Lin S.D. Water and Wastewater Calculations Manual

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2.5 Sludge ﬂow

To estimate every loss in a pipe carrying sludge, the Hazen–Williams

equation with a modified C value and a graphic method based on field

experience are commonly used. The modified C values for various total

solids contents are as follows (Brisbin 1957):

Total solids (%) 0 2 4 6 8.5 10

C 100 81 61 45 32 25

2.6 Dividing-ﬂow manifolds and multiport diffusers

The related subjects of dividing-flow manifolds and multiport diffusers

are discussed in detail by Benefield et al. (1984). They present basic the-

ories and excellent design examples for these two subjects. In addition,

there is a design example of hydraulic analysis for all unit processes of

a wastewater treatment plant.

3 Pumps

3.1 Types of pump

The centrifugal pump and the displacement pump are most commonly

used for water and wastewater works. Centrifugal pumps have a rotating

impeller which imparts energy to the water. Displacement pumps are often

of the reciprocating type in which a piston draws water or slurry into a closed

chamber and then expels it under pressure. Reciprocating pumps are widely

used to transport sludge in wastewater treatment plants. Air-lift pumps,

jet pumps, and hydraulic rams are also used in special applications.

3.2 Pump performance

The Bernoulli equation may be applied to determine the total dynamic

head on the pump. The energy equation expressing the head between

the suction(s) and discharge(d) nozzles of the pumps is as follows:

where H ⫽ total dynamic head, m or ft

, P

⫽ gage pressure at discharge and suction, respectively,

N/m

or lb/in

␥ ⫽ specific weight of water, N/m

or lb/ft

, V

⫽ velocity in discharge and suction nozzles, respectively,

m/s or ft/s

g ⫽ gravitational acceleration, 9.81 m/s

or 32.2 ft/s

, z

⫽ elevation of discharge and suction gage above the datum,

m or ft

H 5

1 z

2 a

1 z

Fundamental and Treatment Plant Hydraulics 265

The power P

required to pump water is a function of the flow Q and the

total head H and can be written as

⫽ jQH (4.28a)

where P

⫽ water power, kW (m ⋅ m

/min) or hp (ft ⋅ gal/min)

j ⫽ constant, at 20⬚C

⫽ 0.163 (for SI units)

⫽ 2.525 ⫻ 10

⫺4

(for British units)

Q ⫽ discharge, m

/min or gpm

H ⫽ total head, m or ft

Example 1: Calculate the water power for a pump system to deliver

3.785 m

/min (1000 gpm) against a total system head of 30.48 m (100 ft) at

a temperature of 20⬚C.

solution:

⫽ jQH

⫽ 0.163 ⫻ 3.785 m

/min ⫻ 30.48 m

⫽ 18.8 kW

or ⫽ 25.3 hp (P

⫽ rQH for SI units)

The power output (P

⫽ QrH for SI units) of a pump is the work

done to lift the water to a higher elevation per unit time. The theo-

retical power output or horsepower (hp) required is a function of a

known discharge and total pump lift. For US customary units, it may

be written as

(4.28b)

For SI units: P

(in N ⋅ m/s or Watt) ⫽ Q␥H

where Q ⫽ discharge, ft

/s (cfs) or m

␥ ⫽ specific weight of water or liquid, lb/ft

or N/m

H ⫽ total lift, ft or m

550 ⫽ conversion factor from ft ⋅ lb/s to hp

Example 2: Determine how many watts equal one horsepower.

solution:

1hp ⫽ 550 ft

⋅ lb/s

⫽ 550 ft ⫻ 0.305 m/ft ⋅ 1 lb ⫻ 4.448 N/(lb ⋅ s)

sin hpd 5

QgH

550

266 Chapter 4

⫽ 746 N ⋅ m/s (J/s ⫽ watt)

⫽ 746 watts (W) or 0.746 kilowatts (kW)

Conversely, 1 kW ⫽ 1.341 hp

The actual horsepower needed is determined by dividing the the-

oretical horsepower by the efficiency of the pump and driving unit.

The efficiencies for centrifugal pumps normally range from 50% to

85%. The efficiency increases with the size and capacity of the

pump.

The design of a pump should consider total dynamic head which

includes differences in elevations. The horsepower which should be

delivered to a pump is determined by dividing Eq. (4.28b) by the pump

efficiency, as follows:

(4.28c)

The efficiency of a pump (e

) is defined as the ratio of the power

output (P

⫽ ␥QH/550) to the input power of the pump (P

⫽ ␻␶). It can

be written as:

for US customary units

(4.28d)

for SI units

(4.28e)

where e

⫽ efficiency of the pump, %

␥ ⫽ specific weight of water, kN/m

or lb/ft

Q ⫽ capacity, m

/s or ft

⫽ brake horsepower

550 ⫽ conversion factor for horsepower to ft ⋅ lb/s

␻ ⫽ angular velocity of the turbo-hydraulic pump

␶ ⫽ torque applied to the pump by a motor

The efficiency of the motor (e

) is defined as the ratio of the power

applied to the pump by the motor (P

) to the power input to the motor

), i.e.

(4.29)e

gQH

3 550

hp 5

gQH

550 3 efficiency

Fundamental and Treatment Plant Hydraulics 267

The overall efficiency of a pump system (e) is combined as

(4.30)

Example 3: A water treatment plant pumps its raw water from a reservoir

next to the plant. The intake is 12 ft below the lake water surface at eleva-

tion 588 ft. The lake water is pumped to the plant influent at elevation 611 ft.

Assume the suction head loss for the pump is 10 ft and the loss of head in the

discharge line is 7 ft. The overall pump effiency is 72%. The plant serves

44,000 persons. The average water consumption is 200 gal per capita per

day (gpcpd). Compute the horsepower output of the motor.

solution:

Step 1. Determine discharge Q

Q ⫽ 44,000 ⫻ 200 gpcpd ⫽ 8.8 Mgal/d (

MGD)

⫽ 8.8 Mgal/d ⫻ 1.547 (ft

/s)/(Mgal/d)

⫽ 13.6 ft

Step 2. Calculate effective head H

H ⫽ (611 ⫺ 588) ⫹ 10 ⫹ 7 ⫽ 40 (ft)

Step 3. Compute overall horsepower output

Using Eq. (4.28c),

Example 4: A water pump discharges at a rate of 0.438 m

/s (10 Mgal/d). The

diameters of suction and discharge nozzles are 35 cm (14 in) and 30 cm (12

in), respectively. The reading of the suction gage located 0.3m (1 ft) above the

pump centerline is 11 kN/m

(1.6 lb/in

). The reading of the discharge gage

located at the pump centerline is 117 kN/m

(17.0 lb/in

). Assume the pump

efficiency is 80% and the motor efficiency is 93%; water temperature is 13⬚C.

Find (a) the power input needed by the pump and (b) the power input to the

motor.

solution:

Step 1. Write the energy equation

H 5 a

1 z

b 2 a

1 z

5 85.7 hp

hp 5

QgH

550e

13.6 ft

/s 3 62.4 lb/ft

3 40 ft

550 ft

lb/hp 3 0.72

e 5 e

268 Chapter 4

Step 2. Calculate each term in the above equation

At T ⫽ 13⬚C, ␥ ⫽ 9800 N/m

, refer to table 4.1a

Let z

⫽ 0 be the datum at the pump centerline

Step 3. Calculate the total head H

H ⫽ [11.94 ⫹ 1.96 ⫹ 0 ⫺ (1.12 ⫹ 1.06 ⫹ 0.30)] m

⫽ 11.42 m or 34.47 ft

Step 4. Compute power input P

by Eq. (4.28e) for question (a)

Step 5. Compute power input to the motor (P

) for question (b)

or 5  88.4 hp

5 65.88 kW

61.27 kW

0.93

5 82.2 hp

5 61.27 kW 3

1.341 hp

1 kW

5 61.27 kW

gQH

s9.8 kNds0.438 m

/sds11.42 md

0.80

510.30 m

s4.55 m/sd

2s9.81 m/s

5 1.06 m

0.438 m

sp/4ds0.35 md

5 4.55 m/s

11,000 N/m

9800 N/m

5 1.12 m

s6.20 m/sd

2s9.81 m/s

5 1.96 m

0.438 m

sp/4ds0.30 md

5 6.20 m/s

117,000 N/m

9800 N/m

5 11.94 m

Fundamental and Treatment Plant Hydraulics 269

3.3 Cost of pumping

The cost of pumping through a pipeline is a function of head loss, flow

rate, power cost, and the total efficiency of the pump system. It can be

expressed as (ductile Iron Pipe Research Association, 1997)

CP ⫽ 1.65 HQ$/E (4.31)

where CP ⫽ pumping cost, $/(yr ⋅ 1000 ft) (based on 24 h/d operation)

H ⫽ head loss, ft/1000 ft

Q ⫽ flow, gal/min

$ ⫽ unit cost of electricity, $/kWh

E ⫽ total efficiency of pump system, %

Velocity is related to flow by the following equation

(4.32a)

where V ⫽ velocity, ft/s

Q ⫽ flow, gal/min

d ⫽ actual inside diameter, in

Head loss is determined by the Hazen–Williams formula

(4.32b)

where C ⫽ a coefficient, and other symbols are as above.

Example: Water is pumped at a rate of 6300 gal/min (0.40 m

/s) through a

24-in pipeline of 10,000 ft (3048 m) length. The actual inside diameters for

ductile iron pipe and PVC pipe are respectively 24.95 and 22.76 in. Assume

the unit power cost is $0.058/kWh; the total efficiency of pump system is

75%; the pump is operated 24 h per day. Estimate: (a) the cost of pumping for

each kind of pipeline, (b) the present value of the difference of pumping cost,

assuming 50-year design pipe life (n), 6.6% annual rate (r) of return on the

initial investment, and 3.5% annual inflation rate of power costs.

solution:

Step 1. Compute the velocity of flow for each pipeline.

Let V

and V

represent velocity for ductile and PVC pipe, respectively, using

Eq. (4.32a)

5 4.13 ft/s

2.448 d

6300 gal/min

2.448s24.95 ind

H 5 1000a

0.115 Cd

0.63

1.852

V 5

2.448d

270 Chapter 4

Step 2. Compute head losses (H

and H

) for each pipe flow using Eq. (4.32b).

Coefficients C for ductile and PVC pipes are 140 and 150, respectively.

Step 3. Compute the costs of pumping, CP

and CP

⫽ 1.65 H

Q$/E ⫽ 1.65 ⫻ 1.89 ⫻ 6300 ⫻ 0.058/0.75

⫽1519 ($/1000 ft/yr)

⫽ 1.65 ⫻ 2.59 ⫻ 6300 ⫻ 0.058/0.75

⫽ 2082 ($/1000 ft/yr)

Step 4. Compute the difference of total cost for 10,000 ft annually (A)

A ⫽ (2082⫺1519) $/(1000 ft/yr) ⫻ (10,000 ft)

⫽ 5630 $/yr

Step 5. Compute the present worth (PW) of A adjusting for inflation using

the appropriate equation below

When g ⫽ r

PW ⫽ An (4.33a)

When g ≠ r

(4.33b)

(4.33c)i 5

r 2 g

1 1 g

PW 5 A c

s1 1 id

2 1

is1 1 id

5 2.59 sft/1000 ftd

5 1000 a

4.96

0.115 3 150 3 22.76

0.63

1.852

5 1.89 sft/1000 ftd

5 1000a

4.13

0.115 3 140 3 24.95

0.63

1.852

5 1000a

0.115Cd

0.63

1.852

5 4.96 ft/s

6300

2.448s22.76d

Fundamental and Treatment Plant Hydraulics 271

where PW ⫽ present worth of annual difference in pumping cost, $

A ⫽ annual difference in pumping cost, $

i ⫽ effective annual investment rate accounting for inflation, %

n ⫽ design life of pipe, yr

g ⫽ inflation (growth) rate of power cost, %

r ⫽ annual rate of return on the initial investment, %

In this example, g ≠ n

4 Water Flow in Open Channels

4.1 Che’zy equation for uniform ﬂow

In 1769, the French engineer Antoine Che’zy proposed an equation

for uniform open channel flow in which the average velocity is a func-

tion of the hydraulic radius and the energy gradient. It can be

written as

(4.34)

where V ⫽ average velocity, ft/s

C ⫽ Che’zy discharge coefficient, ft

1/2

R ⫽ hydraulic radius, ft

⫽ cross-sectional area A(ft

) divided by the wetted perimeter

P (ft)

⫽ A/P

S ⫽ energy gradient (slope of the bed, slope of surface water for

uniform flow)

⫽ head loss (h

) over the length of channel (L) divided by L

⫽ h

The value of C can be determined from

(4.35a)

where g ⫽ gravity constant, ft/s

f ⫽ Darcy–Weisback friction factor

C 5

V 5 C 2RS

5 144,855s$d

PW 5 A c

s1 1 id

2 1

is1 1 id

d 5 5630 c

s1 1 0.03d

2 1

0.03s1 1 0.03d

i 5

r 2 g

1 1 g

0.066 2 0.035

1 1 0.035

5 0.030

272 Chapter 4

Example: A trapezoidal open channel has a bottom width of 20 ft and side

slopes of inclination 1:1.5. Its friction factor f ⫽ 0.056. The depth of the chan-

nel is 4.0 ft. The channel slope is 0.025. Compute the flow rate of the channel.

solution:

Step 1. Determine C value from Eq. (4.34)

Step 2. Compute A, P, and R

Width of water surface ⫽ 20 ft ⫹ 2 ⫻ 6 ft ⫽ 32 ft

Step 3. Determine the flow rate Q

4.2 Manning equation for uniform ﬂow

As in the previous section, Eq. (4.22) was proposed by Robert Manning

in 1889. The well-known Manning formula for uniform flow in open

channel of nonpressure pipe is

(4.22a)

(4.22b)

The flow rate (discharge) Q can be determined by

(4.36)Q 5 AV 5 A

1.486

2/3

1/2

V 5

2/3

1/2

sfor SI unitsd

V 5

1.486

2/3

1/2

sfor British unitsd

5 1937 ft

5 104 ft

3 67.8 ft

1/2

/s 3 23.02 ft 3 0.025

Q 5 AV 5 AC 2RS

5 3.02 ft

R 5 A/P 5 104 ft

/34.42 ft

5 34.42 ft

P 5 2 ft 3 24

1 6

1 20 ft

5 104 ft

A 5

s20 1 32d ft 3 4 ft

C 5

8 3 32.2

0.056

5 67.8 sft

1/2

/sd

Fundamental and Treatment Plant Hydraulics 273

All symbols are the same as in the Che’zy equation. The Manning rough-

ness coefficient (n) is related to the Darcy–Weisback friction factor as

follows:

(4.37)

Manning also derived the relationship of n (in s/ft

1/3

) to the Che’zy coef-

ficient C by the equation

(4.35b)

Typical values of n for various types of channel surface are shown in

Table 4.4.

Example 1: A 10-ft wide (w) rectangular source-water channel has a flow

rate of 980 ft

/s at a uniform depth (d) of 3.3 ft. Assume n ⫽ 0.016 s/ft

1/3

for

concrete. (a) Compute the slope of the channel. (b) Determine the discharge

if the normal depth of the water is 4.5 ft.

solution:

Step 1. Determine A, P, and R for question (a)

A ⫽ wd ⫽ 10 ft ⫻ 3.3 ft ⫽ 33 ft

P ⫽ 2d ⫹ w ⫽ 2 ⫻ 3.3 ft ⫹ 10 ft ⫽ 16.6 ft

R ⫽ A/P ⫽ 33 ft

/16.6 ft ⫽ 1.988 ft

Step 2. Solve S by the Manning formula

Rewrite

Answer for (a), S ⫽ 0.041

Step 3. For (b), determine new A, P, and R

A ⫽ wd ⫽ 10 ft ⫻ 4.5 ft ⫽ 45 ft

P ⫽ 2d ⫹ w ⫽ 2(4.5 ft) ⫹ 10 ft ⫽ 19 ft

R ⫽ A/P ⫽ 45 ft

/19 ft ⫽ 2.368 ft

Step 4. Calculate Q for answer of question (b)

Q 5 A

1.486

2/3

1/2

S 5 a

1.486AR

2/3

5 a

980 cfs 3 0.016 s/ft

1/3

1.486 3 33 ft

3 s1.988 ftd

2/3

5 0.041

Q 5 A

1.486

2/3

1/2

C 5

1/6

n 5 0.093f

1/2

1/6

274 Chapter 4