Analysis and Design of a Simple Surge Tank

RESEARCH NOTEANALYSIS AND DESIGN OF A SIMPLE SURGE TANK M. Azhdary Moghaddam Civil Engineering Department, University of Sistan and Baluchestan Zahedan, Iran, [email protected] (Received: December 9, 2003 – Accepted in Revised Form: October 14, 2004) Abstract In a hydroelectric power plant or in a pumping station in order to avoid sudden large increase of pressure due to instantaneous valve closure sometimes a surge tank is installed. The height of surge tank is designed by the highest possible water level during the operation. The theoretical treatment of oscillation in a surge tank is difficult because of the non-linearity of friction term in the governing differential equation of the system. The present study attempts to find a general solution for the surge oscillation in a simple surge tank in terms of non-dimensional parameters. Equations for the highest and the lowest water level in the tank, which are very important in the design of a surge tank have been found. Key Words Surge Tank, Pressure, Surge Oscillation, Non-Linear Differential Equation            !" # $%& '(%)      ƵŶǀƨģ 36 . . ) *+  !.7 8  . / 1  $ 234 4' '5/ 4 . -&') . ? 4 A. / 9> ? @ . ) :.BC/ 1  . ) 9. . <)  .3.4 . = ' 1. ' .%= . D    )D # ?E4  ?4 . -&') @ / @  F') 4'  . I % ) =) .  )D >H  . @ /  $ 234 1/ 1. with large Depending upon its configuration. INTRODUCTION main functions of a surge tank are [5]: 1. IJE Transactions A: Basics Vol. No. continue indefinitely with the same amplitude. orifice. The maximum amplitude of water level (maximum of water in the surge tank drops [4]. MAIN CONSIDERATIONS IN THE DESIGN tank and water level in the tank at any time during the OF A SURGE TANK oscillation depends on the dimension of the pipeline and tank and on the type of valve movement. because of its small 2.3]. differential. or closed. and so on. suddenly completely closed. the conduit of a hydroelectric power plant for producing a counter-pressure so that water in the preventing the pressure surges entering into it (Figure pipeline flows towards the reservoir and the level 1). 4. surge) can be observed when a full load is suddenly In the absence of damping.D 1/G  /G) .4 ? 1. 17. restricted orifice. The flow into the surge 2.339 . A between pipeline and penstock causes a rise in the simple surge tank is defined as a tank or shaft of water level in the surge tank. The extent of damping is governed by roughness condition. It reduces the amplitude of pressure fluctuations When the valve in a hydroelectric power plant is by reflecting the incoming pressure waves. a surge tank may be inertia retards slowly. The water level rises constant horizontal cross sectional area that connects above the static level of the reservoir water. oscillation would rejected [6]. The difference in flows classified as simple. once [1. November 2004 . The water in the pipeline. The In order to accomplish its mission most effectively. It improves the regulation characteristic of a inertia the water in the penstock stops almost at hydraulic turbine. also the on the following considerations [6]. the surge tank dimensions and location are based between the surge tank and the pipeline. and 4. Figure 1. simplify the derivation of dynamic and continuity 3. The height of a surge tank is governed by the The equation of motion is written as [9]. The surge tank should be located as close to the remains constant [8]. The surge tank must have sufficient cross (iii) the effect of entrance loss in comparison sectional area to ensure stability. Derivation of Governing Equation To prevent overflow for all conditions of operation. discharge line. 2. power or pumping plant as possible. it has been assumed that admit air into the turbine penstock or pumping (i) the conduit walls are rigid. The surge tank should be of sufficient height to 3. and (ii) the water is incompressible. [7]: reservoir is considered so large that its level 1.1. The bottom of surge tank should be low enough equations that describe the oscillations of the water that during its operation the tank is drained out and level in the tank. Simple surge tank system. [12] : highest possible water level that can be anticipated wV wV wy g S0  Sf . with the friction loss has been neglected. January 2005 IJE Transactions A: Basics . 1. These equations describe wV wV wy § wZ wh f · approximate values of peaks and downsurges. No.b) wt wx wx © wx wx ¹ Integration (1. 17. there is very little head loss dt 340 . V g g ¨  ¸ (1.b) with respect to x between the limits 3. since wt wx wx the resistance law flow varies as Reynolds number and relation roughness [2].a) based on a linearized resistance relationship. ANALYSIS OF SURGES IN SIMPLE x=0.Vol. during its operation. All available methods are V g (1. one gets SURGE TANK dV L  gh f  gy 0 (2) In a simple surge tank. x=L (see Figure 1) and simplifying. «¨© R e ¸¹ § 6 16 » · « ¨ Ln§¨ 3.7H  5.5 » for surge oscillation in a simple surge tank.2. Thus 9.a changes to 2 dt dt © DP ¹ dt With initial condition at t=0 fLD T2 dy dy hf (10) 2gD 4P dt dt y h f 0 (5.2.1.b) dt © DT ¹ 3.a) DT L 2gD 2P y V0 sint (7) DP g 2 dy § DP · Equation 7 describes sinusoidal oscillations[10].b) 0. the following (11) particular cases arise: Substituting 10 in 4 one gets 3.a) V ¨¨ ¸¸ (3) 2gD 2P 2gD 4P © dt ¹ © DP ¹ dt With the change of flow direction. the direction of That with use of 3 in 2 the following equation can 2 § dy · be found: friction also changes. Frictionless Flow In this case hf = -hf0 = 0.2.9 ¨ ¸ ¬ © ¹ ¼ on the nature of friction loss. the equation following dimensionless groups are formed: fLV 2 DT g hf (8) y* y (14. Laminar and Turbulent Flow In this case head loss is expressed by Darcy-Weisbach In order to reduce the number of parameters. November 2004 .2.74 ·¸  §¨ 2500 ·¸ ¸ » 3. ¨¨ ¸¸ V0 (13. 5.12 dt © DT ¹ ª º « » «§ 8 » Equation 4 is the governing differential equation f «¨ 64 ·¸  9. Thus Equation 4 changes to d 2 y fD T2 dy dy gD 2P   y 0 (12) d 2 y gD 2P dt 2 2D 3P dt dt D T2 L  y 0 (6) dt 2 D T2 L The initial conditions prescribed on 12 are at t=0 Solving 6 for the initial condition 5.a) 2 In which 8[11] dy § DP · ¨¨ ¸¸ V0 (5.a.a) 2gD 2P D P V0 L IJE Transactions A: Basics Vol. No. Hence ¨ ¸ occurring in © dt ¹ 2 §D · d2y dy dy L¨¨ T ¸¸  gh f  gy 0 (4) 9. Adopted Methods for the Solution Depending « ¨ ¸ ¨ © DP R e ¹ © R e ¹ ¸ » 0 .From continuity condition between tank and pipe. Combining 3 and 8 one gets it can be shown that 2 fLV 2 fLD T2 § dy · § DT 2 · dy hf ¨ ¸ (9.a should be split to .b one gets f 0 LV02 y h f 0  (13.341 . 17. 4. 587 3 3 yt1* -1 (20.c) 4 2 tp1* S 2 (20.02 the following R0 (17) DP Q empirical formula for finding the mentioned values in which R0 is the initial Reynolds number has been suggested: defined as using non-dimensional variables 12 k3 reduces to: ª § h ·k2 º S C «1  ¨¨ f 0* ¸¸ » (20) 2 «¬ © k 1 ¹ »¼ d y* f dy dy * 2  h f 0* *  y* 0 (18) dt * f0 dt * dt * In which the values of S.32  R 00.946 4. k1. tp1*.b) dt * dy * Being non-linear it will be very difficult to find a V V0 (15) dt * closed form solution of 18.362 0.914 -0. A numerical solution.1) 0. No S C k1 k2 k3 Max. k2.35¨¨ ¸¸ ¸ ¨ D © P¹ ¸ dy * © ¹ dt * 1 (19.a) k1  1.85 1  0. The initial conditions for sudden valve closure at t=0 § § H · 0.0 1. Coefficients of Equation 20. using a fourth order Runga Kutta method.2 5 yt01* -1 (20.Error % 1 yp1* 1 (20.9 1.024 . H V0 D P With varying 0 d d 0.h) 5 4 tt1* 3S 2 1.4 6 tp2* 5S 2 3. . can be Substituting 15 in 11 one gets attempted.b) opening are DT L dy * The conduit velocity may be expressed as 1 (19. TABLE 1. From this numerical solution one can see that the various properties of the solution like dy * H Re R0 (16) yp1*.a) (20.653 4.g) (20. and dt * DP hf0*. S.6 DP g and the initial conditions for sudden valve t* t (14.e) 0.f) (20.7 4.b) (20. and tt1* are functions of R0.144 · ¨ 1.49 0. k3 can be found from the Table 1.d) (20. C. yt1*. a) 342 . January 2005 IJE Transactions A: Basics . No.Vol. 1. 17. (20. 041 . 21.h) © DP ¹ k3 0.227¨¨ ¸¸ (20. 20.2 0 1 2 3 4 5 6 y* -0.6R 00.66  1.4 -0.896R 00.a -1 Time Figure 2. k2 1.287 k1 1.4 0.6 0.57 1  1.2 0 -0.011 (20. 0.b) § H · 0. -0.c) 15600  0.8 Eq.01 (20.6 Eq. Surge Oscillation in Simple Surge Tank.946R 00. 037 (20.786  (20.d) k3 0.i) § § H · 0.179 · § H · ¨1  ¨ ¸ ¨ ¨© D P ¸¹ ¸ 10 5  R ¸ 0  . k1 0. 61§¨ H · Equation 18 using fourth order Runga Kutta ¨ ¨D ¸¸ ¸  ¸ 3162  R 0.7 0 .054¨¨ © DP ¸¸ ¹ © ¹ Figure 2 shows a typical plot of surge oscillation 740 k2 1.73  for sudden valve closure obtained by solution § 0. 1  0.65 · ¨1  19. e) equation for h f 0*  1. Analysis of a large number of surge © © P ¹ ¹ oscillation curves suggest the following empirical (20.016 · hf 0* 1.05¨¨ ¸¸ ¸ y* exp Dt* .5 : § § H · 0.36¨1  0. method.  exp Et* . . sinZ*t*  I. f) (21.035 in which k2  0.a) 1. ¨ © DP ¹ ¸ 2 sin I k1 © ¹ (20.024 .92  R 00.5  R 00.462 1. 083 2S 1  0. 17.6¨¨ ¸¸ Z* (21. (20.b) t p 2*  t p1* © DP ¹ IJE Transactions A: Basics Vol. No. November 2004 .343 . 4.g) § H · 0. NOTATIONS I (21.c) 2 t p 2*  t p1* DP Diameter of Conduit DP0 Minmum Conduit Diameter Putting t * t p1* in Equation 21. S 5t p1*  t p 2* 5.a and equating it DT Diameter of Simple Surge Tank to y p1* one gets: DTS Stable Diameter of Tank f Friction Factor f0 Friction Factor for Initial Velocity 2 y p1* sin I exp Dt p1* .  exp Et p1* . a and hf Head Loss equating it to y t1* one gets: hf0 Initial Head Loss h f0* Non-Dimensional Initial Head Loss 2 y t1* sin I H0 Desired Head exp Dt t1* . (22) g Gravitational Acceleration h f 0* h1 Free Board in Tank h2 Cushion Level in Tank Also by putting t * t t1* in Equation 21.  exp E t t1* . y* Non-Dimensional Height of Surge 2. An equation for surge oscillation has been Z0 Level of Reservoir obtained. Equations for minimum downsurge and the corresponding time of occurrence have been yp1 Height of First Peak developed. ZL Bottom Level of Tank 4. The y * R Reynolds Number R0 Initial Reynolds Number versus t * curve can be converted to a y versus t t time curve via using Equations 14. 1. The main t t1* Non-Dimensional Time of First Downsurge conclusions are as follows: V Velocity 1. 17. CONCLUSION First Peak t p2* Non-Dimensional Time of Occurrence In the present study a general solution for surge Second Peak oscillation in a simple surge tank and an optimal tt1 The Occurrence Time of First Downsurge design of system has been discussed. No. (23) HT Height of Tank h f 0* KP Cost Parameter for Conduit KT Cost of Tank per Unit Area The quantities of D and E can be obtained by L Length of Conduit solving Equations 22 and 23. simultaneously [1]. 344 . m Cost Parameter for Conduit Figure 2 shows the result obtained by solving P0 Desired Power Equation 21.a and compares it with the result obtained by numerical solution of differential Q0 Initial Discharge Equation 18 by Runga Kutta method.Vol. Equations for maximum surge height and V0 Initial Velocity corresponding time of occurrence have been y Height of Surge obtained.a and 14. January 2005 IJE Transactions A: Basics . y p1* Non-Dimensional Height of First Peak 3.b. An equation for the occurrence of a second yt1 Height of First Downsurge peak of the surge oscillation has been y t1* Non-Dimensional Height of First Downsurge obtained. t* Non-Dimensional Time Parameter tp1 The Occurrence Time of First Peak t p1* Non-Dimensional Time of Occurrence 4. 345 . W. E. Chaudhary.. J. U. K. "Surge Tanks"..H Roughness of Conduit 3. Van 12. K0 Steady State Efficiency 4..P. A. U. bath. ASCE. (1960)... A. "Engineering Fluid Mechanics". B. B. L. U. Sutton. (1969). Mc Graw-Hill Book Co. H. 6. U.. 446-454. 657-664. ASME. 4. Trans. P Viscosity of Water 5. Div.. (85).K. No.. (1951).. ASCE. John Wiley & Sons Inc. "Explicit Equation for Pipeflow Master of Engineering in Civil Engineering. P. (1955). Hyd. Proc. 225-234. Pickford. (1976). Rich. Jakobsen. E.. 102(5). Mc Graw Hill Nostrand Reinhold Co. Inc. M. J. "Engineering Hydraulics".A. Jain. B. India.S. G. F. (1978). 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