INTERNAL FLOWS / FLOWS IN DUCTS
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1 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
2 Fluid Flow in Tube Systems Pressure balance: g z in Mass balance: 1 1 p in win pin ppump g zout out wout pout loss The pressure must be unambiguous, i.e. several different pressures can NOT exist in one specific point of the system. RoNz 3 Tube connection in series The pressure losses are added. Be aware of that the flow velocity w changes (and hence Re) when the cross-sectional area changes..1 RoNz 4
3 Tube connection in parallel p p p p p p loss,1 pump,1 loss, pump,... loss, n 1 p p g z z i w w loss,i pump,i in,i out,i out,i with the right hand side equal for all branches "i" in,i p in,i pump, n p out,i. RoNz 5 Balance boundaries w in p in z in in.3 RoNz 6
4 Similarities for connections in parallel Shell-and-tube heat exchanger p loss,1 ploss, ploss,3... ploss,n.4 RoNz 7 Branched tube connection g z in 1 1 p in win pin ppump g zout out wout pout loss for each branch of the system a set of equations. Note that pressures p i, heights z i, and/or pressures p i can be different.5 RoNz 8
5 Branched tube connection p ' 1 3 g z3 w 3 p3 g z p p Define: ' 0 p0 p3 ploss,03 ' 3 g z1 p1 ploss,31 ' 3 g z p ploss,3 p p p For a fully developed turbulent flow (horizontal lines in Moody diagram): ' 3 g z0 p0 ploss,03 ' 3 g z1 p1 ploss,31 ' 3 g z p ploss,3 Solve set of 4 expressions for.5 RoNz 9 Example: sprinkler hose In drier regions, large central pivot sprinkler systems are used to irrigate large areas. Consider the simplified schematic of a portion of such a sprinkler (shown below). Water at 10 C is pumped through the spray arm that is constructed of.5-cm diameter galvanized iron. The flow area of each nozzle is 1.5 cm. The pressure at the first nozzle is 50 kpa (gage). Ignoring friction in each nozzle but not in the connecting lengths of pipe, determine the flow rate through the sprinkler (in m 3 /s). Assume four nozzles as shown in the figure. Source: KJ05 RoNz 10
6 Example: sprinkler hose Method: The steady, incompressible flow energy equation can be used. The total flow is such that all the pressure head at location 1 is used to balance losses in the galvanized pipe and the creation of kinetic energy at the nozzle exits. The flow through each nozzle is different and is estimated using the Bernoulli equation. Because the line losses are a function of flow rate and friction factor, an iterative solution is required. If losses in the line and nozzles can be neglected, it holds for nozzle i compared to inlet 1 : where V ji is the velocity through nozzle i. With velocity V 1 ~ 0 (<< V ji ), horizontal lines (all z i the same), and at the nozzle outlet P j 0 (gage pressures!), it is found for nozzle 1: RoNz 11 Example: sprinkler hose For the volume flows it holds that: For the local flow velocities:, etc. For the pressure drop between the nozzles: for hose diameter D and distance L between the nozzles, similar for V 3 and P 3, etc. 11 equations, 11 unknowns: For given case: P P ½ 4 f V 1 1 D L RoNz 1
7 . Pump combinations RoNz 13 Pump (compressor) types Axial flow pumps: similar to a propellor, transforming rotational energy into thrust. Suitable when low head (pressure increase) for large flows is needed. (Compressor = Fan) Radial flow machines: Suitable when high head (pressure increase) for low flow rates are needed. Centrifugal pumps / compressors: pressure increase results from rotation (pressure increases outward) and from the decreasing velocity in the expanding section. (Turbines: extract energy from a moving fluid) pictures: TO6 RoNz 14
8 different impellers Picture: CEWR10 Pump characteristics & performance - examples 15/90 NPSH = net positive suction head = difference in pressure between suction side of the pump and the vapour pressure of the liquid = minimum pressure at intake while avoiding cavitation A well pumping sytem Pump characteristics & performance - examples different impellers Pictures: CEWR10 16/90
9 Pump combinations Produce a combined pump characteristic treat as one pump.6 RoNz 17 Pump combinations Pump network characteristics for two pumps in parallel or in series pic CEWR10.6 RoNz 18
10 Pump combinations A A B dp_pump [kpa] A and B in series A B V [dm 3 /s] dp_pump [kpa] dp_pump [kpa] B A B V [dm 3 /s] A 400 B 300 A and B in parallel V [dm 3 /s] RoNz 19.3 System boundaries, unknown flow direction RoNz 0
11 Pressure losses in T-joints system boundary slightly upstream (similar for outlet) w in p in z in in Neglected since other sources of errors (e.g. estimation of the relative tube wall roughness k) most likely affect the calculations more. Alternatively: simplified geometry: Approximated Approximated losses losses RoNz 1 Unknown flow direction Volume flow m 3 /s, velocity m/s or mass flow kg/s can be < 0 (i.e. in opposite direction) Note that for the pressure drop calculations the velocity squared, w is always > 0! Use product w w?.7 RoNz
12 .4 Cavitation RoNz 3 Cavitation Formation of vapor pockets in a liquid when the local ambient pressure at a point in the system falls below the vapor pressure of the liquid Negative pressure is not possible. Liquid flow may stop. The cross-section of the tube is not filled with liquid with constant density.8 RoNz 4
13 Cavitation PTG / PRC Pictures: CEWR10 Cavitation occurs if fluid pressure is reduced to the vapour pressure (at the given temperature) so that boiling occurs. The formation and collapse of bubbles gives shock waves, noise, and other problematic dynamic effects that can result in reduced performance, failure and damage. Typically occurs at high velocity locations in, for example, pumps or valves, but can damage also tube walls. 5/90.5 Flow or flow velocity measurement RoNz 6
14 Flow or flow velocity measurement Bernoulli s equation allows for designing or assessing the performance of flow meters (since, of course, large pressure drop isn t allowed!) One example is a stagnation tube RoNz 7 Flow or flow velocity measurement Another example of a flow meter, also with negligible (viscous) losses so that Bernoulli s expression can be used, is a Venturi meter More accurately with discharge coefficient C, 0.94 < C < 0.99 correcting for friction losses, non-uniform velocities, construction,... RoNz 8
15 .6 Exercises RoNz 9 Exercises a) Investigate the possibility to add the power losses for tubes connected in series. b) How are the pressure losses calculated in the joints?. Derive the expression for ploss,1 ppump,1 ploss, p pump,... ploss, n ppump, n.3 Calculate the flow velocities and the pressure losses in the tube system below. The total volumetric flow of water (10ºC) is 40 dm 3 /s. The relative roughness k of the pipe wall is 0. mm. RoNz 30
16 Exercises A volumetric flow of 145 dm 3 /s is pumped upwards through a tube heat exchanger, which consists of 60 parallel vertical tubes (l=5.5 m). Each tube has an inner diameter of 15 mm and the relative roughness of each tube wall is 0.05 mm. The water temperature will increase from 0ºC to 90 ºC. The pressure before the pump is 10 kpa, and the pressure in the tube after the exchanger is 35 kpa. The level between the measuring points of the static pressure is 6.75 m. The water flow is led into and out of the heat exchanger through a connection tube (total length 3 m) with an inner diameter of 5 mm (k=0.05 mm), including two 90º bends. How much pump pressure and theoretical pump power is needed to maintain the volumetric flow of 145 dm 3 /s?.5 Water at 0ºC is flowing through the branched connection according to the figure below. The length of each tube is l 0 =10 m, l 1 =6 m and l =1 m. The tubes have the same diameter d=3 cm and the relative roughness of the tube wall is k=0.05 mm. The static pressure is measured at each point, p 0 =150 kpa, p 1 =101 kpa and p =101 kpa. The level of each point is z 0 =5.0 m, z 1 =3.0 m and z =0.0 m. Calculate the volumetric flow in each branch. RoNz 31 Exercise.6.6 The characteristic for two different centrifugal pumps A and B can be described by following values: a) A and B are connected in series. What will the total pump characteristic for this combination be? b) A and B are connected in parallel. What will the total pump characteristic for this combination be? c) Design a connection system with the two pumps and valves resulting in following possibilities to utilize the pumps,1) A alone, ) B alone, 3) A and B connected in series, and 4) A and B connected in parallel RoNz 3
17 Exercises Derive a scheme for flow calculations in the vertical tube below. The scheme should be is independent of the flow direction.?.8 A volumetric flow of 1 dm 3 /s water at 10ºC is to be pumped from a well with a centrifugal pump placed on the ground. The water flows through a tube with a diameter of 4 cm and the relative roughness is mm. a) At what depth will this system still work? b) Can another design enable utilization of an even deeper well? RoNz 33.7 Flow in non-circulars ducts, open channels RoNz 34
18 Flow in non-circular Ducts How to calculate Δp loss and P loss for this duct? w RoNz 35 Assumption Losses are proportional to the inner wall surface b a l w Wetted perimeter: l perimeter = a+b Inner surface: A inner surface =l perimeter l RoNz 36
19 Balances Momentum balance: p 1 p 1 Acs p Acs F p F 0 : shear stress w The friction force (resistance to flow) is also proportional to the inner surface F A inner surface more correctly : F na inner surface RoNz 37 Proposed solution For calculation of losses, use a circular fictive pipe where the same shear stress occurs. The pressures, p 1 and p, at the in- and outlet, the velocity w, the pipe length l, the fluid density ρ, the dynamic viscosity η and the relative roughness of the pipe wall k are all assumed to be the same for the duct and the circular fictive pipe. The hydraulic equivalent diameter of the circular fictive pipe is chosen so that the shear stress is equal in both cases. p 1 l p l p w w p 1 d h RoNz 38
20 Derivation of hydraulic diameter d h nc w p 1 l p p A p A A 0 p1 Acs p1 Acs Ainner surface 0 1 cs,nc 1 cs,nc inner surface,nc nc Momentum balances p1 p Acs, nc l lperimeter,nc nc 0 p1 p Acs,nc nc p p l perimeter,nc l nc d 1 l l A perimeter p p A p p h 1 l perimeter,nc cs,nc l 4 A l cs,nc perimeter,nc 1 p p A l l 0 cs 1 cs perimeter p 1 w d l 4 d p1 p l d h d h p h l h 4 p1 dh p l 4 cs = cross-section nc = non-circular RoNz 39 Pressure losses Note that it is the average velocity w that is assumed to be equal for the non-circular duct and the fictive pipe with the hydraulic equivalent diameter d h. Use d h in the calculations for the pressure losses. Note: errors up to 40% for laminar flow, up to 15% for turbulent flow, must be expected with the hydraulic diameter concept! wd Re k / d log 3.7 h h.51 Re l w ploss d h (or another expression) 3.1 RoNz 40
21 Flow in open channels The adjacent layer of air does not cause any shear stress. The air layer is hence not included when determining the wetted perimeter l perimeter. a b 3., 3.3 RoNz 41 Hydraulic diameter PTG / PRC The ratio A/S (unit: m) is a characteristic dimension of the tube, pipe, duct or channel known as hydraulic radius, while 4 A/S is known as hydraulic diameter D h (see Figure below) with A = cross-sectional area (sv: tvärsnitt); S = perimeter (sv: omkrets) touched by fluid For example for a round tube with diameter D, completely filled with fluid: D h = D; for a square channel with width W, fluid height H: D h = 4 A/S = 4 (H W)/(H+W) 4/84
22 Open channel flows Important characteristics: Three-dimensional velocity distribution Max. velocity typically in the mid-plane, ~0% of depth below the surface Secondary motion, which can be intensified by centrifugal forces Flow = v(x) A(x), with velocity v, cross section A and position x pic: W99 RoNz 43 Open channel flows - Manning Many large channels are rough use fully-rough turbulent flow limit / Moody diagram The hydraulic diameter (or hydraulic radius R h ) concept is seldom used for open channel flows equations by R. Manning for open channel hydraulic are used, with S 0 = tan(angle with horizontal) : Ref: W99 and roughness parameter n see Table next page See W99 chapter 10 for more detail RoNz 44
23 Open channel flows Manning experimental n factor Ref: W99 Note: n can vary with depth, bottom erosion, season (vegitation), etc. For example: Mississippi river: depth 40 ft: n = 0.03 depth 0 ft: n = 0.03 depth 5 ft: n = 0.04 RoNz 45.8 Exercises 3 RoNz 46
24 Exercises How much pressure increase does a fan need to cause for maintaining a flow of 3. kg/s air at 0ºC through an 18 m long bricked channel with a rectangular cross section of 0.35 m x 0.55 m? The channel includes two 90º knees. The inand outlets are not rounded and the relative roughness of the wall is estimated to mm. The pressure in the two rooms that the channel joins is 101 kpa. 3. Derive an expression for d h for following channel. 3.3 Calculate the slope that maintains a volumetric flow of.0 m 3 /s in a rectangular channel with the width (b) of 1.0 m. The relative roughness of the wall is 0.5 mm and the water level (a) is 0.5 m. RoNz 47 Further reading BMH99: Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. Transport phenomena Wiley, nd edition (1999) BSL60: R.B. Bird, W.E. Stewart, E.N. Lightfoot Transport phenomena Wiley (1960) KJ05: D. Kaminski, M. Jensen Introduction to Thermal and Fluids Engineering, Wiley (005) Chapter 4, chapter 9 T06: S.R. Turns Thermal Fluid Sciences, Cambridge Univ. Press (006) Z13: R. Zevenhoven Introduction to process engineering / Processteknikens grunder, Chapter 6: Fluid mechanics: fluid statics, fluid dynamics. Course material ÅA (version 013) available on line: ÖS96: G. Öhman, H. Saxén Värmeteknikens grunder, Åbo Akademi University (1996) Ö96: G. Öhman Teknisk strömningslära Course compendium ÅA (1996) section 1 and (p. 1-7) CEWR10: C.T. Crowe, D.F. Elger, B.C. Williams, J.A. Roberson Engineering Fluid Mechanics, 9th ed., Wiley (010) W99: F.M. White Fluid mechanics, 4th Ed. McGraw-Hill (1999) chapter 10 RoNz 48
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