Ch ap ter 8 St eady In co comp mprr es si sibl ble e Flo low w in Pre ress ssur ure e Con ondu duit its s (Part B)
Outline 8.8 Head loss lo ss duri d uring ng tur t urbul bule ent flo f low w in a pipe pip e 8.9 Solut ol utio ion n of o f Single Sing le Pip Pipe e Flows lo ws 8. 10 L o cal Head Head L osses os ses 8.10.1 Losses at Sudden Enlargement 8.10.2 Losses at Sudden Contraction
Head lo loss ss (Review)
We will wi ll be look lo okin ing g he h ere at the t he flow fl ow of real real fluid flui d in pipe pip es – rea real mea meaning a fluid flu id that that loose loo ses s ene energ rgy y due to fri f ricti ction on as as it inte int eract racts s with wi th the pipe pi pe wall wall as it flows fl ows..
Viscous sublayer
δ 0
=
32 .8d Re λ
Re =
vd
ν
8.8 Head loss in Turbulent Flow
When flow is turbulent, the viscous dissipation effects cannot be derived explicitly as in laminar flow, but the following relation is still valid. 2
h f
= f
l V
d 2 g
l = pipe length d= pipe diameter V = pipe velocity f = friction factor
8.8.1 Nikuradse’s Experiment Nikuradse made a great contribution to the theory of pipe flow by differentiating between rough and smooth pipes. A rough pipe is one where the mean height of roughness is greater than the thickness of the laminar sub-layer. Nikuradse artificially roughened pipe by coating them with sand. He defined a relative roughness value k s/d (mean height of roughness over pipe diameter) and produced graphs of λagainst Re for a range of relative roughness 1/30 to 1/1014.
d /2ks
Figure : Regions on plot of Nikurades’s data
d /2ks
d /2ks
d /2ks
Figure : Regions on plot of Nikurades’s data
d /2ks 15 30.6 60 126 252 507
Laminar Transition
Turbulent
d /2ks 15 30.6 60 12 6 25 2 50 7
Laminar flow
Re < 2000 (lg Re = 3.30) 64 /Re
f Re
d /2ks 15 30.6 60 126 252 507
Transition from laminar to turbulent: 2300< Re < 4000 (3.3 < lgRe < 3.6) Pipe flow normally lies outside this region.
d /2ks 15 30.6 60 126 252 507
Smooth turbulent
The limiting line of turbulent flow.
All value of relative roughness tend toward this as Re decreases.
d /2ks 15 30.6 60 126 252 507
Transitional turbulent
The region which ss
varies with
both Re and relative roughness. Most pipes lie in this region. λ = f (Re,
k s
)
d /2ks 15 30.6 60 126 252 507
Rough turbulent.
remains constant for a given relative
roughness. It is independent of Re.
Nikuradse’s Experiments
In general, friction factor λ = F (Re,
D
)
Function of Re and roughness
Laminar region λ =
k s
λ =
64
k
(Re )
1/ 4
Rough
Blausius
Re
Independent of roughness
Turbulent region
Smooth pipe curve
λ =
64 Re
Rough pipe zone
λ =
All curves coincide @ ~Re=2300 All rough pipe curves flatten out and become independent of Re 0.25
⎡ 5.74 ⎞⎤ ⎛ e ⎢log10 ⎜⎝ 3.7 D + Re0.9 ⎠⎟⎥ ⎣ ⎦
Blausius OK for smooth pipe
2
Laminar
Transition
Turbulent
Smooth
Regions on plot of Nikurades’s data
Turbulent flow in a circular pipe may be classified as: smooth pipe region, rough pipe region and transition region.
8.8.2 Moody chart Colebrook-White Equation
Colebrook and White proposed the following general equation after studying flow in real pipes:
1 f
= −2 lg(
e
3.7 d
+
2.51 Re f
)
The values of friction factor obtained from the equation are plotted on a Moody diagram, which shows a family of curves for f plotted against the relative roughness and Reynolds number. The Moody chart is a graphical method to find the friction factor in pipes.
A good approximate equation for the turbulent region of the Moody chart chart is is given by Haaland’s Haaland’s equation: equation:
Haalan Haaland’s d’s equatio equation n is valid valid for turb turbule ulent nt flow (Re > 2300)
Pip ipe e ro roug ughn hne ess
pipe material
pipe roughness e
glass, drawn brass, copper
0.0015
commercial steel or wrought iron
0.045
asphalted cast iron
0 .1 2
galvanized iron
0.15
cast iron
0.26
concrete
0.18-0.6
rivet steel
0.9-9.0 45 0.12
corrugated metal PVC
(mm )
8.24
Question
If the flow is in turbulent transition region, the Frictional factor f of the industrial pipes _____ with the increase of the Reynolds number
A increases；
B reduces；
C keeps constant.
Question
There are two pipes, one transports oil and the other transports water. If diameter d ,length l and roughness coefficient of the two pipes are all the same, kinematic viscosity υoil is bigger thanυ water , the Reynolds Numbers are equal, then the Frictional Loss______
A. hfo=hfw；
B. hfo>hfw；
C. hfo
D.uncertain.
8.9 Single Pipe Flows
solution basics
Four simultaneous equations:
① continuity ③ Reynolds number For ④
V =
Re =
4Q π D
ρ VD μ
h L
② energy loss
2
=
VD
D 2 g
f = f (Re, e / D)
④
υ 1
Colebrook—White
Haaland
= f
L V 2
f
1 f f =
= −2 lg(
e / D
3.7 d
+
2.51 Re f
)
⎡⎛ e D ⎞ 6.9 ⎤ = − 1.8 log ⎢⎜ ⎟ + ⎥ Re ⎦ ⎣⎝ 3.7 ⎠ 1.11
64ν DV
=
64 Re
3 types of pipe flow problems: 1. Head loss (find h L given D, Q or V ) 2. Discharge (find Q given D and h L) 3. Sizing problem (Find D given Q and h L)
Example ( Laminar flow):
Water, 20oC flows through a 0.6 cm tube, 30 m long, at a flow rate of 0.34 liters/min. If the pipe discharges to the atmosphere, determine the supply pressure if the tube is inclined 10o above the horizontal in the flow direction.
2 10 1
L = 30 m
D = .6 cm
Water Properties:
ρ= 998 kg/m3
ρg = 9790 N/m3
υ= 1.005 E-6 m2/s
Energy Equation (neglecting α) P1
P2 = 0 V =
ρ g Q A
Re =
−3
=
V D
υ
=
Z 2 − Z 1 + h f = L sin 10o + hf
3
0.34 E m / min*1min/ 60 s π (0.3/100 ) m 2
=
0.2*0.006 −6
1.005 E
2
= 1197 →
P = 9790 N/m3*5.75
= 0.2 m / s
laminar flow
56.34 kN/m2 (kPa)
Example (turbulent flow): Oil, = 900 kg/m3, = 1 E-5 m2 /s, flows at 0.2 m3 /s through a 500 m length of 200 mm diameter, cast iron pipe. If the pipe slopes downward 10o in the flow direction, compute hf , pressure drop.
1 o
10 L L = 500 m
D = 200 mm
2
The energy equation for α= 1 can be written as follows
V =
Re =
Q A
3
=
V D
υ
0.2 m / s π (.1) m 2
=
2
6.4*.2 −5 1 E
= 6.4 m / s
cast iron, k s= 0.26 mm
= 128,000 → turbulent flow
Since flow is turbulent, use Haaland’s equation to determine friction factor (check your work using the Moody chart).
Z2 – Z1 + hf = - 500 sin 10 + 116.6 = - 86.8 + 116.6 = 29.8 m
Solution Summary To solve basic pipe flow frictional head loss problem, use the following procedure: 1. Use known flow rate to determine Reynolds number. 2. Identify whether flow is laminar or turbulent. 3. Use correct expression to determine friction factor (with ks/d if necessary). 4. Use definition of hf to determine friction head loss. 5. Use general energy equation to determine total pressure drop.
8.10 Minor Losses in Turbulent Flow
Transitions in pipe systems, such as bends, valves, changes in diameter, entrances and exits, cause head losses in the system. Head losses at transition points are called minor losses. A minor loss is usually a function of the velocity head as follows: 2
h L ( Minor )
=
V k 2g
Where K is minor loss coefficient
Sources of minor losses
• Additional pressure (energy) losses due to: – Fittings, bends, orifice plates, and valves • Losses due to physics – Vena contracta – Abrupt changes in flow area • Losses due to piping networks for fluid distribution
Typical piping elements with minor losses
Minor losses in piping networks
8.10.1 Sudden Enlargement
h j
= ( z1 − z 2 ) + (
p1
ρ g
−
p 2
ρ g
)+(
α 1ν 12 2g
−
α 2ν 22 2g
)
Momentum equation:
p1 A2 − p2 A2 + ρ gA2 ( z1 − z2 ) = ρ Q( β 2v2 − β 1v1) z1 − z 2
h j
h j h j
= =
=
+
( β 2 v 2
(v2 − v1 )2 2g (v2 − v1 ) 2 2g
p1
ρ g
p 2
−
ρ g
− β 1v1 ) v 2
+
g
= (1 −
A1 A2
A2 =( A1
2
)
=
2
− 1)
v1
2g 2
2
v2
2g
( β 2 v 2
− β 1v1 )v 2 g
α 1v1
− α 2 v 2 2
2
2g
2
= K 1
V 1
= K 2
V 2
2g 2
2g
K 1
= (1 −
A1 A2
A2 K 2 = ( A1
)2
− 1) 2
8.10.2 Sudden Contraction A1
A2
A1
v1
A2 K ＝0.5 （1 − ） ⎯⎯ ← h j A1
v2
= K 2
v2
2
2g
8.10.3 Gradual Expansion (Diffusor)
h E
=
K E
(V 1 − V 2 )2 2g
h E
K E
=
0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0
K E
2 2
V
2g
⎛ A 2 ⎞ − 1⎟ ⎜ ⎝ A1
2
Loss due to gradual enlargement 0
20 40 60 diffusor angle (
)
80
8.10.4 Entrance Losses Losses can be reduced by accelerating the flow gradually and eliminating the
he K e
≈ 1 .0
K e
K e
≈
≈
0 .5
0 . 04
=
K e
V 2
2g
8.10.5 Head Loss in Valves •
Function of valve type and valve position
•
The complex flow path through valves can result in high head loss (of course, one of the purposes of a valve is to create head loss when it is not fully open)
hv
=
K v
V
2
2g
Example In a sudden expansion pipe shown in figure, velocity is v1 and v2 respectively. A middle diameter pipe is connected between the two pipes to form a two-sudden-expansion pipe. The interaction of minor resistances is negligible. That is, superposition method can be applied here. Determine: (1) the velocity of the middle pipe when the total minor head loss of the pipe is the least. (2) the total minor head loss, and the comparison with the one sudden expansion pipe.
Solution: (1) The minor head loss of the two-sudden-expansion pipe.
Assume the velocity of middle pipe is v, and make the total minor head loss be the least, so
（2）The total minor head loss is the minor head loss of the one sudden expansion pipe is the total minor head loss of the two-sudden-expansion pipe is half of the minor head loss of the one sudden expansion pipe.
8.11 Branching Pipes Branching pipe systems, such as the one shown by Figure 8.27, can be solved using the following: 1. Q1 = Q2 + Q3. 2 The elevation of P is common to all pipes.
Branching pipe systems
8.12 Pipes in Series Pipes in series, as shown by Figure 8.29, can be solved as follows: Q = Q1 = Q2 = Q3 h L = h L1 + h L2 + h L3
Fig. Pipes in series
The pipe elements of the pipes in series have: A. the same head loss； B. the same total head loss； C. the same hydraulic slope； D. the same discharge through them
Example Consider the two reservoirs shown in figure, connected by a single pipe that changes diameter over its length. The surfaces of the two reservoirs have a difference in level of 9m. The pipe has a diameter of 200mm for the first 15m (from A to C) then a diameter of 250mm for the remaining 45m (from C to B).
For the entrance use k L = 0.5 and the exit k L = 1.0. The join at C is sudden. For both pipes use λ = 0.04.
Total head loss for the system H = height difference of reservoirs
and solve for Q, to give Q = 0.158 m3/s
8.13 Pipes in Parallel Parallel pipes, as shown by Figure 8.30, can be solved as follows: Q = Q1 + Q2 + Q3 h L = h L1 = h L2 = h L3
Pipes in parallel
Question From the figure, we can see that the relationship among the head losses from A to B in the pipes in parallel 1, 2, 3 is:
A. hfAB＝hfl+hf2+hf3； B. hfAB
hfl+hf2；
C. hfAB
hf2+hf3；
＝
＝
D. hfAB
hfl=hf2=hf3.
＝
Example
Two pipes connect two reservoirs (A and B) which have a height difference of 10m. Pipe 1 has diameter 50mm and length 100m. Pipe 2 has diameter 100mm and length 100m. Both have entry loss k L = 0.5 and exit loss k L=1.0 and Darcy f of 0.008.
Calculate: a) rate of flow for each pipe b) the diameter D of a pipe 100m long that could replace the two pipes and provide the same flow.
