Chater 10: Flow in Conduits By Dr Ali Jawarneh Hashemite University 1
Outline In this chater we will: Analyse the shear stress distribution across a ie section. Discuss and analyse the case of laminar flow in ies in articular. Distinguish between laminar and turbulent flows. Discuss and analyse the case of turbulent flow in smooth and rough ies. Analyse the head loss due to ie entrance, elbows and fittings. Analyse flow in non-circular conduits.
10.1: Shear-Stress Distribution across a Pie section Considering the control volume shown in the figure, and assuming that the flow is uniform: The net momentum flow through the control volume is equal to zero. 3
Shear-Stress In Pie Section The momentum equation for the control volume is: F s 0 A d Δ s A Δ W sin α τ ( π r ) Δ s 0 ds d ds dz Δ sa γ A Δ s τ ( π r ) Δ s 0 ds 4
Shear-Stress In Pie Section Re-arranging and simlifying: τ r d( γz) ds Note that the shear stress is zero at the centre of the ie and is maximum (ositive value) at the ie wall. 5
10.: Laminar Flow in Pies For laminar flow, shear stress can be substituted with μ d/dy: d r d( γz) μ dy ds d r d( γz) dr μ ds Solving this differential equation with the condition (r 0 )0: r r 4μ d( ds 0 γ z) Parabolic eqn 6
Laminar Flow in Pies 7
Laminar Flow in Pies The flow rate can be found by integrating the velocity exression over the crosssectional area: Q r 0 r 0 r 4 μ d( γz) π rdr ds 0 Q 4 π r d( 8μμ ds 0 γ z) 8
Laminar Flow in Pies Dividing by the cross sectional area, the mean velocity can be given as: r 0 d ( γ z ) 8μ ds Re-arranging: d( γz) ds 3 μ D 9
Laminar Flow in Pies Integrating between sections 1 and : γ γ 1 z 3 1 z μ L γ D Comaring with the energy equation, an exression for the head loss due to frictional resistance in a ie can be found: h f 3 μ L γ D 10
10.3: criterion for Laminar or Turbulent Flow in a Pie Reynolds related the onset of turbulence in a ie to a non- dimensional number he named after himself: Re ρd μ 11
Laminar s Turbulent Flow He found that: For Re<000 the flow is laminar For Re>3000 the flow is turbulent For 000>Re>3000 the flow is in transition state, changing back and forth between laminar and turbulent. However, he found that under carefully controlled conditions with no vibrations which is and ideal case not existing in engineering g alication laminar flow could be maintained for Re>000. 1
10.4:Turbulent Flow in Pies A universal equation that can be used to calculate late the head loss due to viscous s effects in ies is the Darcy-Weisbach equation: Major loss due to internal friction inside conduits L alid for laminar and turbulent flow h f f D g f : resistance coefficient i or friction i coefficient i For laminar flow it can be easily shown that: f 64 / Re 13
Turbulent Flow in Pies For turbulent flow in smooth ies, an emirical relationshi can be used (develoed by Prandtl): 1 f log ( Re f ) 0.8 For rough ies, relative roughness (k s /D) which is defined as the ratio of the grain size (k s ) to the ie diameter becomes an imortant design factor. alues for grain roughness (k s ) for various ie materials are shown in Table 10.. 14
Table 10. 15
Turbulent Flow in Pies The Colebrook-White equation is an emirical equation that was develoed to evaluate the friction factor for rough ies. The moody diagram (Figure 10.8) was develoed on the basis of this equation. This diagram is widely used in most engineering g alications that incororate flow in conduits. 16
Figure 10.8 17
Turbulent Flow in Pies Instead of using the diagram, an exlicit equation for the friction factor can be used for turbulent flow in case of alying numerical solutions: f 0.5 5.74 log ks 10 0.9 3.7 Re D This equation is also based on the Colebrook-White equation. 18
Turbulent Flow in Pies Also, exlicit equations can be used for either the flow rate through the ie or the diameter of the ie: Q ks / L log 3.7D D 5/ 1.78ν.D ghf 3/ gh f / L D 4.75 5. 0.04 1.5 LQ 9.4 L 0.66 k s ν Q gh f gh f 19
10.5:Flow at Pie Inlets and Losses from Fittings If the inlet to a ie is well rounded, the boundary layer will develo from the inlet and grow in thickness until it extends to the centre of the ie. The length of the boundary layer develoment region can be given aroximately by: Le 0.0505 D Re, for laminar flow. Le 50 D, for turbulent flow. Le entrance Length 0
Flow at Pie Inlets and Fittings If the ie inlet is abrut, searation occurs just downstream of the entrance, this clearly causes relatively higher head loss. Considerable head loss is also roduced in bends and elbows, due to searation that haens downstream of the midsection. 1
Flow at Pie Inlets and Fittings The head loss roduced by inlets, outlets, elbows or fittings is exressed as: h L K g Where K is the loss coefficient. Different values for the loss coefficient (K) are given in Table 10.3.
Table 10.3 3
Note:in The Exansion if the angle is θ 180 0 you have two choices for the head loss: 1- Use the above equation - Use the abrut exansion equation 4
Summary of the Energy equation 1 z 1 1 α1 h z α ht hl γ g γ g Total loss Major loss Minor Loss h L h L, major h L, minor Li i L i j i i j h f K D g g j 5
Examle: Liquid flows downward in a 1-cm, vertical, smooth ie with a mean velocity of.0 m/s. The liquid has a density of 1000 kg/m 3 and a viscosity of 0.06 N.s/m. If the ressure at a given section is 600 ka, what will be the ressure at a section 10 m below that section? Solution: ρd 1000 0. 01 Re 333. 33 μ 0. 06 γ 1 α 1 g 0 0 γ g 1 z1 h z α La min ar -10 0 64 L 10 f 0.19 h.. m Re L f 0 19 39 14 D g 0. 01 g h t h L 314. 1 ka 6
Examle: Kerosene (S0.8 and T68 0 F) flows from the tank shown and through 3/8 inch diameter (ID) tube. Determine the mean velocity in the tube and the discharge. Hint: include the major loss only. 7
Solution: assume a laminar flow 3 g 3., γ 6. 4 0. 8 Ibf/ft, ρ 1. 94 0. 8 slugs/ft μ 4 10 5 Ibf s/ft (Table A. 4 ) 3 L 10 ft, D ft 8 1 0 0 0 0 0 γ 1 g g 1 α1 z1 h α z L h L f D g γ 64 64 f Re ρd μ 8. 45 16. 1 0. 0 1. 60 ft/s h t 3-0.5 0 3 Now check Re 1. 94 1. 60 ( ( ) ρd Re 8 1 194 5 μ 4 10 Q A 1. 60 π ( 1/ 3 ) 4 1. 8 10 3 cfs h L La min ar 8
Examle: Water flows in the ie shown, and the manometer deflects 80 cm. What is the resistance coefficient (friction coefficient) for the ie if 3 m/s? 9
Solution: 10 0 11 0 ( ) L t h h z g α γ h z g α γ 1 1 1 1 L 3 4 Eq.(1) g. f g D L f h L 3 05 0 4 M t ti Manometer equation: a ) y ( ). ( S ). y ( w w w 158 1 8 0 8 0 1 γ γ γ Eq.() a 158 1 From Eq.(1) & () f 0.033 30
Examle: A water turbine is connected to a reservoir as shown. The flow rate in this system is 5 cfs. What ower can be delivered by the turbine if its efficiency is 80%? Assume a temerature of 70 0 F. 31
Solution: 5 6. 369 ft/s π ( 1/ 1 ) Re 4 1 6.369 ( ) D 1 6 10 5 ν 1.0615 10 1 1 α z h 1 1 α z ht hl γ g γ g 5 Turbulent Table A.5 at 70 0 F 0 0 1-100 0 0 0 Eq. (1) L 6. 369 1000 6. 369 hl K e f 0. 5 f g D g 3. 1 / 1 3. Pie entrance at r/d0.0 Shar edge (Table 10.3) Table 10. k s 0. 00" D 1" 0. 00016 f 0.015 5 Re 6 10 Figure 10.8 3
From Eq. (1): W& W& mg & γq W& t W& t mg & γ Q h. ft t 89 6 1 1 h h 550 Ibf.ft/s 745. 7 W h h t η Pum Head fluid γqh wt shaft shaft shaft W shaft Turbine Head W Pum effeciency W W η t γqh Turbine effeciency W t fluid W t Q γ h t η 5 6. 4 89. 6 0. 8 364. 16 ft.ibf f / s P W t 550 364. 16 550 40. 66 ( h ) horseower 33
Examle: Both ies shown have an equivalent sand roughness k s of 0.1 mm and a discharge of o.1 m 3 /s. Also D 1 15 cm, L 1 50 m, D 30 cm, and L 160 m. determine the difference in the watersurface elevation between the two reservoirs. 34
Solution: 6 o ν 10 m / s @ T 0 C Q Q. m/s 15 5 659 30 1. 415 m/s A A30 15 Moody diagram(fig. 10-8): 15D15 5 30D30 Re15 8. 49 10 Re 4. 4 ν f ν 15 0. 0185 k 0.1 k 0.1 s s 0. 00067 0. 00033 D 150 D 300 15 5 30 10 30 f 30 0. 0165 γ 0 i 0 0 i o o α i zi h αo zo ht hl g γ g z i 0 z o Δz h L 0 Total loss Major loss Minor Loss h L h L, major h L, minor Li i L i j i i j h f K D g g j 35
g g g L D g L D g 15 15 30 h L K e K E K E ( f ) 15 ( f ) 30 0.5 (ie entrance) at r/d0 ) 1 1.0 (Exansion) at D 1 /D 0.0 and θ180 o 0.555 (by linear interolation) (Exansion) at D 1 /D 0.5 and θ180 o linear interolation: f(x) f(x o ) f(x1 ) x 1 f(x x o o ) (x x o ) f(x 1 ) f(x) f(x 0 ) x o x x 1 36
h h 1. 787 m L zi zo Δz hl 1. 787 Alternative Solution for Exansion θ180 o g ( g ) 0 ) g 15 15 30 30 L K e ( f ) 15 ( Two Abrut Exansion m L D g ( f L D g ) 30 h 1. 799 m L Note: for Exansion if θ 180 o you should use the table 37
Examle: A tank and iing system is shown. The ie diameter is cm and the total length of ie is 10 m. The two 90 0 elbows are threaded fittings. The vertical distance from the water surface to the ie outlet is 5 m. The velocity of the water in the tank is negligible. Find a) the exit velocity of the water and, b) the height (h) the water jet would rise on exiting the ie. Assume the ie is galvanized iron 38
Solution: Assume a turbulent flow: Energy eq. between (1&): h L K e g K b g f 0 1 z hl g L ( K e K b D g (Eq.1) f L D ) g 0.5: Pie entrance at r/d0.0 Shar edge (Table 10.3) 0.9: 90 0 elbow thread (Table 10.3) From Eq.1: 5 ( 1 K e K b 10 f L D ) g Two unknown, usually we assume the friction factor f f To select close I know: k 3 0. 15 10 s 0. 0075 D 0. 0 Table 10. (galvanized iron) (Eq.) Fig. 10-8 f 0.035 39
. 17 m/s from Eq. Now check Re and new f Re D ν 6. 17 0. 0 4. 34 10 6 10 4 (Turbulent ) 3 k s 0. 15 10 D 0.00 Re 4. 34 10 4 0. 0075 With new f 0. 036 D. 15 0. 0 Re 4. 3 10 6 ν 10 f 0.036036 Fig. 10-8. 15 m/s from Eq. 4 (Turbulent ) This is close to 4.34 x 10 4 so no further iterations are required 0 0 Energy between and 3 to find h: 3 3 Energy between and 3 to find h: hδz0.4 m γ g z γ g z 3 40
Examle: If the um efficiency is 70%, what ower must be sulied to the um in order to um fuel oil (S0.94) at a rate of 1. m 3 /s u to the high reservoir? Assume that the conduit is a steel ie and the viscosity is 5 x 10-5 m /s. 41
γ Solution: 1 Re D ν k s 0. 046 10 0. 00008 D 0. 6 Table 10. (steel ie) Q 4. 15 m/s A 4 5. 1 10 (Turbulent ) Fig. 10-8 f 0. 01 3 1 z 1 h α z ht h h z h g γ g α1 L L hl K e K b f ( K e K g g g D g K e K b 0.19: 90 0 smooth bend at r/d1./0.6 (Table 10.3) h z hl 7. 9 m W abrut 190 b 1 0.5: Pie entrance at r/d0.0 Shar edge (Table 10.3) Q γ h 1. 0. 94 9810 7. 9 441 η 0. 7 kw η L f L D ) g fluid γqh wt shaft shaft shaft W shaft γqh fluid t W W W ηt W 4
10.6: ie Systems When a um is connected to a ieline, it rovides head to the flow. This head sulied by a um is deendant on the flow rate. This imlies that the solution is always straight forward: It is obtained by solving the system equation with the um equation. Every um has its own head s. Discharge curve. 43
When the system curve is lotted against the um curve: the intersection oint is the oerating oint at which the system runs. Finding this oint is through a trial and error rocedure. 44
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10.7: Turbulent Flow in Non- Circular Conduits In many engineering alications, non-circular ducts are encountered. Analysis for flow in such ducts is treated in exactly the same manner as in circular ies with the excetion of using the term D h (hydraulic diameter) instead of the diameter. 4AA P: Wetted erimeter D h A: cross-sectional area P Hydraulic radius R h : D h A Rh Dh 4R P 4 Usually it aears in Darcy-Weisbach eqn, relative roughness, Reynolds number h 50