OE4625 Dredge Pumps and Slurry Transport. Vaclav Matousek October 13, 2004

Transcription

1 OE465 Vaclav Matousek October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

2 OE465 Vaclav Matousek October 13, 004 Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

3 OE465 October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

4 Keywords: Transport (Horizontal, Vertical, Inclined) Pipe (Length, Diameter) Pump (Type, Size) Goals: Design a pipe o appropriate size Design pumps o appropriate size; Determine a number o required pumps October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

5 Objectives: Prediction o energy dissipation in PIPE Prediction o energy production in PUMP Prediction models: Pressure drop vs. mean velocity in PIPE Pressure gain vs. mean velocity in PUMP October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

6 Part I. Principles o Mixture Flow in Pipelines 1. Basic Principles o Flow in a Pipe. Soil-Water Mixture and Its Phases 3. Flow o Mixture in a Pipeline 4. Modeling o Stratiied Mixture Flows 5. Modeling o Non-Stratiied Mixture Flows 6. Special Flow Conditions in Dredging Pipelines October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

7 Part II. Operational Principles o Pump-Pipeline Systems Transporting Mixtures 7. Pump and Pipeline Characteristics 8. Operation Limits o a Pump-Pipeline System 9. Production o Solids in a Pump-Pipeline System 10. Systems with Pumps in Series October 13, Dredge Vermelding Pumps onderdeel and Slurry organisatie Transport

8 1. BASIC PRINCIPLES OF FLOW IN PIPE CONSERVATION OF MASS CONSERVATION OF MOMENTUM CONSERVATION OF ENERGY October 13, 004 8

9 Conservation o Mass Continuity equation or a control volume (CV): ( ) d mass dt = ( q q ) outlet inlet [kg/s] q [kg/s]... Total mass low rate through all boundaries o the CV October 13, 004 9

10 Conservation o Mass Continuity equation in general orm:.( ρv ) 0 t + = ρ For incompressible (ρ = const.) liquid and steady low (ə/ət = 0) the equation is given in its simplest orm vx v v x y z y z + + = 0 October 13,

11 Conservation o Mass The physical explanation o the equation is that the mass low rates q m = ρva [kg/s] or steady low at the inlets and outlets o the control volume are equal. Expressed in terms o the mean values o quantities at the inlet and outlet o the control volume, given by a pipeline length section, the equation is thus q m = ρva = const. [kg/s] (ρva) inlet = (ρva) outlet October 13,

12 Conservation o Mass in 1D-low For a circular pipeline o two dierent diameters D 1 and D V 1 D 1 = V D [m 3 /s] V1 D1 D V October 13, 004 1

13 Conservation o Momentum Newton s second law o motion: ( ) d momentum dt = F external The external orces are - body orces due to external ields (gravity, magnetism, electric potential) which act upon the entire mass o the matter within the control volume, - surace orces due to stresses on the surace o the control volume which are transmitted across the control surace. October 13,

14 Conservation o Momentum In an ininitesimal control volume illed with a substance o density the orce balance between inertial orce, on one side, and pressure orce, body orce, riction orce, on the other side, is given by a dierential linear momentum equation in vector orm DV ρ = ( ρv) + ρv. V = P ρg h. T Dt t October 13,

15 Conservation o Momentum Claude-Louis Navier George Stokes October 13,

16 Conservation o Momentum in 1D-low In a straight piece o pipe o the dierential distance dx (1D-low), quantities in the equation are averaged over the pipeline cross section: V V h P + V + g 4 o + + = 0 t x x x D ρ τ October 13,

17 Conservation o Momentum in 1D-low For additional conditions : - incompressible liquid, - steady and uniorm low in a horizontal straight pipe dp A = τ O o, i.e. dx dp = dx 4τ o D or a pipe o a circular cross section and internal diameter D. October 13,

18 Conservation o Momentum in 1D-low For a straight horizontal circular pipe τ P P 4 1 = o L D P 1 P t 0 V D t 0 L October 13,

19 Liquid Friction in D Pipe Flow The orce-balance equation generalized or D-low gives the shear stress distribution in a cylinder: dp =τ dx r October 13,

20 Liquid Friction in D Pipe Flow Newton s law o liquid viscosity (valid or laminar low): F dv τ µ x = = A dy October 13, 004 0

21 Liquid Friction in D Laminar Flow in Pipe The generalized orce-balance equation or the D-low in a cylinder + Newton s law o liquid viscosity (valid or laminar low) = dp =τ dx r dv τ µ x = dr Velocity distribution in laminar low in a pipe dv dp r x = dr dx µ October 13, 004 1

22 Liquid Friction in 1D Laminar Flow in Pipe The integration o the velocity gradient equation gives a value or mean velocity in pipe and thus a relationship between pressure drop and mean velocity / 1 8 D V = vda= vrdr x x A D A 0 V D dp = 3µ dx which is the required pressure-drop model or laminar low in pipe dp dx = 3µ V D October 13, 004

23 Liquid Friction in 1D Laminar Flow in Pipe A comparison o the pressure-drop model or laminar low in pipe + dp dx = 3µ V D with the general orce balance (driving orce = resistance orce) or pipe low = gives the equation or the shear stress at the pipe wall in laminar low dp dx = τ = µ o 4τ o D 8V D October 13, 004 3

24 Liquid Friction in 1D Turbulent Flow in Pipe The wall shear stress or turbulent low cannot be determined directly rom the orce balance and Newton's law o viscosity (it does not hold or turbulent low). Instead, it is ormulated + by using dimensional analysis. A unction τ 0 = n(ρ, V, µ, D, k) is assumed. The analysis provides the ollowing relationship between dimensionless groups 1 τ o ρ V k = n Re, D October 13, 004 4

25 Liquid Friction in 1D Turbulent Flow in Pipe The dimensionless group Re, Reynolds number, is a ratio o the inertial orces and the viscous orces in the pipeline low Re V Dρ = = µ inertial. orce viscous. orce Remark: The Reynolds number determines a threshold between the laminar and the turbulent lows in a pipe. October 13, 004 5

26 Liquid Friction in 1D Turbulent Flow in Pipe The dimensionless parameter on the let side o the dimensional-analysis equation is called the riction actor. It is the ratio between the wall shear stress and kinetic energy o the liquid in a control volume in a pipeline. Fanning riction actor Darcy-Weisbach riction coeicient τ o 8 = τ λ o = ρ V ρ V 1 λ = 4 October 13, 004 6

27 Liquid Friction in 1D Flow in Pipe A comparison o the Darcy-Weisbach riction coeicient equation + with the linear momentum eq. (driving orce = resistance orce) or pipe low = gives the general pressure-drop equation or the pipe low (Darcy-Weisbach equation, 1850) λ = dp = dx dp = dx ρ 8τ o V 4τ o D λ D ρ V October 13, 004 7

28 Liquid Friction in 1D Laminar Flow in Pipe A comparison o the general pressure-drop equation + with the pressure-drop eq. or laminar low in pipe = gives the pipe-wall riction law or laminar low in pipe dp = dx λ dp = dx λ D ρ V 3µ V D 64µ 64 = = ρ DV Re October 13, 004 8

29 Liquid Friction in 1D Turbulent Flow in Pipe In turbulent lows there is no simple expression linking the velocity distribution with the shear stress (and so with the pressure gradient) in the pipe cross section. The dimensional analysis provides the ollowing relationship between dimensionless groups λ k = n Re, D October 13, 004 9

30 Liquid Friction in 1D Turbulent Flow in Pipe There are three dierent regimes with the dierent wall riction laws: Hydraulically smooth Transitional Hydraulically rough λ =n(re) λ =n(re, k/d) λ =n(k/d) October 13,

31 Liquid Friction in 1D Turbulent Flow in Pipe October 13, Hydraulically smooth Transitional Hydraulically rough λ =n(re) λ =n(re, k/d) λ =n(k/d)

32 Liquid Friction: Moody Diagram October 13, Hydraulically smooth Transitional Hydraulically rough λ =n(re) λ =n(re, k/d) λ =n(k/d)

33 Liquid Friction: Moody Diagram October 13, Hydraulically smooth Transitional Hydraulically rough λ =n(re) λ =n(re, k/d) λ =n(k/d)

34 Conservation o energy P1 V1 P V λ LV + + z 1 = + + z +, ρ g g ρ g g gd where λ is the riction actor, or our case : λ = gd 3 e v R The Colebrook ormula or λ is : 1 λ =, R e 4000,.51 log10 R e λ where R is Reynolds number R e VD = ν e October 13,