Basic Hydraulics of Flow

(Pipe flow, Trench flow, Detention time)

Basic Hydraulics of Flow

(Pipe flow, Trench flow, Detention time)

Math for Water TechnologyMTH 082Lecture 4

Hydraulics Chapter 7 (pgs. 311-319-341)

Math for Water TechnologyMTH 082Lecture 4

Hydraulics Chapter 7 (pgs. 311-319-341)

RULES FOR FLOW RATESRULES FOR FLOW RATES1.DRAW AND LABEL DIAGRAM

2.CONVERT AREA or VELOCITY DIMENSIONS

3 .SOLVE EACH FORMULA INDIVIDUALLY(Velocity and Area)

4. ISOLATE THE FlOW PARAMETERS NECESSARYQ= (Velocity) (Area formula first)

5. USE YOUR UNITS TO GUIDE YOU

6. SOLVE THE PROBLEM

7.CARRY OUT FINAL FLOW RATE CONVERSIONS

1.DRAW AND LABEL DIAGRAM

2.CONVERT AREA or VELOCITY DIMENSIONS

3 .SOLVE EACH FORMULA INDIVIDUALLY(Velocity and Area)

4. ISOLATE THE FlOW PARAMETERS NECESSARYQ= (Velocity) (Area formula first)

5. USE YOUR UNITS TO GUIDE YOU

6. SOLVE THE PROBLEM

7.CARRY OUT FINAL FLOW RATE CONVERSIONS

Types of flow rate?

• Instantaneous flow rate- Flow rate at a particular moment in time. Use cross sectional area and velocity in a pipe or channel

• Average flow rate -Average of instantaneous flow rates over time. Records of time and flow

• Instantaneous flow rate- Flow rate at a particular moment in time. Use cross sectional area and velocity in a pipe or channel

• Average flow rate -Average of instantaneous flow rates over time. Records of time and flow

How do we measure flow rate?Water MeterWater Meter

How do we measure flow rate?

Differential Pressure Metering Devices

• Most common (~50%) units in use today.

• Measure pressure drop across the meter which is proportional to the square of the flow rate.

Differential Pressure Metering Devices

• Most common (~50%) units in use today.

• Measure pressure drop across the meter which is proportional to the square of the flow rate.

How do we measure flow rate?

• Weirs• Weirs

How do we measure flow rate?• Parshall flumes for open channels• Parshall flumes for open channels

How do we measure flow rate?• Orifice meters for closed conduit

• An orifice is simply a flat piece of metal with a specific-sized hole bored in it. Most orifices are of the concentric type, but eccentric, conical (quadrant), and segmental designs are also available.

• Orifice meters for closed conduit

• An orifice is simply a flat piece of metal with a specific-sized hole bored in it. Most orifices are of the concentric type, but eccentric, conical (quadrant), and segmental designs are also available.

How do we measure flow rate?• Venturi meters for closed conduit

• Venturi tubes have the advantage of being able to handle large flow volumes at low pressure drops. A venturi tube is essentially a section of pipe with a tapered entrance and a straight throat.

• Venturi meters for closed conduit

• Venturi tubes have the advantage of being able to handle large flow volumes at low pressure drops. A venturi tube is essentially a section of pipe with a tapered entrance and a straight throat.

Factors that influence flow rate?• Fluid dynamics: typically involves calculation of various

properties of the fluid, such as velocity, pressure, density, and temperature as functions of space and time.

• Viscosity is commonly perceived as "thickness", or resistance to pouring. Viscosity describes a fluids internal resistance to flow and may be thought of as a measure of fluid friction. Water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. Low viscosity =fast moving; high viscosity slow moving

• Density Forces that arise due to fluids of different densities acting differently under gravity.

• Friction of the liquid in contact with the pipe. Friction=slower motion

• Fluid dynamics: typically involves calculation of various properties of the fluid, such as velocity, pressure, density, and temperature as functions of space and time.

• Viscosity is commonly perceived as "thickness", or resistance to pouring. Viscosity describes a fluids internal resistance to flow and may be thought of as a measure of fluid friction. Water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. Low viscosity =fast moving; high viscosity slow moving

• Density Forces that arise due to fluids of different densities acting differently under gravity.

• Friction of the liquid in contact with the pipe. Friction=slower motion

How do we select a flow meter?1. What is the fluid being measured (air, water,etc…)?

2. Do you require rate measurement and/or totalization from the flow meter?

3. If the liquid is not water, what viscosity is the liquid?

4. Is the fluid clean?

5. Do you require a local display on the flow meter or do you need an electronic signal output?

6. What is the minimum and maximum flowrate for the flow meter?

7. What is the minimum and maximum process pressure?

8. What is the minimum and maximum process temperature?

9. Is the fluid chemically compatible with the flowmeter wetted parts?

10. If this is a process application, what is the size of the pipe?

1. What is the fluid being measured (air, water,etc…)?

2. Do you require rate measurement and/or totalization from the flow meter?

3. If the liquid is not water, what viscosity is the liquid?

4. Is the fluid clean?

5. Do you require a local display on the flow meter or do you need an electronic signal output?

6. What is the minimum and maximum flowrate for the flow meter?

7. What is the minimum and maximum process pressure?

8. What is the minimum and maximum process temperature?

9. Is the fluid chemically compatible with the flowmeter wetted parts?

10. If this is a process application, what is the size of the pipe?

How do we quantify flow rate?

Q = V x A where

Q = liquid flow through the pipe/channel (length(ft3)/time)V = average velocity of the flow (length (ft)/time)A = cross-sectional area of the pipe/channel (length ft2)

Q = V x A where

Q = liquid flow through the pipe/channel (length(ft3)/time)V = average velocity of the flow (length (ft)/time)A = cross-sectional area of the pipe/channel (length ft2)

•Because the pipe's cross-sectional area is known and remains constant, the average velocity is an indication of the flow rate.

•Because the pipe's cross-sectional area is known and remains constant, the average velocity is an indication of the flow rate.

Units must match!!! ft3/min, ft3/d, etc.

**MAKE AREA or VELOCITY CONVERSIONS FIRST!****MAKE FLOW RATE CONVERSIONS LAST!!!!******

Units must match!!! ft3/min, ft3/d, etc.

**MAKE AREA or VELOCITY CONVERSIONS FIRST!****MAKE FLOW RATE CONVERSIONS LAST!!!!******

Open Channel Flow Rate (ft3/time)

Open Channel Flow Rate (ft3/time)

V=velocity (ft/time)V=velocity (ft/time)

L=depth (ft)L=depth (ft)

A= W X L =ft2 A= W X L =ft2 W=width (ft)W=width (ft)

V= ft/time V= ft/time

Q (flow rate) = V X A =ft3/time Q (flow rate) = V X A =ft3/time Flow = 7.48 gal or 3.06 X 10-6 acre feet or 1 mgd 1ft3 1 gal 1,000,000 gal

Time = 24 hrs or 1440 min or 86,400 sec 1 day 1 day 1 day

Flow = 7.48 gal or 3.06 X 10-6 acre feet or 1 mgd 1ft3 1 gal 1,000,000 gal

Time = 24 hrs or 1440 min or 86,400 sec 1 day 1 day 1 day

Circular pipe Flowing Full (ft3/time)

Circular pipe Flowing Full (ft3/time)

V=velocity (ft/time)V=velocity (ft/time)

D=diameter (ft)D=diameter (ft)

A= 0.785 (diameter (ft))2 = ft2 A= 0.785 (diameter (ft))2 = ft2

V= ft/timeV= ft/time

Q (flow rate) = V X A =ft3/timeQ (flow rate) = V X A =ft3/time

When pipe is flowing full you can use the full cross sectional area (0.785)When pipe is flowing full you can use the full cross sectional area (0.785)

Pipe Flowing FullPipe Flowing Full

2. Is it flowing full? YES.. I can use 0.785 in equat.3. Formula: Q= A * V4. Substitute and Solve Q= (0.785)(0.67 ft)(0.67 ft)(2.4 fps)Q =0.85 cfs

5. Convert: (0.85 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 380 gpm

2. Is it flowing full? YES.. I can use 0.785 in equat.3. Formula: Q= A * V4. Substitute and Solve Q= (0.785)(0.67 ft)(0.67 ft)(2.4 fps)Q =0.85 cfs

5. Convert: (0.85 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 380 gpm

1. Label Figure!1. Label Figure!

An 8 in transmission main has a flow of 2.4 fps. What is the gpm flow rate through the pipe?

An 8 in transmission main has a flow of 2.4 fps. What is the gpm flow rate through the pipe?

V= 2.4 fpsV= 2.4 fps

D= 8 in=0.67 ftD= 8 in=0.67 ft

Solve for Q!Solve for Q!

Pipe Not Flowing FullPipe Not Flowing Full

V= 3.4 fpsV= 3.4 fps

D= 8 in=0.67 ftD= 8 in=0.67 ft

2. Is it flowing full? NO.. I need d/D ratio 3. d/D= 5”/8”=0.63 (ratio) =_0.5212_ from table4. Formula: Q= A * V5. Substitute and Solve Q= (0.5212)(0.67 ft)(0.67 ft)(3.4 fps)Q =0.8 cfs6. Convert: (0.8 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 359 gpm

2. Is it flowing full? NO.. I need d/D ratio 3. d/D= 5”/8”=0.63 (ratio) =_0.5212_ from table4. Formula: Q= A * V5. Substitute and Solve Q= (0.5212)(0.67 ft)(0.67 ft)(3.4 fps)Q =0.8 cfs6. Convert: (0.8 ft3/sec)(60 sec/ 1 min)(7.48 gal/ft3) = 359 gpm

Solve for Q!Solve for Q!

1. Label Figure!1. Label Figure!

An 8 in transmission main has a flow of 3.4 fps. What is the gpm flow rate through the pipe if the water is flowing at a depth of 5 inches?

An 8 in transmission main has a flow of 3.4 fps. What is the gpm flow rate through the pipe if the water is flowing at a depth of 5 inches?

H2O depth= 5 in=0.41 ftH2O depth= 5 in=0.41 ft

Example 1. Circular pipe Flowing Full (ft3/time)

A 15 in diameter pipe is flowing full. What is the gallons per minute flow rate in the pipe if the velocity is 110 ft/min.

Example 1. Circular pipe Flowing Full (ft3/time)

A 15 in diameter pipe is flowing full. What is the gallons per minute flow rate in the pipe if the velocity is 110 ft/min.

D=diameter (15 inches)Convert! (15in)(1ft/12in)D=1.25 ft

D=diameter (15 inches)Convert! (15in)(1ft/12in)D=1.25 ft

Area (pipe)= 0.785 (diameter)2

A= 0.785 (1.25 ft)2 =1.23 ft2

Area (pipe)= 0.785 (diameter)2

A= 0.785 (1.25 ft)2 =1.23 ft2

V= 110 ft/minV= 110 ft/min

Q (flow rate) = V X A110 ft/min X 1.23 ft2 = 134.92ft3/min

Q (flow rate) = V X A110 ft/min X 1.23 ft2 = 134.92ft3/min

V=110 (ft/min)V=110 (ft/min)

Q (flow rate) = 134.92ft3/min (7.48 gal/ft3)= 1,009 gpmQ (flow rate) = 134.92ft3/min (7.48 gal/ft3)= 1,009 gpm

Q= ?gpmQ= ?gpm

D=diameter (1.5ft) D= 18 inchesD=diameter (1.5ft) D= 18 inches

V = 2 ft/secV = 2 ft/sec8 hrs

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

A full 18” raw sewage line has broken and has been leaking raw sewage into Arcade creek for 8 hours. What is the gpm flow rate

through a pipe-- assume a velocity of 2 ft/sec?

A full 18” raw sewage line has broken and has been leaking raw sewage into Arcade creek for 8 hours. What is the gpm flow rate

through a pipe-- assume a velocity of 2 ft/sec?

158

5 gpm

507

gpm

105

7 gpm

202

929

gpm

60%

0%

10%

30%

Diameter= 18 in,1.5 ft; depth= 18”or 1.5ft,1 ft, V=2 ft/sec= Q?

Q = V X AArea (pipe)= 0.785 (diameter)2

Area (pipe)= 0.785 (diameter)2

A= 0.785 (1.5 ft)2 =1.76 ft2

Q= V X AQ= 2 ft/sec X 1.76 ft2 = 3.56 ft3/sec3.56 ft3 7.48 gal 60sec = 1585 gpm sec 1ft3 1min

Diameter= 18 in,1.5 ft; depth= 18”or 1.5ft,1 ft, V=2 ft/sec= Q?

Q = V X AArea (pipe)= 0.785 (diameter)2

Area (pipe)= 0.785 (diameter)2

A= 0.785 (1.5 ft)2 =1.76 ft2

Q= V X AQ= 2 ft/sec X 1.76 ft2 = 3.56 ft3/sec3.56 ft3 7.48 gal 60sec = 1585 gpm sec 1ft3 1min

1. 1585 gpm

2. 507 gpm

3. 1057 gpm

4. 202929 gpm

1. 1585 gpm

2. 507 gpm

3. 1057 gpm

4. 202929 gpm

How many gallons of raw sewage was released to Arcade Creek after 8 hrs?

How many gallons of raw sewage was released to Arcade Creek after 8 hrs?

951

00 g

allo

ns

760

,800

gal

lon...

105

7 gpm

202

929

gpm

0% 0%0%

100%

1. 95100 gallons

2. 760,800 gallons

3. 1057 gpm

4. 202929 gpm

1. 95100 gallons

2. 760,800 gallons

3. 1057 gpm

4. 202929 gpm

D=diameter (1.5ft) D= 18 inchesD=diameter (1.5ft) D= 18 inches

V = 2 ft/secV = 2 ft/sec

8 hrs

1585 gallons X 60 min X 8 hrs = 760,800 gal min 1 hr1585 gallons X 60 min X 8 hrs = 760,800 gal min 1 hr

Example 2. Channel Flowing Full (ft3/time)

What is the MGD flow rate through a channel that is 3ft wide with water flowing to a depth of 16 in. at a velocity of 2 ft/sec?

Example 2. Channel Flowing Full (ft3/time)

What is the MGD flow rate through a channel that is 3ft wide with water flowing to a depth of 16 in. at a velocity of 2 ft/sec?

L= (16 inches)Convert! (16in)(1ft/12in)D=1.33 ft

L= (16 inches)Convert! (16in)(1ft/12in)D=1.33 ft

V= 2 ft/secV= 2 ft/sec

Q (flow rate) = V X A2 ft/sec X 3.99 ft2 = 7.98ft3/sec

Q (flow rate) = V X A2 ft/sec X 3.99 ft2 = 7.98ft3/sec

V= 2 ft/secV= 2 ft/sec

Depth (L) =16 inDepth (L) =16 in

W= 3ftW= 3ft

Area (rect)= L X WA= (1.33 ft) 3 ft =3.99 ft2

Area (rect)= L X WA= (1.33 ft) 3 ft =3.99 ft2

Q (flow rate)=(7.98ft3/sec) (60sec/1min) (7.48 gal/ft3) (1,440 min/day)= 5,157,251 gpd5,157,251 gpd = 5.16 MGD

Q (flow rate)=(7.98ft3/sec) (60sec/1min) (7.48 gal/ft3) (1,440 min/day)= 5,157,251 gpd5,157,251 gpd = 5.16 MGD

Q= ?MGDQ= ?MGD

Example 4. Water depth in channel (ft)

A channel is 3 ft wide. If the flow in the channel is 7.5 MGD and the velocity of the flow is 185 ft/min, what is the depth (in feet) of water in the channel?

Example 4. Water depth in channel (ft)

A channel is 3 ft wide. If the flow in the channel is 7.5 MGD and the velocity of the flow is 185 ft/min, what is the depth (in feet) of water in the channel?

V= 185 ft/minV= 185 ft/min

Q= V X A where A= L X W Q (flow rate) = V X (L X W) = L= Q÷V(W)

L= 696.3 ft3/min÷185 ft/min (3 ft)

L= 1.25 ft

Q= V X A where A= L X W Q (flow rate) = V X (L X W) = L= Q÷V(W)

L= 696.3 ft3/min÷185 ft/min (3 ft)

L= 1.25 ft

V= 185 ft/minV= 185 ft/min

Depth (L)=? ftDepth (L)=? ft

W= 3 ftW= 3 ftArea (rect)= L X WA= (L=(?ft)) (3ft)Area (rect)= L X WA= (L=(?ft)) (3ft)

Q= 7.5 MGD =7,500,000 gpdQ=7,500,000 gpd(7.48ft3/1gal/)(1day/1440min)Q=696.3 ft3/min

Q= 7.5 MGD =7,500,000 gpdQ=7,500,000 gpd(7.48ft3/1gal/)(1day/1440min)Q=696.3 ft3/min

Example 5. Velocity =rate (length/time)

A float is placed in a channel. It takes 2.5 min to travel 300 ft. What is the flow velocity in feet per minute in the channel?

Example 5. Velocity =rate (length/time)

A float is placed in a channel. It takes 2.5 min to travel 300 ft. What is the flow velocity in feet per minute in the channel?

V= rate (length/time) = 300 ft 2.5 minV= 120 ft/min

V= rate (length/time) = 300 ft 2.5 minV= 120 ft/min

V= 300 ft 2.5minV= 300 ft 2.5min

300 ft300 ft2.5min

Example 7. Velocity in a pipe flowing full (length/time)

A 305 mm diameter pipe flowing full is carrying 35 L/sec. What is the velocity of the water (m/sec) through the pipe?

Example 7. Velocity in a pipe flowing full (length/time)

A 305 mm diameter pipe flowing full is carrying 35 L/sec. What is the velocity of the water (m/sec) through the pipe?

V= ?(m/sec)V= ?(m/sec)

Q= V X A where V= Q/A V= Q÷AV= .035 m3/sec÷(.0703 m2)

V= .49 m/sec

Q= V X A where V= Q/A V= Q÷AV= .035 m3/sec÷(.0703 m2)

V= .49 m/sec

Q=30 L/secQ=30 L/sec

D=diameter (305 mm)Convert! (305mm)(1m/1000mm)D=.305 m

D=diameter (305 mm)Convert! (305mm)(1m/1000mm)D=.305 m

Area (pipe)= 0.785 (.305m)2

A= 0.785 (.305 m)2 =.0703 m2

Area (pipe)= 0.785 (.305m)2

A= 0.785 (.305 m)2 =.0703 m2

Q= 35 L/sec= 35L/sec (1m3/1000L)Q=.035 m3/secQ= 35 L/sec= 35L/sec (1m3/1000L)Q=.035 m3/sec

Velocity???Velocity???

Example 8. Flow rate in channel flowing full (ft3/time)

A channel is 4 ft wide with water flowing to a depth of 2.3 ft. If a float placed in the water takes 3 min to travel a distance of 500 ft, what is the ft3/min flow rate in the channel?

Example 8. Flow rate in channel flowing full (ft3/time)

A channel is 4 ft wide with water flowing to a depth of 2.3 ft. If a float placed in the water takes 3 min to travel a distance of 500 ft, what is the ft3/min flow rate in the channel?

V= 500 ft/3min=166.6 ft/minV= 500 ft/3min=166.6 ft/min

Area (rect)= L X WA= (4 ft) (2.3ft) =9.2 ft2

Area (rect)= L X WA= (4 ft) (2.3ft) =9.2 ft2

Q=? ft3/minQ=? ft3/min

Q (flow rate) = V X A166.6 ft/min X 9.2 ft2 = 1533ft3/min

Q (flow rate) = V X A166.6 ft/min X 9.2 ft2 = 1533ft3/min

V= 500ft/3minV=166.6 ft/minV= 500ft/3minV=166.6 ft/min

D=2.3 ftD=2.3 ft

W= 4 ftW= 4 ft

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

What is the gpm flow rate through a pipe that is 24 inch wide with water flowing to a depth of 12 in. at a velocity of

4 ft/sec?

What is the gpm flow rate through a pipe that is 24 inch wide with water flowing to a depth of 12 in. at a velocity of

4 ft/sec?

12.

56 g

pm

563

7 gpm

452

gpm

285

1gpm

30%

50%

0%

20%

Diameter= 24 in,2 ft; depth= 12”,1 ft, V=4 ft/sec= Q?

Q = V X Ad/D= 12/24=0.5=table 0.3927Area (pipe)= 0.3927 (diameter)2

Area (pipe)= 0.3927(diameter)2

A= 0.3927 (2 ft)2 =1.6 ft2

Q= V X AQ= 4 ft/sec X 1.6 ft2 = 6.35 ft3/sec6.35 ft3 7.48 gal 60sec = 2851 gpm sec 1ft3 1min

Diameter= 24 in,2 ft; depth= 12”,1 ft, V=4 ft/sec= Q?

Q = V X Ad/D= 12/24=0.5=table 0.3927Area (pipe)= 0.3927 (diameter)2

Area (pipe)= 0.3927(diameter)2

A= 0.3927 (2 ft)2 =1.6 ft2

Q= V X AQ= 4 ft/sec X 1.6 ft2 = 6.35 ft3/sec6.35 ft3 7.48 gal 60sec = 2851 gpm sec 1ft3 1min

1. 12.56 gpm

2. 5637 gpm

3. 452 gpm

4. 2851gpm

1. 12.56 gpm

2. 5637 gpm

3. 452 gpm

4. 2851gpm

D=diameter (2 ft) D= 24 inchesD=diameter (2 ft) D= 24 inches

depth=(1 ft)d= 12 inchesdepth=(1 ft)d= 12 inches

Detention TimeDetention Time

Math for Water TechnologyMTH 082Lecture 4

Mathematics Ch 22 (pgs. 193-196)

Math for Water TechnologyMTH 082Lecture 4

Mathematics Ch 22 (pgs. 193-196)

“how long a drop of water or suspended particle remains in a tank or chamber”

“how long a drop of water or suspended particle remains in a tank or chamber”

What is detention time?What is detention time?

Detention time (DT) = volume of tank = MG

flow rate MGD

Detention time (DT) = volume of tank = MG

flow rate MGD

Tank Detention Time

Flash mixing basin

30-60 sec

Flocculation basin

20-60 min

Sedimentation basin

1-12 h

The time it takes for a unit volume of water to pass entirely through a

sedimentation basin is called

Det

entio

n tim

e

Hyd

raulic

load

...

Ove

rflow

tim

e

Wei

r load

ing

r...

25% 25%25%25%1. Detention time

2. Hydraulic loading rate

3. Overflow time

4. Weir loading rate

1. Detention time

2. Hydraulic loading rate

3. Overflow time

4. Weir loading rate

What is the average detention time in a water tank given the following: diameter = 30' depth = 15' flow =

700 gpm

1hr.

34m

in.

1hr.

53m

in.

1hr.

47m

in.

2 h

rs. 3

min

.

25% 25%25%25%

1. 1hr. 34min.

2. 1hr. 53min.

3. 1hr. 47min.

4. 2 hrs. 3 min.

Volume = 0.785 (30 ft)(30ft)(15 ft) = 10597 ft3

10597 ft3 (7.48 gal/1ft3) = 79269 gal

DT= volume/flow = 79269 gal/700 gpm = 113 minutes113 minutes - 60 minutes or 1 hr = 53 minutes or 1 hour and 53 minutes

Volume = 0.785 (30 ft)(30ft)(15 ft) = 10597 ft3

10597 ft3 (7.48 gal/1ft3) = 79269 gal

DT= volume/flow = 79269 gal/700 gpm = 113 minutes113 minutes - 60 minutes or 1 hr = 53 minutes or 1 hour and 53 minutes

What is the average detention time in a water tank given the following: diameter = 80' depth = 12.2' flow =

5 MGD

2.2

hrs

.

1.6

8 hrs

.

2.4

hrs

.

1.7

4 hrs

.

25% 25%25%25%

1. 2.2 hrs.

2. 1.68 hrs.

3. 2.4 hrs.

4. 1.74 hrs.

V= 0.785 (Diameter)(Diameter)(depth)Volume = 0.785 (80 ft)(80ft)(12.2 ft) = 61292 ft3

61292 ft3 (7.48 gal/1ft3) = 458470 gal (1MG/1,000,000 gal) = 0.46 MG

DT= volume/flow = 0.46 MG /5 MGD = .09 days (24 hr/1d) = 2.2 hrs

V= 0.785 (Diameter)(Diameter)(depth)Volume = 0.785 (80 ft)(80ft)(12.2 ft) = 61292 ft3

61292 ft3 (7.48 gal/1ft3) = 458470 gal (1MG/1,000,000 gal) = 0.46 MG

DT= volume/flow = 0.46 MG /5 MGD = .09 days (24 hr/1d) = 2.2 hrs

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

A 10 MG reservoir has a peak of 2.8 MGD. What is the detention time in the tank in

hours?

A 10 MG reservoir has a peak of 2.8 MGD. What is the detention time in the tank in

hours?

3.5

hrs

85.

7 hrs

0.2

8 hrs

4 h

rs

25% 25%25%25%

tank= 10 MG, Flow rate 2.8 mgd

DT= volume of tank/flow rate

DT=VT/FRTime = 10 MG/2.8MGDTime= 3.5 day

3.5 day (24 h/1day)=85.7 hrs

tank= 10 MG, Flow rate 2.8 mgd

DT= volume of tank/flow rate

DT=VT/FRTime = 10 MG/2.8MGDTime= 3.5 day

3.5 day (24 h/1day)=85.7 hrs

10 MG10 MG

2.8 MGD2.8 MGD

1. 3.5 hrs

2. 85.7 hrs

3. 0.28 hrs

4. 4 hrs

1. 3.5 hrs

2. 85.7 hrs

3. 0.28 hrs

4. 4 hrs

DRAW:•Given:•Formula:•Solve:

DRAW:•Given:•Formula:•Solve:

A 36 in transmission main is used for chlorine contact time. If the peak hourly flow is 6 MGD, and the main is

1.8 miles long, what is the contact time in minutes?

A 36 in transmission main is used for chlorine contact time. If the peak hourly flow is 6 MGD, and the main is

1.8 miles long, what is the contact time in minutes?

0.8

3 m

in

.52

min

121

min

250

min

38%

0%

38%

25%

tank= 10 MG, Flow rate 2.8 mgd

Volume = π*r2*hDT= volume of tank/flow rateDT=VT/FRVolume = π*r2*hV= π *(1.5ft)2*(9504 ft)V=6718 ft3Convert to gallonsV=6718 ft3(1 gal/7.48 ft3)V=502,505 gal or .502 MG

DT=VT/FRDT= .502 MG/6 MGDDetention Time =0.83 day0.83 day (24 h/1day)(60 min/1 hr)= 120.6 min

tank= 10 MG, Flow rate 2.8 mgd

Volume = π*r2*hDT= volume of tank/flow rateDT=VT/FRVolume = π*r2*hV= π *(1.5ft)2*(9504 ft)V=6718 ft3Convert to gallonsV=6718 ft3(1 gal/7.48 ft3)V=502,505 gal or .502 MG

DT=VT/FRDT= .502 MG/6 MGDDetention Time =0.83 day0.83 day (24 h/1day)(60 min/1 hr)= 120.6 min

1. 0.83 min

2. .52 min

3. 121 min

4. 250 min

1. 0.83 min

2. .52 min

3. 121 min

4. 250 min

1.8 miles (5280 ft/1mile)= 9504 ft1.8 miles (5280 ft/1mile)= 9504 ft

36 in=3 ft36 in=3 ft

What did you learn?What did you learn?• How is flow measured?

• What equation is used to determine flow rate?

• What are the units for flow rate, velocity?

• What is detention time?

• How is flow measured?

• What equation is used to determine flow rate?

• What are the units for flow rate, velocity?

• What is detention time?

Today’s objective: to become proficient with the concept of basic hydraulic calculations used in the

waterworks industry applications of the fundamental flow equation, Q = A X V, and

hydraulic detention time has been met.

Today’s objective: to become proficient with the concept of basic hydraulic calculations used in the

waterworks industry applications of the fundamental flow equation, Q = A X V, and

hydraulic detention time has been met.

Stro

ngly A

gree

Agre

e

Dis

agre

e

Stro

ngly D

isag

ree

8%0%

15%

77%

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree

1. Strongly Agree

2. Agree

3. Disagree

4. Strongly Disagree