ANSWERS TO EXAMPLE SHEET FOR TOPIC 2 AUTUMN 2013€¦ · ANSWERS TO EXAMPLE SHEET FOR TOPIC 2...

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


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 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



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


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


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.


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.


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.


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.


(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.


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.


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.


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


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.


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.


(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).


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


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.


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


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


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.


Q10.

(a)

L = 3000 m

h = 25 m

ks = 1.0 mm

Q = 0.05 m3 s

–1


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


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

.


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

.


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.


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


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.


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.


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 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 B to J; 66.7 L s–1

from J to C.


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.


(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.


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



(given).

J

A

B

C0.05

Q

J

A

B

C0.05

Q-0.05

Q


(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



.


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


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


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.


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.


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


–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


–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.


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.


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


–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.


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


–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.


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.

ANSWERS TO EXAMPLE SHEET FOR TOPIC 2 AUTUMN 2013€¦ · ANSWERS TO EXAMPLE SHEET FOR TOPIC 2...

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