE 4.12: Development of torque and flow versus speed for a radial flow PAT under constant headIt is now decisive for the magnitude the waterhammer, in what time the PAT would accelerate from the operating point (Qo) to runaway conditions (QR) The faster the change of flow (Delta Q), the more powerful the induced waterhammer will be.Acceleration of the PAT depends on the driving torque and on the rotating masses (PAT impeller, pulleys, flywheel and generator rotor or machine assemblies). Applying Newton's second law of motion on rotating elements we will find:ACCELERATION: dOmega / dt = T(Omega) / J (4. 1 ) where Omega = rotational (angular) speed (rad/s)T (Omega) = accelerating torque (Nm) as a function of rotational speed andJ = moment of inertia (kgm) of all rotating masses (details of calculation see Appendix F)As accelerating torque T(Omega ) we may use the speed - torque relationship developed for constant head in Figure 4.12 a), knowing that this will not be quite correct in many cases since head increases due to pressure surges. Thus, acceleration may be expressed with the following differential equation: dOmega/dt=T0/J *(OmegaR-Omega)/( OmegaR-Omega0 ) ( 4. 2)Integrating this, leads to an exponential law for rotational speed a: versus time. Figure 4.13 below indicates that the runaway speed OmegaR would be reached at infinity. For our first estimate, we will use a straight line with the slope of the curve at the starting point(Omega0). Thus, the accelerating time becomes:Taeff=(OmegaR-Omega0)/Omega0 -Ta where (4 .3)Ta=J*Omega0/T0Ta is called the unit accelerating time.

FIGURE 4.13:Development of PAT speed versus time under a linearly diminishing torque T4.4.5 Penstock ParametersThe main parameter of the penstock is the reflexion time or periodTr = 2*L/a where a = propagation velocity of the pressure wave along the penstock and L = penstock length (see also Appendix A). It represents the time required by the pressure wave to travel from the PAT to the forebay and back. For compound penstocks (varying diameters and/or pipe materials), use the equivalent wave velocity and the equivalent pipe cross-sectional area defined as follows (limits of application see below): a equi= L / (L1/a1 + L2/a2 + L3/a3 ...+Ln/an) (4. 5) where L = total penstock length, a equi = equivalent wave velocity, Li and ai are the length and wave propagation velocities of the individual sections.A equi= L / (L1/A1 + L2/A2 + L3/A3 ...+Ln/An) (4.6) where Aequi = equivalent cross sectional area of the penstock and Li and Ai are the length and cross sections of the individual pipe sections of different diameter.4.4.6 Development of the Graphical MethodFrom the waterhammer computations for valve closure/opening (see Appendix A), we know that the reflection time Tr must be compared with the closing time of the valve which, in the case of load rejection, corresponds to the effective acceleration time of the PAT (formula 4.3). If acceleration of the PAT/generator unit is completed within one period, i.e before the pressure wave arrives back at the PAT, the full pressure rise according to Joukowsky's law occurs:Taeff