pérdidas en fluidos

8/12/2019 Prdidas en fluidos

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CHAPTER 6

FLUID MECHANICS

Fluid mechanics is important in the study and understanding of radiant heat trans-

fer for hydronic systems. Fluid mechanics describes the relationship between pres-sure drops and flow rates through the water conduits of a hydronic system. Thesesame principles can also be applied to airflow through ventilation ducts. This rela-tionship is important to determine pipe diameters, the effect of different tubingmaterials, and the size of the required pump.

The focus of this chapter is to investigate the background information that is nec-essary to predict the pressure drop, flow rate, and pumping power required for agiven piping system.To accomplish this task,we will introduce Bernoullis equation,major and minor head loss terms, and the friction factor.

6.1 BERNOULLIS EQUATION

One of the better known relationships in the field of fluid dynamics is Bernoullisequation. This equation is derived from the conservation of energy equation andrelates fluid velocity and pressure. Bernoullis equation is expressed as:

+ +gz= constant (6.1)

The conditions for using this equation are steady, incompressible flow along a

streamline without the effects of friction.After studying Eq. (6.1), the effects of fric-tion and pipe losses will be introduced.The best application of Bernoullis equation is to calculate the effect of elevation

change. For example, if water is raised from a low level to a high level in a constant-diameter pipe, Bernoullis equation relates the pressure change to the elevationchange. Applying Bernoullis equation between two points in a pipe is:

+ +gz1= + +gz2 (6.2)

This procedure is illustrated in the following example.

EXAMPLE6.1 Water flowing at the rate of 20 gpm in a 1-in-diameter pipe is raisedfrom an elevation of 0 to 30 ft. If the pressure at the lower elevation is 20 psig, calcu-late the pressure at the higher elevation.

V22

2

p2

V12

2

p1

V2

2

p

2.135

Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com)Copyright 2004 The McGraw-Hill Companies. All rights reserved.

Any use is subject to the Terms of Use as given at the website.

Source: RADIANT HEATING AND COOLING HANDBOOK


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Known:

Q=20 P1=20 psi = 999

Q= 0.045 P1= 9.266 104 = 62.366

Z1= 0 ft D= 1 in g=32.174

Z2=20 ft D= 0.083 ft A= 2

Solution:

V1= V2=

C= + +g Z1 C= 1.519 103

P2= C g Z2

Answer: P2= 11.338 psi

EXAMPLE6.2 Water flowing at the rate of 20 gpm in a 1-in pipe at 30 psig enters a0.50-in section of pipe. If the elevation change is zero, calculate the velocity and pres-

sure in the smaller-diameter pipe.

Known:

Q=20 P1=20psi = 999 g=32.174

Q= 0.045 P1= 9.266 104

= 62.366

Z1= 0 ft D1= 1.0 in D1= 0.083 ft

Z2=20 ft D2= 0.5 in D2= 0.042 ft

A1= 2

A2= 2

Solution: Continuity:

0= | V1 A1|+ | V2 A2|

V1= V2=Q

A2

Q

A1

D2

2

D1

2

lb

ft3

lb

fts2ft3

s

ft

s2kg

m3gal

min

(V2)2

2

ft2

s2(V1)

2

2

P1

Q

AQ

A

D

2

ft

s2

lb

ft3lbfts2

ft3

s

kg

m3gal

min

2.136 HEAT TRANSFER AND THERMODYNAMICS



FLUID MECHANICS


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C= + + g Z1 C= 1.519 03

P2= C g Z2

Answers: V2=32.68 P2= 4.599 psi

6.2 PIPE FLOW CALCULATIONS

Pipe flow calculations build on Bernoullis equation,but they remove the restriction

of frictionless flow.The introduced term is the head loss that takes into considerationviscous pipe losses, losses due to valves and bends in the pipe, and other effects.Bernoullis equation is modified as:

+ 1 +gz1 + 2 +gz2= hlT (6.3)The parameter hlT is the head loss term. The terms are velocity profile correc-tions. These can be assumed to be one with little loss in accuracy. The only other dif-ference between Eqs. (6.3) and (6.4) is that the velocity terms have lines over them.

These lines designate that the velocities are average velocities over the cross sectionof the pipe.The average velocity is defined as:

V= = (6.4)

6.2.1 Major Head Losses

The total head loss term is separated into two components: (1) the major-friction

loss term and (2) the minor loss term. By definition, the major head loss is the pres-sure loss through a horizontal constant-area pipe at steady flow. This condition isillustrated in Fig. 6.1. In equation form, this definition is:

= = hl (6.5)p1p2

p

V

A

m

A

V22

2

p2

V12

2

p1

fts

(V2)2

2

ft2

s2(V1)

2

2

P1

FLUID MECHANICS 2.137

A1

V1

A2

V2

FIGURE 6.1 Pressure loss in a horizontal constantarea pipe with steady flow.



FLUID MECHANICS


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Experimental studies show that the major-frictional head loss in a pipe is a func-tion of the pipe roughness, the Reynolds number (Re), and the length-to-diameterratio of the pipe. In equation form, this relationship is:

hl=f , f=fRe, (6.6)The parameter e is the pipe roughness. Figure 6.2 illustrates the roughness of variouspipe materials. For example, drawn tubing is 0.000005 ft and the relative roughnessof 10-in drawn tubing is 0.000006. The parameterfis called the friction factor and isa function of Re and the relative roughness e/D.

eD

LV2

D 2


FIGURE 6.2 Relative roughness of various pipe materials.

The friction factor has been the subject of numerous studies. Arguably the bestset of experimental data was collected and analyzed by Moody (1944). Figure 6.3demonstrates the results of his experiments in the form of a Moody chart.This chartshows the relationship between Re on thex axis, the friction factor on they axis, andthe relative roughness shown as lines on the chart.

EXAMPLE6.3 Determine the friction factor for a 2-gpm flow rate through 0.50-in-diameter drawn tubing.The water flowing through the pipe is 70F. If the pipe is hori-zontal, calculate the pressure drop along a 500-ft section of pipe. Compare this

pressure drop with the pressure drop that would occur in typical hydronic conduit.



FLUID MECHANICS


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2.139

FIGUR

E6.3

Moodyfrictionfactordia

gram.



FLUID MECHANICS


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Known:

Q=2 P1=20 psi = 997 = (968 106) s

Q= 4.456 103 P1= 9.266 104 = 62.241 = 6.505 104

Di= 0.5 in e= 0.000005 g=32.174 L=500 ft

= 0.00012 Df= ft A= 2

V1=

Solution:

Re= Re= 1.303 104

From Moody chart:

f= 0.017 h1= f

Answers:

h1= 1.089 103 =33.857 ft

P0= h1 Q P0= 0.017 hp

EXAMPLE6.4 Calculate the pressure loss in a pipe that has a 20-ft vertical rise over 100ft.The pipe flow rate is 30 gpm,the diameter is 0.50 in,and the water temperature is 90F.

Known:

Q=30 = 995 g=32.174

Q= 0.067 = 62.116

Z1= 0 ft D1= 0.5 in D1= 0.042 ft

Z2=20 ft A1= 2

Solution: Because V1= V2 and Z1= 0, Bernoullis equation reduces to:

= + g Z2

= g Z2

P= g Z2

P2

P1

P2

P1

D1

2

lbft3ft

3

s

ft

s2kg

m3gal

min

h1

g

ft2

s2

(V1)2

2

L

Df

V1 Df

Q

A

Df

2

0.5

12

1

ft

e

Di

ft

s2

lb

fts

lb

ft3lbfts2

ft3

s

N

m2kg

m3gal

min




FLUID MECHANICS


7/16

Answers:

P=3.997 104 P= 8.627 psi

6.2.2 Minor Losses

Minor losses, which can sometimes be quite large, are due to anything other thanstraight sections of pipe. Examples are valves, entrances and exits, pipe elbows,expansions and contractions, and fittings. There are two ways to calculate the effectof minor losses. The first is to introduce a Kvalue and the second is to designate anequivalent length of pipe, which is then used in Eq. (6.6). Each of these methods isrelated to the minor head loss term by:

hlm= K =f (6.7)

Both methods are frequently used and are discussed in the following paragraphs.Table 6.1 illustrates three common inlets and exits from a piping system and their

associated loss coefficients. These become important in the case of a surge tank orpressure volume in a piping system. As shown in the table, the well-roundedentrance exhibits the lowest loss coefficient, whereas the loss term at an exit is inde-pendent of the configuration. Figures 6.4 and 6.5 illustrate the effect of pipe elbowsand miter pipe bends.Table 6.2 provides several loss terms for valves and fittings.Allthese data are from the Crane Industrial Products Group Technical Paper No. 410.

6.2.3 Pipe Solution Techniques

There are four separate cases of pipe system problems that may be encountered.These cases are all functions of the pressure loss equation:

p=f(L/D, V,e/D) (6.8)

V2

2

Leq

D

V2

2

lb

fts2


TABLE 6.1 Inlet and Exit Head Loss Terms

Minor lossEntrance type coefficient, K*

Reentrant 0.78

Square-edged 0.5

Roundedr/D

0.02 0.06

0.15K 0.28 0.15 0.04

* Based on hlm= K(V2/2), where Vis the mean velocity in the pipe.



FLUID MECHANICS


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Of the four variable groups in Eq. (6.8), three need to be specified with the fourthcalculated. Each situation requires a slightly different solution procedure. Each caseis illustrated in the following examples.Each example uses water at 90F and smooth

drawn pipes.EXAMPLE6.5 Flow =20 gpm, D= 1.5 in, L= 100 ft. For the case in which the pres-

sure loss is the unknown, one first obtains the friction factor from the Moody chart


Relative radius, r/D

Dimensionlessequivalentleng

th,

Leq

/D

30r

D

20

100 5 10 15

40

FIGURE 6.4 Head loss due to pipe elbows.

Deflection angle, (degrees)

Dimensionlessequivalentlen

gth,

Leq

/D

40 D

D

D

20

00 3015 45 7560 9

60

FIGURE 6.5 Head loss due to mitered pipe bends.



FLUID MECHANICS


9/16

using Re and the relative roughness.The head loss is then calculated.The pressure lossis calculated by using Eq. (6.5).

Known:

Q=20 P1=20 psi = 995 = (760 106) s

Q= 0.045 P1= 9.266 104 = 62.116 =5.108 104

Di= 1.5 in e= 0.000005 g=32.174 L= 100 ft

= 4 105 Df= ft A= 2

V1=

Solution:

Re= Re=5.52 104

From Moody chart:

f= 0.016

h1= f L

Df

V1 Df

Q

A

Df

2

1.5

12

1

ft

e

Di

ft

s2

lb

fts

lb

ft3lb

fts2ft3

s

N

m2kg

m3gal

min

(V1)2

2


TABLE 6.2 Loss Terms for Valves and Fittings

Equivalent length,*Fitting type Le/D

Valves (fully open)Gate valve 8Globe valve 340Angle valve 150Ball valve 3Lift check valve

Globe lift 600Angle lift 55

Foot valve with strainerPoppet disk 420Hinged disk 75

Standard elbow90 3045 16

Return bend, close pattern 50Standard tee

Flow through run 20Flow through branch 60

*Based on hlm =f .V2

2Le

D



FLUID MECHANICS


10/16

h1= 84.383 =2.623 ft

P= h1

Answers:

P= 1.131 psi

EXAMPLE6.6 p= 10 psi, Q=30 gpm, D= 1 in, L= ? Calculate the total head lossfrom main equation. Getffrom the Moody chart by using Re and e/D. Calculate Lfrom head loss.

Known:

Q=30 P= 10 psi = 997 = (968 106) s

Q= 0.067 P= 4.633 104 = 62.241 = 6.505 104

Di= 1 in e= 0.000005 g=32.174

= 6 105 Df= ft A= 2

V1=

Solution:

h1= h1= 744.379 =23.136 ft

Re= Re= 9.772 104

From Moody chart:

f= 0.016 L=

Answers:

L=51.63 ft

EXAMPLE6.7 p= 10 psi, L=20 ft, D= 1 in, Q= ? This situation requires an itera-tive method,because the calculation offrequires knowledge of the flow rate. Calculatehead loss from the basic equation, guess a high Re number, and then getffrom theMoody chart for the appropriate e/D. Then calculate the velocity and a new head loss

from these estimated values. Correct Re until the two head losses are within 1 percentof each other. Finally, calculate the flow rate.

Known:

P= 10 psi = 997 = (968 106) sN

m2kg

m3

(h1 Df2)

f (V1)2

V1 Df

h1

g

ft2

s2P

Q

A

Df

2

1

12

1

ft

e

Di

ft

s2

lb

fts

lb

ft3lb

fts2ft3

s

N

m

2

kg

m

3

gal

min

h1

g

ft2

s2




FLUID MECHANICS


11/16


P= 4.633 104 = 62.241 = 6.505 104

Di= 1 in e= 0.000005 g=32.174

= 6 105 Df= ft L=20 ft

Solution:

h1= =23.136 ft

Guess:

Re=5 105

From Moody chart:

f= 0.013

V1= V1= 62.706

h1= f = 190.652 ft 23.136< 190.652 Guess lower Re.

Guess:

Re= 1.5 105

From Moody chart:

f= 0.0165

V1= V1= 18.812

h1= f =21.778 ft 23.136>21.778 Guess higher Re.

Guess:

Re= 1.55 105

From Moody chart:

f= 0.0164

V1= V1= 19.439 = 0.099%

h1= f =23.113 ft Close enough!h1

g

(V1)2

2

L

Df

(23.13623.113)

23.136

ft

s

(Re )

Df

h1g

(V1)2

2

LDf

ft

s

(Re )

Df

h1

g

(V1)2

2

L

Df

ft

s

(Re )

Df

h1

g

P

1

12

1

ft

e

Di

ft

s

2

lb

fts

lb

ft3lb

fts2



FLUID MECHANICS


12/16

Answer:

V1= 19.439 A= 2

Q= V1 A Q = 0.106 Q= 47.586

EXAMPLE6.8 p= 10 psi, L=20 ft, Q= 40 gpm, D= ? Obviously, if the flow rate isknown, along with the maximum allowable pressure loss and pipe length, the designerwishes to determine the smallest diameter to keep costs low. This situation requires aniterative solution that starts with a guessed D. From this diameter, e/D can be calcu-lated, and then velocity can be found from the area and flow. Next, fcan be found

from the Moody diagram after calculating Re. With this f, the head loss can be calcu-lated. Use the main equation to calculate the pressure loss and compare with the max-imum allowable pressure drop.

Known:

Q= 40 P= 10 psi = 997 = (968 106) s

Q= 0.089 P= 4.633 104 = 62.241 = 6.505 104

e= 0.000005 g=32.174 L=20 ft

Solution: Guess:

Di= 1 in

Df= Df= 0.083 ft = 6 105

A= 2

A=5.454 103ft2

V1= V1= 16.34

Re= Re= 1.303 105

From Moody chart:

f= 0.017

h1= f h

1=544.665

P= h1 P=3.39 104 P= 7.317 psi 7.317 < 10 Pick

smaller diameter.

lb

fts2

ft2

s2

V12

2

L

Df

( V1 Df)

ft

sQ

A

Df

2

1

ft

e

Di

Di

12i

f

n

t

ft

s2

lb

fts

lb

ft3lb

fts2ft3

s

N

m2kg

m3gal

min

gal

min

ft3

s

Df

2

ft

s




FLUID MECHANICS


13/16

Guess:

Di= 0.75 in

Df= Df= 0.062 ft = 8 105

A= 2

A=3.068 103ft2

V1= V1=29.049

Re= Re= 1.737 105

From Moody chart:

f= 0.016

h1= f h1=2.16 103

P= h1 P= 1.345 105

P=29.02 psi 29.02> 10 Picklarger diameter.

Guess:

Di= 0.94 in

Df= Df= 0.078 ft = 6.383 105

A= 2

A= 4.819 103ft2

V1= V1= 18.492

Re= Re= 1.386 105

From Moody chart:

f= 0.017

h1= f h1= 742.147ft2

s2V1

2

2

L

Df

( V1 Df)

ft

s

Q

A

Df

2

1

ft

e

Di

Di

12i

f

n

t

lb

fts2

ft2

s2V1

2

2

L

Df

( V1 Df)

ft

s

Q

A

Df

2

1

ft

e

Di

Di

12

i

f

n

t




FLUID MECHANICS


14/16

P= h1 P= 4.619 104 P= 9.97 psi

= 0.3% Close enough!

6.2.4 Pump Power Calculations

The head loss equation can be modified to include the power necessary to operate apump.This modification is derived from the fundamental energy conservation equa-tion and is expressed as:

(10 9.97)

10

lb

fts2


Circulating

pump

Supply Discharge To surge tank or

boiler

Radiant Floor loops

FIGURE 6.6 Hydronic piping system using a manifold of fourpipes.

+ 1 +gz1 + 2 +gz2= h1+ hlm+ hP (6.9)The pump head term hP is negative because the power is into the pump. Use of Eq.(6.9) is demonstrated in the following example.

EXAMPLE6.9 The level of a surge tank is 20 ft above the suction of a pump.The pumpdischarges 20 gpm of water into a single-path pipe system that comprises 200 ft of drawntubing.The water is 80F.The pipe system includes one gate valve, and the pipe is 0.50 in

in diameter. The pipe system discharges into a second surge tank that is 40 ft above thepump discharge and is sealed at a pressure of 10 psig. Calculate the pump head andpump power necessary to accomplish this task. Figure 6.6 shows this schematically.

Known:

Q=20 D= 0.5 in = 999

Q= 0.045 D= 0.042 ft = 62.366

Z1=20 ft A= 2

g=32.174

Z2= 40 ft V1= 0 V2= 0ft

s

ft

s

ft

s2D

2

lb

ft3ft3

s

kg

m3gal

min

V22

2

p2

V12

2

p1



FLUID MECHANICS


15/16

Solution:

P1= gZ1= 8.662 psig

P2= 10 psi+ g Z2= 27.324 psig

Power = P Q= 0.218 hp




FLUID MECHANICS


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D l d d f Di it l E i i Lib @ M G Hill ( di it l i i lib )

FLUID MECHANICS

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