PRESSURE FLUCTUATION

1. PRESSURE FLUCTUATION Water hammer

2. e p=po, v=vo p=po, v=vo b f p=po+p, v=0 p=po, v=vo p=po-p, v=0 p=po, v=vo c g p=po+p, v=0 p=po-p, v=0 d h p=po+p, v=0 p=po, v=vo p=po-p, v=0 p=po, v=vo Fig. 4.14 Water hammer stages • Water hammer is phenomenon of pressure fluctuation in closed conduit system, caused by sudden reduction of flow velocity or complete stop of it. (kinetic energy converts into potencial, and conversely)

3. Pressure and velocity fluctuation cycle • Imagine pipeline with valve, which suddenly closes (see. Fig. 4.14). At the first moment after valve closure (t = 0) liquid in the pipeline moves with initial velocity (v > v0), pressure p is initial p0 (stage a). After few moments liquid at valve stops, kinetic energy of it converts into potential, pressure here increases in p. • As a result liquid compresses, pipe expands, velocity drops to zero, pressure increases in p (stage b). • Some moment later zone of stopped liquid and increased pressure reaches beginning of the pipeline (stage c). • Then starts motion of liquid, from pipeline to reservoir or the main, to which pipelines is connected. Pressure drops to initial magnitude, liquid moves with initial velocity (see stage d) until zone with of v = v0 and p = p0 reaches valve (stage e).

4. p1 p3 po p4 p2 p tp tp tp 0 t 0 Fig. 4.15 Pressure at valve – time theoretical relationship • Process of flow stop, liquid compression, pipeline expansion, later liquid expansion and pipe compression repeats again and again, until all initial kinetic energy of the flow converts into thermal energy and dissipates, due to hysteresis of liquid and pipeline deformation.

5. Waves of increased and decreased pressure have decreasing with time amplitude, i.e. p1>p2 >p3>p4… The first, third, fifthwaves of increased pressure are called positive stroke, the second, fourth, sixth - negative strokes. • Duration of all strokes are the same where L is length of pipeline, C is stroke propagation velocity. • Kinetic energy of the flow before stopE converts into potential energy of pipe wall and liquid deformations E1 and E2. Balance of them may be expressed as • E=E1+E2.

6. Next stage starts with sudden drop of pressure at the valve due to tendency of liquid column to divorce from the valve. • Due to vacuum liquid expands, pipe compresses and flow stops(stage f). • After few moments vacuum zone expands along all pipeline, motion of liquid stops everywhere (stage g). • Next moment starts motion of liquid from reservoir, where pressure remains p = p0, to the pipeline, where p=p0-p (stage h). • After few moments pressure restores to p=p0 magnitude, liquid moves along all pipeline with initial direction and velocity v=v0 (stage a).

7. Kinetic energy of the flow may be expressed by velocity head multiplied by mass of moving liquid V and acceleration of gravity force g i.e. Here is volume of liquid. During the first positive stroke the energy converts into energy of elastic deformations of pipewall E1 and liquid E2.

8. F2 F1 Fig. 4.16 Water hammer model L2 L1 • Elastic energy of pipe wall deformation may be expressed as Force of pressure acting pipe wall at the and of deformation (Fig. 4.15) may be expressed as F1= pA1. Here wall internal surface A1 =DL.

9. Distance moved by the wall during deformation depends on tension of the wall Here is pressure force creating tension of the wall FW = pDL and wall area in longitudinal cross section AW = 2L. Thus

10. Deformation of the wall Here  is wall thickness,  is relative wall deformation, D is absolute deformation. Distance moved by the wall now may be written Multiplied by force it expresses work done deforming pipe

11. Energy utilized compressing a liquid Distance moved by imaginary piston L2 = L, where relative deformation of liquid Thus

12. Using expressions (4.29), (4.30) and (4.31) equation (4.28) may be recorded in such way Equalizing Corioli coefficient to unit and solving this equation with respect to p gives result Multiplying and dividing term under square root by El received relationship may by rearranged

13. Part of this formula is known expression of sound propagation velocity Next part of (4.32) let us name by pressure wave propagation velocity correction coefficient and denote by Then (4.32) may be written in such brief form p=vC.

14. This formula is valid for the first positive stroke only. Negative stroke pressure decrement may not be large thanpat. Usually this limit is exceeded. It means that liquid column in the pipe looses continuity and cork of gas (air vapor) appears in the pipe. Interruption of liquid continuity cause additional strokes while they closes during the next positive stroke and which is of smaller strength due to loss of energy during the first two positive and negative strokes. • It happens very often, that wave of pressure increment reaches reservoir and returns to the valve jet being in closure process, i.e phase time tp is shorter closure time tc. Such water hammer is called indirect. First stroke pressure increment for indirect water hammer case is computed by formula

15. c a b Fig. 4.17 Safety valve (a), tear membrane (b) and air damper (c) for pipeline protection from water hammer: • Water hammer stroke propagation velocity in steel pipes is about 1000 – 1200 m/s, the phenomenon proceeds very fast, therefore, in most cases of engineering practice, water hammer belongs to indirect category. It follows from (4.36), that pressure of indirect water hammer may be reduced decreasing velocity of liquid flow (increasing diameter of pipelines) and increasing time of flow stop. • simple measurements for protection of pipelines safety valves are used very often. Some of them are shown in Fig. 4.17.

16. Water hammer may be caused not only by full stop of the flow but also by any sudden decrement of velocity by say partialclosure of valve. All formulas given above valid for this case, but instead of velocityv decrement of velocity vmust be used in computations by (4.35) and (4.36). • Water hammer appears not before valve, another apparatus or obstacle, bur also after it. Water hammer stars here by negativestroke, after which follows positive one, than negative and so on. Computations of pressure here are performed using the same formulas, which are given above.

17. 1 2 Hp 6 7 H 5 4 3 Fig. 4.18 Water hammer pump: 1- dam; 2 - pipe; 3 - impact chamber; 4 -impact valve; 5 - pressurevalve; 6 - air chamber; 7- pressure pipeline Water hammer utilization • Water hammer is utilized in special type pump installed at dam on small river (Fig. 7.18). The pump was invented by J. Mongolfye (1796).

18. impact:valve opens and water start to flow through opening with increasing flow rate. At definite magnitude of flow rate hydrodynamic pressure force acting the valveovercomes gravity force and the valve suddenly closes, causing water hammer in the pipeline. • . Pressure valve opens and water flows from impact to air chamber. At negative shock stage pressure valve closes, impact valve opens and process repeats again and again. • The pump develops pressure exceeding created by the dam pressure headH 10 – 20 times • Capacity of the pump is 15 – 30 times less than flow rate of water passing the pump. Usually capacity of the pump Qp does not exceed 25 l/s, developed pressure head Hp – 40 m, efficiency

19. 2 Fig. 4.19 Collapse of cavitation vapor bubble (1), liquid particles meeting point (2) and solid surface possible erosion place (3) 1 3 Cavitation • Cavitation is phenomenon of water vapor bubbles closure, appearing in boiling water flow, entering zone of higher pressure.

20. Boiling of water may start at rather low temperature, when pressure in a flow drops below boiling point , say in suction pipe at the entrance into pump, or on the vanes of ship or turbine runner • When vapor bubble enters zone of high pressure and collapse, surrounding liquid particles move with high velocity (v>100 m/s) toward the center of the bubble (Fig. 4.19). • . Here the particles meat and suddenly stop. Kinetic energy of them converts into potential, pressure at the point of meeting increases up to 100 MPa • Waves of high pressure spread from the center of collapsing bubbles in all sides and erode walls of a channel limiting the flow. Concrete, cost-iron, usual steel surfaces may be eroded very fast if intensity of cavitations is high.

21. Cavitation happens in centrifugal pumps located too high. Absolute pressure of liquid flow entering a pump may be computed applying Bernoulli equation to cross sections 1–1 and 2–2 with respect data plane 0-0 Elevation head for 1–1 z1 = 0 and for 2–2 z2 =hs – suction height. If absolute pressure is used for 1–1 p1 = pat and for 2–2 p2 = p. Velocities for 1–1 v 0; for 2–2 v2 = v. Bernoulli equation receives expression Equation solved with respect to pressure at the entrance into the pump gives It is evident that pressure at the entrance into the pump is smaller than atmospheric pat, what means vacuum here. If it is large enough cavitation will take place.

22. 1 Fig. 4.20 Cavitation caused by obstacle: 1 – wortex zones; 2 – surface damage zones 2 • Cavitation in pump may be recognized from vibrations and typical noise. Efficiency of hydraulic machines reduces due to cavitation, capacity of pump may drop to zero level. Cavitation appears in turbines, when water level in upstream reach drops below acceptable level. The phenomenon appears also on the vanes of ship propeller, when it is submerged not deep and revolutions of it are too high. • Cavitation phenomena may appear, when open flow of highvelocity meets obstacle (Fig.4.20). In vortex zone pressure may drop to the critical level, boiling water volumes separate from vortex zone, bubbles collapse, walls of obstacle and channel erode.

23. Cavitation may be used for biological purification of water. Passing intensive cavitations zone all micro organisms die.

24. Illustrative of design work Water supply system with pump station

25. 4 4 9 He 8 6 3 0’’ 0’’ 2 2 8 Hs Scheme of water supply system: 0’-0’ and 0’’-0’’ – data planes; 1-1, 2-2, 3-3, 4-4 – cross sections; 5 - water source; 6 – pump; 7 and 8 – suction and pressure pipelines; 9 - reservoir 0’ 1 1 0’ 5 7 • Initial data: suction height Hs = 4.5 m; lift of water Hl = 23.0 m; pressure in the reservoir pm = 85 kPa; • length of suction pipeline Ls = 21.0 m; length of pressure pipeline Lp = 1690 m; water requirement Qd = 570 m3/day; water supply regime No 3.

26. Computation of pumps and reservoir • 1.1 From given flow rate daily distribution according to given water supply regime No 3 I compute hourly flow rates Qh=Qd , where ph andpd are percents of water consumption per hour and day (pd=100 %) correspondingly. • 1.2  Analyzing received magnitudes of hourly flow rates Qh of required water supply (see Table 1) I select 2 pumps with nominal parameters: • -         type NCG-10/65; Qp1n= 10 m3/h; Hp1n = 65 m; p1n = 0.330 and • -         type K45/85; Qp2n= 45 m3/h; Hpn2 = 85 m; p2n = 0.43. • Regime of the pumps is supposed for each hour of day comparing required water flow ratesQh with nominal capacities of selected pumps Qp1n and Qp2n. • Water supply abundance (or deficiency) is computed as differences (Qp–Q). • Volume of water to be accumulated is computed as sum of hourly abundances (Qp–Q). • Volume of the reservoir is computed as sum of modules of maximal andminimal accumulated in reservoir volumes, i.e. Vr=24.2+18.4=42.6 m3. • Notion:Volume of the reservoir should not exceed 10 % from daily water consumption, i.e. 570=57 m3 for the case of given in Table 1. The condition is satisfied: 42.6 m3<57 m3.

27. 2. Computation of pipelines and selection of pipes • Pipes for pipelines are computed from rated flow rate Qr, which is equalized to maximal possible flow rate Qmax of the water in the system under design. In my case maximal possible flow rate I compute as sum of nominal flow rates of both selected pumpsQp1n and Qp2n i.e. • Qmax = 10.00 + 45.0 = 55.0 m3/h • and • Qr = 55.0 m3/h = 15270 cm3/s. • Velocity of water in pipes is freely selected from the range (50150) cm/s for suction pipeline and from range (100  200) cm/s forpressure pipeline. I select the velocities 90 cm/s for suction pipeline and 140 cm/s for pressure pipeline.

28. For computation of pipes diameters I apply such formula • D = 1.129 • Suction pipelines are computed separately for each pump from flow rates Qr1 = 10.00 m3/h = 2780 cm3/s and Qr2 = 45.0 m3/h = 12500 cm3/s receiving the result • Ds1 = 1.129 = 6.27 cm and Ds2 = 1.129 = 13.31 cm. • Pressure pipeline is common for both pumps. Diameter of the pipeline pipes is received as • Dp = 1.129 = 11.79 cm .

29. Pipes for pipelines are selected from nomenclature of standard pipes. From given table dimensions of steel pipes for water supply systems I select such pipes: • -         for suction pipeline of the first pump external diameter De = 75.5 mm; wall thickness  = 4.0 mm and internal diameter D = De - 2 = 75.5 - 2 4.0 67.5 mm; • -         for suction pipeline of the second pump De = 140.0 mm;  = 4.5 mm; D = 131.0 mm; • -         for common pressure pipeline De = 114.0 mm;  = 4.5 mm; D = 105.0 mm.

30. 3. Computation of pipelines characteristics • Capacity of a pumpQ depends on its pressure p. Q – p relationship is called the main pump characteristic. It is given in relative in table Pump characteristics in relative units as Q/Qo – H/Ho numerical relationship and is used to determine actual flow rate of water moving in water supply pipeline. • Developed by the pump pressure p is utilized to lift water into the height He , to overcome pressure in the reservoir pm and to overcome resistance of the pipeline i.e. hydraulic loss in the pipeline.Magnitude of hydraulic loss depends on flow rate Q which is to be determined from Bernoulli equation • The following steps are performed to apply the equation: • data planes 0’-0’ and 0’’-0’’ are accepted at water source free surface and at pumpslevel; cross sections 1-1, 2-2, 3-3 and 4-4 are accepted at water source free surface, at the entrance into pumps, at the outlet from the pumps and at the outlet from pressure pipeline into reservoir. Bernoulli equation we record twice: for cross sections 1-1 and 2-2 with respect 0’-0’ and for cross sections 3-3 and 4-4 with respect data plane 0’’-0’’.

31. Parameters of the cross sections 1-1 with respect 0’-0’ are of the following magnitudes: z1=0; p1=0 and v1=0. Parameters of the cross sections 2-2 with respect 0’-0’ are of the following magnitudes: z2=Hs; p2=ps and v2=vs. Bernoulli equation for suction pipeline takes the following shape 0 + 0 + 0 = Hs + + hs . (1) Parameters of the cross sections 3-3 with respect 0’’-0’’ are of the following magnitudes: z3=0; p3=p and v3=vp. Parameters of the cross sections 4-4 with respect 0’’-0’’ are of the following magnitudes: z4=He - Hs; p4=pm and v4=0. Bernoulli equation for pressure pipeline takes the following shape 0 + + = He- Hs + + 0 + hp (2) Summing equations (1) and (2) the following formula is received . (3)

32. Accepting 2 = 3 = 1 and denoting equation (3) may be solved with respect to H Receiving the following formula H = He + . • Hydraulic loss may be expressed as sum of friction and minor loss, i.e. h = hf + hm . • Assume that in assignment for design work given: reservoir elevation head He = 13.5 m and pressure in the reservoir pm = 125 kPa. Pressure head m. To determine velocity heads let us compute first at all velocities of water motion in suction and pressure pipelines. We design the system with two pumps with separate suction pipelines and common pressure pipeline. Internal diameters of them are as follows: Ds1 = 67.5 mm; Ds2 = 131.0 mm and Dp = 105.0 mm. For computation of the velocities we use nominal flow rates of the pumps: Qs1 = 10.0 m3/h = 2780 cm3/s; Qs2 = 45.0 m3/h = 12500 cm3/s; Qp = 55.0 m3/h = 15270 cm3/s.

33. Dividing indicated flow rates by cross section areas of pipelines of indicated diameters we receive: vs1 = cm/s; vs2 = cm/s; vp = = 176.4 cm/s. Velocity in the second pump suction pipeline is larger, therefore in further computations the velocity vs2 = 0.928 m/s we use to determine pressure head H by (4). Thus And 0.1586 m Friction loss hf we compute by Darcy – Weisbach formula

34. where friction factor  we computing from Reinolds number Re and relative roughness . Accepting water viscosity = 0.010 cm2/s Reinolds number in pressure pipeline is Re = Accepting equivalent roughness for steel pipes of the pipeline e = 0.010 mm relative roughness of the pressure pipeline obtains magnitude 0.000951 From Re = 176500 andfriction factor  = 0.0209 (see Moody diagram).

35. For given length of pressure pipeline, say L = 659 m friction loss in pressure pipeline is =20.8 m. Minor loss in pressure pipeline I compute from friction loss by formula hm = kmhf . Coefficient km for the main is accepted within range 0.05 – 0.15. I accept the coefficient km = 0.10 and receive minor loss magnitude hm = 0.1020.8 = 2.08 m. Hydraulic loss in pressure pipeline hp = 20.8 + 2.08 = 22.9 m. Suction pipelines are short, besides, velocities here are much smaller, therefore hydraulic loss in the pipeline I neglect and accept hs = 0. Pressure head required to lift water, to overcome pressure in the reservoir and resistance of pipelines to water flow I computed by (4) as H = 13.50 + 12.74– 0.0439 + 0.1586 + 22.9 = 49.3 m.

36. Computed pressure head consists of two parts: constant, undependable on flow rate, static pressure head Hs = He + = 13.50 + 12.74 = 26.24 m and variable, dependable on flow rate, dynamic pressure head Hd = = -0.0439 + 0.1586 + 22.9 = 23.0 m. . Computed from nominal flow rate pressure head Hn = Hs + Hdn is used to construct pipeline characteristic H = Hs + Hdn (6) which, to its own turn, is used to determine actual water flow rate moving along pipelines of definite geometric parameters.