BLT2012 Examples

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BSc (Hons) in Civil Engineering BLT2012 Hydraulics 2C Worked Examples

Transcript of BLT2012 Examples

Page 1: BLT2012 Examples

BSc (Hons) in Civil Engineering

BLT2012 Hydraulics 2C

Worked Examples

Page 2: BLT2012 Examples

BSc level 1 Hydraulics revision examples

1) If the pressure in a water main is 2 bar, how high will the water rise in an adjoining block of flats?

P = g h

where P = pressure in N/m2 (1 bar = 100,000 N/m2) is the density of water = 1000 kg/m3 g is the acceleration due to gravity = 9.81 m/s2

h is the height of the water column in m

h = P = 200000 = 20.38 m g 1000x9.81

2) Determine the velocity of flow in a 225 mm diameter pipe when the flow rate of water Q is 3600 litres/min.

Q = A v

Where Q is the flow rate in m3/sec (1000 litres of water in a cubic metre)A is the cross sectional area of the pipe in m2 (d is the pipe diameter in m)v is the velocity of flow of the water in the pipe in m/sec

A = d 2 = 3.142 x 0.2252 = 0.0398 m2

4 4

Q = 3600 = 0.060 m3/sec 1000 x 60

v = Q = 0.060 = 1.507 m/sec A 0.0398

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3) Water flows along a 150mm diameter water main, which has a Darcy friction factor value of 0.025, at a flow rate Q of 10 litres/sec. The pipe is 400m long, has a level of 230m AOD at the start (1) and 235m AOD at the end (2). If the pressure at the start is 3 bar determine the pressure at the end.

Total Energy Line (TEL)

hf

P1/g + v12/2g P2/g + v2

2/2g

2

1

Bernoulli’s equation

hf = L v 2 or hf = L Q 2 Darcy Weisbach equation 2 g d 12.1 d5

A = d 2 = 3.142 x 0.15 2 = 0.0177 m2 Q = 0.010 m3/sec 4 4

v1 = v2 = Q = 0.010 = 0.566 m/sec (note v2/2g = 0.016m) A 0.0177

hf = L v 2 = 0.025 x 400 x 0.566 2 = 1.089 m 2 g d 2 x 9.81 x 0.15

hf = L Q 2 = 0.025 x 400 x 0.01 2 = 1.088 m 12.1 d5 12.1 x 0.155

z1 = 230 m z2 = 235 m Subs into Bernoulli’s equation

230 + 0.566 2 + 300000 = 235 + 0.566 2 + P2 + 1.089 19.62 9810 19.62 9810

P2 = 24.493 x 9810 = 240276.3 N/m2 = 2.403 bar

4 Water is to be discharged from a service reservoir through a 600m long pipeline which connects into a distribution system at a point where the pressure is 3 bar. The water level in the reservoir (1) is 400m AOD, the ground level at the

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distribution system (2) is 362.92 m AOD. If the required flow rate is 50 litres/sec and the pipeline will have a Darcy friction factor value of 0.030, determine the required pipe diameter.

1

2

P1 = 0 (atmospheric), v1 = 0 (large reservoir), z1 = 400 m

P2 = 300,000 N/m2, z2 = 362.92 m , assume v22/2g is negligible

400 = 362.92 + 300000 + hf

9810

hf = 400 – 362.92 – 30.58 = 6.50 m

hf = L Q 2 = 6.50 12.1 d5

0.03 x 600 x 0.05 2 = 6.50 12.1 x d5

d5 = 0.03 x 600 x 0.05 2 = 5.722 x 10-4

12.1 x 6.50

d = (5.722 x 10-4)1/5 = (5.722 x 10-4)0.2 = 0.225m = 225 mm

A = d 2 = 3.142 x 0.225 2 = 0.040 m2 , v2 = 0.05/0.04 = 1.25m/sec, v22/2g = 0.08m

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Pipeflow revision tutorial

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1. The water pressure in a distribution main is 3.6 bars, determine the pressure in kN/m2 and the pressure head in metres. (360 kN/m2, 36.7m)

2. The flow rate of water in a 600mm diameter pipe is 37,322 litres/minute,

determine the velocity of flow. Determine the diameter of pipe in which this flow rate would give a velocity of 3.0m/sec. (2.2 m/sec, 450mm)

3. Water flows between two reservoirs A & B. The water level in A is 384.0 m AOD and in B is 365m AOD. The reservoirs are connected by a 450mm diameter pipe, 1200m in length, which has a Darcy friction factor of 0.025. The lowest point in the pipeline is a distance of 400m from A and has a level of 360m AOD. Determine the rate of flow through the pipeline and the pressure at the lowest point. (376 litres/sec, 170.2 kN/m2)

4. Determine the pipe size required to deliver 30 litres/sec of water between two reservoirs 1300m apart when their water levels differ by 25m. Take the Darcy friction factor as 0.025. For the pipe size chosen, calculate the difference in water levels needed to pass 40 litres/sec. (157mm, 45m)

Laminar pipe flow example

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Oil flows through a 75mm diameter horizontal pipeline, which is 150 m long, at an average flowrate Q of 75 litres/min. The coefficient of dynamic viscosity for the oil is 0.015 kg/ms and the density of the oil is 850 kg/m3. Taking into account only frictional losses determine:

1. The flow type.2. The pressure loss along the pipeline.3. The maximum velocity of flow in the pipe.4. The velocity of flow 5mm from the pipe wall5. The shear stress 0 at the pipe wall.

1. Q = 0.075/60 = 0.00125 m3/sec, A = 3.142 x 0.0752/4 = 0.0044 m2

Average velocity of flow = = Q = 0.283 m/sec A

Re = v d = v d = 850 x 0.283 x 0.075 = 1203 i.e. laminar flow 0.015

2. Apply Bernoulli between two ends of pipeline

, z1 = z2 (pipe horizontal), v1 = v2 = 0.283 m/sec

P1 – P2 = hf therefore pressure change P1 – P2 = g hf

g

hf = 32 L = 32 x 0.015 x 150 x 0.283 = 0.430 m (Hagen-Poiseuille equation Pg 16) g d2 850 x 9.81 x 0.0752

P1 – P2 = dP = 850 x 9.81 x 0.430 = 3585.6 N/m2

3. v = dp (R2 – r2) (Pg 14 in notes) vmax occurs at centreline (r = 0) 4dx

vmax = dp R 2 = 3585.6 x 0.0375 2 = 0.560 m/sec (Note twice average velocity) 4dx 4x 0.015 x 150

4. v32.5 = 3585.6 (0.03752 – 0.03252) = 0.139 m/sec4 x 0.015 x 150

5. 0 = dp R = 3585.6 x 0.0375 = 0.448 N/m2 (Pg 13 in notes) 2 dx 2 x 150

Turbulent pipeflow – velocity profile example

Water flows through 225mm diameter pipeline which is 500m long at a rate of 50 litres/sec. Assuming the pipe has a value of 0.024 determine the frictional head loss

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across the pipeline and determine the actual velocity of flow at the pipe centreline using the universal velocity distribution equation. Take the coefficient of dynamic viscosity for water as 1.14 x 10-3 kg/ms.

hf = L v 2 or hf = L Q 2 Darcy Weisbach equation 2 g d 12.1 d5

A = d 2 = 3.142 x 0.225 2 = 0.040 m2 Q = 0.050 m3/sec 4 4

v1 = v2 = Q = 0.050 = 1.25 m/sec A 0.040

hf = L v 2 = 0.024 x 500 x 1.25 2 = 4.247m 2 g d 19.62 x 0.225

Pipe centreline velocity determination

(Pg 18 in notes)

0 = dp d (Pg 20 in notes) where dp is the pressure drop caused by friction. 4 L

dp = g hf = 9810 x 4.247 = 41663 N/m2

0 = 41663 x 0.225 = 4.687 N/m2

4 x 500

= = 0.069m/sec

= 5.75 log [ 6840 ] + 5.5

vmax = 0.069 ( 22.052 + 5.5 ) = 1.9 m/sec

Barr’s equation exampleDuring a recent network calibration exercise the head loss due to friction over an existing 220m length of 150mm diameter pipe was calculated to be 0.6m when the measured flowrate was 7.5 litres/sec. Estimate the pipe wall roughness ks. Assume the coefficient of dynamic viscosity for water is 1.14 x 10-3 kg/ms.

hf = L v 2 or hf = L Q 2 Darcy Weisbach equation

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2 g d 12.1 d5

= 12.1 d 5 h f = 12.1 x 0.15 5 x 0.6 = 0.045 L Q2 220 x 0.00752

A = d 2 = 3.142 x 0.15 2 = 0.0177 m2 Q = 0.017 m3/sec 4 4

v1 = v2 = Q = 0.0075 = 0.424 m/sec (know v hence we can calculate Re ) A 0.0177

Re = v d = 1000 x 0.424 x 0.15 = 55790 1.14 x 10-3

Barr’s equation (pg 28 in notes)

10-2.357 = ks + 3.059 x 10-4

0.555

4.395 x 10-3 = ks + 3.059 x 10-4

0.555

Ks = 0.555(4.395 x 10-3 - 3.059 x 10-4) = 2.270 x 10-3 m = 2.3mm

Turbulent pipe flow example 1

A 450mm diameter cement lined ductile iron pipe (reason for precision of description to be explained later) 2000m long supplies water from a storage reservoir to a service reservoir. If the difference in water levels in the two reservoirs is 20m determine the flow rate Q between them. Take the pipe wall roughness ks as 0.03mm and the coefficient of kinematic viscosity for water as 1.14 x 10-6 m2/s (at 15°C).

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1

2

Applying Bernoulli between the water surfaces in the reservoirs i.e. between 1 and 2

P1 = P2 = 0 (atmospheric), v1 = v2 = 0 (large reservoirs), z1 - z2 = 20 m

hf = L v 2 (Don’t know Q and hence v, so can’t determine Re - use Colebrook-White) 2 g d

= 2gd hf = 2gd S0 - equation (1) (S0 = hf/L is the friction gradient = 20/2000 = 0.01) L v2 v2

- Colebrook White equation (2) where Re = v d/

Substituting for from equation (1) into equation (2) gives

v = -0.593 log [ 1.802 x 10-5 + 2.14 x 10-5] = -0.593 log [ 3.942 x 10-5]v = 2.612 m/sec

Q = A v = 3.142 x 0.452/4 x 2.612 = 0.416 m3/sec

= 2gd S0 = 19.62 x 0.45 x 0.01 = 0.013 v2 2.6122

Check using Barr’s equation

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Re = v d = 2.612 x 0.45 = 1031053 1.14 x 10-6

= 8.779

= 1 = 0.013 8.7792

Compare with Moody Chart ks = 0.003x10 -3 = 0.000067 D 0.45

= 0.013 (Note in transitional zone)

(* Revisit this example when covering HRS tables)

Turbulent pipe flow example 2

An old tuberculated water main (in normal condition with a slight degree of attack), having a ks value of 1.5mm, has a diameter of 150mm and is 800m in length. When the flow rate Q is 17 litres/sec, the pressure recorded at the inlet is 3 bar and the pressure recorded at the outlet is 1.851 bar. Determine the difference in the pipe levels at the inlet and outlet. Take the coefficient of kinematic viscosity for water as 1.14 x 10-6

m2/sec. Using the universal velocity distribution equation determine the pipe centreline velocity.

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Total Energy Line (TEL)

hf

P1/g + v12/2g P2/g + v2

2/2g

2

1

Bernoulli’s equation

hf = L v 2 or hf = L Q 2 Darcy Weisbach equation 2 g d 12.1 d5

A = d 2 = 3.142 x 0.15 2 = 0.0177 m2 Q = 0.017 m3/sec 4 4

v1 = v2 = Q = 0.017 = 0.962 m/sec (know v hence Re use Barr’s equation) A 0.0177

Re = v d = 0.962 x 0.15 = 126579 1.14 x 10-6

Barr’s equation

= 0.0386

hf = L v 2 = 0.0386 x 800 x 0.962 2 = 9.710m 2 g d 19.62 x 0.150

Subs into Bernoulli’s equation

z1 + 0.962 2 + 300000 = z2 + 0.962 2 + 185100 + 9.710 19.62 9810 19.62 9810

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z2 – z1 = 2.003m (i.e point 2 is 2.003m above point 1)

Check using Moody chart

ks = 1.5 x 10 -3 = 0.01 , Re = 126579, = 0.0385 (Note rough turbulent zone) d 0.15

Pipe centreline velocity determination

(Pg 18 in notes)

0 = dp d (Pg 17 in notes) where dp is the pressure drop caused by friction. 4 L

dp = g hf = 9810 x 9.71 = 95255 N/m2

(Note Because pipe is not horizontal, rises 2.0m, dp ≠ P1 – P2 in this calculation.

P1 – P2 = 300000 – 185100 = 114900 N/m2 which is equivalent to 11.71m head. i.e hf + 2.0 )

0 = 95255 x 0.15 = 4.465 N/m2

4 x 800

= = 0.067m/sec

= 5.75 log [ 4396 ] + 5.5

vmax = 0.067 ( 20.948 + 5.5 ) = 1.772 m/secHRS table examples

For d = 200mm, So = 1/250 (0.004) ks = 1.5mm, = 1.14 x 10-6 Ns/m2

v = -0.251 log [ 2.027 x 10-3 + 1.142 x 10-4] = -0.251 log [ 2.141 x 10-3] = 0.670 m/sec

Q = A v = 3.142 x 0.22/4 x 0.670 = 0.021 m3/sec = 21.0 litres/sec

A 150mm diameter main conveying water at a flow rate of 20 litres/sec rises 1.5m over a distance of 300m. If the pipeline has a roughness value ks of 1.5mm and the pressure at the start is 4 bar determine the pressure at the other end in kN/m2.

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Total Energy Line (TEL)

hf

P1/g + v12/2g P2/g + v2

2/2g

2

1

Bernoulli’s equation and hf = So x L

A = d 2 = 3.142 x 0.15 2 = 0.0177 m2 Q = 0.020 m3/sec 4 4

v1 = v2 = Q = 0.020 = 1.130 m/sec A 0.0177

From HRS tables page (3 in notes) So = 1/60 = 0.0167

hf = 0.0167 x 300 = 5.0m, z2 – z1 = 1.5 m Subs into Bernoulli’s equation

0 + 1.130 2 + 400000 = 1.5 + 1.130 2 + P2 + 5.0 19.62 9810 19.62 9810

P2 = 34.275 x 9810 = 336238 N/m2 = 3.360 barHRS tables examples

1) Determine the frictional head loss in a 375mm diameter uncoated steel pipe in normal condition when it conveys 163 litres/sec of water over a distance of 800m.

2) A water main is to be designed to convey 560 litres/sec over a distance of 700m. If the pipe material is to be coated steel (good condition) and the allowable frictional head loss is 3m determine the required pipe diameter.

3) A 150mm diameter main conveying water at a flowrate of 20 litres/sec rises 1.5m over a distance of 300m. If the pipeline has a roughness value ks of 1.5mm and the pressure at the start is 4 bars, determine the pressure at the other end, in kN/m2.

4) A 300mm diameter concrete sewer (ks = 1.5mm) is required to deliver 88 litres/sec. Determine the minimum gradient at which the sewer should be laid for it not to be surcharged.

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5) A 225 mm diameter concrete sewer (ks = 1.5mm) is 60m long and laid at a gradient of 1/154. What will be the height of surcharge at the upstream manhole when the flowrate is 60 litres/sec.

Parallel Pipelines Example 1

A section of a pipeline consists of two pipes in parallel as shown below. Pipe A has a diameter of 150mm, a length of 150m, a ks value of 0.03mm and a value of 0.0188. Pipe B has a diameter of 225mm, a length of 120m, a ks value of 0.03mm and a value of 0.0163. If the friction head loss HL across the section (between 1 and 2) is 0.6m determine the flow in each pipe.

A

1 2

B

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The 0.6m head loss across the parallel section ( 1 to 2) is the same in each pipe i.e.

HLA = HLB = 0.6m

= = 0.6m and

SOA LA = SOB LB = 0.6m

= 0.01398 m3/sec = 13.98 litres/sec

= 0.04626 m3/sec = 46.26 litres/sec

Using HRS table 3 (ks = 0.03mm pg 11 HRS tables handout)

SOA = 0.6 = 0.004 ( 1 in 250) 150

SOB = 0.6 = 0.005 ( 1 in 200) 120

For pipe A 150mm diameter QA = 13.952 litres/sec

For pipe B 225mm diameter QB = 46.253 litres/sec

Parallel Pipelines Example 2

A service reservoir is to be constructed to provide a water supply for a town 1 km distant. The initial population is expected to be 86,000 and the average rate of demand is 200 litres/head/day.

Design the supply pipeline which connects the reservoir to the distribution system. The pipe material is to be metal with a concrete lining, which has a Darcy friction factor of 0.013. Allow a peaking factor of 2. The top water level in the reservoir is 415.6m AOD and the pipe level at the start of the distribution system is 376.3m AOD. The pressure at the start of the distribution system is to be 2.94 bar.

After 25 years the population increases to 125,000. If the water level in the service reservoir and the pressure in the distribution system remains unchanged determine:

(a) the length of a parallel pipeline, of the same diameter and material as the original pipeline, required to ensure that the demand for water is met.

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(b) the length of a parallel pipeline, of 525mm diameter and the same material as the original pipeline (with a Darcy friction factor value of 0.011), required to ensure that the demand for water is met.

(c) the diameter of a lined ductile iron parallel pipeline, of length 850m and Darcy friction factor of 0.014, required to ensure that the demand for water is met.

Parallel pipeline example – solution

A 415.6m AOD

B 376.3m AOD

Original flowrate Qo = 2 x 86000 x 200 = 398 litres/sec = 0.398 m3/sec 24 x 3600

Pressure in distribution system = 2.94 bar = 294,000 N/m2

Pressure head H in distribution system = PB = 294,000 = 30m g 9810

Apply Bernoulli between A & B (ignore residual K.E. at B)

hf = zA – zB – 30 = 415.6 - 376.3 - 30 = 9.3m

hf = L Q 2 = 0.013 x 1000 x 0.398 2 = 9.3m 12.1 d5 12.1 x d5

d = ( 0.013 x 1000 x 0.398 2 )1/5 = 0.45m = 450mm. 12.1 x 9.3

By HRS tables: Ks = 0.03mm, So = 9.3/1000 = 0.0093 (1 in 108)

(Table 14 page 13 in HRS handout) Q = 0.398 m3/sec, d = 450mm

(a) Population increases to 125,000 identical parallel pipe diameter specified, find length.

QN = 2 x 125,000 x 200 = 579 litres/sec = 0.579 m3/sec

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24 x 3600

Insert parallel pipeline along bottom section of original pipeline to accommodate new flow and maintain pressure in distribution system.

A

1 3

2 B

Total Head loss between A & B = HL = HL1 + HL2 = HL1 + HL3 i.e. HL2 = HL3

Thus 2 L2 Q22 = 3 L3 Q3

2 12.1 d2

5 12.1 d35

Identical pipes, L2 = L3 , d2 = d3 , 2 = 3 Therefore Q2 = Q3

( note if any of the above relationships are not equal then Q2 ≠ Q3 and the relationship between Q2 and Q3 has to be determined by calculation.)

Q1 = Q2 + Q3 = 0.579 m3/sec

As Q2 = Q3 = 0.579/2 = 0.290 m3/sec

1000 = L1 + L3, therefore L1 = 1000 - L3

HL = 9.3m = HL1 + HL3

1 L1 Q12 + 3 L3 Q3

2 = 9.3 12.1 d1

5 12.1 d35

0.013 x (1000 – L 3) x 0.579 2 + 0.013 x L3 x 0.290 2 = 9.312.1 x 0.455 12.1 x 0.455

19.519 - 0.0195 L3 + 0.0049 L3 = 9.3

10.219 = 0.0146 L3

L3 = 700m

By HRS table 14 (pg 13 ): d1= d3 = 450mm, Q1= 0.579 m3/sec, Q3 = 0.290 m3/sec, HL = 9.3m

Ks = 0.03mm (all pipes), So1 = 1 in 54 (0.0185) and So3 = 1 in 198 (0.0051)

HL1 + HL3 = So1 x L1 + So3 x L3 = 0.0185 (1000 - L3) + 0.0051 L3 = 9.3

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18.518 - 0.0185 L3 + 0.0051 L3 = 9.3

9.218 = 0.0134L3

L3 = 688m

(b) Population increases to 125,000 different parallel pipe diameter specified, find length.

QN = 0.579 m3/sec A

1 3

2 B

Total Head loss between A & B = HL = HL1 + HL2 = HL1 + HL3 i.e. HL2 = HL3

Thus 2 L2 Q22 = 3 L3 Q3

2 12.1 d2

5 12.1 d35

L2 = L3 , d2 = 450mm , d3 = 525mm 2 = 0.013, 3 = 0.011

0.013 x Q 22 = 0.011 x Q3

2

0.455 0. 5255

0.704 Q22 = 0.276 Q3

2

Q32 = 2.551 Q2

2

Q3 = 1.597 Q2

Q1 = Q2 + Q3 = 0.579 m3/sec

Q2 + 1.597 Q2 = 2.597 Q2 = 0 .579

Q2 = 0.223 m3/sec and Q3 = 0.356m3/sec

1000 = L1 + L3, therefore L1 = 1000 - L3

HL = 9.3m = HL1 + HL3

1 L1 Q12 + 3 L3 Q3

2 = 9.3 12.1 d1

5 12.1 d35

0.013 x (1000 – L 3) x 0.579 2 + 0.011 x L3 x 0.356 2 = 9.312.1 x 0.455 12.1 x 0.5255

19.519 - 0.0195 L3 + 0.0029 L3 = 9.3

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10.219 = 0.0166L3

L3 = 615.6m

By HRS tables (pg 12 &13) Ks = 0.03mm (all pipes): d1= d2 = 450mm, d3 = 525mm,

Q1= 0.579 m3/sec = Q2 + Q3

So1 = 1 in 54 (0.0185), So2 = So3

Need to look down HRS table to find friction gradient (So) where Q2 + Q3 = 0.579 m3/sec

For So = 1 in 303 (0.0033) Q2 = 0.231 m3/sec Q3 = 0.348 m3/sec

HL1 + HL3 = So1 x L1 + So3 x L3 = 0.0185 (1000 - L3) + 0.0033 L3 = 9.3

18.519 - 0.0185 L3 + 0.0033 L3 = 9.3

9.219 = 0.0152 L3

L3 = 606.5m

(c) Length of parallel pipe defined, diameter required

L3 = 850m, L1 = 150m, 3 = 0.014, 1 = 0.013, d1 = 450mm

HL = 9.3m = HL1 + HL3

HL1 = 1 L1 Q12 = 0.013 x 150 x 0.579 2 = 2.928m

12.1 d15 12.1 x 0.455

HL3 = HL2 = 9.3 – 2.928 = 6.372m

Determine flowrate Q2 going down existing pipe (2)

HL2 = 2 L2 Q22 = 0.013 x 850 x Q2

2 = 6.372m 12.1 d2

5 12.1 x 0.455

Q2 = ( 6.372 x 12.1 x 0.45 5 )1/2 = 0.359 m3/sec 0.013 x 850

Q1 = Q2 + Q3 = 0.579 m3/sec

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Therefore Q3 = 0.579 - 0.359 = 0.220 m3/sec

HL3 = 6.732m = 3 L3 Q32 = 0.014 x 850 x 0.220 2

12.1 d35 12.1 x d3

5

d3 = ( 0.014 x 850 x 0.220 2 )1/5 = 0.375m 12.1 x 6.732

By HRS tables (pg 13) d1 = d2 = 450mm, Q1= 0.579 m3/sec, L1= 150m, L3 = 850m, HL = 9.3m

Ks = 0.03mm (all pipes), from HRS table So1 = 0.0185 (1 in 54)

HL1 = So1 x L1 = 0.0185 x 150 = 2.778m

HL2 = HL3 = 9.3 - 2.778 = 6.522m

So2 = So3 = 6.552 = 0.0077 (1 in 130) 850

From HRS table, Q2 = 0.363 m3/sec

Therefore Q3 = 0.579 – 0.363 = 0.216 m3/sec

From HRS table, d3 = 375mmHydraulics 2C – Flow Balancing Example

The reservoir A at a treatment plant feeds two service reservoirs B and C as shown below. Use the Flow Balancing Method to approximately determine the flows entering the service reservoirs.

A

J

B

C

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Reservoir Water Level m AODA 95B 80C 76

Pipe Diameter (mm) Length (m) Ks(mm)A-J 300 900 1.5J-B 225 1100 1.5J-C 150 800 1.5

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1st estimate Hj = 90.5m

2nd estimate Hj = 91.5m

3rd estimate Hj = 90.9m

Pipe hf across

Pipe (m)

SO

(1 in)

Q

(litre/s)

hf across

Pipe (m)

SO

(1 in)

Q

(litre/s)

hf across

Pipe (m)

SO

(1 in)

Q

(litre/s)

A 4.5 200 69.2 3.5 257 61.0 4.1 220 65.9

B 10.5 105 -44.5 11.5 96 -46.8 10.9 101 -45.4

C 14.5 55 -20.9 15.5 52 -21.6 14.9 54 -21.1

Error in Q = 3.8l/sec Error in Q = -7.4l/sec Error in Q = 0.6l/sec

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Hardy Cross/Synergee exampleSee diagram on page 24.

Based on the draw-off from the network, the reservoir supplies 340 litres/sec to node A

From HRS table 14, ks = 0.03mm for 450mm pipe carrying 340 litres/sec then hydraulic gradient So = 0.00681 ( 1 in 147).

Head loss from reservoir to A = 0.00681 x 900 = 6.13m

Total head at A = 136.13 – 6.13 = 130m (AOD)

By inspection of diagram, decide on flow directions and quantities (remember more water will pass down bigger, shorter and smoother pipes). Because the pipe A-C is common to two loops it is shown as if it is split into two in the diagram below. In reality of course it is just a single pipe and there can only be one value for the flow passing down it common to both loops.

Note clockwise flows are deemed +ve and anti-clockwise flows –ve as are their respective friction head losses. The basic logic is that when the flows are correctly allotted the head losses in a loop will sum to zero.

i.e. in the top loop: hf A-B + hf B-C + hf A-C = 0

And in the bottom loop: hf A-D + hf D-C + hf A-C = 0

Calculate head loss hf for each pipe:

For pipe A-B, Q = 130 litres/sec (clockwise), dia = 300mm, L = 550m

HRS table 3 gives So = 0.00833 and thus hf = 0.00833 x 550 = +4.58m

Complete table on page 25.

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Pipe Dia(mm)

Length(m)

1st Adjustment 2nd Adjustment 3rd Adjustment Total Head (m)

PressureHead (m)

Q(m3/sec)

hf

(m)hf/Q Q

(m3/sec)hf

(m)hf/Q Q

(m3/sec)hf

(m)

A-B 300 550 +0.130 4.58 35 +0.127 4.57 34 +0.129 4.50A

130 30

B-C 200 450 +0.040 3.10 78 +0.037 2.68 72 +0.039 2.93B

125.5 27.5

A-C 250 800 -0.080 -6.64 83 -0.088 -7.92 90 -0.085 -7.40C

122.6 21.6

Q for AC = (-0.003 + -0.005) = - 0.008m3/s

+1.04 196 -0.87 196 0.03

Q = +0.002m3/s

AC = 0.003m3/s

A-C 250 800 +0.080 6.64 83 +0.008 7.92 90 +0.085 +7.50A

130

D-C 250 250 -0.060 -1.21 20 -0.055 -1.05 19 -0.056 -1.06C

122.3

A-D 300 850 -0.130 -6.94 54 -0.126 -6.59 53 -0.126 -6.70D

123.3 27.3

Q for AC = (+0.005 + +0.003) = + 0.008m3/s

-1.57 157 +0.28 162 -0.26

Q = -0.001m3/s

AC = -0.003m3/s

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Page 26: BLT2012 Examples

Foul Sewer Design ExampleDesign the foul sewer system shown below. Assume 3 persons per dwelling, 250 litres/head/day water consumption and 40 litres/head/day infiltration. The new system is to connect into a existing combined sewer at A4 which has an invert level of 95.00m AOD. There is an existing 600mm diameter surface water sewer midway between A2 and A3 of invert level 95.02m AOD and a water main midway between C1 and C2 invert level 98.10m AOD.

A1 C2A2 A3

C1

B1

A4

Pipe run Length(m)

Number of houses

Ground level upstream Mh

Ground level downstream Mh

A1-A2 90 250 100.00 98.94B1-A2 80 220 100.27 98.94A2-A3 85 300 98.94 97.86C1-C2 120 180 100.15 98.49C2-A3 50 150 98.49 97.86A3-A4 100 280 97.86 97.19

Sewer run Cumulative houses

Cumulative flow (litres/sec)

Ground gradient

Pipe diameter(mm)

Min Pipe gradient(1 in)

Chosen pipe gradient(1 in)

A1-A2 250 15.1 1 in 85 150 1 in 100 1 in 100

B1-A2 220 13.3 1 in 60 150 1 in 133 1 in 100

A2-A3 770 46.5 1 in 79 225 1 in 96 1 in 96

C1-C2 180 10.9 1 in 72 150 1 in 133 1 in 100

C2-A3 330 20.0 1 in 79 150 1 in 59 1 in 59

A3-A4 1380 83.4 1 in 149 300 1 in 138 1 in 138

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Page 27: BLT2012 Examples

Foul Sewer Design Example

Check overall fall, say invert level @ A4 = 95.150m AOD and @ A1, B1 & C1 invert level is 2.0m below ground.

Fall between A1 – A4 = 98.0 – 95.15 = 1 in 96 which is OK 275

Fall between: B1 – A4 = 1 in 85 and C1 – A4 = 1 in 90.

Complete table

a) A1 – A2 cumulative houses = 250

b) Cumulative flow = 6 x 250 x 3 x 290 = 15.1 l/sec 24 x 3600

c) Ground gradient = 100 – 98.94 = 1 in 85 90

d) 1st pipe in system 150mm dia minimum gradient 1 in 100

e) A2 – A3 cumulative houses = 770

f) Cumulative flow = 6 x 770 x 3 x 290 = 46.5 l/sec 24 x 3600

g) Ground gradient = 1 in 79

h) Pipe dia = 225mm, gradient = 1 in 96

Manhole Invert levels & depths

A3 = 95.15 + 100 = 95.88m AOD 138

Depth = 97.86 – 95.88 = 1.98m

A2 = 95.88 + 0.075 + 85 = 96.84m AOD, depth = 2.10m 96

A1 = 96.84 + 0.075 + 90 = 97.82m AOD, depth = 2.18m 100

B1 = 96.84 + 0.075 + 80 = 97.72m AOD, depth = 2.55m 100

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Page 28: BLT2012 Examples

C2 = 95.88 + 0.15 + 50 = 96.88m AOD, depth = 1.61m 59

C1 = 96.88 + 120 = 98.08m AOD, depth = 2.07m 100

Check service clearance A2 – A3

A2 A3 clearance

A2 – A3 invert level midway = 96.84 + 95.88 = 96.36m AOD 2

SW sewer top of pipe level = 95.02 + 0.6 + 0.1 = 95.72m AOD

Clearance = 0.64m – 0.05m (pipe wall) = 0.59m

Check service clearance C1 – C2

Water main invert level = 98.10

C1– C2 invert level midway = 98.08 + 96.88 = 97.48m AOD 2

Clearance = 98.1 – (97.48 + 0.15 + 0.05) = 0.42m

28

Page 29: BLT2012 Examples

Lloyd Davies Rational method example

Analyse the above storm sewer system using the Rational method for a 1 in 1 year event taking the time of entry Te as 2.0 mins. ks for all pipes is 0.6mm. Rainfall table provided. See Table below.

Pipe length 1.00Difference in level = 1.10m, length = 63.1m,

Gradient = 1.10 = 1 in 57, choose pipe dia = 150mm 63.1V = 1.33m/sec from HRS tables (pg 7 in HRS handout)

Time of flow = TF = 63.1 = 47.4 secs = 0.79 mins 1.33Time of entry = TE = 2.0 mins.

For 1st pipe in branch: Time of concentration = TC = TE + TF

TC1.00 = 2.0 + 0.79 = 2.79 mins

From table (pg 31) rainfall intensity = i = 70.5 mm/hr. AP = impermeable area (ha)

Q1.00 = 2.778 x AP x i = 2.778 x 0.102 x 70.5 = 20.0 litres/sec

Note capacity of 150mm dia @ 1 in 57 is 23.6 litres/sec – OK

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Page 30: BLT2012 Examples

Pipe length 1.01Choose 225 mm dia pipe, from HRS tables V = 1.7m/sec

TF = 66.1 = 38.9 secs = 0.65 mins 1.7TC = TC1.00 + TF1.01 = 2.79 + 0.65 = 3.44 mins

i = 63.0 mm/hr, AP = 0.102 + 0.226 = 0.328 ha

Q1.01 = 2.778 x 0.328 x 63.0 = 57.4 litres/sec

Check capacity 225 @ 1 in 59 = 67.9 litres/sec – OK

Pipe length 2.02Choose 300mm dia pipe, V = 2.74 m/sec TF2.02 = 0.33 mins

Note when calculating TC2.02 use the LONGER of the two upstream times of concentration.

i.e. TC2.01 = 2.78 mins and TC3.00 = 2.56 mins. Therefore TC2.02 = TC2.01 + TF2.02 = 2.78 + 0.33 = 3.11 mins

AP = 0.647 ha and i = 65.5 mm/hr and Q2.02 = 2.778 x 0.647 x 65.6 = 117.7 litres/sec

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Page 31: BLT2012 Examples

RAINFALL TABLE - Rates of rainfall in mm/h for a range of durations for given return periods

RETURN PERIOD (YEARS)

DURATION 1 2 5 10 20 50 100

2.0 MINS 85.6 93.4 120.5 138.3 158 187 2132.5 MINS 76.5 87.5 113.4 130.4 149 177 2023.0 MINS 66.3 82.3 107.2 123.4 141 168 1923.5 MINS 62.8 77.8 101.7 117.3 135 161 1844.0 MINS 59.6 73.8 96.8 111.8 128 154 1764.1 MINS 59.1 73.1 95.9 110.8 127 152 1744.2 MINS 58.5 72.3 95.0 109.8 126 151 1734.3 MINS 57.9 71.6 94.1 108.8 125 150 1724.4 MINS 57.4 71.0 93.2 107.9 124 149 1704.5 MINS 56.9 70.3 92.4 106.9 123 148 1694.6 MINS 56.3 69.6 91.6 106.0 122 146 1684.7 MINS 55.8 69.0 90.8 105.1 121 145 1664.8 MINS 55.3 68.3 90.0 104.2 120 144 1654.9 MINS 54.8 67.7 89.2 103.4 119 143 1645.0 MINS 54.3 67.1 88.5 102.5 118 142 1635.1 MINS 53.9 66.5 87.7 101.7 117 141 1625.2 MINS 53.4 65.9 87.0 100.9 116 140 1605.3 MINS 53.0 65.4 86.3 100.1 115 139 1595.4 MINS 52.5 64.8 85.6 99.3 115 138 1585.5 MINS 52.1 64.3 84.9 98.5 114 137 1575.6 MINS 51.7 63.7 84.2 97.8 113 136 1565.7 MINS 51.2 63.2 83.5 97.0 112 135 1555.8 MINS 50.8 627 82.9 96.3 111 134 1545.9 MINS 50.4 62.2 82.3 95.6 110 133 1536.0 MINS 50.0 61.7 81.6 94.9 110 132 1526.2 MINS 49.3 60.7 80.4 93.5 108 130 1506.4 MINS 48.5 59.8 79.2 92.2 107 129 1486.6 MINS 47.8 58.9 78.1 90.9 105 127 1466.8 MINS 47.1 58.0 77.0 89.6 104 125 1447.0 MINS 46.4 57.2 75.9 88.4 102 124 1437.2 MINS 45.8 56.4 74.9 87.3 101 122 1417.4 MINS 45.2 55.6 73.9 86.1 100 121 1397.6 MINS 44.5 54.8 72.9 85.0 99 119 1387.8 MINS 44.0 54.1 71.9 84.0 97 118 1368.0 MINS 43.4 56.4 71.0 82.9 96 117 1358.2 MINS 42.8 52.7 70.1 81.9 95 115 1338.4 MINS 42.3 52.0 69.3 81.0 94 114 1328.6 MINS 41.8 51.4 68.4 80.0 93 113 1318.8 MINS 41.2 50.7 67.6 79.1 92 112 1299.0 MINS 40.8 50.1 66.8 78.2 91 110 1259.2 MINS 40.3 49.5 66.0 77.3 90 109 1279.4 MINS 39.9 49.0 65.3 76.4 89 108 1259.6 MINS 39.4 48.4 64.6 75.6 88 107 1249.8 MINS 39.0 47.9 63.8 74.8 87 106 12310.0 MINS 38.6 47.4 63.1 74.0 86 105 12110.5 MINS 37.6 46.1 61.5 72.1 84 102 11811.0 MINS 36.7 44.9 59.9 70.2 82 100 11611.5 MINS 35.8 43.8 58.4 68.5 80 97 11312.0 MINS 35.0 42.8 57.0 66.9 78 95 11112.5 MINS 34.2 41.8 55.7 65.4 76 93 10813.0 MINS 33.4 40.8 54.4 64.0 75 91 10613.5 MINS 32.7 39.9 53.3 63.6 73 89 10414.0 MINS 32.0 39.1 52.1 61.3 72 87 10214.5 MINS 31.4 38.3 51.0 60.0 70 86 10015.0 MINS 30.8 37.5 50.0 58.8 69 84 9816.0 MINS 29.36 36.1 48.1 56.6 65 81 9417.0 MINS 28.6 34.8 46.3 54.6 64 78 9118.0 MINS 27.6 33.5 44.7 52.7 62 76 8819.0 MINS 26.7 32.4 43.2 51.0 60 73 85

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Page 32: BLT2012 Examples

20.0 MINS 25.9 31.4 41.8 49.3 58 71 83

32

Page 33: BLT2012 Examples

Lloyd Davies method: Time of entry = 2 mins, ks = 0.6mm, 1 in 1 year storm 1 Pipelengthref No

2Differencein level

(m)

3 Pipelength

(m)

4Pipegradient (1 in )

5 Vel

(m/s)

6 Time of flow (min)

7 Time of Conc. (min)

8 Rate ofrainfall i(mm/hr)

9 Imp. Area

(ha)

10Cum Imp. Area AP

(ha)

11 Flow Q

(m3/s)

12Pipedia.

(mm)

1.00 1.10 63.1 57 1.33 0.79 2.79 70.5 0.102 0.102 20.0 150

1.01 1.12 66.1 59 1.70 0.65 3.44 63.0 0.226 0.328 57.4 225

1.02 0.73 84.7 116 0.081 300

2.00 1.40 44.8 32 2.32 0.32 2.32 82.5 0.194 0.194 44.5 225

2.01 0.61 49.1 80 0.150 300

3.00 0.98 48.5 49 0.129 150

2.02 1.65 54.3 33 2.74 0.33 3.11 65.5 0.174 0.647 117.7 300

1.03 1.22 27.7 23 0.356 300

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Page 34: BLT2012 Examples

Table 1 Lloyd Davies method: Time of entry = 2 mins, ks = 0.6mm, 1 in 1 year storm 1 Pipelengthref No

2Differencein level

(m)

3 Pipelength

(m)

4Pipegradient (1 in )

5 Vel

(m/s)

6 Time of flow (min)

7 Time of Conc. (min)

8 Rate ofrainfall i(mm/hr)

9 Imp. Area

(ha)

10Cum Imp. Area AP

(ha)

11 Flow Q

(m3/s)

12Pipedia.

(mm)

1.00 1.10 63.1 57 1.33 0.79 2.79 70.5 0.102 0.102 20.0 150

1.01 1.12 66.1 59 1.70 0.65 3.44 63.0 0.226 0.328 57.4 225

1.02 0.73 84.7 116 1.46 0.97 4.41 57.4 0.081 0.409 65.2 300

2.00 1.40 44.8 32 2.32 0.32 2.32 82.5 0.194 0.194 44.5 225

2.01 0.61 49.1 80 1.77 0.46 2.78 70.0 0.150 0.344 66.9 300

3.00 0.98 48.5 49 1.43 0.56 2.56 76.0 0.129 0.129 27.2 150

2.02 1.65 54.3 33 2.74 0.33 3.11 65.5 0.174 0.647 117.7 300

1.03 1.22 27.7 23 3.29 0.14 4.55 56.9 0.356 1.412 223.2 300

34