DEPARTMENT OF CIVIL ENGINEERING
CVE 372 HYDROMECHANICS
__________________________________________________________________________________________
SOLVED PROBLEMS - 3
FLOW IN CLOSED CONDUITS
Parallel – Series Pipes
Problem 1: Given a three-pipe series system as shown below. The total pressure drop is pA-pB = 150
kPa, and the elevation drop is zA-zB = 5 m. The pipe data are
pipe
1
2
3
L (m)
100
150
80
ε (mm)
0.24
0.12
0.20
D (cm)
8
6
4
The fluid is water (ρ= 1000 kg/m3 , ν= 1.02x10-6 m2/s). Calculate the flow rate Q (m3/hr). Neglect minor
losses.
PA/γ
D1
D2
A
D3
PB/γ
B
zB
zA
L1= 100 m
Solution:
Energy eqn b/w A and B
L3= 80 m
0
H A = HB + ∑ h f + ∑ h m
2
L2= 150 m
2
V3
V1
2g
2g
2
2
2
2
2
L 3 V3
L 2 V2
VA
PA
VB
PB
L 1 V1
+
+ zA =
+
+ z B + f1
+ f2
+ f3
D 3 2g
γ
γ
D 2 2g
2g
2g
D1 2g
2
 L3
 V3 2
 V12
 L1
 p A pB 
L 2 V2
 + (z A − z B ) =  f1

+ f2
+  f3
+ 1
− 1
−
γ 
D 2 2g  D 3
 γ
 2g
 D1
 2g
from continuity: V1 A 1 = V2 A 2 = V3 A 3
2
2
8
8
A 1 =   A 2 =   A 3 and
6
 4
Page 1 of 7
A 1 = 1.78 A 2 = 4 A 3
V2 = 1.78 V1 and V3 = 4 V1
2
 L
 V2
 p A pB 
L V
2
−  + (z A − zB ) =  f1 1 − 1 1 + (1.78 ) f2 2 1

D2 2g
γ 
 γ
 D1  2g
 V12
 L3
+ (4 )  f3
+ 1
 2g
 D3
2
2
V
20.3 = (1250 f1 + 7920 f 2 + 32000 f3 + 15 ) 1
2g
V1 and fi ’s are not known. Assume hydraulically rough flow !
pipe
1
2
3
ε (m)
0.00024
0.00012
0.00020
ε/D
0.003
0.002
0.005
f (use S-J)
0.0262
0.0234
0.0304
S-J: f =
Then,
1.325
  ε / D 
ln 3.7 

 
2
V
20.3 = (1250 ⋅ 0.0262 + 7920 ⋅ 0.0234 + 32000 ⋅ 0.0304 + 15 ) 1
2g
V1 = 0.57 m/s
pipe
1
2
3
V
0.57
1.02
2.29
Re
45976
61378
91953
ε/D
0.003
0.002
0.005
f (use S-J)
0.0288
0.0260
0.0314
Then,
2
V
20.3 = (1250 ⋅ 0.0288 + 7920 ⋅ 0.0260 + 32000 ⋅ 0.0314 + 15 ) 1
2g
V1 = 0.562 m/s
pipe
1
2
3
V
0.562
1.00
2.25
Re
44960
60022
89920
ε/D
0.003
0.002
0.005
f (use S-J)
0.0288
0.0260
0.0314
Since the agreement b/w the last two iterations is acceptable
V1 = 0.562 m/s
π 2
Q = V1 ⋅ D1 = 0.00282 m3 /s
4
Q = 10.167 m3/hr
Page 2 of 7
2
Q = 0.00282 ⋅ 3600 = 10.167 m3 /hr
⇒ Solution using equivalent pipe approach:
Based on the assumptions that all
L eq
D eq
5
=∑
i =1
f ’s are of the same magnitude
Li
n
Di
5
hence one may either choose an equivalent Leq and determine Deq or choose an equivalent Deq and
find Leq to proceed.
for L eq = ∑ L = 330
then
330
D eq
5
=
100
150
80
+
+
= 0.1⋅ 10 10 and D eq = 0.0505 m
5
5
5
0.08
0.06
0.04
PA/γ
Deq = 5.05 cm
A
zA
B
zB
L= 330 m
HA = HB + f
PB/γ
L V2
D 2g
2
2
VA
P
V
P
L V2
+ A + z A = B + B + zB + f
γ
γ
D 2g
2g
2g
L V2
20.3 = f
D 2g
Assume ε=0.2 mm and hydraulically rough flow. Then
ε / D = 0.004 ⇒
20.3 = 0.028 ⋅
f=
1.325
  0.004 
ln 3.7 

 
2
= 0.028
330 V 2
⋅
0.0505 2g
Page 3 of 7
V = 1.46 m/s ;
Re = 72284
f = 0.0301
V = 1.42 m/s ;
⇒
;
ε
= 0.004
D
;
f=
R e = 70304
;
ε
= 0.004
D
f = 0.0302
V = 1.42 m/s
π
2
Q = 1.42 ⋅ (0.0505 ) = 0.0028 m 3 /s
4
1.325
5.74 
  0.004
ln 3.7 + 722840.9 


Q = 10.24 m 3 /hr
For D eq = 0.06 m
L eq
D eq
5
= 0.1⋅ 1010
then H A = HB + f
L eq = 778 m
L V2
D 2g
2
2
VA
PA
VB
PB
L V2
+
+ zA =
+
+ zB + f
2g
2g
D 2g
γ
γ
20.3 = f
L V2
D 2g
Assume ε=0.2 mm and hydraulically rough flow. Then
ε / D = 0.0033
f = 0.0269
f = 0.0289
f = 0.0291
⇒
⇒
⇒
⇒
f=
1.325
  ε / D 
ln 3.7 

 
2
778 V
20.3 = 0.0269 ⋅
0.06 2g
V = 1.14 m/s ;
V = 1.06 m/s ;
V = 1.06 m/s ;
2
= 0.0269
Re = 67059
R e = 62353
Re = 62353
π
(0.06 )2 = 0.0030 m 3 /s
4
Q = 10.8 m 3 /hr
;
;
;
f = 0.0289
f = 0.0291
f = 0.0291
Q = 1.06 ⋅
Page 4 of 7
Q = 10.8 m 3 /hr
2
Problem 2: Assume that the same three pipes of previous example are now in parallel with the same
total loss of 20.3 m. Compute the total rate Q(m3/hr), neglecting the minor losses.
D1=8 cm
A
B
D2=6 cm
D3=4 cm
Solution-I:
Energy equation b/w A and B:
0
H A = H B + hL = H B + hf + hm
no matter which route is followed b/w A and B
L V2
V12
L V2
= f 2 2 2 = f3 3 3
D1 2g
D 2 2g
D 3 2g
L
HA – HB = 20.3 = f1 1
Substituting the known L’s and D’s ,
20.3 = 1250 f1
V32
V12
V2
=2500 f 2 2 =2000 f3
2g
2g
2g
Since Vi’s and fi’s are not known, assume hydraulically rough regime.
First iteration:
Pipe
ε/D
1
0.003
2
0.002
3
0.005
Second iteration:
Pipe
ε/D
1
0.003
2
0.002
3
0.005
Solution:
Pipe
1
2
3
TOTAL
V(m/s)
3.46
2.55
2.52
f0
0.0262
0.0234
0.0304
V
3.49
2.61
2.56
Re
273726
153529
100392
f1
0.268
0.247
0.315
V
3.46
2.55
2.52
Re
271373
150000
98823
Q(m3/s)
0.0174
0.0072
0.0032
Q(m3/hr)
62.6
26.0
11.4
100
f1
0.268
0.247
0.315
f2
0.268
0.247
0.315
Q= 100 m3/hr
Page 5 of 7
f’s have converged
Solution-II: Using equivalent pipe approach
Based on the assumption that f’s are of the same magnitude
 D 5eq 


 L eq 


1/ 2
 D5 
= Σ i 
 Li 
1/ 2
hence one may either choose an equivalent length Leq and determine Deq or vice versa to proceed.
For Leq =
 D 5eq 


 L eq 


ΣL i
=110 m
3
1/ 2
 0.08 5 

=
 100 


1/ 2
 0.06 5 

+
 150 


1/ 2
 0.04 5 

+
 80 


1/ 2
=2.88x10-4
Deq = 0.098 m
Then
H A = H B + hf
L V2
HA - HB = 20.3 = f
D 2g
; L = 110 m, D = 0.098m
Since both f and V are not known assume ε=0.2 mm and hydraulically rough flow such that
ε/D = 0.002
20.3 = 1122.45 f
i
0
1
ε/D
0.002
0.002
Therefore,
V2
2g
fi
0.0234
0.024
V
3.89
3.84
Re
373745
368941
fi+1
0.024
0.024
V = 3.84m/s
Q = 0.029 m3/s
Q = 104.3 m3/hr
Page 6 of 7
f1= f2
For
Deq =0.08 m
Leq = 39.3 m
Then
H A = H B + hf
HA - HB = 20.3 = f
L V2
D 2g
; L = 39.3 m
, D = 0.08 m
Since both f and V are not known assume ε=0.2 mm and hydraulically rough flow such that
ε/D = 0.0025
20.3 = 491.25 f
i
0
1
ε/D
0.0025
0.0025
Therefore,
V2
2g
fi
0.0249
0.0253
V
5.71
5.66
Re
447843
443922
fi+1
0.0253
0.0253
V = 5.66 m/s
Q = 0.0284 m3/s
Q = 102.4 m3/hr
Page 7 of 7
f1= f2