1. Energy principle ( 에너지 법칙 ) The 1 st law of thermodynamics ( 열역학 제 1 법칙 ) – “conservation of energy” (no appearance or disappearance of energy in a system, but conversion to different energy: heat to mechanical, mechanical to heat) Q: heat transferred to the system W: work done by the system E p : potential energy E k : kinetic energy E u : internal energy, associated with motion of molecules (structure of atoms, chemical E, electrical E) By Reynolds transport theorem,

2. Classification of work Flow work – work done by pressure forces as the system moves through space Flow work of a CV i System flow work Shaft work – work other than flow work pump – work on the flow, E of flow increases, negative work (-) turbine – work done by the flow, E of flow decreases, positive work (+) For steady flow (E accumulation = “0”) and uniform velocity across section in a pipe

3. Energy equation for steady flow/an incompressible fluid in a pipe  = 1, for the uniform velocity across the section  > 1, for nonuniform velocity distribution  = 2, parabolic velocity distribution   1, for most turbulent flows

4. Energy equation Shaft work – pump work on the flow (-) and turbine work given by the flow (+) Pump power Turbine power Power delivered by the pump to the flow Actual power delivered by the turbine

5. Class quiz) What is the pressure in the pipe at L=2000m? Head loss = 0.02 (L/D)V 2 /(2g), D=0.2m, Q=0.06m 3 /s,  =1

6. Class quiz) How much of power is required to have a pressure of 350kPa (gage pressure) at section 2? Q=0.5m 3 /s, D=0.5m, z 1 =30m, z 2 =40m, p 1 =70 kPa (gage), Head loss = 3m,  1 =  2 =1

7. Class quiz) How much of power is generated from a hydroelectric turbine? Q=141m 3 /s, Head loss = 1.52m,  1 =  2 =1

8. Bernoulli equation - Energy equation Bernoulli equation – for a steady, incompressible, and inviscid flow along a streamline Energy equation – for viscous, incompressible flow in a pipe (with pumps and/or turbines) For the flow of inviscid (no head loss) in a small stream tube (uniform velocity distribution) and no pump and turbine, the Energy equation = the Bernoulli equation.

9. Class practice – Abrupt expansion Energy equation in the sections 1 and 2 (assumption,  =1) Momentum equation in the sections 1 and 2 Continuity equation in the sections 1 and 2 If a pipe discharges liquid into a reservoir, V 2 =0, and then

10. Class quiz) How much of force is required to hold the transition in place? Q=0.707m 3 /s, Head loss = 0.1V 2 2 /(2g), p 1 =250 kPa,  1 =  2 =1

11. Class quiz) What is the head loss bwt. the reservoir and outlet and how much is the pressure at point B if the ¾ of the head loss occurs btw. the reservoir and point B? Assume a turbulent flow in the pipe (  1 =1) Given: Discharge – Q Pipe diameter - D Note that cavitation occurs if P B < P vapor (1.23 kPa abs., -100.07 kPa gage) A

12. Class quiz) Prob 7.56 What is the max. discharge before cavitation occurs at the throat of the venturi meter? Assume a turbulent flow in the pipe (  1 =1) and no head loss around venturi meter inlet. D = 0.3 m, d = 0.15 m, P atoms = 100 kPa (abs), H = 5 m, water temp = 20 o C

13. HGL and EGL Each term in the energy equation represents “energy per unit weight of the flowing liquid (N  m/N)” in length unit (m)→ “Head” Hydraulic grade line (HGL) – piezometric head (P/  +z) Energy grade line (EGL) – total head (P/  +z+  V 2 /2g) Remarks in EGL and HGL for a pipe EGL-HGL =  V 2 /2g EGL=HGL at V=0 (reservoir surface) The downward slope in the direction of flow due to headloss The constant slope (=  h L /  L, head loss per unit length) of EGL for steady flow in a pipe of which physical characteristics are uniform (diameter, roughness, shape, etc.) Abrupt rise given by a pump and abrupt drop given by a turbine in the EGL HGL = system line at p=0 in a pipe or channel (P/  =0) – outlet (to the atmosphere) and reservoir surface) Changes in pipe diameter → changes in EGL/HGL slopes If the HGL were below the pipe, there’s subatmospheric pressure in a pipe.

14. HGL and EGL

16. Class quiz) How much of power is required? Q=7.85cfs, head loss = 0.01 (L/D)V 2 /(2g), pipe length (L)=5000ft, D=1ft

17. Class quiz) HGL and EGL Ex 7.3 Ex 7.5

18. Class quiz) HGL and EGL (a) What is the flow direction? (b) What kind of machine is A? (c) Pipe CA and pipe AB, which has a larger diameter? Or the same? (d) Is there a vacuum along the pipe? (e) How about EGL?

19. Class quiz) Prob. 7.59 HGL and EGL h L = 0.025(L/D)(V 2 /2g)

20. Dimensional Analysis and Similitude ( 차원해석과 상사 ) The analytical solutions directly describing fluid behavior → very limited Verification of the fluid model solutions → experiments using physical models How to perform experiments? - determine variables (parameters) - set up appropriate relationships (governing equations) for the fluid behavior using variables - set up / perform physical model as a function of variables - determine the correlations among the variables in the relationships  Possibly too many experiments required depending on the complexity of geometrical fluid characteristics and the number of variables involved By reducing the number of variables via dimensional analysis, we can save our efforts, time, and cost significantly. Example, Analysis of an fluid passing an orifice,  Reduction in the correlating parameters from 5 to 2

21. Buckingham  theorem The number of independent dimensionless groups of variables (dimensionless parameters) = n – m n: the number of variables involved m: the number of basic dimensions included in the variables The dimensionless parameters – “  groups” The equation describing a physical system that has n dimensional variables Expressed in terms of dimensionless parameters (  groups)

22. Dimensional analysis Determination of  groups 1. Identification of all independent variables describing physical phenomena (geometry, fluid characteristics, impacts on the fluid, etc.– the number of variables = n 2. Determination of basic dimensions (typically, M, L,T) – the number of basic dimensions = m 3. Expression of variables with the basic dimensions 4. Determination of  groups by combining the variables to form dimensionless parameters = the number of dimensionless groups (  groups) = n - m Step-by-step method (Ipsen’s method) - Elimination of basic dimensions step-by-step - Always yields the correct number of parameters Exponent method - Combine variables as the product - Set up a set of algebraic simultaneous equations consist of the exponents of the variables (based on dimensional homogeneity) - Solve the algebraic simultaneous equations – select the most frequently shown exponents as variables (repeating parameters) which will not be solved - Express the relationship in a dimensionless form Remark - Identification of significant variables → based on experience - Double-inclusion of variables - Omission of significant variables

23. Common  groups in fluid mechanics A hypothetical flow Determination of pressure difference btw. two points, P = f(V, L, , , E v, ,  ) V: velocity, L: characteristic length,  :density,  :viscosity, E v :bulk modulus of elasticity,  :surface tension,  :specific weight → important when free-surface phenomena (waves) exist (open channel, ship model, etc.) General form Step-by-step method (Ipsen’s method) Exponent method 

24. Forces involved in a general flow situation Pressure force Inertial force Viscous force Compressibility force Surface (tension) force Gravity force Dimensionless numbers defined by the ratios of forces involved, Euler number (C p ) = Reynolds number = Mach number = Weber number = Froude number =

25. Experiment in fluid dynamics Prototype – original target structures for the fluid dynamic analysis (real-scale) Model – made for the analysis of performance of prototype via experiments, usually smaller than prototype Similitude – relationships of physical properties btw prototype and model, required for the acceptable prediction of prototype performance from model observations Model design/construction Performance test Analysis of experimental results not considered in the theoretical aspects 1. Geometric similitude ( 기하학적 상사 ) – the most obvious/basic requirement in similitude, match of geometric scales (dimensions, sizes) btw prototype and model 2. Kinematic similitude ( 운동학적 상사 ) – the ratios of physical terms related to the flow (v, a, Q, etc.) must be constant time ratio, velocity ratio, acceleration ratio, flowrate ratio,

26. Satisfaction of similitude If couple of significant  groups were equal, the dynamic (and kinematic) similitude will be met. All involved  groups should NOT necessarily be equal since all forces are neither acting on nor significant (negligible, or sometimes canceled by each other) in particular fluid motions. In general, the equality of 1 or 2 significant  groups satisfy the dynamic similitude 3. Dynamic similitude ( 역학적 상사 ) – the ratios of the forces acting on corresponding masses in the model and prototype must be constant throughout the flow field  similarity in flow pattern

27. Similitude in forces represented by  groups Ratio of pressure forces and inertial forces Ratio of inertial forces and viscous forces Ratio of inertial forces and compressibility forces Ratio of inertial forces and surface forces Ratio of inertial forces and gravity forces

28. Significance of the common  groups Euler number – flows in which pressure drop is important (most flow situations; e.g., closed conduits-no impacts on gravity, surface tension, compressibility → no need for the consideration of Fr, We, and M) Reynolds number – flows that are influenced by viscous effects: internal flows, boundary layer flows etc. (most non-ideal flow situations) Mach number – compressibility is important in these flows, usually V>0.3c (sound velocity) Weber number – surface tension influences the flow, flow with an interface (liquid flow facing gases or solids, bubbles, surface of orifice/weir of which head is very small, etc.) Froude number – flows that are influenced by gravity, primarily free-surface flows in which waves are involved

29. Dimensional analysis for mass transport (reactions) of substances Changes in the concentrations of substances in fluids caused by transport/reaction processes Transport of substances  Macro transport – advection, dispersion  Micro transport – diffusion Advection ( 이송 / 이류 ) – bulk movement of substance mass as a result of cocurrent movement of the fluid elements → substance mass moves along with fluid mass Dispersion ( 분산 ) – mixing of substances driven by exchange of momentum btw fluid elements in a turbulent flow field or different velocity gradient (fluid drag forces) or velocity fluctuation due to irregular flow pattern (in porous media) Diffusion ( 확산 ) – molecular level mass transport, driven by chemical potential (concentration gradient) or Brownian motion

30. Dimensional analysis for mass transport (reactions) of substances Fick’s 1 st law Describes the spatial mass transport (diffusion) of substances Diffusive flux, J x, is proportional to the concentration gradient J x : mass transported per unit area per unit time (ML -2 T -1 ) XiXi X i+1 cici c i+1 Flux, J x

31. Dimensional analysis for mass transport (reactions) of substances Fick’s 2 nd law Describes the spatial and temporal mass transport (diffusion) of substances From mass balance around the cv, dx A

32. Dimensional analysis for mass transport (reactions) of substances The diffusional flux General form After the dimensional analysis (by Ipsen or exponent method)

33. Review of  groups in mass transport Sherwood number = Peclet number = Schmidt number = Dispersion number = Damkohler number = Find correlation coefficients ( , a, b) via experiments Ex) Mass transfer of substances bwt fluid and single sphere