answers to example sheet for topic 2 autumn 2013€¦ · answers to example sheet for topic 2...
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Hydraulics 2 A2-1 David Apsley
ANSWERS TO EXAMPLE SHEET FOR TOPIC 2 AUTUMN 2013
(Full worked answers follow on later pages.)
Q1. (a) 0.166 m s–1
and 4.6810–6
m3 s
–1
(b) 1.32 m s–1
and 3.7210–5
m3 s
–1
Q2. A to B ; λ = 0.0129
Q3. DJB = 0.334 m; DJC = 0.443 m
Q4. (a) 10 m
(b) λ = 0.0181; Q = 0.0517 m3 s
–1
(c) 0.138 m
(d) 149 m
Q5. (a) 260 mm
(b) 24.2 kW
Q6. (a) 155 L s–1
(b) 65.7 m
Q7. (a) 48.5 m
(b) 95.2 kW
(c) 5.73 bar
Q8. (a) 67.5 kPa
(b) 27.6 m
(c) 159 L s–1
(d) 19.3 m
(e) 12.9 kW
Q9. (a) 0.401 m
(b) 0.01503; 71.3 m
Q10. (a) 0.235 m
(b) 31.8 L s–1
Q11. 6.62 L s–1
Q12. 2.79 m
Q13. (a) QC = 0.2 m3 s
–1
(b) QBC = QAB, QAD = 0.6 – QAB, QDC = 0.2 – QAB
(c) QAB = 0.185 m3 s
–1, QBC = 0.185 m
3 s
–1, QAD = 0.415 m
3 s
–1, QDC = 0.015 m
3 s
–1
(d) QAB = 0.341 m3 s
–1, QBC = 0.341 m
3 s
–1, QAD = 0.259 m
3 s
–1, QDC = –0.141 m
3 s
–1
Q14. (a) 171 m
(b) QAJ = 0.154 m3 s
–1, QBJ = 0.161 m
3 s
–1, QJC = 0.314 m
3 s
–1
Hydraulics 2 A2-2 David Apsley
Q15. (a) 81.9 L s–1
from A to J; 11.4 L s–1
from J to B; 70.4 L s–1
from J to C
(b) 97.5 L s–1
from A to J; 9.2 L s–1
from B to J; 66.7 L s–1
from J to C
Q16. (a) 28263 JAJA QHz , 252881 JFFJ QzH , 239661 JGGJ QzH
(b) 22.1 L s–1
; 6.5 m
(c) QAJ = 37.6 L s–1
; QJF = 18.6 L s–1
; QJG = 19.0 L s–1
; fountain height = 4.6 m
Q17. (a) From A to J: 156 L s–1
; from J to B: 106 L s–1
; from J to C: 50 L s–1
(given)
(b) From A to J: 284 L s–1
; from J to B: 82 L s–1
; from J to C: 202 L s–1
Q18. (All flows in L s–1
)
4
3
2
3
1.937
2.125 0.875
1.9370.062
Q19. QAC = 0.103 m3 s
–1; QBC = 0.069 m
3 s
–1; QCD = 0.172 m
3 s
–1; QDE = 0.075 m
3 s
–1;
QDF = 0.097 m3 s
–1
HC = 293 m; HD = 145 m
Q20. (a) 1.30 m3 s
–1
(b) 0.587 m
Q21. (a) 0.853 m3 s
–1
(b) 0.909 m3 s
–1
Q22. 188 mm
Q23. (b) 1.27 m3 s
–1
(c) 1.16 m
Q24. (a) 0.426 m3 s
–1
(b) 0.535 m
Q25. (a) 0.0152
(b) 229 mm
Hydraulics 2 A2-3 David Apsley
Q1.
The Reynolds number determines the regime of flow:
“small” Re laminar;
“large” Re turbulent.
“Small” or “large” depends on what flow is being considered and the particular velocity and
length scales used to define the Reynolds number.
In pipe flow the critical Reynolds number (based on pipe diameter and average velocity) for
transition in smooth-walled circular pipes is about 2300.
(a)
hf = 0.03 m
D = 0.006 m
L = 2 m
(ν = 1.010–6
m2 s
–1)
For this small head, guess the flow regime to be laminar. (Check later.) Then:
VD
ν64
Re
64λ
Substituting in the head-loss formula:
2
22 ν32
2
ν64)
2(λ
gD
LV
g
V
D
L
VDg
V
D
Lh f
Rearrange for V:
1
6
22
s m 1655.00.2100.132
03.0006.081.9
ν32
L
hgDV
f
[Check that the flow regime is indeed laminar:
993100.1
006.01655.0
νRe
6
VD
This is much less than 2300, so the assumption of laminar flow is justified.]
The discharge is:
13 622
s m 10679.44
006.0π1655.0
4
π
D
VVAQ
Answer: mean velocity V = 0.166 m s–1
; discharge Q = 4.6810–6
m3 s
–1.
(b) Data as above, except that hf = 1 m.
For this much larger head guess that the flow is turbulent; (check later). Then:
4/14/1
4/1
4/1
ν32.0
Re
32.0λ
DV
Substituting in the head-loss formula:
4/5
4/74/12
4/14/1
4/12 ν16.0
2
ν32.0)
2(λ
gD
LV
g
V
D
L
DVg
V
D
Lh f
Hydraulics 2 A2-4 David Apsley
Rearrange for V:
1
7/4
4/16
4/57/4
4/1
4/5
s m 317.10.2)100.1(16.0
0.1006.081.9
ν16.0
L
hgDV
f
[Check that the flow regime is indeed turbulent:
7900100.1
006.0317.1
νRe
6
VD
This is much greater than 2300, so the assumption of fully turbulent flow is justified.]
The discharge is:
13 522
s m 10724.34
006.0π317.1
4
π
D
VVAQ
Answer: mean velocity V = 1.32 m s–1
, discharge Q = 3.7210–5
m3 s
–1.
Hydraulics 2 A2-5 David Apsley
Q2.
D = 0.75 m
Q = 0.6 m3 s
–1
There are changes in both pressure and elevation. The overall change in head ΔH = HB – HA
is determined by the change in piezometric pressure p* = p + ρgz:
m 432.2)4034(81.91000
10)75.11.2(
ρ
5
AB
ABAB zz
g
ppHH
The head at B is lower. Therefore, the flow is from A to B. The frictional head loss is
hf = 2.432 m.
The flow velocity is
1
22s m 358.1
4/75.0π
6.0
4/π
D
Q
A
QV
The head-loss formula
)2
(λ2
g
V
D
Lh f
can be rearranged for the friction factor λ:
01294.0358.11500
432.275.081.922λ
22
LV
gDh f
Answer: friction factor λ = 0.0129.
Hydraulics 2 A2-6 David Apsley
Q3.
The heads at the reservoirs can be found from the slopes of the pipes (since the final water
levels are all the same distance above the respective pipe outfalls). The head at the junction
can be deduced from the head losses in pipe AJ (since the flow in this pipe is the sum of
flows in the two branch pipes). This gives the head loss in the branch pipes and hence, from
the head-loss vs discharge relationship, the diameters of the branch pipes.
Refer all heads to the water level in A (say). Then, using the length and slopes of the pipes:
0AH
m5.32005.015000025.010000 AB HH
m30001.050000025.010000 AC HH
The flow in AJ is, by continuity, the sum of the flows in the branch pipes:
13 sm30.015.015.0 JCJBAJ QQQ
The relationship between head loss and flow rate in each pipe is given by:
g
V
D
Lh
2λ
2
where 2π
4
D
Q
A
QV
52
2
π
λ8
gD
LQh
The head at the junction is then
m13.196.081.9π
3.01000002.080
π
λ852
2
52
2
AJ
AJAJ
AJgD
QLHH
Inverting the head-loss formula for diameter:
5/1
2
2
π
λ8
gh
LQD
Applying this to each branch pipe:
m334.0)5.3213.19(81.9π
15.0150002.08
)(π
λ85/1
2
25/1
2
2
BJ
JBJB
JBHHg
QLD
m443.0)3013.19(81.9π
15.0500002.08
)(π
λ85/1
2
25/1
2
2
CJ
JCJC
JCHHg
QLD
Answer: DJB = 0.334 m; DJC = 0.443 m.
Hydraulics 2 A2-7 David Apsley
Q4.
(a)
Static lift = 150 – 120 = 30 m.
Head supplied by the pump = 40 m.
Hence, the head lost to pipe friction is 40 – 30, or 10 m.
Answer: 10 m.
(b)
hf = 10 m
L = 800 m
D = 0.2 m
ks = 10–4
m
ν = 10–6
m2 s
–1
Darcy-Weisbach head-loss equation:
g
V
D
Lh f
2λ
2
Rearranging for the unknowns:
212 )sm(04905.0800
102.081.922λ
L
gDhV
f
From the Colebrook-White equation,
2
10 )λRe
51.2
7.3(log0.2
1λ
D
k s
But
2λ
ν51.2
λ
ν51.2
λRe
51.2
VDVD
Hence,
01809.0
)04905.02.0
1051.2
2.07.3
10(log0.2
1λ
264
10
Since we know λ and V2 we can find the bulk velocity:
12
sm647.101809.0
04905.0
λ
λ V
V
The flow rate is then given by (velocity area):
1322
sm05174.04
2.0π647.1
4
π
D
VVAQ
Answer: λ = 0.0181; Q = 0.0517 m3 s
–1.
Hydraulics 2 A2-8 David Apsley
(c) The dynamic head is
m1383.081.92
647.1
2
22
g
V
Answer: 0.138 m.
(d) Total head drops toward the pump, so the worst case will be at the pump itself. The total
head at the entrance to the pump is the head in the river, minus the frictional head loss over
distance x:
g
V
D
xH
g
Vz
g
p pump
2λ
2ρ
2
0
2
Relative to the initial still-water level, H0 = 0, z = –2 m. If the gauge pressure ppump is 0 then
zg
V
D
x
2)λ1(
2
Substituting values:
21383.0)2.0
01809.01( x
which solves to give
m8.148x
Answer: 149 m.
Hydraulics 2 A2-9 David Apsley
Q5.
(a)
L = 30 000 m
hf = 150 m
13 s m05787.0606024
5000
Q
ks = 310–4
m
From the Darcy-Weisbach head-loss equation:
2
2
π
4),
2(λ
D
Q
A
QV
g
V
D
Lh f
52
2
π
λ8
gD
LQh f
5/1
2
25/1
2
2
λ15081.9π
05787.0300008
π
λ8
fgh
LQD
Hence:
5/1)λ05534.0(D (*)
An expression for the Reynolds number is required in the Colebrook-White equation:
DDD
QD
D
QVD 736801
10π
05787.041
πν
4
νπ
4
νRe
62
Substituting in the Colebrook-White equation and rearranging,
255
10 )λ
10407.310108.8(log0.2
1λ
D
D
(**)
Starting iteration with initial guess λ = 0.01 and iterating (*) and (**) in succession:
λ D (m)
0.01000 0.2232
0.02218 0.2617
0.02122 0.2594
0.02127 0.2595
0.02127 0.2595
Answer: D = 260 mm.
(b) Head available to drive the turbine will be 150 m minus the frictional head loss.
L = 30 000 m
13 s m03472.0606024
3000
Q
ks = 310–4
m
D = 0.2595 m
Find hf. First need to iterate for a new λ.
Hydraulics 2 A2-10 David Apsley
1
22s m 6565.0
2595.0π
03472.04
π
4
D
Q
A
QV
17040010
2595.06565.0
νRe
6
VD
Hence
2
54
10 )λ
10473.110125.3(log2
1λ
Successive approximations for λ starting from the value in part (a):
λ
0.02127
0.02184
0.02182
0.02182
Then
m 41.5581.92
6565.0
2595.0
3000002182.0
2λ
22
g
V
D
Lh f
Hence the head available for the turbine is
m 59.9441.55150 turbineh
Hence,
W24160
59.9403472.081.9100075.0
ρoutputpower
turbinegQhefficiency
Answer: power = 24.2 kW.
Hydraulics 2 A2-11 David Apsley
Q6.
(a)
L = 3000 m
D = 0.3 m
ks = 1.010–4
m
ν = 1.010–6
m2 s
–1
At maximum discharge, the head lost to friction is equal to the available gravitational head:
hf = 40 m
Inspect the head-loss equation:
g
V
D
Lh f
2λ
2
The only unknowns are λ and V. Collecting these:
212 )sm(07848.03000
403.081.922λ
L
gDhV
f
Rearranging the Colebrook-White equation and expanding the Reynolds number:
01626.0
)07848.03.0
1051.2
3.07.3
10(log0.2
1
)λ
ν51.2
7.3(log0.2
1λ
264
10
2
210
VDD
k s
Solving for V:
12
sm197.201626.0
07848.0
λ
λ V
V
Finally,
1322
sm1553.04
3.0π197.2
4
π
D
VVAQ
Answer: Q = 155 L s–1
.
(b) With the given flow rate, find the frictional head loss and subtract the available
gravitational head to find the amount which must be made up by a pump.
L = 3000 m
D = 0.3 m
ks = 1.010–4
m
ν = 1.010–6
m2 s
–1
Q = 0.2553 m3 s
–1
From these we can deduce
1
22sm612.3
3.0π
2553.04
π
4
D
Q
A
QV
Hydraulics 2 A2-12 David Apsley
6
610085.1
100.1
3.0612.3
νRe
VD
Inspect the head-loss equation:
g
V
D
Lh f
2λ
2
hf will be known if we can find λ.
Rearranging the Colebrook-White equation:
2
6
4
10
2
10 )λ10085.1
51.2
3.07.3
10(log0.2
1
)λRe
51.2
7.3(log0.2
1λ
D
ks
Hence,
2
65
10 )λ
10313.210009.9(log0.2
1λ
Iterating from initial value 0.01626 gives λ = 0.01590.
Hence,
m7.10581.92
612.3
3.0
300001590.0
2λ
22
g
V
D
Lh f
40 m of this is available from gravity. The rest must be made up from a pump:
Hpump = 105.7 – 40 = 65.7 m
Answer: Hpump = 65.7 m.
Hydraulics 2 A2-13 David Apsley
Q7.
(a) The total head required from the pump is the static head (30 m) plus the frictional head
loss, hf. The latter can be found from the head-loss equation and Colebrook-White equations.
Known parameters:
L = 5000 m
D = 0.3 m
Q = 0.08 m3 s
–1
ks = 1.010–4
m
ν = 1.010–6
m2 s
–1
From these we can calculate the average velocity and Reynolds number:
339600ν
Re
sm132.1π
4 1
2
VD
D
Q
A
QV
Inspect the head-loss formula:
g
V
D
Lh f
2λ
2
The frictional head loss will be known if we can find λ.
From the Colebrook-White formula:
λRe
51.2
7.3log0.2
λ
110
D
ks
Either rearrange for λ and iterate, or, for a change, write λ/1x . Substituting values:
xx 65
10 10391.710009.9log0.2
Iterating from x = 10, gives, successively,
x = 10, 7.570, 7.671, 7.667, 7.667
01701.01
λ2
x
m52.1881.92
132.1
3.0
500001701.0
2λ
22
g
V
D
Lh f
The total head which must be provided by the pump is then
m52.4852.1830 H
Answer: total head = 48.5 m.
(b) The output power (ρgQH) is the efficiency times the input power Pin; i.e.
inPgQH ηρ
W952004.0
52.4808.081.91000
η
ρ
gQHPin
Answer: input power = 95.2 kW.
Hydraulics 2 A2-14 David Apsley
(c) The maximum head (and pressure) occur immediately downstream of the pump. At this
point, relative to the top-water level of the reservoir, the total head H = 48.52 m, the elevation
is z = –10 m and the velocity is V = 1.132 m s–1
. Then
g
Vz
g
pH
2ρ
2
Pa10734.5)81.92
132.11052.48(81.91000)
2(ρ 5
22
g
VzHgp
Answer: 5.73105 Pa (= 5.73 bar).
Hydraulics 2 A2-15 David Apsley
Q8.
Density of oil = 0.86 1000 = 860 kg m–3
.
(a) By hydrostatics,
Pa67490881.9860Δρ zgp
Answer: at the entrance to the tank, p = 67.5 kPa.
(b) Pressure loss
Pa23251067490300000Δ p
Corresponding head loss:
m56.2781.9860
232510
ρ
Δ
g
ph
Answer: head loss along the pipeline = 27.6 m
(c)
D = 0.25 m
L = 400 m
ks = 110–4
m
hf = 27.56 m
ν = 910–5
m2 s
–1
Assume that the head loss is entirely due to pipe friction. The frictional head-loss equation is
g
V
D
Lh f
2λ
2
Collecting the unknowns onto one side:
212 )sm(3380.0400
56.2725.081.922λ
L
gDhV
f
Rearrange the Colebrook-White equation for λ, expanding the Reynolds number ν
ReVD
:
03237.0
3380.025.0
10951.2
25.07.3
10log0.4
1
λ
ν51.2
7.3log0.4
1
λRe
51.2
7.3log0.4
1λ
254
10
2
210
2
10
VDD
k
D
kss
Then
44.1003237.0λ
3380.0λ 2
2
V
V
Hydraulics 2 A2-16 David Apsley
1sm231.3 V
1322
sm1586.04
25.0π231.3
4
π
D
VVAQ
Answer: Q = 159 L s–1
.
(d) The head loss at the valve will be the overall head loss minus the frictional head loss hf
(which must be found). If the flow is reduced by half then so is the velocity:
1sm616.1231.32
1 V
4489109
25.0616.1
νRe
5
VD
(The Reynolds number is still just sufficient for turbulent flow).
Inspect the head-loss formula:
g
V
D
Lh f
2λ
2
The only unknown is λ.
Rearrange the Colebrook-White equation:
2
4
10
2
10
λ4489
51.2
25.07.3
10log2
1
λRe
51.2
7.3log2
1λ
D
ks
Substituting values, an iteration formula is
2
44
10λ
10591.510081.1log2
1λ
Starting from the value from part (c),
λ = 0.03237 → 0.04023 → 0.03880 → 0.03904 → 0.03900 → 0.03900
The frictional head loss is then
m306.881.92
616.1
25.0
40003900.0
2λ
22
g
V
D
Lh f
The remainder of the available head (27.56 m) is lost across the valve; i.e.
m25.19306.856.27 valveH
Answer: head loss at the valve =19.3 m.
(e) Power loss at the valve is
W1290025.19)1586.0(81.9860ρ21 valvegQH
Answer: power loss = 12.9 kW.
Hydraulics 2 A2-17 David Apsley
Q9.
(a)
Q = 0.2 m3 s
–1
hf = 25 m
L = 5000 m
ks = 1.010–4
m
ν = 1.010–6
m2 s
–1
From the Darcy-Weisbach head-loss equation:
2
2
π
4),
2(λ
D
Q
A
QV
g
V
D
Lh f
52
2
π
λ8
gD
LQh f
5/1
2
25/1
2
2
λ2581.9π
2.050008
π
λ8
fgh
LQD
5/1)λ6610.0(D (*)
An expression for the Reynolds number is required in the Colebrook-White equation:
DDD
QD
D
QVD 2546001
10π
2.041
πν
4
νπ
4
νRe
62
Substituting in the Colebrook-White equation and rearranging,
265
10 )λ
10859.910703.2(log0.2
1λ
D
D
(**)
Starting iteration with initial guess λ = 0.01 and iterating (*) and (**) alternately:
λ D (m)
0.01000 0.3665
0.01595 0.4023
0.01558 0.4005
0.01560 0.4006
0.01561 0.4006
0.01561 0.4006
Answer: D = 0.401 m.
(b) If the flow rate is increased then the frictional head loss must also increase; the excess
above 25 m must be supplied by pumps.
Q = 0.4 m3 s
–1
D = 0.4006 m
L = 5000 m
ks = 1.010–4
m
ν = 1.010–6
m2 s
–1
Hydraulics 2 A2-18 David Apsley
Find hf. First need to iterate for a new λ.
1
22s m 174.3
4006.0π
4.04
π
4
D
Q
A
QV
6
610272.1
100.1
4006.0174.3
νRe
VD
Hence
2
65
10 )λ
10973.110747.6(log2
1λ
Successive approximations for λ starting from the value in part (a):
λ
0.01561
0.01502
0.01503
0.01503
Then
m 32.9681.92
174.3
4006.0
500001503.0
2λ
22
g
V
D
Lh f
25 m of this head is supplied by the difference in water levels; the rest must be supplied by
pumps:
m32.712532.96 pumpH
Answer: friction factor = 0.01503; pumping head = 71.3 m.
Hydraulics 2 A2-19 David Apsley
Q10.
(a)
L = 3000 m
h = 25 m
ks = 1.0 mm
Q = 0.05 m3 s
–1
From the Darcy-Weisbach head-loss equation:
g
V
D
Lh
2λ
2
where 2π
4
D
Q
A
QV
52
2
π
λ8
gD
LQh
5/1
2
2
λπ
8
gh
LQD
Substituting values:
5/1)λ02479.0(D (*)
From the Colebrook-White equation for fully-rough walls (i.e., no Re-dependent term):
)7.3
(log0.2λ
110
sk
D
Substituting values and rearranging:
2
10 )3700(log0.2
1λ
D (**)
Normally, one would iterate equations like (*) and (**) in turn to find the pipe diameter.
However, because the friction-factor formula is simplified here by the absence of Reynolds-
number terms, and there are no minor losses, it is easier in this particular problem to
substitute (*) in (**) to obtain a single iterative equation for λ:
25/1
10 )λ1766(log0.2
1λ (***)
Iteration of (***), from e.g. λ = 0.01 or λ = 0.1, gives:
λ = 0.02894
The diameter is then, from (*),
D = 0.2351 m
Answer: diameter = 0.235 m.
(b) Let the flow in the last third of the pipe be Q (in m3 s
–1). Then, by continuity at the
junctions, the flow in the middle third is Q + 0.02 (m3 s
–1) and the flow in the first third is
Q + 0.03 (m3 s
–1). Adding the head losses in each section (which have common length
L1 = 1000 m, diameter D and friction factor λ):
hgD
QL
gD
QL
gD
QL
52
2
1
52
2
1
52
2
1
π
λ8
π
)02.0(λ8
π
)03.0(λ8
Hydraulics 2 A2-20 David Apsley
1
52222
λ8
π)02.0()03.0(
L
hgDQQQ
007509.00013.01.03 2 QQ
0006209.01.03 2 QQ
Solving this quadratic equation (noting that the negative root is meaningless):
132
sm03178.06
006209.0341.01.0
Q
Answer: discharge to the lower tank = 31.8 L s–1
.
Hydraulics 2 A2-21 David Apsley
Q11.
The pipes are in series so the same quantity of flow Q passes through each and the velocity in
each pipe can be found from
4/π 2D
Q
A
QV
where D is the relevant diameter.
We can use either the total head loss or the total pressure loss; here it is easier to use the total
pressure loss, so avoiding using g.
Equate total pressure loss to the sum of the individual losses (friction + junction + friction):
)ρ(λ)ρ(5.0)ρ(λΔ 2
221
2
22
2
1212
121
1
11 V
D
LVV
D
Lp
Substitute for V1 and V2 in terms of the single unknown, Q:
2
2
2
21
2
22
2
2
1
21
1
11
π
4ρλ
π
4ρ)5.0λ(Δ
D
Q
D
L
D
Q
D
Lp
Substituting |Δp| = 2000 Pa, ρ = 900 kg m–3
, λ1 = 0.014, λ2 = 0.012, D1 = 0.2 m, D2 = 0.1 m,
L1 = L2 = 50 m gives
2710559.42000 Q
Q = 6.6210–3
m3 s
–1
Answer: quantity of flow Q = 6.62 L s–1
.
Hydraulics 2 A2-22 David Apsley
Q12.
Because the pipes are in parallel, the head loss h is the same along each pipe. Hence,
2
33
2
22
2
11 ααα QQQh
3
3
2
2
1
1α
,α
,α
hQ
hQ
hQ
The total flow is
)
α
1
α
1
α
1(
321
321
h
QQQQ
Substituting values Q = 0.4 m3 s
–1, α1 = 100, α 2 = 150, α 3 = 300 gives
2394.04.0 h
m79.2h
Answer: head loss = 2.79 m.
Hydraulics 2 A2-23 David Apsley
Q13.
(a) Continuity for the whole network: total flow in = total flow out
CQ 4.06.0
Answer: QC = 0.2 m3 s
–1.
(b) Continuity for each individual junction: total flow in = total flow out.
Answer: QBC = QAB, QAD = 0.6 – QAB, QDC = 0.2 – QAB.
(c) Head changes along a pipe are of the form:
negativeisifα
positiveisifα2
2
ABABAB
ABABAB
BAQQ
QQHH
and similarly for the other pipes.
Head change along ABC = head change along ADC; i.e.
)()()()( CDDACBBA HHHHHHHH
Assume first the most likely scenario that QAB, QBC, QAD, QBC are all positive (check later!).
2222 αααα DCDCADADBCBCABAB QQQQ
2222 )2.0(100)6.0(100200300 ABABABAB QQQQ
)4.004.0(100)2.136.0(100500 222
ABABABABAB QQQQQ
040160300 2 ABAB QQ
02815 2 ABAB QQ
30
215488 2 ABQ (only the positive root is desired)
This gives
QAB = 0.185 m3 s
–1, QBC = 0.185 m
3 s
–1, QAD = 0.415 m
3 s
–1, QDC = 0.015 m
3 s
–1.
These are all positive, so consistent with the head-loss equation.
Answer: QAB = 0.185 m3 s
–1, QBC = 0.185 m
3 s
–1, QAD = 0.415 m
3 s
–1, QDC = 0.015 m
3 s
–1.
(d) If we make the same assumptions about direction, but take αAD = 900 instead, then the
same analysis leads to
2222 )2.0(100)6.0(900200300 ABABABAB QQQQ
)4.004.0(100)2.136.0(900500 222
ABABABABAB QQQQQ
03281120500 2 ABAB QQ
1000
328500411201120 2 ABQ (both roots are positive)
This gives either solution
QAB = 1.894 m3 s
–1, QBC = 1.894 m
3 s
–1, QAD = –1.294 m
3 s
–1, QDC = –1.694 m
3 s
–1
QAB = 0.346 m3 s
–1, QBC = 0.346 m
3 s
–1, QAD = 0.254 m
3 s
–1, QDC = –0.146 m
3 s
–1
Hydraulics 2 A2-24 David Apsley
In both cases there is flow of the wrong direction in at least one pipe … so the assumed head
changes are wrong.
Instead, since we increased the resistance in AD we might anticipate that the flow is actually
from C to D; that is, QDC is negative. The head changes along the two paths are then:
2222 )2.0(100)6.0(900200300 ABABABAB QQQQ
)4.004.0(100)2.136.0(900500 222
ABABABABAB QQQQQ
03201040300 2 ABAB QQ
0165215 2 ABAB QQ
30
161545252 2 ABQ (both roots are positive)
This gives either solution
QAB = 3.125 m3 s
–1, QBC = 3.125 m
3 s
–1, QAD = –2.525 m
3 s
–1, QDC = –2.925 m
3 s
–1
QAB = 0.341 m3 s
–1, QBC = 0.341 m
3 s
–1, QAD = 0.259 m
3 s
–1, QDC = –0.141 m
3 s
–1.
In the first case the direction of QAD is wrong. However, the last solution is OK, consistent
with actual flow from C to D.
Answer: QAB = 0.341 m3 s
–1, QBC = 0.341 m
3 s
–1, QAD = 0.259 m
3 s
–1, QDC = –0.141 m
3 s
–1.
Hydraulics 2 A2-25 David Apsley
Q14.
(a) The minimum top-water level in B is that at which the heads at J and B are equal and
there is no flow in JB. The head at J is then that arising from flow from A to C.
First find the head-loss vs discharge relation for each pipe. From the head-loss formula:
)2
(λ2
g
V
D
Lh where
2π
4
D
Q
A
QV
52
2
π
λ8
gD
LQh
With the given data the head losses (in m) against the flow rates (in m3 s
–1) for each pipe are:
222 323,1088,2031 JCCJJBBJJAAJ QHHQHHQHH
or
323
,1088
,2031
CJ
JC
BJ
JB
AJ
JA
HHQ
HHQ
HHQ
(*)
When B and J have the same head the same flow, Q say, goes from A to J and then J to C as
for pipes in series. Thus,
2
2
323
2031
QHH
QHH
CJ
JA
Adding these head losses to get the total head loss from A to C (which is known):
22354QHH CA
2132 )s (m 03398.02354
160240
2354
CA HHQ
The head at J (in this part, the same as that at B) is
m 17103398.020312402031 2 QHH AJ
Answer: minimum water level = 171 m.
(b) Use the head-discharge relations (*) and a succession of trial values of HJ until there is
continuity at junction J. With the given heads at the reservoirs:
323
160,
1088
220,
2031
240
J
JC
J
JB
J
JA
HQ
HQ
HQ
Trial values and the net outflow are given in the following table. After the first two guesses,
better values can be found by interpolation/extrapolation.
HJ (m) QJA (m3 s
–1) QJB (m
3 s
–1) QJC (m
3 s
–1) Net outflow (QJA + QJB + QJC)
190 –0.1569 –0.1661 0.3048 –0.0182 (too much in; raise HJ)
200 –0.1403 –0.1356 0.3519 +0.0760 (too much in; lower HJ)
191.93 –0.1538 –0.1606 0.3144 0.0000
Answer: QAJ = 0.154 m3 s
–1, QBJ = 0.161 m
3 s
–1, QJC = 0.314 m
3 s
–1.
Hydraulics 2 A2-26 David Apsley
Q15.
(a) First find the resistance laws for each pipe:
g
V
D
Lh
2λ
2
where 2π
4
D
Q
A
QV
52
2
π
λ8
gD
LQh
Substituting values gives the following for the three pipes:
22380 JAAJ QHH or 2380
AJ
JA
HHQ
230985 JBJ QHBH or 30985
BJ
JB
HHQ
212910 JCCJ QHH or 12910
CJ
JC
HHQ
The direction of flow is from high head to low head. Heads are in m, discharges in m3 s
–1.
Vary HJ until the net (signed) flow out of the junction is zero. This is best set out in a table.
Flow directions in pipes JA and JC are obvious, but that in JB must be determined by the
heads at each end: a first guess HJ = 280 allows one to determine the direction of flow in pipe
JB. (Entries for HJ after the first two are by interpolation to target net outflow = 0.)
HJ 2380
300 J
JA
HQ
30985
280
J
JB
HQ
12910
220 J
JC
HQ JCJBJA QQQ
280 –0.09167 0 +0.06817 –0.02350
290 –0.06482 +0.01797 +0.07363 +0.02678
284.67 –0.08025 +0.01228 +0.07078 +0.00281
284.17 –0.08155 +0.01160 +0.07050 +0.00055
284.05 –0.08186 +0.01143 +0.07044 +0.00001
Answer: discharges: 81.9 L s–1
from A to J; 11.4 L s–1
from J to B; 70.4 L s–1
from J to C.
(b) The only difference here is an additional outflow of 0.04 m3 s
–1 from the junction.
HJ 2380
300 J
JA
HQ
30985
280
J
JB
HQ
12910
220 J
JC
HQ 04.0 JCJBJA QQQ
280 –0.09167 0 +0.06817 +0.01650
270 –0.11227 –0.01796 +0.06223 –0.02800
276.29 –0.09981 –0.01094 +0.06603 –0.00472
277.12 –0.09804 –0.00964 +0.06652 –0.00117
277.39 –0.09746 –0.00918 +0.06667 +0.00003
Note that the direction of flow in JB is now reversed.
Answer: discharge: 97.5 L s–1
from A to J; 9.2 L s–1
from B to J; 66.7 L s–1
from J to C.
Hydraulics 2 A2-27 David Apsley
Q16.
(a) Since there is no dynamic head at A:
AA zH
whereas at F and G the total head includes the dynamic head at exit:
g
VzH JF
FF2
2
g
VzH JG
GG2
2
Applying the head-loss equations for each pipe (the direction of flow being obvious):
AJ
JAg
V
D
LHz
2
λ 2
JFJF
FJg
V
D
L
g
VzH
2
λ)
2(
22
JF
FJg
V
D
LzH
2)
λ1(
2
JGJG
GJg
V
D
L
g
VzH
2
λ)
2(
22
JG
GJg
V
D
LzH
2)
λ1(
2
(Note that, as far as the piezometric head is concerned, the exit dynamic head looks like an
effective “loss”, with loss coefficient 1.0; i.e. this kinetic energy is not available as potential
energy).
Noting that, for each pipe,
2π
4
D
Q
A
QV
the head vs discharge conditions for each pipe are:
AJ: 28263 JAJA QHz
JF: 252881 JFFJ QzH
JG: 239661 JGGJ QzH
where H is in m, Q in m3 s
–1.
(b) In this case there is no JG; the same Q goes through AJ and JF. Adding head changes:
261144QzHHz FJJA
26114430 Q
21342 )sm(10906.461144
30 Q
13 sm02215.0 Q
The velocity head at exit is
m486.605.081.9π
10906.48
π
8
2 42
4
42
22
gD
Q
g
V
and this is the height to which the fountain rises (in the absence of losses).
Answer: Q = 22.1 L s–1
; fountain height = 6.5 m.
Hydraulics 2 A2-28 David Apsley
(c) In this case we require repeated trials for HJ. We have a free choice, but it is convenient to
take the height datum as the exit point F. Note the sign convention for flow (here: positive
out of the junction).
HJ 8263
30 J
JA
HQ
52881
0 J
JF
HQ
39661
4 J
JG
HQ
Net outflow
JGJFJA QQQ
15 –0.04261 0.01684 0.01665 –0.00912
20 –0.03479 0.01945 0.02009 +0.00475
18.29 –0.03765 0.01860 0.01898 –0.00007
18.31 –0.03761 0.01861 0.01899 -0.00001
The last row gives the flow rates. The velocity head at exit from F is now:
m579.405.081.9π
01861.08
π
8
2 42
2
42
22
gD
Q
g
V
and this is the height to which the fountain rises (in the absence of losses).
Answer: QAJ = 37.6 L s–1
; QJF = 18.6 L s–1
; QJG = 19.0 L s–1
; fountain height at F = 4.6 m.
Hydraulics 2 A2-29 David Apsley
Q17.
(a) First find the loss coefficients for each pipe:
g
V
D
Lh
2λ
2
where 2π
4
D
Q
A
QV
2
52π
λ8Q
gD
Lh
Applying this for each pipe in turn (noting that the direction of flow in JB is not obvious):
22.605 JAJA QHH
27615 JBJB QHH
22720 JCCJ QHH (when not valve-regulated)
where heads are measured in m, flow rates in m3 s
–1.
In this part, QJC ≡ 0.05.
Let the flow rate from A to J be Q.
To determine the direction of flow suppose HJ = 200. Then there is no
flow in JB but the flow in AJ is
13 sm4065.02.605
200300
2.605
JA HHQ
This is too large; hence HJ must be raised and the flow is from J to B.
By continuity the flow in JB is Q – 0.05 and we have the situation
shown right.
Considering the head losses along AJ and JB:
22.605300 QH J
2)05.0(7615200 QH J
Adding gives
04.195.7618220)0025.01.0(76152.605100 222 QQQQQ
096.805.7618220 2 QQ
0632.0or1558.082202
96.80822045.7615.761 2
Q
We require the positive root: Q = 0.1558 m3 s
–1.
The flow from J to B is then
13 sm1058.005.01558.005.0 Q
Answer: flow from A to J: 156 L s–1
; from J to B: 106 L s–1
; from J to C: 50 L s–1
(given).
J
A
B
C0.05
Q
J
A
B
C0.05
Q-0.05
Q
Hydraulics 2 A2-30 David Apsley
(b) Use the head-discharge relations and a succession of trial values of HJ until there is
continuity at junction J. With the given heads at the reservoirs:
2720
140,
7615
200,
2.605
300
J
JC
J
JB
J
JA
HQ
HQ
HQ
Trial values and the net outflow are given in the following table. After the first two guesses,
better values can be found by interpolation/extrapolation.
HJ (m) QJA (m3 s
–1) QJB (m
3 s
–1) QJC (m
3 s
–1) Net outflow (QJA + QJB + QJC)
200 –0.4065 0.0 +0.0221 –0.3844 (too much in; raise HJ)
250 –0.2874 +0.0810 +0.2011 –0.0053 (too much in; raise HJ)
250.7 –0.2854 +0.0816 +0.2017 –0.0021 (too much in; raise HJ)
251.16 –0.2841 +0.0820 +0.2022 +0.0001
Answer: flow from A to J: 284 L s–1
; from J to B: 82 L s–1
; from J to C: 202 L s–1
.
Hydraulics 2 A2-31 David Apsley
Q18.
Divide the network into upper and lower loops. The flow increment in each loop is
Q
QsQ
α2
αδ
2
or, since in this example α is the same for each pipe,
Q
sQQ
2δ
2
Initial guess satisfying continuity:
Update of upper loop:
7.1)014(2
)014(δ
222
Q
Update of lower loop:
15.0)027.1(2
)027.1(δ
222
Q
Iteration continues by alternating upper and lower loops until convergence.
Final network:
4
3
2
30
4 1
20
4
3
2
3
1.7
2.3 0.7
20
4
3
2
3
1.85
2.3 0.7
1.850.15
4
3
2
3
1.937
2.125 0.875
1.9370.062
Hydraulics 2 A2-32 David Apsley
Q19.
Head-loss vs discharge relations:
2αQh
The relationship between small changes in head and discharge (with correct sign) is then
QQh δα2δ or Q
hQ
α2
δδ
Using these, flow rates in the various pipes and their response to small changes are given by:
DF
DDF
DDF
DE
DDE
DDE
CD
DC
CD
DC
CD
BC
C
BC
C
BC
AC
C
AC
C
AC
Q
HQ
HQ
Q
HQ
HQ
Q
HHQ
HHQ
Q
HQ
HQ
Q
HQ
HQ
20000
δδ
10000
50
16000
δδ
8000
100
10000
δδδ
5000
24000
δδ
12000
350
20000
δδ
10000
400
Starting from HC = 300 and HD = 150 gives
D
DDFDF
DDEDE
CCDCD
CBCBC
CACAC
H
HQQ
HQQ
HQQ
HQQ
HQQ
δ
δ00050.0δ1000.0
δ00079.0δ0791.0
00058.0δ00058.0δ1732.0
δ00064.0δ0645.0
δ00050.0δ1000.0
The net flow out of junction C is
0087.0 BCACCD QQQ
Hence, we should make changes of head such that
0087.0δδδ BCACCD QQQ
or
0087.0δ00058.0δ00172.0 DC HH
870δ58δ172 DC HH (*)
Similarly, the net flow out of junction D is
0059.0 CDDFDE QQQ
Hence we should make changes of head such that
0059.0δδδ CDDFDE QQQ
or
0059.0δ00187.0δ00058.0 DC HH
59018758 DC HH (**)
Solving (*) and (**) simultaneously gives δHC = –6.8 m, δHD = –5.3 m
Hydraulics 2 A2-33 David Apsley
Hence the next guess uses junction heads
7.1442.293 DC HH
This gives flow rates
0973.0
0748.0
1723.0
0688.0
1033.0
DF
DE
CD
BC
AC
Q
Q
Q
Q
Q
Check continuity. At junction C:
0002.0 BCACCD QQQ
whilst at junction D:
0002.0 CDDFDE QQQ
This is close enough (accuracy 0.0002/0.1723 = 0.0012 or about 0.1%).
Answer: Flow rates:
QAC = 0.103 m3 s
–1;
QBC = 0.069 m3 s
–1;
QCD = 0.172 m3 s
–1;
QDE = 0.075 m3 s
–1;
QDF = 0.097 m3 s
–1.
Heads at the junctions:
HC = 293 m;
HD = 145 m.
Hydraulics 2 A2-34 David Apsley
Q20.
S = 0.01
n = 0.012
The flow cross-section can be composed of two right triangles of height h, width
330tan/ hh and hypotenuse hh 230sin/ . Hence, for any depth of water h the wetted
perimeter and cross-sectional area of flow are given by
hhP 422
2
21 3)3(2 hhhA
Then
hP
ARh
4
3
3/22/1
3/2
2/13/2 770.401.04
3
012.0
11hhSR
nV h
3/823/2 261.83770.4 hhhVAQ
(a) If h = 0.5 m then
133/8 sm30.15.0261.8 Q
Answer: discharge = 1.30 m3 s
–1.
(b) If Q = 2 m3 s
–1 then
m587.0261.8
2
261.8
8/38/3
Qh
Answer: depth = 0.587 m.
Hydraulics 2 A2-35 David Apsley
Q21.
S = 0.005
n = 0.012
R = 0.375 m
(a) When the pipe is full:
2πRA
RP π2
m 1875.02// RPARh
From Manning’s formula:
12/13/22/13/2 s m 930.1005.01875.0012.0
11 SRn
V h
1322 s m 8526.0375.0π930.1)π( RVQ
Answer: discharge = 0.853 m3 s
–1.
(b)
radians498.2)8.0(cos)1(cos)(cosθ 111
R
h
R
hR
2
212
212 m 4188.0))498.22sin(498.2(375.0)θ2sinθ( RA
m 874.1498.2375.02θ2 RP
m 2235.0874.1
4188.0
P
ARh
From Manning’s formula:
12/13/22/13/2 s m 170.2005.02235.0012.0
11 SRn
V h
13 s m 9088.04188.0170.2 VAQ
Answer: discharge = 0.909 m3 s
–1.
The answer in (b) exceeds that in (a) because in going from 90% of maximum depth to full
the additional cross-section through which there is flow increases only very slightly whereas
the wetted perimeter, over which there is friction, increases substantially.
Hydraulics 2 A2-36 David Apsley
Q22.
S = 0.0025
n = 0.015 m–1/3
s
R = 0.15 m
Q = 0.03 m3 s
–1
Find the answer first in terms of , then convert to a height.
θ (radians) A (m2)
)θ2sinθ(212 R
P (m)
θ2R
Rh (m)
PA/
V (m s–1
)
2/13/21SR
nh
Q (m3 s
–1)
VA
1.5 0.03216 0.45 0.07147 0.5740 0.01846
2 0.05351 0.6 0.08918 0.6653 0.03560
1.84 0.04717 0.552 0.08545 0.6466 0.03050
1.826 0.04658 0.5478 0.08503 0.6445 0.03002
The first two values of θ are guesses. The following values are obtained systematically by
interpolation.
θ = 1.826 radians. Then
m 188.0θcos RRh
Answer: depth = 0.188 m.
Hydraulics 2 A2-37 David Apsley
Q23.
(a)
h D/2
Use the semi-angle θ such that
)θcos1( Rh
)2
1(cos)1(cosθ 11
D
h
R
h
Then,
)θcosθsinθ()
2(
)θcosθsinθ()θcosθsinθ(2
2
2
212
21
D
RRRA
and
θ
θ2
D
RP
h D/2
Adding a semi-circle and a rectangle:
)2
(8
π 2 DhD
DA
)2
(22
π Dh
DP
(b) When D = 0.6 m and h = 0.8 m we are clearly in the second regime, and
222
m4414.05.06.08
6.0π)
2(
8
π
DhD
DA
m942.15.022
6.0π)
2(2
2
π
Dh
DP
m2273.0942.1
4414.0
P
ARh
12/13/22/13/2 sm865.201.02273.0013.0
11 SRn
V h
13 sm265.14414.0865.2 VAQ
Answer: discharge = 1.27 m3 s
–1.
(c) Set out a table, with calculations and formulae.
h
(m)
2),
2(
8
2π
2),θcosθsinθ(2)
2(
Dh
DhD
D
Dh
D
A
(m2)
2),
2(2
2
π2
,θ
Dh
Dh
D
DhD
P
(m)
P
ARh
(m)
2/13/21SR
nV h
(m s–1
)
VAQ
(m3 s
–1)
0.8 0.4414 1.942 0.2272 2.864 1.264
1 0.5614 2.342 0.2396 2.968 1.666
1.166 0.6610 2.674 0.2471 3.029 2.002
1.165 0.6604 2.672 0.2471 3.029 2.000
Answer: 1.16 m.
Hydraulics 2 A2-38 David Apsley
Q24.
(a)
b = 0.6 m
2h2h
h 5h 5h h
For a given depth h, the fluid cross-sectional area, A, and wetted perimeter, P, are given by
2222
12 hbhhhbhA
hbP 52
where b = 0.6 m.
For the given depth the sequence of calculations is:
m4.0h
22 m56.02 hbhA
m389.252 hbP
m2344.0P
ARh
13/2
2/1
3/22/13/2 sm7603.02400
1
025.0
11
hhh RRSR
nV
13 sm4258.0 VAQ
Answer: discharge = 0.426 m3 s
–1.
(b) Need h when Q = 0.8 m3 s
–1. Follow the same sequence of calculations as above to
assemble a general expression for Q(h). After the first two trials, subsequent test values of h
are determined by extrapolation/interpolation. Metre-second units are assumed throughout.
3/2
3/52/1
3/21
P
A
n
SAS
P
A
nVAQ
With the given values of S and n, and the formulae for A and P:
3/2
3/52
)526.0(
)26.0(2
h
hhQ
h (m) Q (m3 s
–1)
0.4 0.4258
0.8 1.9781
0.5 0.6882
0.55 0.8484
0.535 0.7982
The required discharge is obtained to within 0.2%.
Answer: h = 0.535 m.
Hydraulics 2 A2-39 David Apsley
Q25.
(a)
2m12.03.04.0 A
m0.13.024.0 P
m12.00.1
12.0
P
ARh
2/12/13/22/13/2 27.2012.0012.0
11SSSR
nV h
2/12/1 432.212.027.20 SSVAQ
01522.0432.2
3.0
432.2
22
QS
Answer: S = 0.0152.
(b) The new slope is
03044.001522.02 S
Write geometric parameters in terms of h:
hA 4.0
hP 24.0
Using Manning’s formula:
3/2
3/52/1
3/21
P
A
n
SAS
P
A
nVAQ
With the given values of S and n, and the formulae for A and P:
3/2
3/5
)24.0(
)4.0(54.14
h
hQ
Use successive trials for h (guided by interpolation/extrapolation) to aim for Q = 0.3.
h (m) Q (m3 s
–1)
0.15 0.1696
0.2 0.2506
0.230 0.3014
0.229 0.2997
Answer: h = 229 mm.