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Analysis of Pipe Networks Applications: Water supply system, Pressure sewerage system, Water treatment plant Theory:

 A1V  1 =   A2V 2  

1.  Continuity equation 2.  Energy equation

2

V 1

2g  p1

Energy line

 

V 22

h p

Hydraulic grade line

 

γ  

2g

 p 2

V 1  

 

γ  

V 2  

P

 

 z1  

datum

 z 2   2

2

 z1 +  p1 + V 1 + h p  =  z 2 +  p 2 + V 2 + h L   γ   γ   2g 2g 2

 L V 

h L =  f     D 2 g Friction factor   f   is a function of Reynolds number ( R  R N ) and relative roughness ( k s / D ). It ca can n be be ffou ound nd fr from om Moo oody dy’s ’s di diag agra ram mo orr Col Coleb ebro rook ok eq equa uati tion on::

⎛ k s / D 2.51  ⎞ ⎜ ⎟  = −2 log⎜ + ⎟  f  ⎝  3.7  R N   f  ⎠

1 Jain's approximation :

 ⎞ ⎛  s 1 = −2 log⎜ k  / D + 5.74 0.9 ⎟   3 . 7  R  f   N   ⎠ ⎝  Table Tab le 1 1.. Eq Equiv uivale alent nt ssand and roug roughnes hness, s, k s in n new ew pipe pipes, s, (M (Mood oody, y, 1944 1944)) Material

Equivalent sand roughness, k s (mm)

Riveted steel Concrete Ductile and cast iron Galvanized iron Asphalt-dipped ductile/cast iron Commercial steel or wrought iron

0.9-9.0 0.3-3.0 0.26 0.15 0.12 0.046

Copperplastic or brass tubing Glass, (PVC)

0.0015 ≈ 0 

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Example: Water from a treatment plant is pumped into a distribution system at a rate 3 o of 4.38 m /s, a pressure of 480kPa, and a temperature of 20 C. The diameter of the  pipe is 750mm and is made of ductile ductile iron. (a) Calculate the friction factor by Colebrook equation (b) Estimate the pressure 200m downstream of the treatment plant if the pipeline remains horizontal. Solution:

 f  = 0.016  

 p2   = 270 kPa

  Designing a pipe for a given flowrate and head loss 1.  Assume f   2.  Calculate diameter from rearranged Darcy-Weisbach

⎛  8 LQ 2  ⎞  D = ⎜⎜   2 f  ⎟⎟ ⎝ h L g  π   ⎠

equation:

1/ 5

 

3.  Calculate Reynolds number,  R N   =

VD ν 

 

4.  Calculate relative roughness, k s / D 5.  Use R N  and k  /D  to calculate f  from  from Colebrook equation. s 6.  Using new value of f  repeat  repeat the procedure until new f  agrees  agrees with the old one. Example: A galvanized iron service pipe from a water main is required to deliver 200 L/s during a fire. If the length of the service pipe is 35m and the head loss in the  pipe is not to exceed 50m, calculate calculate the minimum pipe diamet diameter er that can be used. Solution: D= 136 mm Three-reservoir problem

 

A B V 1 , L 1 , D1   V 2 , L 2 , D2  

zA

zB V 3 , L 3 , D3  

datum

C zC

Unknowns: V 1 , V  2 , V 3 Equations: One continuity and two energy equations

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Other friction formulas: Hazen-Williams formula (SI units)

V  = 0.85C  H    R 0.63 S 0.54

Manning’s formula (SI units)

1 V   =  R 2 / 3 S 1 / 2   n

Table 2. Pipe roughness coefficients C 

 n

Pipe material Ductile and cast iron:  New, unlined Old, unlined Cement lined and seal coated

130 80 120

0.013 0.013

Steel: Welded and seamless Riveted Concrete Vitrified clay Polyvinyl chloride (PVC)

120 110 110 -

0.015 0.015 0.013 0.009

 H 

The head loss can be expressed as:

h L  =   KQ x   Manning’s formula:  x

=2

Darcy-Weisbach equation:  x

Hazen-Williams formula:  x

= 1.75

(smooth pipe) to  x

= 1.85   = 2  (rough pipe)

Design of water distribution system

A municipal water distribution system is used to deliver water to the consumer. Water is withdrawn from along the pipes in a pipe network system, for computational  purposes all demands on the system are assumed to occur at the junction junction nodes. Pressure is the main concern in a water distribution system. At no time should the water pressure in the system be so low that contaminated groundwater could enter the system at points of leakage. The total water demand at each node is estimated from residential, industrial and commercial water demands at that node. The fire flow is added to account for emergency water demand.

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Water Demand

Total water demand at a node is equal to the average water demand per capita multiplied by the population served by that node. The population may be estimated by any of the available models of population growth. These models use the previously available population data for future  projections. For example, according to Arithmetic model:

P(t ) =  P0 + kt   P(t) is the population at time t  and  and P0 is the reference population. Similarly a higher order polynomial function can be obtained as follows:

P(t ) = P0 + at    + bt 2 + ct 3 + L   and the population can be estimated at any time t . Example: You are in the process of designing a water-supply system for a town, and the design life of your system is to end in the year 2020. The population in the town has been measured every 10 years since 1920 and is given below. Estimate the  population in the town using graphical extension and arithmetic growth projection. Year 1920 1930 1940 1950 1960 1970 1980 1990

Population 125,000 150,000 150,000 185,000 185,000 210,000 280,000 320,000

Solution: 500000 400000 300000 200000 100000 0 1920

1940

1960

1980

2000

2020

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  From the graph, if the line is extended to year 2020, we get, the population as:

P = 440,000   By regression on the data for 1970 to 1990, we get the following expression:

P (t ) = −60000   + 5500t   where, t   is is the time in years starting from year 1920. So, for year 2020, t= 100, and we get the population,

P = 490,000  

Variations in demand Maximum daily demand, Maximum hourly demand Table 3. Typical demand factors Condition Range of demand factors Daily average in maximum month 1.1-1.5 Daily average in maximum week 1.2-1.6 Maximum daily demand 1.5-3.0 Maximum hourly demand 2.0-4.0

Minimum hourly demand

0.2-0.6

Typical value 1.2 1.4 1.8 3.25

0.3

Fire demand

Insurance Services Office, Inc. (ISO, 1980) formula:

 NFF i = C i O i ( X    + P)i   where,  NFF i is needed fire flow at location i, C i is the construction factor based on size and type of construction of the building, Oi is the occupancy factor reflecting the

  + P) i   kinds of material stored in the building (value range from 0.75 to 1.25), and ( X  is the sum of the exposure factor and communication factor that reflects the  proximity and exposure of other buildings (value range from 1.0 to 1.75). 1.75).

C i ( L / min)    = 220 F  Ai

 

 Ai is the effective floor area in square meters, typically equal to the area of the largest

floor in the building plus 50% of the area of all other floors, F  is   is a coefficient based on the class of construction. The maximum value of C i and typical F   values are given below: Class of Construction Description Max. C i(L/min) F 1 Frame 1.5 30,000 2 Joisted masonry 1.0 30,000 3 4 5

Noncombustible Masonry, noncombustible Modified fire resistive

0.8 0.8 0.6

23,000 23,000 23,000 5

 

6 Fire resistive 0.6 Occupancy factors: Combustibility class Examples C-1 Noncombustible Steel or concrete products storage C-2 Limited combustible Apartments, mosques, offices C-3 Combustible Department stores, supermarkets C-4 Free-burning Auditoriums, warehouses C-5 Rapid burning

Paint shops, upholstering shops

23,000 Oi 0.75 0.85 1.0 1.15

1.25

  + P) i is 1.4. The NFF should be rounded to the nearest Average value of ( X  1000L/min if less than 9000 L/min and to the nearest 2000L/min if greater than 9000L/min. Required fire flow durations: Required fire flow (L/min) Duration (h) 140 kPa

Maximum pressure (not strict) : 650kPa. Storage facilities 1.  20% to 25% of the maximum daily demand volume 2.  Fire demand 3.  Emergency storage (minimum storage equal to average daily system denmand) 4.  Minimum height of water in the elevated storage tank based on minimum  piezometric head in the service area 5.  Normal operating range 4.5 to 6.0 m. Access manhole

Roof vent

 Normal operating range Fire/emergency range Overflow  pipe

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  Example: A service reservoir is to be designed for a water-supply system serving 250,000 people with an average demand of 600L/d/capita and a needed fire flow of 37000 L/min. Estimate the required volume of service storage. 3

Solution: 237600m . Example: water-supply system is tosystem be designed in anAarea whereanalysis the minimum allowable A pressure in the distribution is 300kPa. hydraulic of the distribution network under average daily demand conditions indicates that the head loss between the low-pressure service location, which has a pipeline elevation of 5.4m and the location of the elevated storage tank is 10m. Under maximum hourly demand conditions, the head loss between the low-pressure service location and the elevated storage tank is 12m. Determine the normal operating range for the water stored in the elevated tank. Solution: From 48m to 46 m. Pipe Network Systems T-2  T-1  P-

P-

C2 

1  J-1 

4  J-2  P-3 

P-2 

J-3  C3 

P-

C1 

5  J-4 

C4 

P-8 

P-

P-

6 

7  J-5 

C5 

P-9 

J-6  C6 

 N eq =  N  j +    N     p + N f  − 1    N eq eq= Number of equations required  N  j= number of junction nodes  N l= number of loops  N  f = number of Fixed Grade Nodes (e.g. elevated reservoirs)

In the above figure

 N eq = 6 + 2 +  2 − 1 = 9   8

 

  Conditions to satisfy: 1.  The algebraic sum of the pressure drops around any closed loop must be zero 2.  The flow entering a junction must equal the flow leaving it. Hardy Cross Method

Q=Q   a + ∆   Q = Actual flow rate

Qa = Assumed flow rate

∆ = Correction

∑

K (Qa +  ∆ ) = 0    

∑ KQ + ∑ xK ∆Q

 x −1 a

 x a

For small values of correction

∑

 + 

∆=−

∑  

 ∑  

KQa + ∆   x

x

 x − 1 2  x − 2  xK ∆ Qa + L = 0   2  x −1

KxQa

∑ ∑  xKQ

= 0 

 x

KQa

 x −1   a

Example: Determine the discharge in each line and the pressure at each junction node for the pipe network shown below using Hardy Cross method. The pipe and junction data is given in the table below.

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 A  B 

C2 

P-

J-2 

P-

P-2 

P-5 

J-1 

J-4 

PP-3 

P-6 

C1 

J-3 

C4 

C3 

Line 1 2

Nodes A-1 1-2

Length(m) 300 250

Diameter(cm) 30 20

3 4 5 6 7

1-3 2-3 2-4 3-4 4-B

200 220 180 250 270

20 15 20 20 25

 

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K ((ss /m ) 153.0 1226.5

0.019 0.02 0.019 0.019 0.017

981.2 4787.6 883.1 1226.5 388.4

 Node A B 1 2

Elevation(m) 126 123 96 99

Demand(L/s) 56 112

3 4

93 90

28 85

From Darcy-Weisbach formula,

2

f 0.015 0.019

h L   = KQ 2 , where

K  =

8 fL π 

2

gD 5

 

HGL elevation at the end of pipe = HGL elevation at the beginning-Head loss

 p = γ  ( HGL   . − G.El.)      El

10

 

 

 A  B 

P1 

J-2 

P-

P-2 

P-5 

J-1 

7  J-4 

PP-3  4 

P-6  J-3 

Pipe 1 2 3 4 5 6 7  Node A 1 2 3 4 B

Discharge(L/s) 167.9 58.4 53.5 16.4 37.2 9.1 113.1

Elevation(m) 126 96 99 93 90 123

Head loss (m) 5.61 4.78 3.22 1.55 1.42 0.13 5.97

HGL (m)

 p(kPa)

120.39 115.61 117.17 117.03

239.3 163.0 237.1 265.2

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Linear Method Hardy Cross method requires an initial estimate of the flow that should be close to the final solution. Linear method is a very stable method in which all the equations are solved simultaneously. Each nonlinear head loss term is linearized using first two terms in Taylor series:

 f (Q ) =  f ( q  ) +

∂ f 

(Q − q )

∂q

 

q= flow rate from previous iteration Q= unknown flow rate.

So,  

KQ = Kq + nKq n

n 

n−1

(Q − q ) =  DQ + D '  

where  

 D = nKq n

−1

 D ' = (1 − n) Kq n  

For loops with pumps, nonlinear pump characteristic equation (LHS) is replaced with linear equation (RHS):

 AQ 2 + BQ + H  =  Aq 2 + Bq + H    + (2 Aq + B )(Q − q ) =  EQ + E '   where

 E  = 2 Aq + B  

   E ' =  H  − Aq 2  

Example: Write down the equations to solve so lve the following network by linear method.  A  B 

P-

C2 

J-2 

P-2  J-1 

P-5  J-4 

PP-3 

C1 

P-

P-6  J-3 

C4 

C3 

Solution: First of all we assume the direction of flow in each pipe.

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 A  B 

P-1 

C2 

P-7 

J-2 

P-2 

P-5  J-4 

P-4 

J-1 

P-6 

P-3  C1 

C4  J-3 C3 

Continuity equations:

Q1 − Q2 − Q3 − C 1 = 0 Q2 + Q4 − Q5 − C 2 = 0 Q3 − Q4 + Q6 − C 3 = 0  

− Q5 − Q6 + Q7 − C 4 = 0 Loop equations (clockwise positive): 2 2 2 K 2Q2 − K 3Q3 − K 4Q4 = 0

K 4Q4 − K 5Q5 + K 6Q6 = 0 2

2

2

2

2

 

2

2  B

 A

Q − K  Q + K  Q + K  Q =  z −  z − K continuity The equations are already linear, so linearizing the loop equations we get:  D2Q2 − D3Q3 − D4Q4 = − D2′ + D3′ + D4′ = C 5  D4Q4 − D5Q5 + D6Q6 = − D4′ + D5′ − D6′ = C 6   − D1Q1 − D2Q2 + D5Q5 + D7Q7 =  z B − z A + D1′ + D   2′ − D5′ − D7′ = C 7 1

1

2

2

5

5

7

7

In matrix form the above equations can be written as follows:

[ A][][Q  ] = [C ]   Where [ A] is coefficient   matrix and [C ]  is a column matrix of constants. 13

 

 

⎡ 1 ⎢ 0 ⎢ ⎢ 0 ⎢ ⎢ 0 ⎢ 0 ⎢ ⎢ 0 ⎢− D ⎣ 1

−1

0

0

0

1

0

1

−1

0

1

−1

0

0

0

0

−1

 D2

− D3

− D4

0

0

0

 D4

− D5

− D2

0

0

 D5

0 ⎤ ⎡ Q1 ⎤

⎡ C 1 ⎤ ⎥ ⎢ ⎥ ⎢C  ⎥ 0 0 Q2 ⎥⎢ ⎥ ⎢ 2 ⎥ 1 0 ⎥ ⎢Q3 ⎥ ⎢C 3 ⎥ ⎥⎢ ⎥ ⎢ ⎥ − 1 1 ⎥ ⎢Q4 ⎥ = ⎢C 4 ⎥ 0 0 ⎥ ⎢Q5 ⎥ ⎢C 5 ⎥   ⎥⎢ ⎥ ⎢ ⎥ 0 ⎥ ⎢Q6 ⎥ ⎢C 6 ⎥  D6 0  D7 ⎥⎦ ⎢⎣Q7 ⎥⎦ ⎢⎣C 7 ⎥⎦ 0

Solving for the flow rates:

[Q ] = [ A]−1[C ]  For the initial iteration a velocity of 1m/s may be assumed in each pipe. The computed flow rates (Q) become estimated flow rates (q) for the next iteration. The iterative process is continued until the equations converge to a solution. In other words when the value of  N 

∑Q −q i

i

i =1  N 

∑

Qi

 

i =1

is within a specified limit.

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