Ch-10 Flow in Conduits - ch-1,2,3,4,5,7,8,10 PDF

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Chapter 10: Flow in Conduits By Dr Ali Jawarneh Hashemite University 1 Outline In this chapter we will: • Analyse the shear stress distribution across a pipe section. • Discuss and analyse the case of laminar flow in pipes in particular. • Distinguish between laminar and turbulent flows. • Discuss D...

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Chapter 10: Flow in Conduits By

Dr Ali Jawarneh Hashemite University

1

Outline In this chapter we will: • Analyse the shear stress distribution across a pipe section. • Discuss and analyse the case of laminar flow in pipes in particular. • Distinguish between laminar and turbulent flows. • Discuss and analyse the case of turbulent flow in smooth and rough pipes. • Analyse the head loss due to pipe entrance entrance, elbows and fittings. • Analyse flow in non-circular conduits. 2

10.1: Shear-Stress Distribution across a Pipe 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 Pipe Section • The momentum equation for the control volume is:

∑F

s

=0

dp ⎞ ⎛ pA − ⎜ p + Δ s ⎟ A − ΔW sin α − τ (2πr ) Δs = 0 ds ⎠ ⎝

−

dp dz ΔsA − γ A Δs − τ (2πr ) Δs = 0 ds ds

4

Shear-Stress In Pipe Section • Re-arranging and simplifying: τ=

r ⎡ d ( p + γz ) ⎤ − 2 ⎢⎣ ds ⎥⎦

• Note that the shear stress is zero at the centre of the pipe and is maximum (positive value) at the pipe wall. 5

10.2: Laminar Flow in Pipes • For laminar flow, shear stress can be substituted with μ dV/dy: μ

dV r ⎡ d ( p + γz ) ⎤ = − ⎥ dy 2 ⎢⎣ ds ⎦

dV r =− d dr 2μ

⎡ d ( p + γz ) ⎤ ⎢⎣ − ⎥⎦ d ds

• Solving this differential equation with the condition V(r0)=0: r02 − r 2 V= 4μ

⎡ d ( p + γz ) ⎤ ⎢− ⎥ ds ⎣ ⎦

Parabolic eqn 6

Laminar Flow in Pipes

7

Laminar Flow in Pipes • The flow rate can be found by integrating the velocity expression over the crosssectional area: r0

r02 − r 2 Q=∫ 4μ 0

π r04 Q= 8μ

⎡ d ( p + γz ) ⎤ − ⎥ 2π rdr ds ⎣⎢ ⎦

⎡ d ( p + γ z) ⎤ ⎢⎣ − ⎥⎦ ds

8

Laminar Flow in Pipes • Dividing by the cross sectional area, the mean velocity can be given as: r02 V = 8μ

⎡ d ( p + γz ) ⎤ ⎢⎣ − ⎥⎦ ds

• Re-arranging: Re arranging: 32 μV d ( p + γ z) =− ds D2

9

Laminar Flow in Pipes • Integrating between sections 1 and 2: p1

γ

+ z1 =

p2

γ

+ z2 +

32 μ LV γ D2

• Comparing with the energy equation, an expression for the head loss due to frictional resistance in a pipe can be found: hf =

32 μ LV γ D2 10

10.3: criterion for Laminar or Turbulent Flow in a Pipe • Reynolds related the onset of turbulence in a pipe to a nondimensional number he named after himself: Re =

ρ VD μ

11

Laminar Vs Turbulent Flow • He found that: – For Re3000 the flow is turbulent – For 2000>Re>3000 the flow is is in transition state state, changing back and forth between laminar and turbulent.

• However However, he found that under carefully controlled conditions with no vibrations –which is and ideal case not existing in engineering application – laminar flow could be maintained for Re>2000. 12

10.4:Turbulent Flow in Pipes • A universal equation that can be used to calc late the head loss due calculate d e to viscous isco s effects in pipes is the Darcy-Weisbach equation: Major loss due to internal friction inside conduits

L V2 hf = f D 2g

Valid for laminar and turbulent flow

• f : resistance coefficient or friction coefficient • For laminar flow it can be easily shown that: f = 64 / Re

13

Turbulent Flow in Pipes • For turbulent flow in smooth pipes, an empirical relationship can be used (developed by Prandtl): 1 = 2 log ( Re f

f ) − 0.8

• For rough pipes, relative roughness (ks/D) which is defined as the ratio of the grain size ( ks) to the pipe diameter becomes an important design factor. factor • Values for grain roughness (ks) for various pipe materials are shown in Table 10 10.2. 2 14

Table 10.2

15

Turbulent Flow in Pipes • The Colebrook-White equation is an empirical equation that was developed to evaluate the friction factor for rough pipes. • The moody diagram (Figure 10.8) was developed on the basis of this equation. This diagram is widely used in most engineering applications that incorporate flow in conduits. 16

Figure 10.2

17

Turbulent Flow in Pipes • Instead of using the diagram, an explicit equation for the friction factor can be used for turbulent flow in case of applying numerical solutions: 0.25 f =

⎡ 5.74 ⎞ ⎤ ⎛ ks + log ⎜ ⎢ 10 3.7D Re0.9 ⎟ ⎥ ⎝ ⎠⎦ ⎣

2

• This equation is also based on the Colebrook-White equation. 18

Turbulent Flow in Pipes • Also, explicit equations can be used for either the flow rate through the pipe or the diameter of the pipe: Q = −2.22 D

5/ 2

⎛ k ⎞ 1.78ν s ⎜ ⎟ + 3/ 2 ghf / L log ⎜ 3.7 D D gh f / L ⎟⎠ ⎝

⎡ 2 ⎛ 1. 25 ⎜ LQ D = 0 .66 ⎢k s ⎜ ⎢ ⎝ gh f ⎣

⎞ ⎟ ⎟ ⎠

4.75

⎛ L ⎞ ⎟ ⎟ ⎝ gh f ⎠

+ν Q 9.4 ⎜ ⎜

5.2

⎤ ⎥ ⎥ ⎦

0.04

19

10.5:Flow at Pipe Inlets and Losses from Fittings • If the inlet to a pipe is well rounded, the boundary layer will develop from the inlet and grow in thickness until it extends to the centre of the pipe. • The length of the boundary layer d development l t region i can be given approximately by: Le= 0.05 0 05 D Re, Re for laminar flow. Le= 50 D, for turbulent flow.

Le = entrance Length

20

Flow at Pipe Inlets and Fittings • If the pipe inlet is abrupt, separation ti occurs jjustt downstream of the entrance, this clearly causes relatively higher head loss. • Considerable head loss is also produced in bends and elbows, due to separation that happens downstream of the midsection. 21

Flow at Pipe Inlets and Fittings • The head loss produced by inlets, outlets, elbows or fittings is expressed as: V2 hL = K 2g

– Where K is the loss coefficient coefficient.

• Different values for the loss coefficient (K) are given in Table 10.3. 22

Table 10.3

23

Note:in The Expansion if the angle is θ =1800 you have two choices for the head loss: 1- Use the above equation 2- Use the abrupt expansion equation

24

Summary of the Energy equation V 12 p2 V 22 + z 1 + α1 + hp = + z 2 + α2 + ht + h L γ γ 2g 2g

p1

Total loss= Major loss + Minor Loss h L=h L, major + h L, minor hL =

∑ i

Li V i 2 + fi D i 2g

∑K j

V j2 j

2g

25

Example: Liquid flows downward in a 1-cm, vertical, smooth pipe with a mean velocity of 2.0 m/s. The liquid has a density of 1000 kg/m3 and a viscosity of 0.06 N.s/m2. If the pressure at a given section is 600 kpa, what will be the pressure at a section 10 m below that section?

Solution: Re =

ρVD VD 1000 × 2 × 0 .01 = = 333.33 La min ar μ 0 .06

-10 0 0 p1 V1 2 0 p2 V22 + α1 + z1 + h p = + α2 + z 2 + ht + hL γ γ 2g 2g

64 = 00. 192 f= Re

LV2 10 2 2 = 0 .192× = 39 .14 m hL = f D 2g 0 .01 2 g

p2 = 314 .1 kpa 26

Example: Kerosene (S=0.8 and T=68 0F) 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.

27

Solution: assume a laminar flow g = 32. 2, γ = 62.4 × 0.8 = Ibf/ft 3, ρ = 1.94 × 0.8 slugs/ft 3 μ = 4 × 10− 5 Ibf − s/ft 2 (Table A.4 ) 3 L = 10 ft, D = ft 8 × 12 0

0

0

2

-0.5 0 0 V12 p2 V22 p1 + α1 + z1 + hp = + α2 + z 2 + ht + hL γ γ 2g 2g

L V2 hL = f D 2g

64 64 = Re ρVD μ

f =

V 2 + 8 .45V − 16 .1 = 0.0 → V = 1.602 ft/s Now check Re Re =

ρVD = μ

Q = VA = 1.602 ×

3 ) 8× 12 = 1942 ⇒ La min ar 4 × 10− 5

1.94 × 1.602 × (

π ( 1 / 32 )2 = 1.228 × 10− 3 cfs 4

28

Example: Water flows in the pipe shown, and the manometer deflects 80 cm. What is the resistance coefficient (friction coefficient) for the pipe if V=3 m/s?

29

Solution: 2

0

10

2 2

11

0

p1 V p V + α 1 1 + z1 + hp = 2 + α2 + z 2 + ht + hL Eq.(1) γ γ 2g 2g L V2 4 32 hL = f = f D 2g 0.05 2g M Manometer t equation: ti

p1 + γ w ( y + 0.8 ) − Sγ w ( 0.8 ) − γ w ( y + 1 ) = p 2

Eq.(2)

⇒ p1 − p2 = 21582 pa From Eq.(1) & (2)

f = 0.033 30

Example: A water turbine is connected to a reservoir as shown shown. The flow rate in this system is 5 cfs. What power can be delivered by the turbine if its efficiency is 80%? Assume a temperature of 70 0F.

31

Solution: V =

5 π ( 12 / 12 ) 2 4

0

0

= 6.369 ft/s

12 ) VD 12 = = = 6× 105 ⇒ Turbulent Re −5 1.0615 × 10 μ 6.369 × (

Table A.5 at 70 0F 1 0 0 0 2

V = V2

-100

p1 V12 p V + α1 + z1 + h p = 2 + α 2 2 + z2 + ht + ∑ hL γ γ 2g 2g

Eq. (1)

V2 LV2 6. 3692 1000 6. 3692 ∑ hL = K e 2 g + f D 2 g = 0.5 2 × 32.2 + f 12 / 12 2 × 32.2 Pipe entrance at r/d=0.0 Sharp edge (Table 10.3) Table 10.2

k s 0.002" = = 0.00016 12" D

Re = 6 × 10 5

f = 0.015 Figure 10 10.8 8 32

From Eq. (1):

ht = 89.6 ft

hp = ht =

1 hp = 550 Ibf.ft/s 1 hp = 745 .7 W

W& p

=

& mg W& t

=

& mg

W& p

Pump Head

γQ

W&t γQ

Turbine Head

•

ηp =

W

fluid

•

W

=

shaft

•

ηt =

W

w T shaft

Pump effeciency

•

shaft

•

W

γ Qh p

fluid

=

W

shaft

γ Qht Turbine effeciency

•

W t = Q γ ht η = 5× 62. 4× 89. 6× 0. 8 = 22364. 16 ft .Ibf / s •

P=

W t 22364.16 = = 40.66 ( hp ) horsepower 550 550 33

Example: Both pipes shown have an equivalent sand d roughness h k s off 0 0.1 1 mm and d a discharge di h off o.1 m3/s. Also D1=15 cm, L1=50 cm, D2=30 cm, and L2=160 m. m determine the difference in the water water-surface elevation between the two reservoirs.

34

Solution: ν = 10 −6 m2 / s @ T = 20 oC Q V15 = / = 5 .659 m/s A15

V30 =

Q = 1.415 m/s A30

Moody diagram(Fig. 10-8): V15 D15 V D = 8. 49 × 10 5 Re 30 = 30 30 = 4. 24 × 105 ν ν f15 = 0 .0185 f 30 = 0.0165 ks 0.1 ks 0.1 = = 0.00033 = 0.00067 = D30 300 D15 150

Re15 =

0

0 2

0

0

0

pi Vi po Vo2 + αi + zi + hp = + αo + z o + ht + ∑ hL γ γ 2g 2g

z i − zo = Δ z = ∑ hL Total loss= Major loss + Minor Loss h L=h L, major + h L, minor

hL =

∑ i

Li V i 2 fi + Di 2 g

∑K j

V j2 j

2g 35

V152 V152 V302 LV2 L V2 ∑ hL = K e 2 g + K E 1 2 g + K E 2 2 g + ( f D 2 g )15 + ( f D 2 g )30 0.5 (pipe entrance) at r/d=0

1.0 (Expansion) at D1/D2≅0.0 and θ=180o 0.555 (by linear interpolation) (Expansion) at D1/D2=0.5 and θ=180o

linear interpolation:

f(x) = f(xo ) +

f(x1 ) − f(xo ) (x − xo ) x1 − xo

f(x1) f( ) f(x) f(x0)

xo

x

x1

36

∑h

L

= 12.787 m

z i − zo = Δz = ∑ hL = 12.787 m

Alternative Solution for Expansion θ=180o V152 ( V15 − V30 ) 2 ( V30 − 0 ) 2 L V2 LV2 ∑ hL = K e 2 g + 2 g + 2 g + ( f D 2 g )15 + ( f D 2 g )30

Two Abrupt Expansion

∑h

L

= 12.799 m

Note: for Expansion if θ≠180o you should use the table

37

Example: A tank and piping system is shown. The pipe diameter is 2 cm and the total length of pipe is 10 m m. The two 900 elbows are threaded fittings. The vertical distance from the water surface to the pipe outlet id 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 pipe. Assume the pipe is galvanized iron

38

Solution: Assume a turbulent flow: 2 2

V + z 2 + ∑ hL (Eq.1) 2g V2 V2 LV2 L V2 ∑ hL = K e 2g + 2 Kb 2 g + f D 2g = ( K e + 2K b + f D ) 2 g Energy eq. between (1&2):

0.5: Pipe entrance at r/d=0.0

0=1

Sharp edge (Table 10.3)

0.9: 900 elbow thread (Table 10.3)

10

L V2 ) 5 = ( 1 + K e + 2K b + f D 2g

From Eq.1:

Two unknown, usually we assume the friction factor

k s 0.15 × 10 = 0. 02 D Table 10.2 (galvanized iron)

To select close

f I know:

V2 = V

−3

= 0.0075

(Eq.2)

f

Fig. 10-8 f = 0.035 39

V = 2 .17 m/s

from Eq.2

f

Now check Re and new VD 2.17 × 0.02 Re = = = 4. 34 × 104 ( Turbulent ) −6 ν 10 k s 0.15 × 10− 3 = = 0.0075 00. 02 D

f = 0.036 Fig. Fig 10 10-8 8

Re = 4.34 × 10 4

With new f = 0.036

Re =

V = 2 .15 m/s from Eq.2

VD 2.15× 0.02 4 = × = . ( Turbulent ) 4 3 10 −6 ν 10

This is close to 4.34 x 104 so no further iterations are required 0 0 2 2 p V p V 2 2 3 3 Energy between 2 and 3 to find h:

h=Δ Δz=0.24 m

γ

+

2g

+ z2 =

γ

+

2g

+ z3 40

Example: If the pump efficiency is 70%, what power must be supplied to the pump in order to pump fuel oil (S=0 (S=0.94) 94) at a rate of 1.2 m3/s up to the high reservoir? Assume that the conduit is a steel pipe and the viscosity is 5 x 10 -5 m2/s.

41

Solution: Re =

V=

Q = 4.15 m/s A

VD = 5.1 × 10 4 ( Turbulent ) ν −3

Fig. 10-8 f = 0.021

ks 0.046 × 10 − = = 0. 00008 D 0. 6 Table 10.2 (steel pipe)

p1 V1 2 p2 V22 + α1 + z1 + hp = + α2 + z 2 + ht + ∑ hL h p = z2 + ∑ hL γ γ 2g 2g abrupt 190

2

2

2

2

V V V LV L V2 ∑ hL = K e 2g + 2K b 2 g + 2 g + f D 2 g = ( K e + 2K b + 1 + f D ) 2 g K e = 0. 5 : Pipe entrance at r/d=0.0 Sharp edge (Table 10.3) • γ Qh p W fluid K b = 0. 19 : 900 smooth bend at r/d=1.2/0.6=2 (Table 10.3) = ηp = • w T shaft W shaft h p = z 2 + ∑ hL = 27.9 m •

•

Wp=

Q γ hp η

=

1. 2× 0. 94× 9810× 27.9 = 441 kW 0.7

ηt =

W

•

shaft

•

W

fluid

=

W

shaft

γ Qht 42

10.6: pipe Systems • When a pump is connected to a pipeline, it provides head to the flow. • This head supplied by a pump is dependant on the flow rate. • This implies that the solution is always straight forward: It is obtained by solving the system equation with the pump equation. • Every pump has its own head Vs. Discharge curve. 43

• When the system curve is plotted against the pump curve: the intersection point is the operating point at which the system runs. • Finding this point is through a trial and error procedure. See example 10.11 44

10.7: Turbulent Flow in NonCircular Conduits • In many engineering applications, non-circular ducts are encountered. t d • Analysis for flow in such ducts is treated in exactly the same manner as in circular pipes with the exception of using the term Dh (hydraulic diameter) instead of the diameter. 4 A P: Wetted perimeter Dh = P A: cross-sectional area A Dh = ⇒ D h = 4R h Hydraulic radius Rh : Rh = P 4 Usually it appears in Darcy-Weisbach eqn, relative roughness, Reynolds number 45...