DYNAMIC FLOWS / COMPRESSIBLE FLOWS / MOMENTUM BALANCES

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1 Fluid and Particulate systems 4451 /018 Pages added DYNAMIC FLOWS / COMPRESSIBLE FLOWS / MOMENTUM BALANCES 4 Ron Zevenhoven ÅA Thermal and Flow Engineering ron.zevenhoven@abo.fi 4.1 Dynamic, time-dependent flow april 018 RoNz

2 Dynamic Flow t Two processes Fast process Slow process seconds minutes april 018 RoNz 4 t

3 Slow process Fast process Lowering of level in storage vessel; Lowers tube inlet pressure seconds Slow process minutes Calculate the pseudo-stationary state at several points during the process. Integrate the result to achieve the value of the searched variable. dx change of property X : f ( X ) t t1 dt X X dx 1 f ( X ) 4.1 april 018 RoNz 5 Fast process Fast process Acceleration of the water mass in the of tube: F = ma = Δp/L seconds Slow process minutes The state is changing within the control volume; time derivates of energy need to be accounted for. april 018 RoNz 6

4 Time derivates of energy Potential energy: Internal energy: du dt d m g z dt d m u dt Neglected at non-compressible flow when assuming that the liquid mass in a pipe is constant. Partial energy balance for internal energy Use enthalpy h instead Kinetic energy: d m dt 1 w l A w dw dt 4. 7 Time-variant pressure balance for fast processes g 1 z z w w p p p loss l dw dt 4.3, 4.4 april 018 RoNz 8

5 4. Example: water hammer april 018 RoNz 9 Water hammer l What happens if the valve is suddenly closed? Theoretically: Deceleration of an incompressible fluid and a non-expandable pipe would cause infinite high pressure before the valve. Practically: The kinetic energy of the liquid in motion will compress the fluid and expand the pipe (a slightly bigger perimeter). This results in a high pressure shock wave bouncing back and forth, which is experienced as a loud bang resembling a hammering noise. Friction will, however, decay the wave fast. april 018 RoNz 10

6 Compressibility of a liquid Compressibility For water at 0ºC dv 1 V dp Pa -1 For ideal gases (isentropic compression) c c p v p 4.5, Pressure rise by water hammer Transformation of energy w p 0 c E kinetic W liquid,kompression + W pipe,expansion E internal d E (Unit term.. = s/m) d E w 0 d E speed of sound in the liquid in tube m/s 1 velocity of the liquid before the valve closes m/s Density kg/m 3 For example Compressibility Pa -1 steel pipe pipe diameter m E = Pa coefficient of elasticity Pa at w o = 10 m/s: pipe wall thickness m Δp ~ 10 MPa p w0 c speed of sound in free flow m/s c 1 1

7 Water hammer period Length l What will happen if the valve is suddenly closed? april 018 RoNz 13 Water hammer period FLOW 0 c FLOW 0 l Consider water flowing to the right as a spring. At t = 0, the water is hitting the instantaneously closed valve. c (velocity of the phenomena) t=0 t=½l/c The water flow at the valve will stop and water is compressed, at the velocity of sound c in water. At t = ½l /c, half of the water has stopped flowing and is compressed (kinetic energy flow energy = enthalpy). The rest is still moving to the right. april 018 RoNz 14

8 Water hammer period FLOW 0 c FLOW 0 All the water has stopped flowing at t = l /c and the pressure in the pipe is very high. The water is beginning to flow backwards into the tank, since the pressure is lower there. c t=l/c t=1½l/c Half of the water in the pipe is flowing back into the tank and pressure is reduced (flow energy (enthalpy) kinetic energy). The rest has still a high pressure, which will soon will be released. april 018 RoNz 15 Water hammer period FLOW 0 c FLOW 0 All the water is moving to the left and the high pressure is released at t = l /c. c t=l/c t=½l/c At t = ½ l /c, half of the water has stopped flowing to the left and is stretched, which means that the pressure will become low. april 018 RoNz 16

9 Water hammer period FLOW 0 c FLOW 0 t=3l/c The flow backwards has stopped completely and the pressure in the pipe is very low at t = 3l /c. Water will start to flow forwards into the pipe again. c t=3½l/c Half of the pipe has water flowing to the right, while the other half has a low pressure causing the water to flow at t = 3½l /c. april 018 RoNz 17 Water hammer period FLOW 0 c t=4l/c All the water is flowing towards the valve at t = 4l/c. The situation is equal to t = 0. The phenomenon will decay fast because of friction. dead link april 018 RoNz 18

Critical closing time Water hammer occurs in a pipe if a

com/1-d_water_hammer video # 14 (checked April 018) 4.

10 Critical closing time Water hammer occurs in a pipe if a valve is closed faster than t critical l c More risk for water hammer with long tubes!! See for a simulation of water hammer: video # 14 (checked April 018) 4.7 april 018 RoNz 19 Water hammer arrester Hydraulic ram Hydropower application / april 018 RoNz 0

11 Pressure surges during valve closure Pipe diameter D Cross section A Fluid volume = AL; Fluid mass = ρ A L ; Gradual closure of valve: Decelaration = u/t; t= time for reducing velocity to 0. Inertial force required = F = m a = ρ A L u/t pressure rise Δp = F/A = ρ L u/t april 018 RoNz 1 Pressure surges during valve closure Pipe diameter D Cross section A Fluid volume = A L; Fluid mass = ρ A L ; Sudden closure of valve, with compressibility κ: a shock wave with velocity c = 1/ (κρ) The change in volume ~ change in pressure Average pressure rise is ½Δp; storing compression energy ½Δp ΔV, which is equal to kinetic energy change ½m u pressure rise Δp = cuρ april 018 RoNz

Pressure surges during valve closure Wall (material volume = πdδl Δp may cause pipe swelling.an elastic wall (elasticity modulus E) will lower the pressure rise.

12 Pressure surges during valve closure Wall (material volume = πdδl Δp may cause pipe swelling.an elastic wall (elasticity modulus E) will lower the pressure rise. Kinetic energy will be absorbed by the wall. With wall thickness δ the stress is σ = ΔpD/δ;giving strain energy S = σ /E material volume = Δp DAL/δE Wall strain energy + fluid strain (compression) energy is equal to the kinetic energy lost by the flow which is ½mu = ½ρALu. Fluid strain energy ½ΔpΔV = ½Δp ALκ Δp DAL/δE + ½ΔpΔV= ½ρALu p u D april 018 E RoNz Excercises 4 april 018 RoNz 4

13 Exercises 4 marked** will not be discussed; answers will be distributed 4.1 Two containers, A and B, with cylindrical cross sections have the diameters d A = 4 m and d B = 3 m. They are joined together with a horizontal pipe (l = 0 m, d = 75 mm, wall roughness k = 0.5 mm, Σζ = 4.5). The water level z A in container A is 6.5 m over the pipe, while the water level in container B has the same level (z B = 0m) as the pipe at the time t 0. Calculate the time needed for transporting 8 m 3 water (10ºC) from A to B, under the conditions that only the mentioned pipe is utilized and the storing capacity of B is big enough. 4. Derive the expression for the time derivate of kinetic energy. 4.3 Derive the expression for the time-variant pressure balance for fast processes. ** 4.4 Water (ρ = 998 kg/m 3, kinematic viscosity ν =10-6 m /s) is flowing through a horizontal pipe (l = 100 m, d =100 mm, k = 0. mm) with the velocity w = 3.0 m/s. The static pressure before the inlet is p 1 = 0 kpa and the static pressure after outlet is p = 100 kpa. A valve near the outlet is instantly closed at the time t 0 and the ζ valve value changes then from 1 to 50. Calculate and illustrate the velocity w during the first 4 seconds after the valve was closed. Calculate the pressure in front of the valve at the time of closing. april 018 RoNz 5 Exercises A volume of 1 m 3 water at 0ºC and 100 kpa is compressed by %. a) What is the pressure at the end of the compression? b) How much theoretical compression work is needed per kg of water? c) How much does the specific enthalpy of the water increase? **4.6 To what value will the pressure rise in a 400 m long pipe (d=100 mm) where water at 0ºC is flowing with a velocity of 5 m/s if a valve in the far en is closed immediately? Assume that all kinetic energy of the moving water is used for compression of the liquid. 4.7 a) What is the speed of sound in water at 0ºC in a 400 m long steel pipe (d =100 mm, δ = 5 mm, E = Pa)? b) What is the critical closing time for a valve in the far end? c) What pressure increase can be expected if the valve is closed too fast and the water is flowing with a velocity of 5 m/s? april 018 RoNz 6

14 4.4 Compressible flow april 018 RoNz 7 Compressible Flow static pressure static pressure Non-compressible flow velocity velocity

15 Compressible fluid p is high specific volume v is low (density high) p is low specific volume v is high (density low) Concerns gases. Liquids are usually approximated as non-compressible. april 018 RoNz 9 Gas flow in a pipe calculations static pressure velocity 1. p decreases v increases V increases velocity w increases The normal pressure balance can be used if Δp = p 1 p < 0%. Use then w and ρ pertaining at p mean ; gas behaves like liquid Calculate as a connection in series. Constant values of w and ρ can be assumed for each section. Temperature changes due to heat transfer can also be accounted for. april 018 RoNz 30

16 Calculations 3 Consider the case in which the temperature is constant and the changes in the potential and kinetic energies between the in- and outlet can be neglected. The decrease in static pressure dp for a thin slice dl of the pipe (duct) is equal to the loss of pressure due to fluid friction. d w dp dl d The velocity w of a flowing gas can anywhere in a pipe be expressed as a function of the pressure p and the state at the inlet p 1, w 1 (alternatively the state at outlet p, w ), using the ideal gas law, for molar mass M kg/kmol:. p V M n n R T V M p R T The density ρ of a flowing gas can likewise anywhere in a pipe be expressed as a function of the pressure p and the state at the inlet p 1, ρ 1 (alternatively the state at the outlet p, ρ ), according to the ideal gas law. 31 Calculations *see pages continued d w Integration of dp dl and expressing w=f(w d 1,p 1,p), and ρ = f(ρ 1,p 1,p), the pressure at the outlet p can be calculated if the state of the inlet 1 is known according to p p1 p ( p1 loss,1) where p loss,1 is the fictive pressure loss if the state at the inlet would pertain for the whole pipe. 3

17 Calculations 3 continued d w Alternatively, integration of dp dl and expressing d w = f(w, p, p) and ρ = f(ρ, p, p), the pressure at the inlet p 1 can be calculated if the state of the outlet is known according to p 1 p p ( p loss,) where p loss, is the fictive pressure loss if the state at the outlet would pertain for the whole pipe. 33 Mass flow always constant

18 Temperature constant? An isentropic expansion of an ideal gas would decrease the temperature in an isolated system. An isentropic compression of an ideal gas would increase the temperature in an isolated system. Isentropic changes from a beginning state 1 to an end state can be described by 1 1 p 1 V1 p V T p 1 / 1 / 1 T p T V T v c c p V Temperature constant? The expansion of a flowing ideal gas in a pipe is not isentropic since the pressure loss due to fluid friction involves an entropy generation. Temperature decrease due to expansion Temperature increase due to fluid friction Changes in potential and kinetic energies are neglected. temperature remains constant The temperature increase for a real gas, e.g. air, can also be neglected. 5.3, 5.4 april 018 RoNz 36

19 4.5 Excercises 5 april 018 RoNz 37 Exercises 5 marked** will not be discussed; answers will be distributed, 5.1 a) Derive the expressions for p 1 ( 1 ploss,1 p p ) p p p ) 1 ( ploss, b) Can it be assumed that the values of Re and, for a compressible gas, are constant although the pressure decreases? c) In what cases can the change in kinetic and potential energies, between the inand outlet of the control volume, be neglected? 5. Explain qualitatively what temperature and pressure changes occur after the valve has been opened. A p A,1 = p B,1 B T A,1 =T B,1 april 018 RoNz 38

20 Exercises What is special with the enthalpy of a flowing gas in a pipe? **5.4 Air with a pressure of 600 kpa and a temperature of 00ºC is transported through a pipe (l=500 m, d=50 mm, roughness k=0.05 mm). The pressure at the outlet is 100 kpa. a) Calculate the mass flow rate under the assumption that the process is isothermal at 00ºC. b) Calculate the mass flow rate in a case where the surrounding air at 0ºC has a cooling influence; the heat transfer coefficient related to the inner surface of the pipe is assumed to be 3.0 W/(m K). Calculate the pressure distribution along the pipe. Pressure losses due to inlet, outlet and valves can be neglected. april 018 RoNz Momentum balances, the ejector april 018 RoNz 40

The pump medium b is sucked into the mixing space due to the high momentum flux provided by the drive medium a.

21 Applied Momentum Balances The Ejector a b nozzle mixing space diffuser The drive medium a is supplied at high pressure and flow through the nozzle at high speed into the mixing space. The pump medium b is sucked into the mixing space due to the high momentum flux provided by the drive medium a. The drive medium a and pump medium b are mixed (s) in the mixing space and the pressure increases in the flow direction. A further increase in the pressure is achieved in the diffuser, where kinetic energy is partly transformed into flow energy. s april 018 RoNz 4

22 a The Ejector 0 0 b nozzle 1 mixing space diffuser If the drive medium a is steam, which condensates in the mixing space, the outlet pressure p 3 might reach values higher than the inlet pressure p a0 of the steam. This kind of ejector is called injector, which can be used as feed pump for steam boilers. This is an easy and inexpensive pump to manufacture, and the operation is safe since it does not contain any moving parts. It has, however a low efficiency and it might be a negative aspect that the drive medium a is mixed together with the pump medium b. 3 s april 018 RoNz 43 The Ejector Energy and momentum balances can be utilized as tools for picturing the working boundaries and the efficiency of the ejector when put into practice. Following ideal conditions are assumed: - densities are constant for the drive medium ρ a and the pump medium ρ b - no losses due to fluid friction in the mixing space - kinetic energy and momentum ξ coefficients are assumed to be 1 - ideal inflow; no losses when flow energy is transformed to kinetic energy at boundary 1 - constant diffuser efficiency (80%) april 018 RoNz 44

23 The Ejector a 0 0 nozzle 1 mixing space diffuser 3 s b The Ejector The following equations can be utilized for calculations of ejectors. w w a1 b1 w 7. pa0 p a b 1 pb0 p 1 b1 s wa1 1 w p s A A nozzle mixing space w a1 s a 1 w w b1 w p1 a wa1 1 b b1 s ws p A A n cs 3 p. 4 ws e w 0 w a1 (m /m ) s a0 s p3 pb0 p p b b0 46

24 The Ejector (Injector) If the drive or pump medium (or both) are gaseous (steam), changes in the specific volume (density) have to be accounted for. In such case it might be appropriate to utilize energy and entropy balances for rough estimates of the ejector performance. 7.3 april 018 RoNz Momentum balances, propellors and wind turbines see also ÅA course New Energy Technologies and REN1 annual reports on renewable energy for wind energy status REN1 april 018 RoNz 48

25 Wind energy Betz Law Theoretically (Betz Law) a wind turbine can exploit 60% (16/7) of incoming wind power. In practice it is ~40%, due to (exergy) losses, mainly as heat, for example friction between rotor shaft and bearings, cooling fluids in gear box and generator. Pics: april /64 Power input P > 0 (boats, airplanes) 1 w 1 undisturbed input velocity Input power P Input axle force F w velocity at a distance of ~ D p after the propeller w p velocity through the propeller D Sweep area of Ap the propeller: 4 p D p 1 april 018 RoNz 50

26 Power output P < 0 (wind power plants) Output power -P Output axle force -F w 1 undisturbed input velocity 1 w p velocity through the propeller w velocity at a distance of ~ D p after the propeller : mass flow if fluid through the propeller 1 april 018 RoNz 51 Energy and momentum balances The simplified theory is based on the assumption that the propeller power is used for changing the kinetic energy (without losses) of the mass flow going through the sweep area of the propeller. Idealizations: -The propeller is considered as a pump (η =100%) in a pipe with the diameter D p and the length 0 - The kinetic energy coefficients ξ are assumed to be 1 - Neglecting that the propeller jet is spinning around its axis (= power loss) - Neglecting the influence of the number of propeller wings - The pressure is constant along the balance boundaries. However, there is a pressure difference Δp on both sides of the propeller distributed over the whole sweep area 7.4 april 018 RoNz 5

27 Velocity w p through the propeller 53 Betz Law (196) P is power extracted from the wind P ρ 4 air w1 w (w1 w ) Ap ½ρair w1 w ½(w1 w ) Ap Power of undisturbed flow through area A p ρair 3 P0 w1 Ap F w Δp A w Efficiency of power exraction from wind P P 1 1 w 0 w1 w1 with maximum 16/7 at w /w 1 =1/3 w (1 ) april 018 RoNz 54

28 Propeller pitch s (sv: stigning) w p s n n: rotational speed 7.5 april 018 RoNz 55 Recent developments See S Langton Fitting a pitch Mechanical Engineering (ASME) December 009, p april 018 RoNz 56

Savonius turbines @ Vaasan yliopisto http://www.savonius.net/project-background.html Efficiency etc.

In practice, constant 16/7 0.36 Angular frequency ω v/r (depending on turbine). For example, r = 0.

info/savonius-tuulivoimala-saa-lisaa-tehoa/ (dr thesis S Marmutova, May 016) https://www.researchgate.

http://vaasalaisia.info/wp-content/uploads/016/05/savonius_tuuliv-oimala_vaasan_yliopisto.

shipping with one rotor on a 9700 DWT were.6%, assuming potential of ~0% with several rotors.

29 Savonius Vaasan yliopisto Efficiency etc. follow from Betz law assessment, power : with fluid density ρ, rotor height h, radius r and fluid flow velocity v. In practice, constant 16/ Angular frequency ω v/r (depending on turbine). For example, r = 0.5 m, h = 1 m, v = 10 m/s, ω 0 rad/s ~ 190 rpm, gives P ~ 180W SOURCES: (dr thesis S Marmutova, May 016) april 018 RoNz 57 Rotor sails, rotor ships Principle of operation: Magnus effect Norsepower : Fuel savings for shipping with one rotor on a 9700 DWT were.6%, assuming potential of ~0% with several rotors. SOURCES: ArticleID/1059/Shipping-Company-Takes-Rotor-Sail-for-a-Spin.aspx april 018 RoNz 58

30 4.8 Excercises 7 april 018 RoNz 59 Exercises 7 marked** will not be discussed; answers will be distributed 7.1 Derive equations for evaluating the ejector pump in practice. a 0 0 b nozzle 1 mixing space diffuser 7. An ejector pump is used for emptying a flooded basement floor. The drive medium consists of tap water with a pressure p a0 of 900 kpa. A suitable inlet pressure p b0 of the pump medium would be 80 kpa, and the outlet pressure p 3 is expected to be 130 kpa. The tap water flow is 1 l/s. Design the ejector (calculate A cs and A cs,a ) that maximizes the mass flow rate of the pump medium. 3 s april 018 RoNz 60

31 Exercises 7 **7.3 A steam injector is consuming kg/s saturated drive medium steam with a pressure of 40 kpa for compression of 0.10 kg/s wet steam (containing % water) with a temperature of 15ºC to the pressure of 4.0 kpa. a) Calculate the temperature of the out coming steam at 4.0 kpa b) Calculate the efficiency of the injector by comparing the turbine power, which the drive medium steam (0.105 kg/s) would theoretically give off at an isentropic expansion to 4.0 kpa, with the compressor power, which would be needed for isentropic compression of 0.10 kg/s steam to 4.0 kpa. 7.4 Derive momentum balances for a) 1 w 1 undisturbed input velocity Input power P Input axle force F w velocity at a distance of ~ D p after the propeller april 018 RoNz 61 Exercises Derive momentum balances for a) b) Output power -P Output axle force -F w 1 undisturbed input velocity 1 w p velocity through the propeller w velocity at a distance of ~ D p after the propeller 1 c) What is the relationship between the power (input or output) P and the pressure difference Δp on both sides of the propeller? april 018 RoNz 6

32 Exercises 7 **7.5 A wind motor has a propeller diameter of 6 m. a) How much axle power can theoretically be achieved if the wind flow (0 ºC) is w 1 8 m/s and the wind velocity w (ca 6 m after the propeller) is m/s. b) What will the rotational speed be if the propeller pitch is 15 m. c) Is w = m/s optimal? What w would be optimal for reaching maximum power output? d) Show that the maximum power output is 16/7 (=59%) of the kinetic energy that an air flow is possessing, that undisturbed with the velocity w 1 is flowing through a ring with the propeller diameter D p. april 018 RoNz 63 Further reading DWIA09 Proof of Betz Law Danish Wind Industry Association KJ05: D. Kaminski, M. Jensen Introduction to Thermal and Fluids Engineering, Wiley (005) Chapter 4, chapter 10 L09: S Langton Fitting a pitch Mechanical Engineering (ASME) December 009, p (as pdf at course material www) SWE = magazine Sun & Wind Energy T06: S.R. Turns Thermal Fluid Sciences, Cambridge Univ. Press (006) Z13: R. Zevenhoven Introduction to process engineering / Processteknikens grunder (PTG), # 6: Fluid mechanics: fluid statics, fluid dynamics. Course material ÅA (version 013) available on line ÖS96: G. Öhman, H. Saxén Värmeteknikens grunder, Course compendium ÅA (1996) Ö96: G. Öhman Teknisk strömningslära Course compendium ÅA (1996) sections 3, 4, 7 (p. 8-14, -6) CEWR10: C.T. Crowe, D.F. Elger, B.C. Williams, J.A. Roberson Engineering Fluid Mechanics, 9th ed., Wiley (010) REN1: Renewables 017 Global Status Report (GSR 017) (30 pages) (links to Figures etc.) april 018 RoNz 64

33 Add: compressible flow pressure drop april 018 RoNz 65 Add: compressible flow pressure drop april 018 RoNz 66

34 Add: compressible flow pressure drop april 018 RoNz 67 Add: compressible flow pressure drop april 018 RoNz 68

INTERNAL FLOWS / FLOWS IN DUCTS

Fluid and Particulate systems 4451 /018 INTERNAL FLOWS / FLOWS IN DUCTS Ron Zevenhoven ÅA Thermal and Flow Engineering ron.zevenhoven@abo.fi.1 Flow tube sections in series, in parallel, or branched RoNz