19.14 WATER HAMMER IN PIPES Introduction Consider the flow of water between two points A and B in a long pipe as shown in Fig 9.12. When the valve is suddenly closed, the momentum of flowing water is destroyed and a wave of high pressure is set up which is transmitted along the pipeline. This creates noise called knocking and the wave has the effect of hammering action on the walls of the pipe, hence known as water hammer or hammer blow. H Valve A B V Fig 9.12 The pressure rise due to water hammer depends upon; (i) The velocity of flow of water in the pipe (ii) The speed at which the valve is closed (iii) The length of the pipe (iv) The elastic properties of pipe material as well as that of flowing fluid. The rise in pressure in some cases may be so large that the pipe may even burst. Hence it is essential to take this pressure into account in the design of pipes. Time Taken by Pressure Wave to Travel From Valve to Tank and Back Let: t = time taken by pressure wave L = length of pipe C = velocity of pressure wave Total distance= L+L=2L dis tan ce travelled from valve to tan k and back ∴ t= velocity of pressure wave 2L Or t= …(9.1) C Valve closure is said to be; 2L (i) Gradual (slow) if t > C 2L (ii) Sudden if t < C Cases of water hammer in Pipes: The following cases of water hammer in pipes are to be considered; (i) Gradual closure of valve (ii) Instantaneous closure of valve in rigid pipes (iii) Instantaneous closure of valve in elastic pipes h.m.m. V ρAL × = p ×A t ρLV ∴ p= …(9. (ii) Pressure Due to Instantaneous Closure in Rigid Pipes Consider pipe AB shown in Fig 9. the liquid is compressed to some extent.E.. If t = 0. increase in pressure seems infinite. Loss of K. = mean rise in pressure × volumetric strain of water in pipe × volume h. = pressure × area of pipe = p ×A …(ii) Equating equations (i) and (ii). Let A = area of cross-section of the pipe L = length of pipe V= velocity of flow through pipe t = time in seconds required to close valve p= intensity of pressure wave produced Due to gradual closure.3) gt Equation (1) gives the relation between increase of pressure due to water hammer in the pipe and time required to close valve. . ∴Retardation of water = change of velocity/time V−0 V = = t t Axial force producing retardation = mass of water in pipe AB × retardation V = ρAL × …(i) t Force due to pressure wave.) of water (if friction is neglected).E. of water.E.m. p ρLV 1 ρLV 1 ρLV H= = × = × = w t w t ρg ρgt LV or H= …(9. 1 1 = mV 2 = × ρAL × V 2 2 2 Gain of S.E. and the pipe material is also stretched. and rigid pipes. 2 (i) Pressure Due to Gradual Closure of Valve Consider a long pipe conveying a liquid (Fig 9.) of the flowing water is converted to strain energy (S. The equation is thus valid for incompressible fluids. Kinetic energy (K.m. this is finite.12 when the valve is closed suddenly.12) and provided with a valve which is closed gradually. In the case of very high pressure. water is brought from initial velocity V to zero velocity in time t seconds.2) t Pressure head. but from experiments. x .m..E.E. like other moduli of elasticity may also be thought of as stress/strain i. of fluid. of fluid will be converted partly into S. of pipe. The above theory is closely related to Mach’s number which is given as.25 to 0. Consider a length of pipe.m. mV 2 = ⋅ ρ × d2 x ⋅ V 2 2 2 4 S.E.34 for mild steel) p = increase in pressure due to water hammer d = internal diameter of the pipe t = thickness of pipe wall (small compared to outside diameter D of the pipe) σC= circumferential stress in pipe σL= longitudinal stress in pipe Sudden closure of valve produces a wave of high pressure of intensity p. 3 p p 1 p2 = × × AL = AL .4) ρV Where C is the propagation velocity (or celerity of a pressure wave).E. which causes circumferential and longitudinal stresses in the pipe wall.12 when the valve is closed suddenly. and is also called the coefficient of compressibility. and is the same as velocity of sound in a liquid of density ρ. σL σ Longitudinal strain of pipe = −µ C E E h.E. .E. 1 1 p2 ρALV 2 = AL 2 2 K 1 2K or p 2 = ρALV 2 × = ρKV 2 2 AL ∴ p = ρKV 2 = ρKV 2 × ρ = ρV K = ρVC ∵ K = C ρ ρ ρ p or C= …(9. p/(dV/V). Original K. Equating K. to S. V M= Kρ (iii) Pressure Due to Instantaneous Closure in Elastic Pipes Consider pipe AB shown in Fig 9.e. where K= bulk modulus of water 2 K 2 K It may be noted that K. and varies from 0. of the fluid. Let: E = modulus of elasticity of the pipe material µ = Poison’s ratio for the pipe material (µ =1/4. π 2 Volume of fluid in length x = d x 4 1 1 π Original K. of fluid = mean rise in pressure ×volumetric strain× volume p p π 2 1 p2 π 2 = ⋅ ⋅ d x= ⋅ ⋅ d x 2 K 4 2 K 4 From knowledge of strength of materials.E. and partly into S. . of fluid = S. = S.E.m.m.E. of fluid + S.E. 1 = ∑ average stress × strain = ∑ 2 stress × strain 1 σL σ 1 σ σ = σL − µ C + σC C − µ L 2 E E 2 E E 1 = 2E (2 2 σ L + σ C − 2µσ C σ L ) For a thin walled pipe.E.E. stored in the pipe material per unit volume of pipe wall. 4 σC σ Circumferential strain in pipe = −µ L E E ∴ S. it can be shown that: pd pd σL = and σ C = 4t 2t ∴ Total S. stored in pipe material.5) 1 d + K tE h. per unit volume of pipe material × total volume of pipe material 1 pd pd pd 2 2 pd = + − 2µ ⋅ × πdtx 2E 4 t 2t 2 t 4 t 1 p2 d2 p2 d2 p2 d2 = ⋅ πdx + − 2µ ⋅ 2E 16t 4t 8t πdx p2 d2 1 2µ = ⋅ + 1− 2E 4t 4 2 πd2 x p2 d = ⋅ (1 + 4 − 4µ ) 4 2 × 4E πd2 x p2 d = ⋅ ( 5 − 4µ ) 4 2 × 4E Original K.E. of pipe 1 π 2 1 p2 π 2 πd 2 x p 2 d ⋅ d x × ρ ⋅ V2 = ⋅ ⋅ d x+ ⋅ (5 − 4µ ) 2 4 2 K 4 4 2 × 4Et 1 p2 p 2d or ρV 2 = + (5 − 4µ ) 2 2K 2 × 4Et 1 = p2 + d (5 − 4µ ) 2K 2 × 4Et 1 2 ρV 2 ρ ρ ∴ p=V =V =V 1 ( 5 − 4µ ) 1 d d 1 d + ( 5 − 4µ ) 2 + + ( 5 − 4µ ) 2K 2 × 4Et 2K 2 × 4Et K 4Et 1 Since µ = (for mild steel pipes) 4 ρ ∴ p=V …(9. 5 Derivation of equations for longitudinal stress (σL). = pressure × area = p×xd …(iii) This force must be balanced by upward force due to circumferential σC (or hoop stress σH). σH×2(x×t) = p×xd pd ∴Hoop stress. σL: Consider a thin walled pipe acted upon by pressure due to water hammer as shown in Fig 9. p is the gauge pressure.m.m.13 (b). there is a longitudinal stress in the cylinder wall. Total force on the end of the cylinder wall due to fluid pressure p.7) 2t In the above equations. h. σC: Now consider equilibrium of wall and contents of a length x of half the cylinder as shown in Fig 9. = pressure × area π = p × d2 …(i) 4 Force due to longitudinal stress σL in the pipe wall to counteract fluid pressure. . equate forces in equations (i) and (ii). = stress × area = σH×2(x×t) …(iv) Equating equations (iii) and (iv). π 2 p⋅ d x 4 σH d d σL t (a) (b) Fig 9. there is a downward force. and hoop stress (σC): Longitudinal stress. = stress × mean circumferential thickness = σ L × πd × t (approximately) …(ii) (If D is taken as mean diameter. then area of wall tube is πDt exactly) For equilibrium. σ L = …(9. σC = σH = …(9.13 Due to pressure on the ends of the cylinder. π ∴ σ L × πd × t = p × d2 4 pd ∴Longitudinal stress.6) 4t Hoop stress.13 (a). Due to fluid pressure. through which a pressure wave is being transmitted from right to left as shown in Fig 9. V+dV = a(v+dv) …(ii) But a =1 and V = v Hence.14 Force on fluid between sections A and B. pressure increases by dp. (∵ area 'a' is unity ) …(i) At section B. having velocity V. (v+dv) and (V+dV). dV = dv (comparing the two equations above) A B unit area.m. At section A. Now bring the wave to rest by imagining the fluid to have a velocity V in the opposite direction. V= a×v = v. increase in pressure Bulk mod ulus. let p. Consider a tube of fluid of unit cross-sectional area. with the maximum pressure occurring at B. …(iv) dV V h. volume compressed/s. . 6 Velocity of Pressure Wave in a Fluid: The velocity of pressure wave (same as velocity of sound) transmitted through a fluid depends on the density and the bulk modulus of the fluid. velocity and the volume respectively. = change of momentum per second = mass/s ×change of velocity ∴ [p − (p + dp )]a = ρaV × dv or dp = -ρVdv But dv = dV Hence dp = -ρVdv dp or = −ρV …(iii) dV The bulk modulus of a fluid is given by.14. K = decrease in volume / original volume * [* (decrease in volume/ original volume) = volumetric strain] dp dp ∴ K= = -V -dV/V dV dp K or =. Between sections A and B. v. and V be the intensity of pressure. and the volume compressed per second increases by dV. At section A. Let the corresponding values at section B be (p+dp). a V V+dV p p +dp p +dp Fig 9.m. Take Kwater= 2×109N/m2. assuming the pipe to be rigid. h.1 is closed in 2 seconds.m.and Kwater= 2×109N/m2. . determine: (a) the rise in pressure (b) the circumferential and longitudinal stress developed in the pipe wall.1: Water flows at a velocity of 1. 7 From equations (iii) and (iv). K ρV = V But V = v. then K K ρV = .5m/s. K p = ρV == ρVC = 103 ×1.54s C 1414 2L ∴ t< . Diameter =0. the thickness of the pipe is 10mm and the valve is suddenly closed.3: If in Example 9.5m/s in a pipe of length 2500m and diameter 500mm Determine the rise in pressure if a valve at the end of the pipe is closed in 25 seconds. Solution K 2 × 10 9 C= = = 1414m / s ρ 10 3 2L 2 × 2500 The ratio t = = = 3. Take C= 1460m/s Solution Given: V = 1.m.42s C 1460 2L Since t > . t=25s.5m.2: If the valve in Example 9.8) ρ Example 9. ρLV 10 3 × 2500 × 1. or v = v ρ The pressure wave is often denoted by C. hence closure of valve is rapid C For rapid closure of valve in rigid pipe. K ∴ C= …(9.5 ×1414 = 2101kN/m 2 ρ Example 9.5 p= = = 150kN/m 2 t 25 Example 9. C=1460m/s 2L 2 × 2500 The ratio t = = = 3.1. closure of valve is gradual C For gradual closure. L =2500m. determine the pressure exerted at the valve. Take E =200×109N/m2. 65 × 10 6 N / m 2 = 21650kN/m .01 h. 8 Solution ρ (a) p=V 1 d + K tE 10 3 = 1 . 4t 4 × 0.732 × 10 6 × 0.732 × 10 6 × 0. 2t 2 × 0.5 9 + 2 × 10 0.m.3 × 10 6 N / m 2 = 43300kN/m . .01 × 200 × 10 9 = 1732 kN/m2 (b) (i) Circumferential stress: pd 1.5 =1.5 2 σC = = = 43.5 2 σL = = = 21.m.732×106N/m2 1 0.01 (ii) Longitudinal stress: pd 1.
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