Brad Peterson, P.E.

New Website: New Website:http://njut2009fall.weebly.com

M P ’ E il Add Mr. Peterson’s Email Address: bradpeterson@engineer.com

If 6 m3 of oil weighs 47 kN calculate itsIf 6 m3 of oil weighs 47 kN, calculate its specific weight Ƴ and specific gravity.

specific weight Ƴ = 47 kN = 7 833 kN/m3specific weight Ƴ = 47 kN = 7.833 kN/m3

6 m3

specific gravity = Ƴoil = 7.833 kN/m3 = 0.800Ƴwater 9 79 kN/m3Ƴwater 9.79 kN/m

If 1 m3 of concrete has a mass 2 4 Tons If 1 m3 of concrete has a mass 2.4 Tons, calculate its specific weight Ƴ and specific gravitygravity.

39.81 /ifi i ht 2300 22 56 /N kgk kN 33specific weight 2300 22.56 /

1gkg kN m

m

322.56 /specific gravity 2.30con kN m 3specific gravity 2.30

9.79 /water kn m

V i h l h h i Vacuum = space with less that atmospheric pressureA h i f ili Atmospheric pressure refers to prevailing pressure in the air around us

l l d d h At sea level, standard atmospheric pressure is:

101.3 , orp kPa760 of mercury, or

1 atmosphereh mm1 atmosphere

ph

101 3p kPa

Atmospheric pressure 101.3p kPaAtmospheric pressure

2101 3 103 3 /p kPa kN m 101.3 103.3 /p kPa kN m

:Compute :13.6

Compute Specific gravity of mercury = R b

Remember,

weight of substance specific gravity = =specific gravity = = weight of equal amount of water

= specific gravity weight of equal amount of water

313.6 9.79 /kN m = 3133.1 /kN m

ph

2101 3 /p kN m

3

101.3 /

133.1 /

p kN m

kN m 2101.3 / / 0 76 760kN mh m mm

3 0.76 760133.1 /

h m mmkN m

1.1m

12 34kN3

12.34Weight of glycerin kNm

12.34P A 1 1

mkNh 3

.3Pr essure at A 1.1kNp h mm

2

13.57 13.57kN kPa2 13.57kPam

Determine the gage pressure in kPa at a depthDetermine the gage pressure in kPa at a depth of 10.0 meters below the free surface of a body of waterbody of water.

39.79 /kN m Weight of water Ng

p h

9.79 97.910 0 97 9kN kNp m kPa 3 210.0 97.9p m kPam m

Find the pressure at the bottom of a tankFind the pressure at the bottom of a tank containing glycerin under pressure as shown in following figurein following figure.

5050

p hkN

pressure at bottom =

25050 kNkPam

312.34kN

mweight of glycerin =

m

2 3 3

50 12.34 74.682 74.68kN kN kNp m kPam m m

m m m

(a) Find the elevation of the liquid surface in Piezometer(a) Find the elevation of the liquid surface in Piezometer A(b) The elevation of mercury in Piezometer B(b) The elevation of mercury in Piezometer B(c) The pressure at the bottom, Elevation 0

(a) The liquid in Piezometer A will rise to the same elevation as the top of the tankas the top of the tank.

9 79kN3

9.79( ) (0.72 )(1.7 )AkNb p h m

m

2

11.98 11.98kN kPa2

211.98 // 0 519

mkN mh p 3 / 0.519

(2.36 9.79 / )0 3 0 519 0 819

Ah pkN m

h

0.3 0.519 0.819TOTALh m

0( ) A Bc p p p 311.98 (2.36 9.79 / )(0.3 )kPa kN m m

18.9kPa

specific weight of liquidcgF h A

specific weight of liquiddepth of the center of gravitycgh

depth of the center of gravity

Areacgh

A

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3m

Example

Total force (FBC) on the bottom of the tankof the tank

Total weight (W) of the waterTotal weight (W) of the waterExplain the difference

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3.0m

Example

( )F A h A( )BCF pA h A

9.8 (6 ) (7 3 )kN3 (6 ) (7 3 )m m m

m

1235BCF kN 1235BCF kN

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3m

Example

( )W Volume

9 8kN 33

9.8 [(7 2 3 (4 0.1 )]kNW m m m m mm

k416W kN

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3m

Example

The Force on the bottom of the tank is:The Force on the bottom of the tank is:

1235 BCF kN

B h l i h f h i l

N

But the total weight of the water is only:

416W kN416W kN

What is the source of the additional force?

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3m

Example

( )ADF pA h A

29.8 ( ) ( )kN 23

9.8 (4 ) (7 3 0.1 )kN m m m mm

819ADF kN 819ADF kN

20.1A m4.0m

2 02.0mWater

7.0m

Tank Width 3m

Example

1235 416 819d th f

kN kN Kn and therefore:

BC ADF W F BC ADF W F

The large forces that can be developed with a The large forces that can be developed with a small amount of liquid (the liquid in the tube) acting over a much larger surfaceacting over a much larger surface.

Hydraulic lift Hydraulic lift

A stone weighs 90N in air When immersed inA stone weighs 90N in air. When immersed in water it weighs 50N. Compute the volume of the stone and its specific gravitythe stone and its specific gravity.

0Y 50T N

W t0BW T F

F W T

90W N

Water

90 50B

B

F W TF N N

90W N

90 50 40BF N N N BFBF

buoyant force 40BF N

3

y

40 9.8 kNN v 3

3

40 9.8

40

N vmN 3

3

40 0.0041 4.1 9800 /

Nv m litersN m

weight of the stonespecific gravity = weight of an equal volume of waterweight of an equal volume of water

90specific gravity = 2.2540

NN

An object that is 0 2m wide by 0 2m thick by An object that is 0.2m wide by 0.2m thick by 0.4m wide is found to weigh 50N in water at a depth of 0 6m What is its weight in air anda depth of 0.6m. What is its weight in air and what is its specific gravity.

50T N50 T N

0Y 00B

YW T F

50B

B

W T FW N F

W50 BW N F

BF

buoyant force weight of displaced liquidBF

3

9.8 (0.2 0.2 0.4 ) 0.157 157BkNF m m m kN N

m

50 50 157 207B

m

W N F N N N 50 50 157 207BW N F N N N

weight of the stonespecific gravity = weight of an equal volume of water207specific gravity = 1.31157

NN

157N

A hydrometer is an A hydrometer is an instrument used to measure the specific gravity of liquids.

Remember, specific gravity is the ratio of the density of thethe density of the liquid to the density of water.water.

A hydrometer weighs 0 0216N and has a stemA hydrometer weighs 0.0216N and has a stem at the upper end that is cylindrical and 2 8mm in diameter How much deeper will it2.8mm in diameter. How much deeper will it float in oil with sp gr 0.780 than in alchohol of sp gr 0.821?of sp gr 0.821?

h

alcohol0 821

oil0 780sp gr 0.821 sp gr 0.780

1For position 1, in alcohol, compute weight of hydrometer = weight of displaced liquid

v

1

weight of hydrometer weight of displaced liquid9.80 0216 0 821 kNN v 13

3

0.0216 0.821

0 0216

N vm

N

31 3

0.0216 0.000002680.821 9800 /

Nv mN m

6 31 2.68 10 v m

For position 2 in oilFor position 2, in oilweight of hydrometer = weight of displaced liquid

13

9.80.0216 0.780 ( )kNN v Ah 3

3 3

( )

0.0216( ) 0 00000283

mNAh 3 3

1

6 3

( ) 0.000002830.780 9800 /

( ) 2 83 10

v Ah m mN

Ah

6 31( ) 2.83 10 v Ah m

6 3( )h 6 31

6 3 6 3

( ) 2.83 10 2.83 10 2.68 10

v Ah mm mh

hA

2 6 2

6 3 6 3

(0.0028 ) 6.16 10 4

A m m

6 3 6 3

6 2

2.83 10 2.68 10 0.024 246.16 10

m mh m mmm

A rectangular tank 6 4m long by 2 0m deep A rectangular tank 6.4m long by 2.0m deep by 2.5m wide contains 1.0m of water. If horizontal acceleration is 2 45m/s2 then:horizontal acceleration is 2.45m/s , then:◦ Compute the total force due to water acting on each

end of the tank◦ Show that the difference between these two forces

is equal to the force needed to accelerate the mass.

2

1.0m

22.45m / s

6.4m

T k i 2 0 dTank is 2.0 deepm

2

2

( , / )tan a acceleration of vessel m s 2

2

( , / )2.45 / 0 25

g acceleration of gravity m sm s 2

. 5 /tan 0.259.8 /

tan 0 25

m sm s

tan 0.25

tan

0 2 /slope

2

0.25 /m m

1.0m

22.45m / s

6.4m

T k i 2 0 dTank is 2.0 deepm

3.2 tan 3.2 0.25 0.8y m m m 1.0 0.8 0.2CD

yd m m m

1.0 0.8 1.8ABd m m m

2

1.0m

22.45m / s

6.4m

T k i 2 0 dTank is 2.0 deepm

3

9.8 1.8 (1.8 2.5 ) 39.72

AB cgkN mF h A m m kN

m

3

29.8 0.2 (0.2 2.5 ) 0.5

2CD cg

mkN mF h A m m kN 3 ( )

2m

2

1.0m

22.45m / s

39 7ABF kN 0 5CDF kN39.7ABF kN 0.5CDF kN

6.4m

T k i 2 0 dTank is 2.0 deepm

Force needed to accelerate =Force needed to accelerate = mass of water X acceleration =

3

2 2

6.4 2.5 1.0 9.8 / 2.45 39.2m m m kN m m kN 2 2 39.2

9.8 /:

kNm s s

check39.7 0.5 39.2AB CDF F kN kN kN

Force needed to accelerate =Force needed to accelerate = mass of water X acceleration =

36.4 2.5 1.0 1000 / 16,000mass m m m kg m kg

2 2

2.45 39,20016,000 m kg mforce kgs s

s s

2 2

2.45 39,20016,000 m kg mforce kg

2 2,f gs s

kg m2

kg mnewtons

39.2force kN

1.0m

22.45m / s

39 7ABF kN 0 5CDF kN39 2force kN 39.7ABF kN 0.5CDF kN39.2force kN

6.4m

T k i 2 0 dTank is 2.0 deepm

:check39.7 0.5 39.2AB CDF F kN kN kN

A similar tank filled with water and A similar tank filled with water and accelerated at 1.5m/s2.◦ Compute how many liters of water are spilled◦ Compute how many liters of water are spilled.

drop in watersurface

2.0m

7.0m

2( / )a acceleration of vessel m s2

( , / )tan( , / )

a acceleration of vessel m sg acceleration of gravity m s

2

2

1.5 /tan 0.1539 8 /

m sm s

9.8 /

tan 0.153 slope of water surfacem s

drop in surface = 7 0.153 1.07m

1.07m

2.0m

7.0m

37.0 1.072.5 ( ) 9.362

m mVolume m m

29360Volume liters

A 1 5m cubic tank is filled with oil with sp gr A 1.5m cubic tank is filled with oil with sp gr 0.752.◦ Find the force acting on the side of the tank with an◦ Find the force acting on the side of the tank with an

acceleration of 4.9m/s2 up and with 4.9m/s2 down.

for acceleration up:for acceleration up:

(1 )Bap h

23 4.9 /(0.752 9.8 / )(1.5 )(1 )B

gm sp kN m m 2

2

(0.752 9.8 / )(1.5 )(1 )9.8 /

16.58 / 16.58

B

B

p kN m mm s

p kN m kPa

21area of loading diagram ( 16.58 / 1.5 )(1.5 )2

ABF kN m m m

18.65ABF kN

for acceleration up:2

32

for acceleration up:4.9 /(0.752 9.8 / )(0.75 )(1 )(1.5 1.5 )9 8 /

ABm sF kN m m m m 29.8 /

18.65AB

m sF kN

for acceleration down:2

32

for acceleration down:4.9 /(0.752 9.8 / )(0.75 )(1 )(1.5 1.5 )9 8 /

ABm sF kN m m m m 29.8 /

6.22AB

m sF kN

When 0 03m3/s flows through a 300mm pipe When 0.03m3/s flows through a 300mm pipe that reduces to 150mm, calculate the average velocities in the two pipesvelocities in the two pipes.

300 300 150 150Q A V A V 3

3002

0.03 / 0.42 /Q m sV m sA

2300

3

0.3004

A m

3

1502150

0.03 / 1.70 /0 150

Q m sV m sA

150 0.150

4m

If the velocity in a 300mm pipe is 0.50m/s, h h l d fwhat is the velocity on a 75mm-dia jet from a

nozzle attached to the pipe?

300 300 75 75

2 2

Q A V A V

2 2300 75

2 2

0.300 0.0754 4

m V m V

2 275

2

0.300 0.50 / 0.075

0 300 0 50 /

m m s m V

75 2

0.300 0.50 /8.00 /

0.075

m m sV m s

m

Oil of sp gr 0 75 is flowing through a 150mm Oil of sp gr 0.75 is flowing through a 150mm pipe under a pressure of 103kPa. If the total energy relative to a datum plane 2 4m belowenergy relative to a datum plane 2.4m below the center of the pipe is 17.9m, determine the flow of oil.the flow of oil.

103p kPap

2.40z m 150pipe dia mm

2.40z m0.750specific gravity

E PE KE FE 2

E PE KE FEV pE z

2

2103

E zg

V kP

2

2 3

10317.9 2.402 9.8 / 0.750 9.8 /

V kPam mm s kN m

2 217.9 2.40 14.0 19.6 /5 4 /

V m m m m sV

5.4 /V m s

103p kPap

5 4 /V m s5.4 /V m s

2.40z m 150pipe dia mm

2.40z m0.750specific gravity

2(0 150 )Q AV

m

2(0.150 ) 0.0184

mA m

2

5.4 /0 018 5 4 /

V m sQ m m s

3

0.018 5.4 /0.097 /

Q m m sQ m s

In the following figure water flows from A to In the following figure, water flows from A to B at the rate of 0.40m3/s and the pressure head at A is 6 7m Considering no loss inhead at A is 6.7m. Considering no loss in energy from A to B, find the pressure head at B. Draw the energy line.B. Draw the energy line.

2

????2

BVg2

????2

AVg

????Bp

2g

p

600Dia

mm

6.70Apm

Dia

600mm

8.00Bz m

3 00300Dia

mm3.00Az m

30.40 /Q m s

Use the Bernoulli theorem from A to B: Use the Bernoulli theorem, from A to B:

energy at + energy energy energy = energy atenergy at + energy – energy – energy = energy atsection 1 added lost extracted section 2

2 2 2 2

2 2A A B B

A A L E Bp V p V

z H H H zg g

2 2 0A

g gH

00

LH 0EH

2

????2

BVg2

????2

AVg

????Bp

2g

p

600Dia

mm

6.70Apm

Dia

600mm

8.00Bz m

3 00300Dia

mm3.00Az m

30.40 /Q m s

2 2A A B Bp V p V

z z

3

2 2A Bz zg g

3

2

0.40 / 5.66 /( 300 ) / 4A

Q m sV m sA m

3

(.300 ) / 40.40 / 1 41 /

AA mQ m sV m s

2 1.41 /(.600 ) / 4B

B

V m sA m

2

????2

BVg2

????2

AVg

????Bp

2g

p

600Dia

mm

6.70Apm

Dia

600mm

8.00Bz m

3 00300Dia

mm3.00Az m

30.40 /Q m s

2 25.66 1 41m mp

5.66 1.416.7 3.0 8.0

2 2B

m mpm m m

g g

6.7 1.6 3.0 0.1 8.0Bpm m m m m

Bp

11.3 8.1Bpm m

3.2 Bpm water

2

0.12

BVm

g2

1.62

AVm

g

3.2Bpm

2g

p

600Dia

mm

6.7Apm

Dia

600mm

8.0Bz m

3 0300Dia

mm3.0Az m

30.40 /Q m s

The energy line The energy line The hydraulic grade line

2

0.12

BVm

g2

1.62

AVm

gKE

KE3.2Bp

m

2g

p

FE

600Dia

mm

6.7Apm

FE

Dia

600mm

8.0Bz m

3 0PEPE

300Dia

mm3.0Az mPE

30.40 /Q m s

A pipe carrying oil of sp gr 0 877 changes in A pipe carrying oil of sp gr 0.877 changes in size from 150mm at section E to 450mm at section R Section E is 3 66m lower than Rsection R. Section E is 3.66m lower than R and the pressures are 91.0kPa and 60.3kPa, respectively. If the discharge is 0.146m3/s,respectively. If the discharge is 0.146m /s, determine the lost head and the direction of flow.flow.

Draw a diagram to illustrate the problem Draw a diagram to illustrate the problem

Calculate average velocity at each section:g yQ AV

Q

QVA

3

150 2

0.0146 / 8.26 /(0 150 ) / 4

m sV m s 2

3

(0.150 ) / 40.0146 / 0 92 /

mm sV

450 2 0.92 /(0.450 ) / 4

V m sm

Using lower section, E, as datum:2

150

Using lower section, E, as datum:

E VpE z

2 2

2

91 0 / (8 26 / )

E EE zg

kN

2 2

3 2

91.0 / (8.26 / ) 00.877 9.8 / 2 9.8 /E

kN m m sEkN m m s

14.1EE m

2 2150

2R

R RVp

E zg

2 260.3 / (0.92 / ) 3 66

g

kN m m sE

3 2 3.660.877 9.8 / 2 9.8 /

10 7

REkN m m s

E

10.7RE m

14 1E 14.110.7

E

R

E mE m

Flow will occur from E to R because

R

the energy head at E is greater

14 1 10 7 3 4L H d 14.1 10.7 3.4Lost Head m m m

A 0 15m pipe 180m long carries water from AA 0.15m pipe 180m long carries water from A at elevation 24.0m to B at elevation 36.0m. The frictional stress between the liquid andThe frictional stress between the liquid and the pipe walls is 0.26N/m2. Determine the lost head and the pressure change.lost head and the pressure change.

First draw a sketch of the problemFirst, draw a sketch of the problem

2 2

Use Bernoulli's Theorum:

p V p V 2 2

A A B BA L B

p V p Vz H z

g g

First, calculate loss due to friction LHL

An equation for loss due to friction HL

LH L

LHR

h

shear stressL length g

areaR Hydraulic Radiusd

ywetted perimeter

For a round pipe flowing full:2

For a round pipe flowing full:/ 4A d dR

4

:

RP d

So

4L

L LHR d

24 0.26 / 180 14 7

L R dN m mH

3

4 0.26 / 180 14.79800 / 0.15L

N m mH mN m m

2 2

14 7A A B Bp V p V 14.7

2 2Velocity V is the same at both ends of the pipe

A A B BA B

p pz m z

g g

A B

Velocity V is the same at both ends of the pipeand z =24m, z =36m, so:

14.7 12A Bp pm m

26.7A Bp pm

23

9.8 26.7 262 / 262A BkNp p m kN m kPa

m

A 1m diameter new cast iron pipe (C=130) is A 1m diameter new cast iron pipe (C=130) is 845m long and has a head loss of 1.11m. Find the discharge capacity of the pipeFind the discharge capacity of the pipe according to the Hazen-Williams formula.

0.63 0.540.8492V CR S130C

1 4 4d mR hydraulic radius 4 4

1.11head loss mS hydraulic grade line 845

S hydraulic grade linelength

0.63 0.54

0.63 0.54

0.8492

1 1 11

V CR S

0.63 0.541 1.110.8492(130) 1.281 /

4 845m mV m s

2

31

1 281 / 1 01 /m

Q AV m s m s

1.281 / 1.01 /4

Q AV m s m s

Solve Problem 8 28 using the Manning formulaSolve Problem 8.28 using the Manning formula.2/3 1/2R SV

0 012

Vn

n

0.0121

nd mR hydraulic radius

4 4

1.11

R hydraulic radius

head loss mS hydraulic grade line 845

S hydraulic grade linelength

2/3 1/2R SV

2/3 1/21 1 11

Vn

m m

1 1.114 845 1.199 /

0 012

m m

V m s

2

0.0121m

31.199 / 0.94 /

4Q AV m s m s

m/s m3/sHazen‐Williams 1.281 1.01Manning 1.199 0.94

A 0 9m diameter concrete pipe (C=120) is A 0.9m diameter concrete pipe (C=120) is 1220m long and has a head loss of 3.9m. Find the discharge capacity of the pipe usingFind the discharge capacity of the pipe using Hazen-Williams.

0.63 0.540.8492V CR S120C

0.9 4 4d mR hydraulic radius 4 4

3.9head loss mS hydraulic grade line 1220

S hydraulic grade linelength

0.63 0.54

0.63 0.54

0.8492

0 9 3 9

V CR S

0.63 0.540.9 3.90.8492(120) 1.789 /

4 1220m mV m s

2

30.9

1 789 / 1 14 /m

Q AV m s m s

1.789 / 1.14 /4

Q AV m s m s

Solve Problem 8 32 using the Manning formulaSolve Problem 8.32 using the Manning formula.2/3 1/2R SV

0 013

Vn

n

0.0130.9

nd mR hydraulic radius

4 4

3.9

R hydraulic radius

head loss mS hydraulic grade line 1220

S hydraulic grade linelength

2/3 1/2S2/3 1/2R SVn

2/3 1/20.9 3.94 1220m m

2

4 1220 1.609 /0.013

V m s

2

30.9

1.609 / 1.02 /4

mQ AV m s m s

4

m/s m3/sm/s m /sHazen‐Williams 1.789 1.14M i 1 609 1 02Manning 1.609 1.02

What size square concrete conduit is needed What size square concrete conduit is needed to carry 4.0m3/s of water a distance of 45m with a head loss of 1 8m? Use Hazen-with a head loss of 1.8m? Use HazenWilliams.

find square dimension a0.63 0.54

3

0.84924 0 /

V CR SQ

3

2

4.0 /Q m sVA a

2

120 (for concrete from Problem 8.32)Ca aR

4 41 8

Ra

head loss m

1.8 45

head loss mS hydraulic grade linelength m

0.63 0.54

0.63 0.54

0.8492

4 0 1 8

V CR S

a

2

4.0 1.80.8492 1204 45a

a

0.63 0.542.63

0.54

4.0 4 450.75

0 8492 120 1 8a m

0.8492 120 1.8Specify 0.80m by 0.80m conduit

Water is flowing in a 500mm diameter new Water is flowing in a 500mm diameter new cast iron pipe (C=130) at a velocity of 2 0m/s Find the pipe friction loss per 100m2.0m/s. Find the pipe friction loss per 100m of pipe. Use Hazen-Williams.

0 63 0 540.63 0.540.84922 0 /

V CR SV m s 2.0 /

130V m sC

0.50 0.1254

mR m 4

0.63 0.540.8492V CR S

0.63 0.54

0.8492

2.0 / 0.8492 130 0.125

V CR S

m s m S 0.54

0 63

2.0 /m sS 0.630.8492 130 0.125

0 0067 /m

S m m 0.0067 /Total head loss for 100m pipe: S m m

p p0.0067 / 100 0.67m m m m

To express the loss as pressure:9 8kN

3

9.8 0.67kNp h mm

2

6.6 6.6kNp kPam

m

For a lost head of 5 0m/1000m and C=100 For a lost head of 5.0m/1000m and C=100 for all pipes, how many 20cm pipes are equivalent to one 40cm pipe? How may areequivalent to one 40cm pipe? How may are equivalent to a 60cm pipe?

Out of curiousity, let's compare the cross-sectional areas:2

2020A = 1004

2

4040A = 4004

2

6060A = 12004

If area was all that mattered:it would take 20cm pipes to equal 1-40cm pipe, and4 it would take 20cm pipes to equal 1-60c12 m pipe

But we must consider head loss But we must consider head loss. Let’s use two equations:

Q AVQand

0.63 0.540.8492V CR S

Q AV2

4dA

0.63 0.540.8492V CR SdR

4

5 0 005 /

R

mS m m

0.005 /1000100

S m mm

C 0.63

2 0.540.8492 100 0.0054dd

Q

4

Q

0.632 0 540 20 2 0.54

320

0.200.20 0.8492 100 0.0054 0.023 /

4Q m s

0.63

4

0 40 2 0.54

340

0.400.40 0.8492 100 0.0054 0.143 /

4Q m s

0.63

4

0 60 2 0.54

360

0.600.60 0.8492 100 0.0054 0.416 /

4Q m s

4

340

320

0.143 / 6.2 20cm pipes equivalent to a 40cm pipe0.023 /

Q m sQ m s

360 0.416 / 18 1 20 i i l t t 60 i

Q m s603

20

18.1 20cm pipes equivalent to 60cm pipe0.023 /

QQ m s

A 300mm pipe that is 225m long and a 500m A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series Find the diameter of a 625m longseries. Find the diameter of a 625m long equivalent pipe. Assume all pipes are concrete.concrete.

2d 20.63 0.54; ; 0.8492

4dQ AV A V CR S

; 120 for concrete4dR C

31

4

; assume 0.1 /h

S Q m s ; QL

0.540.632 10.8492 120

4hddL

40.14

L

for the 300mm dia 225m long pipe:0.540.63

2 1

for the 300mm dia, 225m long pipe:

0.300 30 0 8492 120h

2 10.300.30 0.8492 120

4 2250.1

0.544

h 10.1 1.4087225h

1 1.678h m

for the 500mm dia, 400m long pipe:0.540.63

2 2

for the 500mm dia, 400m long pipe:

0.500 50 0 8492 120h

20.50 0.8492 120

4 4000.14

0.542

4h 20.1 5.3985

400h

2 0.248h m

total head loss 1.678 0.248 1.926m

Th fTherefore:for a 625m long equivalent pipe:

0.63 0.542

for a 625m long equivalent pipe:

1.9260 8492 120 dd 0.8492 1204 6250.1

4

d 4

360d mm

A 300mm pipe that is 225m long and a 500m A 300mm pipe that is 225m long and a 500m pipe that is 400m long are connected in series Find the diameter of a 625m longseries. Find the diameter of a 625m long equivalent pipe. Assume all pipes are concrete.concrete.

Use Diagram B-3 to solve Use Diagram B-3 to solve.

3Assume a flow rate, 0.1 /Q m sAssume a flow rate, 0.1 /From Diagram B-3,

Q m s

1

2

for 300mm pipe, 0.0074 /for 500mm pipe, 0.00064 /

h m mh m m2p p ,

total head loss =

1

0.0074 / 225 0.00064 / 400 1.921for a 625m long pipe, 1.921 / 625 0.00

m m m m m m mh m m

307m1g p p ,

from diagram B-3, 360d m

Water flows at a rate of 0 05 m3/s from Water flows at a rate of 0.05 m3/s from reservoir A to reservoir B through three concrete pipes connected in series as shownconcrete pipes connected in series, as shown on the following slide. Find the difference in water surface elevations in the reservoirs.water surface elevations in the reservoirs. Neglect all minor losses.

Use Diagram B-3 Use Diagram B-3

30 05 /Q m s 0.05 /from Diagram B-3:Q m s

1

gfor the 400mm pipe, 0.00051 /h m m

1for the 300mm pipe, 0.0020 /h m m

1for the 200mm pipe, 0.015 /h m m

total head loss0 00051 / 26000.00051 / 2600

0 0020 / 1850m m m

m m m 0.0020 / 1850

0.015 / 970m m m

m m m

19.58m

Use Hazen-Williams Formula Use Hazen-Williams Formula2

0 63 0 540 8492dQ AV A V CR S 0.63 0.54; ; 0.84924

Q AV A V CR S

d

; 120 for concrete4dR C

3; 0.05 /hS Q m sL

L

0.63 0.542 0.8492 120

4d hd

L

40.054

L

0 540 63

for the 400mm dia, 2600m long pipe:0.540.63

2 4000.400.40 0.8492 1204 2600

h 4 26000.05

41 32h m

300 1.32h m

0 540 63

for the 300mm dia, 1850m long pipe:0.540.63

2 3000.300.30 0.8492 1204 1850

h 4 18500.05

43 82h m

300 3.82h m

0.540.63

for the 200mm dia, 970m long pipe:

0 20 h 2 2000.200.20 0.8492 1204 9700.05

h

300

0.054

14.44h m300

1.32 3.82 14.44 19.58total head m m m m

The flow in pipes AB and EF is 0 850m3/s All The flow in pipes AB and EF is 0.850m3/s. All pipes are concrete. Find the flow rate in pipes BCE and BDEpipes BCE and BDE.

Assume head loss from B to E 1.00m

11.00for pipe BCE, 0.00043 /2340

mh m m

3

2340from Diagram B-3, 0.133 /BCE

mQ m so g 3, 0. 33 /BCEQ m s

11.00for pipe BDE, 0.00031 /3200

mh m m 1

3

p p ,3200

from Diagram B-3 0 038 /m

Q m sfrom Diagram B 3, 0.038 /BDEQ m s

if head loss from B to E = 1 00m is correct thenif head loss from B to E = 1.00m is correct, thensum of the flow rates through BCE and BDE will

3

gequal 0.850m /s

but 0 133 0 038 0 171 0 850 but, 0.133 0.038 0.171 0.850

head loss of 1.00m is not correct, however,actual flow rates through BCE and BDE willbe at the same proportion Sobe at the same proportion. So,

3BCE

0.133Q = 0.850 0.661 /m s BCE

3

Q171

0.038Q 0 850 0 189 /3BDEQ = 0.850 0.189 /

171m s

C t th

1500m

Compute the flow in each 1500

120m

C

30 51 /Q

branch.

W Z

30.51 /Q m s

900120m

C

2d 20.63 0.54; ; 0.8492

4dQ AV A V CR S

; 120 for concrete; 4d hR C S

L

4 L

0.63 0.542 0.8492 120

4d hd

L

44

LQ

C t th

1500m

Compute the flow in each 1500

120m

C

30 51 /Q

branch.

W Z

30.51 /Q m s

900120m

C

h d l f 10 f W t Z

0 63 0 54

assume a head loss of 10m from W to Zthen, for the 300mm pipe:

0.63 0.54

2

3300

0.3 100.3 0.8492 1204 1500 0.09 /

4Q m s

0.63 0.54

4and, for the 400mm pipe:

0 4 10 2

400

0.4 100.4 0.8492 1204 900

4Q

30.26 /m s4

1500 , 300m mm

30.09 /Q m s1500 , 300

120m mm

C

30 51 /QW Z

30.51 /Q m s

900 , 400120m mm

C 30 26 /Q m s 3 3 30 09 / 0 26 / 0 51 /m s m s m s 0.26 /Q m s 0.09 / 0.26 / 0.51 /m s m s m s

3 3 3 33 3 3 3

3300

0.09 / 0.26 / 0.35 / 0.51 /0.09 0.51 0.13 /0 35

m s m s m s m s

Q m s

300

3400

0.350.26 0.51 0.38 /0 35

Q

Q m s 0.35

1500 , 300m mm

30.13 /Q m s1500 , 300

120m mm

C

30 51 /QW Z

30.51 /Q m s

900 , 400120m mm

C 30 38 /Q m s 3 3 30 13 / 0 38 / 0 51 /m s m s m s 0.38 /Q m s 0.13 / 0.38 / 0.51 /m s m s m s

Next let’s solve using Hardy Cross Method Next, let s solve using Hardy Cross Method

C t th

1500 , 300m mm

Compute the flow in each 1500 , 300

120m mm

C

30 51 /Q

branch using Hardy‐Cross 

W Z

30.51 /Q m s yMethod.

900 , 400120m mm

C

Many pipes connected in a complex manner Many pipes connected in a complex manner with many entry and exit points.

Analyze using the Hardy Cross Method• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.

2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.

– 3. find the sum of head losses in each loop.p– 4. remember, head loss between two joints is the

same for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.

7 dj t th d fl t f ll i d– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

For this problem we are looking at just one For this problem, we are looking at just one loop:

Q

Q

C t th

1500 , 300m mm

Compute the flow in each 1500 , 300

120m mm

C

30 51 /Q

branch using Hardy‐Cross 

W Z

30.51 /Q m s yMethod.

30 51 /Q900 , 400120m mm

C

30.51 /Q m s

Analyze using the Hardy Cross Method• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.

2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.

– 3. find the sum of head losses in each loop.p– 4. remember, head loss between two joints is the

same for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.

7 dj t th d fl t f ll i d– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

For correction use: For correction, use:

LH 01.85 ( / )

LHLH Q

3, /flow adjustment m s

LH lost head L S

0 Q assumed initial flow

For Consistency: For Consistency:◦ Clockwise, Q and LH are positive◦ Counterclockwise Q and LH are negativeCounterclockwise, Q and LH are negative◦ So, in:

LH

01.85 ( / )LH Q

◦ Sign (+ or -) is important in numerator◦ But, denominator is always positive

In the table that follows:

- :S is from the Hazen Williams formula

0.63 0.54 0.8492V CR S

l

:also use

Q VA Q VA

C t th

1500 , 300m mm

Compute the flow in each 1500 , 300

120m mm

C

30 51 /Q

branch using the Hardy Cross 

W Z

30.51 /Q m s yMethod.

900 , 400120m mm

C

See Excel Spreadsheet See Excel Spreadsheet Or

Pdf Version Pdf Version

Analyze using the Hardy Cross Method• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.

2 calculate the head loss thru each pipe using– 2. calculate the head loss thru each pipe using Hazen-Williams.

– 3. find the sum of head losses in each loop.p– 4. remember, head loss between two joints is the

same for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.

7 dj t th d fl t f ll i d– 7. adjust the assumed flow rates for all pipes and repeat the process until all corrections are zero.

30.4 /m s600m 600mBA C300 mm dia 300 mm dia

400250

mmm dia

400250

mmm dia

400250

mmm dia

600300

mmm dia

600300

mdiEF D

30.4 /m s300 mm dia 300 mm diaEF D

• Analyze using the Hardy Cross Method– 1. assume flows for each individual pipe.– 2. calculate the head loss thru each pipe using

Hazen WilliamsHazen-Williams.– 3. find the sum of head losses in each loop.– 4 remember head loss between two joints is the4. remember, head loss between two joints is the

same for each branch.– 5. sum of head losses in each loop must be zero.– 6. compute a flow rate correction.– 7. adjust the assumed flow rates for all pipes and

h il ll irepeat the process until all corrections are zero.

For this problem we are looking at only For this problem, we are looking at only loops:

Q

Q

Solve using Excel Solve using Excel .pdf Version

0.63 0.54For , use Hazen-Williams: 0.8492S V CR S

LH total head loss in pipe L S

LH

1.85 /LH Q

30.4 /m s600m 600mBA C300 mm dia 300 mm dia

30 0.200 /Q m s 3

0 0.100 /Q m s

400250

mmm dia 400

250 mmm dia 400

250 mmm dia

30 0.100 /Q m s

30 0.100 /Q m s

30 0.200 /Q m s

600300

mmm dia 600

300mmm diaEF D

0Q

30.4 /m s300 mm diaEF D3

0 0.300 /Q m s3

0 0.200 /Q m s

30.4 /m s600m 600mBA C300 mm dia 300 mm dia

3

30

0 0.2410.200 /

/Q sQ s

mm

3

030.15 /

.100 /9

0Q m sQ m s

400250

mmm dia 400

250 mmm dia 400

250 mmm dia

3

30

0 0820.100 /

/QQ sm

30

30 16 /.200 /

00Q m s

Q m s 3

0 .100 /0Q m s

600300

mmm dia 600

300mmm diaEF D

30 0.082 /Q sm0.16 /0Q m s 0

30.15 /9QQ m s

30.4 /m s300 mm diaEF D3

030.16 /

.200 /0

0Q m sQ m s

30

30.24 /.300 /

10Q m s

Q m s

Water flows in a rectangular concrete open Water flows in a rectangular concrete open channel that is 12.0m wide at a depth of 2 5m The channel slope is 0 028 Find the2.5m. The channel slope is 0.028. Find the water velocity and the flow rate.

using the Manning equation:2/3 1/2

g g qRV= S

2.5 12.0 1 765

nareaR m

2/3 1/2

1.765 2.5 12.0 2.5

1 765 0 0028

R mwetted perimeter

3

1.765 0.0028V= 5.945 /0.013

m s

32.5 12.0 5.945 178 /Q AV m s

Water is to flow at a rate of 30

3m3/s in the concrete channel shown on the following slide. What slope of channel will be required?

2/3 1/2RQ SV

21.6 3.6

VA n

m m

2

2

1.6 3.6(3.6 2.0 ) 2.0 12.402

m mA m m m m

2

2 2

12.40 1.2363.6 1.6 2.0 2.0 2.0

mR mm m m m m

3 2/3 1/2

2

30 / 1.236 0.000746 /0 01312 40

m s m S m mm

0.01312.40

or, 0.746 /m

m kilometer

On what slope should a 600mm concrete On what slope should a 600mm concrete sewer pipe be laid in order that 0.17m3/s will flow when the sewer is one-half full? Whatflow when the sewer is one half full? What slope if the sewer flows full? (use n=0.013)

2V 2

2/3: =RVnuse S

R

21 d 2 4 0.6 0 15d mR m

1/2 0.15

1 4 4R m

d

2

0 6

d

d

0.6 0.154 4Fulld mR m 4 4

0 17Q1/2 2

0.17 1.206 /1 0.6

QV m sA

2 4

2 2

1/2 2/3 2/3

1.206 0.013 0.0031 /VnS m m

1/2 2/3 2/30.15R

0 17Q2

0.17 0.601 /0.6Full

QV m sA

42 2

2/3 2/3

0.601 0.013 0.00077 /0 15Full

VnS m mR

2/3 2/30.15Full R

A 600mm diameter concrete pipe on a 1/400 A 600mm diameter concrete pipe on a 1/400 slope carries water at a depth of 240mm. Find the flow rate QFind the flow rate, Q.

240depth mm600

1 / 400

pdiameter mmslope

p

240depth mm600

1 / 400

pdiameter mmslope

p

300r mm

240300 240 60depth mm

mm mm mm

300r mm 60mm0 060 1 0.060cos 1.3690.30

0 30 sin(1 369 ) 0 294

rad

rad m

0.294m

0.30 sin(1.369 ) 0.294rad m

Area of triangles g1 0.060 (0.294 2) 0.01762

f i l ti300r mm

60mm2

area of circle section1.369 2 0.60 0.123

2 4

2 4

0.123 0.018 0.105A

20 105A m

0.294m 0.294m

0.123 0.018 0.105A 0.105A m

tt d i twetted perimeter1.369 2 0.60

2

2

1.369 0.60 0.821

300r mm60mm

20.105A m0.105R 0.1280.821

0.128R

2/3 1/2RV= S

1/22/3 10.128

400

n 400V=

0.0130.977V

300r mm60mm

0.977V

Q AV 20.105A m

30.105 0.977 0.10 /Q

m s 0.128R

After a flood had passed an observation After a flood had passed an observation station on a river, an engineer visited the site.

By locating flood marks performing By locating flood marks, performing appropriate survey and doing necessary computations she determined that at thecomputations, she determined that, at the time of peak flooding :◦ The cross-sectional area was 2960m2The cross sectional area was 2960m◦ The wetted perimeter was 341m◦ The water surface slope was 0.00076m/m p /

The engineer also noted that the channel The engineer also noted that the channel bottom was “earth with grass and weeds,” for which a handbook gave a Manning n value ofwhich a handbook gave a Manning n value of 0.030.

Estimate the peak flood discharge Estimate the peak flood discharge.

2/3

1/22960(2960) 0 00076 2/3 1/2 (2960) 0.00076

3410.030

AR SQn

311,500 /Q m s

A rectangular channel 6 1m wide carries A rectangular channel 6.1m wide carries 11.3m3/s and discharges onto a 6.1m wide apron with no slope at a mean velocity ofapron with no slope at a mean velocity of 6.1m/s. What is the height of the hydraulic jump? What energy is absorbed (lost) in thejump? What energy is absorbed (lost) in the jump?

21 2

1 2y yq y y 1 2 2

y yg

311.3 /Q m s 11.3 /Q m s

1 6.1 /v m s

21 2

1 2y yq y y 1 2 2

y yg

311.3/ 1.85 /6.1

q flow unit width m s

11.85 0.3036 1

qy mV

1

22

6.10.3031.85 0 303

Vy

y 20.303

9.8 21 37

y

y m

2 1.37y m

21 2

1 2y yq y y 1 2 2

y yg

311.3 /Q m s

311.3 /1.85 /

Q m sq m s

1 6.1 /V m s

0 303y m 2 1.37y m1 0.303y m

1.37 0.303 1.07jump m m m

critical depth:32 23

critical depth:

/ 1.85 / 9.8 0.70Cy q g m Cy q g

Therefore:◦ Flow depth before the jump (0.303m) is < 0.70 and

i iti lis supercritical◦ Flow depth after the jump (1.37m) is > 0.70 and is

subcriticalsubcritical

21 2

1 2y yq y y 1 2 2

y yg

311.3 /Q m s

311.3 /1.85 /

Q m sq m s

1 6.1 /V m s

0 303y m 2 1.37y m0.70cy m1 0.303y m c

1.37 0.303 1.07jump m m m

Lost energy: Lost energy:

26 1 / 21 1 1

6.1 // 2 0.303 2.20

2 9.8m s

E y V g m m

2

22 2 2

1.85 / 1.37 // 2 1.37 1.46

sE y V g m m 2 2 2 2 9.8

2.20 1.46 0.74

y g

lost head m m m

2.20 0.303 1.90m m m 1.46 1.37 0.09m m

1 6.1 /V m s 1.90m0.09m0.74m

0 303y m 2 1.37y m0.70cy m1 0.303y m c

1.37 0.303 1.07jump m m m

Measures stagnation pressure (at B) which Measures stagnation pressure (at B), which exceeds the local static pressure (at A), to determine velocity headdetermine velocity head.

Ah Bh

Velocity (V) atVelocity (V) at Point B is zerozero.

A l th Ah

Bh

Apply the Bernoulli

tiequation, next slide

2 2V Vl 2 2

2 2A A B B

A Bp V p Vno lossz z

g assumed g

0; zB A B

g gV z

2

,so2

2A A Bp V p

g

2g

2A A Bp V p

2A A Bp p

g

2 B Ap pV g g

p p

B AB A

p ph h d

With no friction:2V gd

h h dB Ah h d

Ah Bh

A small amount of friction normally occurs, ff f l ( dso a coefficient of velocity cV (see discussion

on following slides) is sometimes used:

V

actual velocityc

2

V theoretical velocity

V d2VV c gd

1 id ffi ito assume 1 provides sufficient accuracy for most engineering problems involving Pitot tubes

vc

involving Pitot tubes.

The ratio of the actual velocity in a stream to h h l l h ldthe theoretical velocity that would occur

without friction.

actual velocity V

actual velocityctheoretical velocity

A Pitot tube having a coefficient of 0 98 is A Pitot tube having a coefficient of 0.98 is used to measure the velocity of water at the center of a pipe The stagnation pressurecenter of a pipe. The stagnation pressure head is 5.67m and the static pressure head in the pipe is 4.73m. What is the velocity?the pipe is 4.73m. What is the velocity?

4.74m5.67m

B Ah h d 5.67 4.74 0.94

B Ah h dm m m

4.74m 5.67m

2V d25 67 4 73 0 94

VV c gdd m m m

5.67 4.73 0.940.98V

d m m mc

29.8 /V

g m s20.98 2 9.8 / 0.94 4.21 /V m s m m s

A 100mm A 100mm diameter standard orificestandard orifice discharges water under a 6.1munder a 6.1m head. What is the flow?flow?

2area; 0 6

Q cA gHA c

area; 0.6total head causing flow

A cH

22

g0.10 6 2 9 8 / 6 1mQ m s m

3

0.6 2 9.8 / 6.14

0 05 /

Q m s m

Q

30.05 /Q m s

The tank in problem 12 9 is closed ant the air The tank in problem 12.9 is closed ant the air space above the water is under pressure, causing to flow to increase to 0 075m3/scausing to flow to increase to 0.075m3/s. Find the pressure in the air space.

2

2

0 1

Q cA gH

m

3 20.10.075 / 0.6 2 9.8 /

4m

m s m s H

12.912.9 6.1 6.8P Z

H mh H h m m m

3 2

12.9 6.1 6.8

9.8 / 6.8 70 / 70P Z

P

h H h m m m

p h kN m m kN m kPa

During a test on a 2 4m suppressed weir that During a test on a 2.4m suppressed weir that was 0.9m high, the head was maintained constant at 0 3m In 38 seconds 29m3 ofconstant at 0.3m. In 38 seconds, 29m of water were collected. Find the weir factor m using equations A and B.using equations A and B.

0 3H m0.3H m

0 9Z 0.9Z m

329m 329 0.763 /38

0 9 0 3 1 2

mQ m ss

fl d h

3

0.9 0.3 1.20.763 / 0 265 /

flow depth m m mQ m sV

0.265 /2.4 1.2

QV m sA m m

3/2 3/22 2

using Eq. A:

V VQ b H

3/2

2 2Q b H

gm

g

3/2 3/22 23

2 2

0.265 0.2650.763 2.4 0.32 9.8 2 9.8

mm

3/2 3/230.763 2.4 0.3 0.00358 0.00358mm

1.90m

i E B3/2

using Eq. B:Q bHm

3/20.763 2.4 0.3Q

mbHm

1.93m

1.90 1.93 Equation B is OK for weirsl d hi h

placed high

During a test on a 2 4m suppressed weir that During a test on a 2.4m suppressed weir that was 0.0m high, the head was maintained constant at 0 3m In 38 seconds 29m3 ofconstant at 0.3m. In 38 seconds, 29m of water were collected. Find the weir factor m using equations A and B.using equations A and B.

0 3H m0.3H m

0 0Z 0.0Z m

b Hb

Z=0

From: http://www.lmnoeng.com/Weirs/cipoletti.htm

329m 329 0.763 /38

0 3 0 30 0

mQ m ss

fl d h

3

0.3 0.30.763 / 1 06 /

0.0flow depth m mQ m sV

m

1.06 /2.4 0.3

QV m sA m m

3/2 3/22 2

using Eq. A:

V VQ b H

3/2

2 2Q b H

g gm

3/2 3/22 23

2 2

1.06 1.060.763 2.4 0.32 9.8 2 9.8

m m

3/2 3/230.763 2.4 0.3 0.0573 0.0573mm

1.53m

using Eq. B:3/2

3/2

Q bHm3/20.763 2.4 0.3

1 93m

m

1.93m

1.53 1.93, Equation A must be used , q for shallow weirs