chapter 5(color)

7/14/2019 Chapter 5(Color)

1/56

CHAPTER 5:

PIPING SYSTEM


2/56

EL AND HGL

Energy line (EL) shows totalhead for a certain crosssection in system.

p V2

EL 1 1+ +z 2 g 1

Hydraulic Grade Line (HGL)shows piezometric headfora certain cross section in

system.p1

HGL +z

1

3


3/56

EL AND HGL# Example 1

EL & HG L

V22/2g

P2/g

z1 z2

Water flows from tank to atmosphere4


4/56

EL AND HGL# Example 2EL & HGL

EL

HGL

z

V22/2g

hturbine

turbine

drop immediately in total energy head

5


5/56

EL AND HGL # Example 3

EL

HGLV22/2g

Velocity

increased when

there is a

contraction in

diameter of pipe

Fluid flows through a nozzle

6


6/56

EL AND HGL# Example 4

Water flows through 2 types of diameter pipes7


7/56

EL AND HGL# Example 5EL

Energy equationHGL

Apam

zA

B

turbin

zB

from A to B:

DatumEnergyhead

2 2 lossesp V p VA A+ +z +( h Bh )= B+ +z +h

2 gA pam turbin B L

2 g

h hf hmmachine

8


8/56

EL AND HGL Conclusion

EL always higher or same level with HGL. The difference between 2 lines is the value ofkinetic energy head, V2/2g.

Major losses - cause EL and HGL to decreasegradually.

Minor losses - cause EL and HGL to decreaserapidly.

Use of pump - Lines increased immediately Use of turbine - Lines decrease immediately

9


9/56

FLOW TO ATMOSPHEREEL and HGL

Or z

A

V22/2gEL

HGL Hs= zA- zB

to atm

B

Energy equation from A to B:

p

AV

+2

2

A + zg A

2p VB B= + +

2gz +h +

B fh

m

10


10/56

FLOW TO ATMOSPHERE

p V2 2p V

A

+ A +z

2gA

B B= + +z +hB L

2g

zA

z =B

2

V

2

2

B + h + hf mg

2 2

hS

V=

2B +g

fLV 0.5V+

2gD 2g

hS=2VB

2g

4fL1+ + 0.5

D

11


11/56

FLOW TO ATMOSPHERE

V 2 fL 2 ghBh = 1+ s+ 0.5 V =S

2g DB

Let assume

as K

1.5+fLD

2ghQ =AV =A

k

s V =B2ghs

k

12


12/56

EXAMPLE 5.1

Determine the discharge of water flows in the system.

A

Hs10 m

f = 0.005ke atm

L=100 m, d = 0.5 cm B

13


13/56

EXAMPLE 5.1: SOLUTION

Taken from previous slide2V =

gh

ks

In test, the formula must be derivedfrom Bernoulli Equation

hS = zA+ zB= 10 cm , f = 0.005 , L = 100 m , D = 0.5 cmk = 1.5 + fL/D = 1.5 +(0.005x100)/0.005 = 101.5

V=2(9.81)(10)

101.5

=1.39m/s

5Q =AV =4

(0.005)2x1.39= 2.73x10 m3 /s

14


14/56

FLOW WHICH CONNECTING 2

RESERVOIRSA

Hs

Energy equation

p 0 V

0 02

p V

B

02

from A to B:

Head loss atA A+ +z B B= + + z +h +h entrance and

2gA

2gB m f

submergeddischarge

15


15/56

FLOW WHICH CONNECTING 2

RESERVOIRS

zA

z =B

0.5V2 g

2

2 2

fLV+ +2 gD

2

V2g

2ghV B fL sV =hS

= 0.5 +2g

+1D

Let assumeas K

B

1.5+fLD

Q =AV =A 2gh s Vk

2ghB =k

s

16


16/56

EXAMPLE 5.2

Determine the difference level between 2 reservoirswhich is connected by single pipe of cast iron for 1km. Discharge and diameter is given as 0.01 m3/sand 5 cm respectively. ( =1.14 x 10-6m2/s)

17


17/56


V= QA

=

4

0.01

(0.05)

VD

=5.09m/s2

5.09x0.05 5

e

NR= =

v0.25

61.14x10

=2.23x10

=D

= = 0.00550

From Moody Chart , f =0.03

k=1.5+fL

D

0.03x1000= 1.5+

0.05=601 .5

18


18/56

EXAMPLE 5.2 :SOLUTION

2V =

gh

ks

hs =kV

2g

2

601.5(5.09)2

hs

= = 794.28m2x9.81

19


19/56

EXAMPLE 5.3 (Use Pump)

The tanks, pump and pipelines in the figure havethe characteristics noted. The suction lineentrance from the pressure tank is flush and thedischarge into the open tank is submerged. If the

pump P puts 2.0 hp into the liquid. (a) Determinethe flow rate (b) find the pressure in the pipe onthe suction side of the pump.

20


20/56

EXAMPLE 5.3 (Use Pump)

21


21/56

EXAMPLE 5.3a: SOLUTION

Qh 52QhP= p =2.0=

550

p Thus h550

p

21.2=

Q

Energy Equation, 1 to 3

p V2 2

p V6 6+ +z 8+h = + 8 +z + h + h 2g

5(144) 21.2

6 p

2V6

8

2g

50

m f

2

V62 2200 V V

8 8

52+ =10+0.5

Q 2 g

Q

+0.0256/12 2g

+0.03 8/12 2g 2g

QV=6 and V=80.1963 0.349

22


22/56

EXAMPLE 5.3a: SOLUTION

Substituting for hp, V6, V8

23.8+21.2

Q

22.48Q = 2.48Q0

3 23.8Q21.2=0

Solving this by trial and error or equation polynomialsolver

Q=3.47cfs

23


23/56

EXAMPLE 5.3b: SOLUTION

To obtain the pressure P2at the suction side of the pump

V=63.47

0.1963=17.68

Energy equation, 1 to 2

2 2 25(144) P=

17.682

+15+17.68

+0.550 17.68

+0.02552

P

2(32.2) 2(32.2)

52

6/12 2(32.2)

2

=20.6ft P2 =20.6 =7.43 psi

144


24/56

NONRIGOROUS HEADLOSS EQNS

The empirical equation which are Hazen-William,Manning and Darcy Weisbach were rearranged intothe form of head loss equation.

Darcy - Weisbach2

Hazen - William Manning

hf 8fLQ= 2 5 10.67LQh=

1.85

2 210 .29LnQ gD

8fL

f 4.87C DH

hf

=16 .3D

210 .29Lnk=

, n = 22 5gD

10.67Lk=

4.87, n = 1.85 k=

D 16.3CHD, n = 2

As we can see that hf= kQn

25


25/56

BRANCHING PIPE

Let us consider 3 pipes connected to 3 reservoirsand branching together at common junction point J.

We shall assume that all the pipesare sufficient long that we can

neglect minor losses and velocityhead so hL= hfwhich we shalldesignate as h

26


26/56

BRANCHING PIPE

As there are no pumps, the elevation of P must liebetween the surfaces of reservoirs A and C.

If P is below the surface of reservoir B then h2andQ2are both zero.

If P is above the surface of reservoir B then watermust flow into B and Q1= Q2+Q3.

If P is below the surface of reservoir B then watermust be out of B and Q1+Q2 = Q3.

There are several different method of solutions

27


27/56

EXAMPLE 5.4

The elevations of water surfaces in reservoirs A andC are 250 ft and 160 ft, respectively and thedischarge Q2into reservoir B is 3.3 cfs. Find thesurface elevation of reservoir B.

28


28/56

EXAMPLE 5.4

29


29/56


30


30/56


31

EXAMPLE 5 5


31/56

EXAMPLE 5.5

With the sizes, length and material of pipes given asexample 5.4, suppose that the surface elevation ofreservoirs A, B and C are 525 ft, 500 ft and 430 ftrespectively.

(a) Does water enter or leaves reservoir B?(b) Find the flow rates of 60oF water in each pipe.

32


32/56


33


33/56


34


34/56


35


35/56

EXAMPLE 5.6

The surface elevation of reservoirs A, B and C are160 m, 150 m and 120 m respectively.

(a) Does water enter or leaves reservoir B?(b) Find the flow rates in each pipe.

36

EXAMPLE 5 6


36/56

EXAMPLE 5.6

ZMA/S20607 37


37/56


38


38/56


39


39/56

PIPES IN SERIESIf a pipeline is made up of lengths of differentdiameters, conditions must satisfy the continuity andenergy equations namely,

If given the rate ofdischarge Q, we findthat hL= Q

Using the nonrigorous eqn,h =h =kQ

n n n+kQ +kQ +L f 1 1 2 2 3 3

40


40/56

PIPES IN SERIESBut since all the Qs are equal, this becomes

h =( k + k+k + ....) Qn n=(k)QL 1 2 3 3If substituting from Darcy eqn into hL(for series) and

including minor losses if we wish (usually L/D


41/56

EXAMPLE 5.7

Referring to figure (previous slide) and data givenfor new cast iron pipes, find the rate of flow from Ato B in conveying 15oC water. (z or hs= 10 m andneglect minor losses)

42

EXAMPLE 5 7 :SOLU ION


42/56

EXAMPLE 5.7 :SOLU ION

43

EXAMPLE 5 7 :SOLU ION


43/56

EXAMPLE 5.7 :SOLU ION(Sample Problem 8.17)

44


44/56

EXAMPLE 5.8

2 reservoirs were connected by a pipeline 1 (150mm diameter,6 m length) and pipeline 2 (255 mmdiameter and 15 m length). Inlet and outlet ofpipes are categories as a sharp and z is 6 m. Tableall losses, calculate the rate of flow and sketch EL

and HGL for the system. Take f = 0.01 for bothpipes.

45


45/56


Continuity Equation:Q = A1V1= A2V2 V1= (A2/A1)2V2 = 2.89V 2

46


46/56


Pers. Bernoulli diantara 1 dan 2,

p

1V

+2

2

1 +zg

1

2p V2 2= + +

2g

12 .94V

z +h2 L

22 V =3.02m/s6=

2g

(0.255 )

2

2

3Q =AV =2 2

x3.02 =0.154m /s4

V1= 2.89V 2 = 2.89 (3.02) = 8.73 m/s

47


47/56


EL and HGL

A

EL

HGL

d1= 0.15 m

L1= 6 m

d2 = 0.255 m

L2= 15 m

The sketches of EL and HGL

6 m

B

48


48/56

PIPES IN PARALLEL

If a pipeline is made up of lengths of differentdiameters, conditions must satisfy the continuity andenergy equations namely,

If we are given the total flow Q and want to find hL

1/n 1/nhh=kQ n Q

hf1 f2= + +

fk k

1 2

49


49/56

PIPES IN PARALLEL

But since all the hfs (= hLs) are equal, this becomes1/n 1/ n 1/n

1/n 1 1 1/ n 1Q=(h) + + =(h)

fk k

f

k 1 2

To find hLand the individual Qs, and including minor losses,

can write the head loss eqn as

hL

=L

f +kD

2V

2 g

Where k is the sum of the minor

loss coefficient

We usually neglect if the pipe is longer than 1000

diameters. Solving for V and then Q for pipe 1.

50


50/56

PIPES IN PARALLEL

2ghLQ =AV =A =C h Where C1=constant1 1 1 1f(L/D) +

1 Lk

1 1 1 1 for given pipe

We can similarly express the flows in the other pipesusing assumed values of f. Finally becomes

Q =C h +C h +C h =(C +C +C) h1 L 2 L 3 L 1 2 3 L

This enables us to find a first estimate of hLand distribution offlows and velocities in the pipes. Using these, improvements to

the f values can be made. If necessary repeat them until wefinally obtain a correct determination of hLand distribution of

flows.51


51/56

EXAMPLE 5.9

There pipes A, B and C are interconnected as in figurebelow. The pipe characteristics are as follows:

Find the rate at whichwater will flow in each

pipe. Find also the

pressure at point P. Allpipe lengths are much

greater than 1000diameters, so neglect

minor losses

52

EXAMPLE 5 9 S OLUTION


52/56

EXAMPLE 5.9: S OLUTION

53

EXAMPLE 5 9: SOLUTION


53/56

EXAMPLE 5.9: SOLUTION(Sample Problem 8.18)

54



54/56

EXAMPLE 5.9: SOLUTION(Sample Problem 8.18)

55


55/56

EXAMPLE 5.10

Pipe 2 has a length 4 times longer thanpipe 1. Both pipe has a same diameterwhich is 305 mm. If discharge in pipe 2 is1 m3/s, find discharge for pipe 1. Calculate

flow rate of system if f equal to 0.01 andneglect minor losses.

2

1

A B

56


56/56

EXAMPLE 5.10

V Q 1= =13.69m/s2=A

24 (0.305)

2

In parallel pipe:

hL1= hL2 and h f1 = h f2

2 22 2 f LV f LVV1

V1

=1x4x1x13.69=27.3m/s

1 1

2gD1 =1

2 2 2

2gD2

where L2/L1= 4

Q1=A1V1=

(0.305 )

4

2

x27.30=3

1.99m / s

Qsystem = Q1+ Q2= 1.99 + 1 = 2.99 m3/s57

chapter 5(color)

Documents