Closed Conduit Systems

CLOSED CONDUIT SYSTEMS
Minor Losses
Sudden Expansion: using the momentum equation it was
2 2
2
v1   D1  
established that: hl  k ; where k  1    
2g   D2  
D1
Thus, if a pipe discharges into reservoir, D  0 , k = 1.
2
Conical Expansions: From Streeter Figure 6.22:
2
v22  1 
h
Sudden Contraction: l  k ; where k  
C  1
2g  c 
A2/A1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Cc 0.62 0.63 0.64 0.66 0.68 0.72 0.76 0.81 0.89 1.0
From reservoir into a pipe, k = 0.5 for square-edged entry,
k = 0.01~0.05 for rounded entry, and k = 0.8~1.0 for
pipe extending into reservoir (re-entrant).
Fittings: Manufacturers provide loss coefficients; e.g.:
Equivalent Length: Minor losses can be incorporated into
friction loss by adding an equivalent length of pipe, i.e.
Le v 2 v2 KD
from f  k we get Le  f . Thus we substitute
D 2g 2g
D
Ltotal  L   K into L developed in earlier procedures.
f
Hydraulic Grade Line and Energy Grade Line:
p
The hydraulic grade line is the plot of  z i.e. the locus of

points to which the water surface will rise up to when a
piezometer is attached anywhere in the pipe system. The
v2 p
energy grade line is higher than  z by an amount .
 2g
Negative Pressure: pipe can be above hydraulic grade line
Pmin =  (Patm  Pv); Pv = vapor pressure
= 0.44 m. above absolute zero pressure
=  (10.34  0.44)
Pmin =  9.9 m. (gage pressure)
Particular attention given for systems with pumps, e.g.:
Shallow Well Pump Deep Well Pump
Min. neg.
pressure
P Q Q
Pump must
be primed No neg.
before pressure
starting
H (suction pipe H
completely
filled)
Water Table Water Table
Drawdown
during pumping
P
Set H < 0.9 m. to avoid cavitation.
Pipes in Series: best illustrated by an example

1 Given: H=6 m
H 2 L1=300 m; D1=600 mm
L2=240 m; D2=900 mm
D1 D2 ε1=2 mm; ε2=0.3 mm
υwater = 0.8x10-6 m2/s
L1, ε1 kexp L2, ε2 Find Q.
ken kex
Energy equation from 1 to 2:
P1 v12 P2 v22
 z1  1  hP   z2   2  hT  hL
 2g  2g
 v12 L1 v12 v12 L2 v22 v22 
z1  z2  6   ken  f1  kexp  f2  kex 
 2 g D1 2 g 2 g D2 2 g 2 g 
Assuming square-edged entry and exit (ken=0.5, kex=1.0),
 2  2
and using continuity, i.e. 1 v D1  v2 D2 :
4 4
v12  240  6   6  
2 2
300   6  
4 4
6 0.5  f1  1      f 2     
2g  0.6   9   0.9  9   9  
 
6
v12
1.006  500 f1  52.86 f 2  ; 1  1 ;  2  1
2g D1 300 D2 3000
From Moody: f1=0.026; f2=0.015 (fully turbulent rough pipe)
Solving for v: v1 = 2.82 m/s; v2 = 1.25 m/s
Solving for R: R1 = 2.1x106; R2 = 1.41x106
From Moody: f1=0.026; f2=0.0155 ok!
Answer: v1 = 2.82 m/s Q = 0.8 m3/s
Notice that Q1 = Q2 and hf = hf1 + hf2 + hl, i.e. for pipes in
series, discharge is constant and head losses are additive.
Equivalent Pipes: For two pipes of different diameters to be
L1 L2
equivalent, hf1 = hf2 for Q1 = Q2  f1 D5  f 2 D 5 , thus
1 2
5
f D 
L2  L1 1  2 
f 2  D1 
Solving above example using equivalent lengths for minor

losses and equivalent pipes: For minor losses:
2
  6 2   
k1  0.5  1      0.81  Le  0.81 0.6  18.66 m.
  9   1
0.026
10.9
k2  1  Le2   58.1 m. ;  L1 = 318.66 m.
0.0155
5
0.0155  0.6 
L2 = 298.1 m.  e
L  298.1    23.4 m.
0.026  0.9 
Problem is reduced to: L=342.1 m; D=0.6 m; ε=2 mm. or
342.1 v 2
6 f  ; try f=0.026; v=2.816 m/s; R=2.1x10 6
,
0.6 2 g
3
same as above.  Q  0.8 m
s
Pipes in Parallel: Q=Q1+Q2+…Qn and hf1=hf2=…=hfn

1 L1,D1,ε1
A B
2 L2,D2,ε2
Q
3 L3,D3,ε3
Problem Type I: With pipe properties D, L,  and hA and hB

known, find Q. Use methods in simple pipe problem to
find Q's in each pipe since hf=hA-hB. Then Q = Q1+Q2+Q3.
Problem Type II: With Q and pipe properties known, find
hA-hB and Q1, Q2, and Q3.
1. Try a value of Q1'  find hf1'. Then hf1'=hf2'=hf3'.
2. Find Q2' and Q3' using DW eqn. Then Q'=Q1'+Q2'+Q3'.
Qi
3. Update Q's using Q  Q  compute hfi of each pipe.
i
Q 
4. If each is within 5% of hf1', then hA-hB=hfi and Q's are ok.
Otherwise, repeat from 1. using new Q1 as Q1'.
The 3-reservoir Problem: (applicable to >3 reservoirs)
z1
Pj

1 2 z3
j
z3
3 zj
Datum
Known: z1, z2, z3 and D, L, ε of each pipe. Find Q1, Q2, Q3.
pj
If z   z2 then Q1  Q2  Q3 ; otherwise Q1  Q2  Q3 .
j

pj
1. Assume j z 
 then compute Q1, Q2, and Q3.
2. If Qj = 0 then problem is solved.
pj
3. If Qj > 0, raise  , otherwise lower it; repeat from 1.
Try this problem: L1=2 km, D1=900 mm, ε1=0.2 mm;

L2=600 m, D2=450 mm, ε2=0.2 mm;
L3=1 km, D3=600 mm, ε3=0.2 mm;
z1=30 m, z2=20 m, z3=10 m.
Pipe Networks:
QA
2 QB
Principles:
1. hf around any loop = 0.
Going downstream, add hf. I
Going upstream, subtract hf. 1 5 3
2. Q=0 for each junction (node).
3. hf at each pipe follows DW or II
similar equation (minor losses
QD QC
are converted to equivalent
4
lengths).
Solution: by successive approximation, e.g. Hardy-Cross
Balancing Method
1. Assume Q' of each pipe while satisfying continuity at
each node. For convenience, the head loss equation is
simplified to exponential form h f  rQ , e.g. for the DW
n
2
 Q 
L  A 8 fL
h
equation f  f so that n=2 and r
D 2g g 2 D 5 . In
the water supply industry, the Hazen-Williams equation
is commonly used: V  0.85CH R S f , resulting in
0.63 0.54
1.85
 Q  L
hf  S f L  
A 10.65L
 1.85 4.87 Q1.85 .
0.85CH  R
1.85 1.17
CH D
2. For each loop, compute corrections to balance head

m
  rQin
Q  m
i 1
; m  no. of pipes in loop
losses:
 nr Q
n 1
i
i 1
3. Update Qi  Qi  Q of each pipe i in the loop.
4. Repeat 2. and 3. for all loops until all ΔQ become very

small, say <5% of the smallest Q.
Try the problem: r1=4, r2=5, r3=1, r4=2, r5=3;
QA=100, QB=20, QC=50; QD=30
Hazen Williams Roughness Coefficients:
Pipe Material CH
Brick Sewers 100
Cast Iron
New 130
5 years old 120
20 years old 100
Concrete 130
PVC 140
Riveted Steel 110
Vitrified Clay 110
Welded Steel 120
Wood Stave 120
Examples of Design Considerations for water supply

networks: Maintain pressure at junctions at 14 m. above
ground level for 1st class development, 7 m. for economical
design, e.g. low-cost, low-rise housing, or provincial water
district systems. Maximum pressure should be < 60 m. to
protect bathroom fixtures. As much as possible, maintain
velocity between 0.3 m/s and 10 m/s to minimize scaling or
sediment deposition on the one hand, and water hammer
risk or pipe vibration on the other.
Software, commercial or freeware, are available for pipe

network analysis or design. All have computational
engines that are essentially steady-state network analysis.
Extended Period Simulation: For Contaminant Propagation

problems, the same software are used, operating at steady
state on an hourly basis, e.g., flows are updated every
hour. Within each steady state interval, contaminant
concentration are evaluated (traced at check points) at
shorter time intervals, e.g., 30 sec. Filling up and drawing
from Distribution Reservoirs are evaluated in the same
manner.

Closed Conduit Systems

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CLOSED CONDUIT SYSTEMS

Conical Expansions: From Streeter Figure 6.22:

Pipes in Series: best illustrated by an example

Solving above example using equivalent lengths for minor

Pipes in Parallel: Q=Q1+Q2+…Qn and hf1=hf2=…=hfn

Problem Type I: With pipe properties D, L,  and hA and hB

Try this problem: L1=2 km, D1=900 mm, ε1=0.2 mm;

2. For each loop, compute corrections to balance head

3. Update Qi  Qi  Q of each pipe i in the loop.

4. Repeat 2. and 3. for all loops until all ΔQ become very

Examples of Design Considerations for water supply

Software, commercial or freeware, are available for pipe

Extended Period Simulation: For Contaminant Propagation

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