DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING
1ST YEAR FLUID MECHANICS
Solutions to problem sheet 1: Flow in pipes and pipelines
2. A pump is used to deliver water through a pipe 150mm in diameter. A certain section
C of the pipe is 5m above the pump outlet B, which is also 150mm in diameter. The
pump outlet is 600mm above the inlet A, which is 225mm in diameter. At A the mean
velocity is 1.35m/s, the pressure 150mm Hg vacuum and the power required to drive the
pump is 12.7kW. If friction in the pipe BC dissipates energy at the rate 2.5kW and the
gauge pressure at C is 35kPa, calculate the overall efficiency of the pump. (For mercury
σ = 13.65)
[ Ans: 68.0% ]
5m
B
ZC
C
PL
600mm
A
PS
Solution:
By definition, the vacuum pressure is the negative gauge pressure. Then, the gauge pressure at
A is:
kg
m
PA = −σ ρ g hHg = −13.56 × 1000 3 × 9.81 2 × 0.15m = −19953P a .
m
s
Note the negative pressure value, which means that the pressure is lower then the atmospheric
one.
Flow rate:
Q = AA UA =
π d2A
π
m
m3
UA = × 0.2252 m2 × 1.35 = 0.054
.
4
4
s
s
Mean velocity in the pipe BC:
UB = UC = UA
d2
m 2252
AA
m
= 3.038 .
= UA 2A = 1.35 ×
2
AC
s
150
s
dC
Steady flow energy equation for AC:
HA − HC = −
PW
PL
+
,
ρgQ ρgQ
where PW is the power supplied to the flow by the pump, and PL is the dissipation rate in the
pipe. Then
PC − PA UA2 − UC2
PW = ρ g Q (HC − HA ) + PL = ρ g Q ZC − ZA +
+
+ PL
ρg
2g
1
PW = 1000
kg
m
m3
× 9.81 2 × 0.054
×
3
m
s
s
×
5.6m +
35000P a + 19953P a
kg
m
1000 m
3 × 9.81 s2
2
2
3.0382 m
− 1.352 m
s2
s2
+
2 × 9.81 sm2
+ 2500W = 8634W
Efficiency is the ratio of the useful power PW to the shaft power PS :
η=
8634W
PW
=
= 68.0% .
PS
12700W
3. Water flows through a pipe 230m long and 20mm diameter with mean velocity 0.11m/s.
If the kinematic viscosity of water is 1.32mm2 /s, determine the pipe Reynolds number. Is
the flow laminar or turbulent?
Solution:
Re = Red =
0.02m × 0.11m/s
Ud
=
= 1667
ν
1.32 × 10−6 m2 /s
< 2000
⇒
laminar.
4. For fully developed flow of a liquid of dynamic viscosity 2.9 × 10−3 Pa s and relative density 0.9 in a horizontal pipe of diameter 12mm, determine the wall shear stress, pressure
gradient and head loss over a 200mm length of the pipe for the two conditions (i) mean
velocity 0.5m/s and (ii) mean velocity 4.6m/s. For laminar flow f = 16/Red , for turbulent
flow f = 0.079 Re−0.25
if Red < 105 .
d
[ Ans: (i) 0.968Pa; 322.6Pa/m; 7.3mm (ii) 65.7Pa; 21900Pa/m; 496mm ]
Solution:
Re =
ρU d
:
µ
(i)
(ii)
kg
m
900 m
3 0.012m 0.5 s
= 1862
0.0029P a s
kg
m
900 m
3 0.012m 4.6 s
= 17131
0.0029P a s
Re < 2000
τ =f
ρ U2
:
2
⇒
laminar
Re >> 2000
f = 0.079 Re−0.25 = 0.0069
0.0086
kg
m2
900 3 0.52 2 = 0.968P a
2
m
s
0.0069
kg
m2
900 3 4.62 2 = 65.7P a
2
m
s
d U2
200mm 0.52 m
s2
= 0.0073m
: 2 × 0.0086
L 2g
12mm 9.81 sm2
∆P
ρ g hf
=
:
l
l
turbulent
f = 16/Re = 0.0086
2
hf = 4 f
⇒
kg
m
900 m
Pa
3 9.81 s2 0.0073m
= 322.3
0.2m
m
2
2
2 × 0.0069
200mm 4.62 m
s2
= 0.496m
12mm 9.81 sm2
kg
m
900 m
Pa
3 9.81 s2 0.496m
= 21896
0.2m
m
5. Two points 700mm apart are joined by a pipe. If this pipe divides at mid length into two
parallel pipes, the same diameter being used through, calculate the percentage increase of
flow obtained with this system compared with that for a single pipe of the same diameter
used over the whole distance. (Neglect all losses except pipe friction and take f to be
constant)
[ Ans: 26.5% ]
Case B
Case A
QB / 2
H1
H2
QA
H1
QB
QB / 2
L
H2
L/2
L/2
Solution:
Single pipe (case A):
hf A = 4 f
L Q2A
.
d 2 A2 g
Divided pipes (case B):
hf B = 4 f
L/2 (QB /2)2
5
L Q2B
L/2 Q2B
+
4
f
=
4
f
.
d 2 A2 g
d
2 A2 g
8
d 2 A2 g
The head loss is the same for both cases. Then we have:
r
5
Q
8
B
=
Q2A = Q2B ⇒
= 1.26 .
8
QA
5
The increases of flow rate is 26%.
6. A pipe 600m long and 200mm diameter discharges water to atmosphere at a point 6m
below the level of the inlet. The pressure at the inlet is 40kPa and conditions are such
that the pipe is always full. At a point half way along the pipe water is withdrawn at the
rate 30 liters a second. Neglecting all losses of head except that due to pipe friction, and
taking the friction factor f as 0.006 throughout, determine the rate of discharge from the
end of the pipe.
[ Ans: 34.85 liters/s ]
Solution:
Steady flow energy equation for 1–2 and for 2–3:
H1 − H2 = 4 f
L/2 Q21
;
d 2 A2 g
H2 − H3 = 4 f
Summing up these two equations we have:
H1 − H3 =
f L 2
(Q1 + Q23 ) .
A2 g d
3
L/2 Q23
.
d 2 A2 g
L
L/2
1
Q1
P1
2
Z
Q3
3 P3 = 0
Q2
Heads at points 1 and 3 (gauge pressure at 3 is 0):
H1 =
Q21
P1
+
+ Z;
ρ g 2 A2 g
H2 =
Q23
.
2 A2 g
After substitution and rearrangement we have:
(Q21
+
Q23 )
A2 g d
1d 2
−
(Q1 − Q23 ) −
2 l
f l
P1
+Z
ρg
= 0.
The second therm of this equation is much smaller then the two other terms (d/(2 l) = 0.00017)
and can be neglected. By continuity we have Q1 = Q2 + Q3 , and the final equation for Q3
becomes:
Q2 − C
Q23 + Q2 Q3 + 2
= 0,
2
where we use the notation
A2 g d
P1
C=
+Z .
f l
ρg
Solutions of the quadratic equation are
q
1
Q3 = ( ± 2C − Q22 − Q2 ) ,
2
and we should take the solution with plus (Q3 is positive). Substituting the values:
C=
π 2 0.24 m4 × 9.81 sm2
0.2m
40 × 103 P a
m6
×
×
(
+
6m
)
=
0.00542
kg
m
42 × 0.006
600m
s2
1000 m
3 × 9.81 s2
r
m6
m6
m3
m3
1
Q3 = ( 2 × 0.0054 2 − 302 × 10−6 2 − 30 × 10−3
) = 0.03485
.
2
s
s
s
s
7. A tank 900mm square in plain is drained via a 60mm diameter pipe, 6m in length. The
pipe has two bends along its length with head loss coefficients k = 0.6 each, and the entry
of the pipe is not rounded. The outlet of the pipe is 230mm below the level of the base of
the tank. What is the depth of water in the tank when the rate of flow through the pipe
is 7 liters per second? Starting from this condition, how much water will drain from the
tank in the next tree minutes? (Assume f=0.008)
[ Ans: 1.613m; 994 liters ]
4
1
.
Z(t)
h
A
230 mm
K = 0.5 ( e.g. Massey )
K = 0.6
K = 0.6
a
2
Solution:
Steady flow energy equation along 1–2:
Z(t) −
Q2
=
2 a2 g
4f
L
+ KΣ
d
Q2
,
2 a2 g
P
where a is the pipe cross section area, KΣ = i Ki is the sum of all loss coefficients, and Z(t)
is the the height of the water surface in the tank above the outlet at time t. For Z(t) we have
L
Q2
Z(t) = 1 + KΣ + 4 f
,
d
2 a2 g
and when Q = 7 l/s (we chose this moment as time reference t = 0) the value of Z is:
6
72 × 10−6 m
6m
42
s2
Z(0) = (1 + 0.6 + 0.6 + 0.5 + 4 × 0.008
)×
= 1.843m ,
m
0.06m
2 × 9.81 s2 π 2 0.064 m4
and the corresponding depth h = 1.843 − 0.230 = 1.613m .
The flow rate can be expressed via the rate of change of water level as:
Q = −A
dZ
,
dt
where A is the area of the water surface in the tank. Substituting into the equation for Z we
have
s
p
a
2g
dZ
=−
Z(t) .
dt
A 1 + KΣ + 4 f L/d
After the integration
ZZ
dZ
a
√ =−
A
Z
s
2g
1 + KΣ + 4 f L/d
Zt
t dt
0
Z0
and using the initial condition Z = Z(0) as t = 0, we obtain
s
p
p
1 a
2g
Z(t) = Z(0) −
t.
2 A 1 + KΣ + 4 f L/d
5
After 3 minutes (t = 180s), we have:
Z(180s) =
√
π 0.062 m2
1.843m −
8 × 0.92 m2
s
2 × 9.81 sm2
× 180s
2.7 + 4 × 0.008 × 6m/0.06m
!2
and the volume of water drained from the tank is
V = A (Z(0) − Z(180s)) = 0.92 m2 (1.843m − 0.616m) = 0.994m3 .
6
= 0.616m ,