CVEN 3323 Hydraulic Engineering
Homework 3
Fall 2021
19. Water is flowing through the pipe shown in Figure P19 at a rate of 0.350 m 3/s. Manometers
are placed near the pipe entrance and downstream of the valve, diffusor, and confusor. Near
the pipe entrance, the static head, d, (d = P/+z) is 0.641 m relative to the centerline of the
pipe. Calculate the static head (d = P/+z) and total head (h = P/+z+V2/2g) at points 1, 2, and
3. Neglect pipe friction losses (i.e., the only losses are minor losses). Answer: d2 = 0.739 m
Figure P19
Given: Q = 0.350 m3/s, Dlarge = 0.400 m, Dsmall = 0.360 m, d0 = 0.641 m (since it is
measured relative to the centerline of the pipe, z = 0).
Needed: d1, d2, d3, ho, h1, h2, h3
Solution: The di represent pressure head at each point in the pipe. Calculate the minor
losses through the pipe sections, starting with the valve, and use Bernoulli equation to
calculate static head and total head. With the Bernoulli equation, total head is
Pi
Vi 2
hi = + zi +
= di + hVi

2g
where hVi is shorthand for velocity head at point i.
First calculate the velocities and velocity heads:
4 0.350m3 /s
Q
4Q
Vsmall =
=
=
= 3.44m/s = Vsmall = V0 = V1 = V3
2
2
Asmall  Dsmall
π ( 0.360m )
(
Vlarge
)
4 ( 0.350m3 /s )
Q
4Q
=
=
=
= 2.79m/s = Vlarge = V2
2
2
Alarge  Dlarge
π ( 0.400m )
( 3.44m/s ) = 0.603m = h
V2
= small =
V − small = hV 0 = hV 1 = hV 3
2g
2 ( 9.81m/s 2 )
2
hV − small
hV −l arg e =
Vl 2arg e
2g
=
( 2.79m/s )
2
2 ( 9.81m/s 2 )
= 0.395m = hV −l arg e = hV 2
Between points 0 and 1, the Bernoulli equation is
d0 + hV 0 = d1 + hV 1 + KV hV 1
For a gate valve, KV = 0.15 (from Table 3.6). Using hV0 = hV1, we can solve for d1 and h1
as
d 0 = d1 + KV hV 1  d1 = d 0 − KV hV 1 = 0.641m-0.15 ( 0.603m ) = 0.551m = d1
h1 = d1 + hV 1 = 0.551m + 0.603m = 1.154 m = h1
Between points 1 and 2, the Bernoulli equation is
2
2
' (V1 − V2 )
' (V1 − V2 )
d1 + hV 1 = d 2 + hV 2 + K E
 d 2 = d1 + hV 1 − hV 2 − K E
2g
2g
where K’E is the minor loss coefficient for the diffusor. From Figure 3.14, with a diffusor
angle of 40o, K’E = 0.95. Using this value in the previous equation, we can solve for d2
and h2 as
d 2 = d1 + hV 1 − hV 2 − K
'
E
(V1 − V2 )
2
2g
( 3.44 m/s − 2.79 m/s )
= 0.551m + 0.603m − 0.395m − 0.95
2 ( 9.81m/s 2 )
2
= 0.739 = d 2
h2 = d 2 + hV 2 = 0.739m + 0.395m = 1.134 m = h2
Between points 2 and 3, the Bernoulli equation is
d 2 + hV 2 = d3 + hV 3 + KC' hV 3  d3 = d 2 + hV 2 − hV 3 − KC' hV 3
where K’C is the minor loss coefficient for the confusor. For the confusor, the loss
coefficient depends on the angle and the ratio of the areas, A3/A2. Since area is
proportional to D2, A3/A2 = (D3/D2)2 =(0.360 m/0.400 m)2 = 0.81. For A3/A2 = 0.81 and
angle of 40o, K C' =0.01 (from Figure 3.11, reading below the line for A2/A1 = 0.8, about
1/10 of the way between the lines for A2/A1 = 0.8 and A2/A1 = 0.9). Using this value in the
previous equation, we can solve for d2 and h2 as
d3 = d 2 + hV 2 − hV 3 − K C' hV 3 = 0.739m + 0.395m − 0.603m − ( 0.01) 0.603m = 0.525m = d3
h3 = d3 + hV 3 = 0.525m + 0.603m = 1.128 m = h3
NOTE: Depending on when you round the answers, the heads may vary by up to 0.004
m.
20. In Problem 19, the flow rate is measured by measuring the time for 3 m3 of water to discharge
from the pipe, which took 8.57 seconds for the trial reported in Problem 19. Suppose the
uncertainty of this duration is t = 0.05 seconds and the uncertainty in the measured static head
at point 0 is d = 0.002 m. Estimate the uncertainty in the calculated static head at point 1.
Given: Q = 0.350 m3/s, D = 0.360 m, t = 0.05 s, Vol = 3 m3, t = 8.57 s, d0 = 0.002 m
Needed: d2
Solution: Use error propagation formula. First, write the equation for d2 in terms of the
unknown variables. We can write Q = (Vol)/ t, and V2 = Q/A2:
d1 = d 0 − KV hV 1 = d 0 − KV
8KV (Vol )
d1 = d 0 −
g 2 D14 t 2
V12
Q2
= d 0 − KV
2g
2 gA12
2

(1)
The next step is to write the error propagation formula for this equation. In this case, there
are two uncertain values, d0 and t, so there error propagation formula is
2
 d1 = 
2
d0
 d1 
2  d1 

 +t 

 t 
 d 0 
2
(2)
Next we obtain expressions for the partial derivatives in (2) using the equation in the last
line of (1):
d1
=1
d 0
(3)
 8K (Vol )2  16 K (Vol )2
d1
V
= −2  − 2 V 4
=
 g D1 t 3  g 2 D14 t 3
t


Finally, we substitute these expressions into (2) and then substitute the numerical values for
the symbols to obtain
2
 d 
 d 
 d 1 =  d20  1  +  t2  1 
 t 
 d 0 
2
 16 K (Vol )2 
2
2
=  d 0 (1) +  t  2 V 4

 g D1 t 3 


2
2
2

( 3m3 )  = 0.0023m = 
16 ( 0.15 )

d1
 ( 9.81m/s 2 )  2 ( 0.360m )4 ( 8.57 s )3 


2
=
( 0.002 m ) (1) + ( 0.05s )
2
2
2
21. Water flows at a rate of 0.750 m3/s through a reducing pipe bend and is deflected 30o in the
horizontal plane as shown in Figure P21. Water flows through a reducing pipe bend and is
deflected 25o in the horizontal plane as shown in Figure P21. Calculate the force of the bend
wall on the water as it passes through the bend. Be sure to draw a free body diagram showing
the forces acting on the water inside the bend. Also, sketch the vector of the calculated force
on Figure P21.
Figure P21
Given: Q = 0.75 m3/s, V1 = 3.00 m/s, A1 = 0.250 m2, P1 = 95.2 kN/m2
V2 = 6.00 m/s, A2 = 0.125 m2, P2 = 70.1 kN/m2
Deflection of 25o
Needed: force of the bend wall on the water in the bend.
Solution: Use conservation of momentum, along with  = 1000 kg/m3. The conservation
of momentum equations are
 F x = Q V x−out − V x−in and  F y = Q V y−out − V y−in
(
)
(
)
First draw a free body diagram. Forces are shown in red. Velocity vectors are shown in
blue.
The pressure force, P1A1, only has a component in the +x direction, while p2A2 has a
component in the -x direction (with a magnitude of P2A2 cos 25o) and in the +y direction
(with a magnitude of P2A2 sin 25o). V1 is directed in the +x direction, and V2 has components
in the +x direction (V2 cos 30o) and in the -y direction (magnitude of V2 sin 25o).
Using those components and directions, the conservation of momentum equations are
F
x
(
=  Q V x −out − V x −in
)

P1 A1 − P2 A2 cos 25o + Rx =  Q ( +V2 cos 25o − V1 )
 Rx =  Q ( +V2 cos 25o − V1 ) − P1 A1 + P2 A2 cos 25o
 1N 
 Rx = (1000kg/m3 )( 0.750m3 /s ) ( + ( 6.00m/s ) cos25o − 3.00m/s ) 
+
2 
 1kg-m/s 
1000N 
 − ( 95.2kN/m 2 )( 0.25m 2 ) + ( 70.1kN/m 2 )( 0.125m 2 ) cos25o  

  1kN 
= −14, 030N = Rx = 14, 000N 
F
y
(
= Q V y −out − V y −in
)

P2 A2 sin 25o + Ry =  Q ( −V2 sin 25o − 0 )
 Ry = −  QV2 sin 25o − P2 A2 sin 25o
 1N 
= − (1000kg/m3 )( 0.750m3 /s ) ( 6.00m/s ) sin25o 
2 
 1kg-m/s 
 1000N 
− ( 70.1kN/m 2 )( 0.125m 2 ) sin25o 

 1kN 
= −5605N = Ry = 5600N 
R = -14,000 N i - 5600N j.
Force is shown with the thick green arrow on the sketch.
22. (Synthesis Problem) In the reducing bend shown in Figure P22, water enters the bend flowing
at a rate of 7.00 ft3/s at an angle of -60o relative to the x axis and exits the bend at an angle 
relative to the x axis. Calculate the force of the wall on the water for angles of  = -90o, 80o, -70o, ... 70o, 80o, 90o. Show all equations used, and show one set of sample calculations.
Make plots of the magnitude of the resultant force vs.  and the direction of the resultant force
(in degrees, relative to the x axis, with positive angles in the counterclockwise direction) vs. 
. Use proper plot formatting. Discuss how the magnitude and direction of the resultant force
vary with  , and provide physical justification. The example problem in Lecture 7 was
identical to this scenario with  =0o, so you should be able to evaluate your answer at least for
=
Figure P22
Given: Q = 7.00 ft3/s, V1 = 14 ft/s, A1 = 0.50 ft2, P1 = 5320 lb/ft2
V2 = 70 ft/s, A2 = 0.10 ft2, P2 = 1040 lb/ft2
Flow direction at section 1, 1 = -60o relative to the x axis
Needed: Force of the bend wall on the water in the bend for = -90o, -80o, -70o, ... 70o,
80o, 90o.
Solution: Use conservation of momentum, along with  = 1.94 slug/ft3. The next page
shows a free body diagram with a generic outflow angle, .
Using this generic outflow angle, the conservation of momentum equations are
 F x = Q V x−out − V x−in
(
)
Rx + P1 A1 cos 60o − P2 A2 cos  =  Q (V2 cos  − V1 cos 60o )
Rx =  Q (V2 cos  − V1 cos 60o ) − P1 A1 cos 60o + P2 A2 cos 
(
)
− P A sin  =  Q (V sin  −  −V sin 60  )
F
Ry − P1 A1 sin 60o
y
=  Q V y −out − V y −in
o
2
2
2
1
Ry =  Q (V2 sin  + V1 sin 60o ) + P1 A1 sin 60o + P2 A2 sin 
These equations hold regardless of the sign on . Note that P1A1 and V1 are pointing in
the positive x direction and in the negative y direction, while P2A2 is pointing in the negative x
and y directions, and V2 is pointing in the positive x and y directions. This explains the
difference in signs on the P1A1 terms in the equations for Rx and Ry, and the difference in signs
on the V1 terms in the two equations.
Solving for Rx and Ry is done by substituting in the numerical values. The results are shown in
the Table on the next page, with sample calculations here for = -30o.
Rx =  Q (V2 cos  − V1 cos 60o ) − P1 A1 cos 60o + P2 A2 cos 


1lb
= (1.94slug/ft 3 )( 7.0 ft 3 /s ) ( 70 ft/s ) cos ( −30o ) − (14 ft/s ) cos 60o 
2 
 1slug-ft/s 
(
)
− ( 5320 lb/ft 2 )( 0.50 ft 2 ) cos 60o + (1040 lb/ft 2 )( 0.10 ft 2 ) cos ( −30o ) = −512 lb
Ry =  Q (V2 sin  + V1 sin 60o ) + P1 A1 sin 60o + P2 A2 sin 


1lb
= (1.94slug/ft 3 )( 7.0 ft 3 /s ) ( 70 ft/s ) sin ( −30o ) + (14 ft/s ) sin 60o 
2 
 1slug-ft/s 
(
)
+ ( 5320 lb/ft 2 )( 0.50 ft 2 ) sin 60o + (1040 lb/ft 2 )( 0.10 ft 2 ) sin ( −30o ) = 1941lb
Rx (lb)
Ry (lb)
|R| (lb)
-90
-80
-70
-60
-50
-40
-30
-20
-10
0
10
20
30
40
50
60
70
80
90
-1.57
-1.40
-1.22
-1.05
-0.87
-0.70
-0.52
-0.35
-0.17
0.00
0.17
0.35
0.52
0.70
0.87
1.05
1.22
1.40
1.57
-1425
-1242
-1064
-898
-747
-617
-512
-434
-386
-370
-386
-434
-512
-617
-747
-898
-1064
-1242
-1425
1414
1430
1477
1555
1660
1790
1941
2108
2285
2468
2651
2829
2996
3146
3276
3382
3459
3507
3523
2007
1894
1821
1796
1821
1894
2007
2152
2318
2496
2679
2862
3039
3206
3360
3499
3619
3720
3800
135
131
126
120
114
109
105
102
100
99
98
99
100
101
103
105
107
110
112
4000
160
3500
140
3000
120
2500
100
2000
80
1500
60
1000
40
500
20
Direction of R (deg)
 (rad)
|R| (lb)
 (deg)
Angle
(deg)
0
0
-90
-60
-30
0
30
60
90
 (deg)
The results show that the lowest |R| occurs for  = -60o, which is the same angle as the inflow to
the bend. When the outflow and inflow have the same angle, there is no deflection of the flow;
there is only a change in the velocity through the reduction. Thus, the force exerted by the wall
on the water is lowest. The magnitude of the force increases as the outflow angle moves away
from the inflow angle. The same force is required to deflect the flow an equal angle in either
direction, i.e.,  = -90o and  = -30o both require a 30-degree deflection, and both have the same
magnitude of the force.
When the outflow angle is equal to the inflow angle ( = -60o), the direction of the force is in the
opposite direction of the flow ( = 120o), indicating that there is no component of force in the
direction transverse to flow, which is consistent with the change in flow direction. For  < -60o,
the angle of the force increases, indicating that the component of force in the negative x direction
is increasing, to account for the deflection of flow toward the negative x direction. The opposite
occurs for  > 60o, with the component of force in the negative x direction decreasing.
23. Water is flowing through an 8000-m long, 0.90-m diameter commercial steel pipe in
its best condition. The total head at the upstream end of the pipe is 12.3 m higher than
at the downstream end. Use the Hazen Williams equation to solve for the flow rate in
this pipe. The Hazen-Williams coefficient for commercial steel pipe is 140 – 150.
Answer: Q = 0.96 m3/s.
Given: L = 8000 m, D = 0.90 m, hL = 12.3 m; CHW = 140 -150
Needed: water velocity from Hazen-Williams equation, and flow rate from continuity
Solution: Use Hazen Williams equation given by
V = 0.85 CHW Rh0.63 S 0.54
where Rh = D/4 = (0.90 m)/4 = 0.225 m is the hydraulic radius and S = hL/L. For
commercial steel pipe in its best condition, use the high end of the range of CHW, i.e., CHW
= 150. Using these values in the Hazen Williams equations produces
V = 0.85 CHW R
0.63
h
S
0.54
= ( 0.85)(150 )( 0.225)
0.63
 12.3 


 8000 
0.54
= 1.51m/s = V
( 0.90m ) = 0.96m3 /s = Q
D2
Q = VA = V 
= (1.51m/s ) 
4
4
2
24. Water (20oC) flows through a 0.05-m-diameter horizontal copper pipe at a rate of
0.0085 m3/s. Use the Hazen-Williams equation to calculate the pressure drop along a
25-m-long section of pipe. The Hazen-Williams coefficient for copper pipe is 130.
Given: L = 25 m, D = 0.05 m, Q = 0.0085 m3/s, CHW = 130, T = 20oC
Needed: pressure drop from Hazen-Williams equation
Solution: Since the pipe is horizontal, hL= P/ Use Hazen-Williams equation for SI units
and rearrange for head loss, and substitute hL= P/
Q = VA = 0.85 CHW R
0.63
h
hL0.54 =
QL0.54
0.2787CHW D 2.63
S
D2
D

= 0.85  CHW  
4
4
0.54
 hL =
P

=
0.63
 hL 
 
L
0.54
D 2 0.2787CHW D 2.63hL0.54
=
4
L0.54
10.7Q1.85 L
10.7Q1.85 L


P
=
1.85 4.87
1.85 4.87
CHW
D
CHW
D
At T = 20oC,  = 9790 N/m3. Substituting this and the given values leads to
1.85
10.7 Q1.85 L (10.7 ) ( 0.0085) ( 25 )( 9790 )
P =
=
= 1.0 105 N/m 2 = P
1.85
4.87
1.85 4.87
CHW
D
(130 ) ( 0.05)
25. (FE Question) A concrete sanitary sewer is 150 m long and has a pipe diameter of 1.25
m. The inlet elevation is 50.0 m, and the outlet elevation is 49.2 m. The HazenWilliams coefficient is 130. During heavy rainfalls, the sewer pipe flows full with no
surcharge. During heavy rainfalls, the capacity of the sewer is most nearly
a. 3.1 m3/s
b. 3.8 m3/s
c. 4.7 m3/s
d. 5.7 m3/s
Given: L = 150 m, D = 1.25 m, zinlet = 50.0m, zoutlet = 49.2 m, CHW =130
Solution: Use Hazen-Williams equation given by
V = 0.85 CHW Rh0.63 S 0.54
where Rh = D/4 = (1.25 m)/4 = 0.3125 m is the hydraulic radius and S = hL/L, and hL = zinlet
– zoulet = 50.0 m – 49.2 m = 0.8 m. Using these values in the Hazen-Williams equations
produces
0.63
0.54
V = 0.85 CHW Rh0.63 S 0.54 = ( 0.85 )(130 )( 0.3125 ) ( 0.8 ) = 3.15m/s = V
(1.25m ) = 3.9 m3 /s = Q
D2
Q = VA = V 
= ( 3.15m/s ) 
4
4
2