6 flow of fluids

7/29/2019 6 Flow of Fluids

1/9

Chapter 6Flow of Fluids

Page 1 of 9

6_001 Water at 68 F flows from a lake through 500 ft of 4-inch i.d. cast-iron pipe to a water turbine located 250 ftbelow the surface of the lake. After flowing through the turbine, the water is discharged into the atmospherethrough a horizontal 50-ft section of the same pipe. The turbine power output is 10 hp when the water in thedischarge pipe is flowing at 5 ft/sec. What is the turbine efficiency, defined as the actual power output of theturbine divided by the power output that would be withdrawn if there were no friction within the turbine?

Solution:

D = 4 inch = 1/3 ft

@ 68 F, = 2.43 lb/hr-ft = 6.75 x 10-4

lb/sec-ft

m

= 0.016056 cu ft / lb

p1

at the lake = 0 gauge pressure

p2

before turbine

p3

after turbine

p4

at discharge = 0 gauge pressure

( )

m

12

chc

1m2

m

21

zz

g

g

r2g

LGf

pp

+

=

where:z

2- z

1= -250 ft

L1

= 500 ft

g = 32.17 ft/(sec)(sec)g

c= 32.2 ft x pounds matter/(sec)(sec)(pounds force)

G = V / m

= 5 / 0.016056 = 311.41 lb/sec-sq ft

Solving for fm:

( )( )153,783

106.75

311.41

DGN

4

31

Re =

==

Equation 6-8e.

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f153,7833.2logf

110 +=

f = 0.00502

( )( ) ( )( )( )( )

( )( )

( )( )0.016056

500

32.2

32.17

32.22

5000.016056311.410.00502p0

121

2

2 +=

p2 = 14,828 lb/sq ft

( ) ( )( ) ( )( )( )( ) 3121

2

hc

m2

m43 p

32.22

500.016056311.410.00502

r2g

LGfpp ==

=

2

p

3= 73 lb/sq ft

p2

- p3

= 14,828 - 73 = 14,755 lb/sq ft


2/9


Page 2 of 9

Volumetric rate of flow = q

q = AV = (/4)(1/3)2(5) = 0.43633 cu ft/sec

( ) ( )( )550

0.436314,755

550

qpphp 32 =

=

hp = 11.70 hp

Turbine Efficiency = (10 hp / 11.70 hp) (100 %)Turbine Efficiency = 85.5 % . . . Ans.

6_002 If the turbine in Prob. 1 were by-passed, what would be the mass flow rate of water through the 550 ft of 4-inch pipe?

Solution:

If the turbine is by-passed:

( )0

zz

g

g

r2g

LLGf

pp m

12

chc

21m2

m

41 =

+

+

=

( ) ( )( )( )( )

( )( )

( )( )

00.016056

250-

32.2

32.17

32.22

5500.016056G0.00502

121

2

=+

G = 1,372.3 lb/sec-ft

w = GA

w = (1,372.3)(/4)(1/3)2w = 119.76 lb/sec . . Ans.

6_003 Air is to be delivered from a compressor to a distribution line by means of standard 1-1/2-in. steel pipe, 300 ftlong. The maximum rate of consumption by the equipment connected to this distribution system is 600 cu ftof free air (measured at 68 F) per minute. In order for the pressure in the distribution line to be always 70

lb/sq in. gauge or higher, what must be the pressure rating of the compressor? Assume that the air flow isisothermal.

Solution:

Air at 600 cu ft per minute at 68 Fg

c= 32.2 ft x pounds matter/(sec)(sec)(pounds force)

L = 300 ftD = 1 1/2 in = 1/8 ft

p

TRG=

T = 68 F + 460 F = 528 F

RG = 53.34rh

= D/4 = 1/32 ft

p2

= 70 lb/sq in gauge = 84.696 lb/sq in abs.

p2

= 12,196 lb/sq ft abs.

A = (/4)D2 = (/4)(1/8)2A = 0.012272 sq ft


3/9


Page 3 of 9

( )( )12196

52853.34=2

2= 2.309 cu ft /lb

Air viscosity at 68 F

= 0.0440 lb/hr-ft = 1.2222 x 10-5

lb/sec-ft

( )( )5

81

Re101.2222

353

DGN

==

3,610,293NRe =

Therefore, = 1Steel pipe:

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f3,610,2933.2logf

110 +=

f = 0.003036

( )( )

111

Gp

28,164

p

52853.34

p

TR===1

( )212

1m +=

1.1545p

14,0822.309

p

28,164

1121

m +=

+=

Use Equation (6-9d) assuming

2/

1> 2.

hc

2m

c

2

mG

22

21

r2g

LGfln

g

G

T2R

pp+

=

1

2

12,196

p

p

28,164

2.309 1

1

=

=

1

2

( )( )( )

( )( )( )

( )( )( )( )( )321

21

2221

32.22

3533000.003036

12,196

pln

32.21

353

52853.342

12,196p+

=

( )56,395

12,196

p3,870ln

56,327

12,196p 122

1+

=

By Trial and Error:p

1= 60,621 lb/sq ft abs.

p1

= 406.31 lb/sq in gauge. . . Ans.

6_004 Glycerol at 68 F, specific gravity 1.26, is pumped at a rate of 28,000 cu ft/hr by a single pump through twohorizontal pipes connected in parallel. Both pipes have a length of 100 ft. One is a standard 4-in. steel pipe,the other a standard 9-in. steel pipe. It is agreed to neglect pressure drop due to fittings. What is the velocityin the smaller pipe? What is the pressure drop through the lines? If the lines discharge into an open tank,what head must be developed by the pump?

Solution:


4/9


Page 4 of 9

Use Line a for 4 in pipe and line b for 9 in pipe

Glycerol at 68 F, = 7 cp = 4.704 x 10-3

lb/sec-ft

q = volumetric flow rate = 28,000 cu ft/hr = 7.777778 cu ft/secq = q

a+ q

b

= (1.26)(62.4) = 78.624 lb/cu ft

Line a:

r2g

LGf

r2g

LGfpp

hac

2aa

hc

m2

aa21 =

=

rha

= Da/4

For 4 in dia. pipe, D = 4.026 in = 0.3355 ftrha

= Da/ 4 = 0.083875 ft

gc

= 32.2 ft x pounds matter/(sec)(sec)(pounds force)

( )( )( )( )78.6240.08387532.22100Gfpp

2

aa21 =

2

aa21 G0.235465fpp = Line b:

r2g

LGf

r2g

LGfpp

hbc

2bb

hc

m2

bb21 =

=

rhb

= Db/4

For 9 in dia. pipe, D = 8.941 in = 0.7451 ftrhb

= Db/ 4 = 0.186275 ft

gc

= 32.2 ft x pounds matter/(sec)(sec)(pounds force)

( )( )( )( )78.6240.18627532.22

100Gfpp2bb

21 =

2

bb21 G0.106024fpp =

Then:

2bb

2aa G0.106024fG0.235465f =

2aa

2bb G2.220865fGf =

2D

4q

A

qG ==

( )

( )2aa

0.3355

q78.6244G =

G

a= 889.4q

a

( )

( )2b

b0.7451

q78.6244G =

G

b= 180.3q

b


5/9


Page 5 of 9

Then:

( ) ( )2aa2

bb 889.4q2.220865f180.3qf =

b

aab

f

f7.351272qq =

Equation 6-8e.

( ) 1.2fN3.2logf

1Re10 +=

Then:

( )( ) 1.2fN3.2log

1.2fN3.2log

f

f

aRea10

bReb10

b

a

+

+=

( )( )

a3

aaaRea 63,434q

104.704

889.4q0.3355

GDN =

==

( )( ) b3abb

Reb q104.704

180.3q0.7451GDN 559,28=

==

( )( ) 1.2f63,474q3.2log

1.2f28,559q3.2log

f

f

aa10

bb10

b

a

+

+=

( ) 1.2f63,474q3.2log1.2f209,945q3.2log

f

f

aa10

aa10

b

a

+

+=

By trial and error, try

13.1=b

a

f

f

q

b= (7.351272)(1.13)q

a

qb = 8.306937qa

7.777778 = qa

+ 8.306937qb

qa

= 0.8357 cu ft/sec

( ) 53,0120.835763,434NRea ==

( ) 1.2f53,0123.2logf

1a10

a

+=

0.07826fa = fa

= 0.006125

( )( )( )[ ]( )( )( )[ ]

1.131.20.078260.835763,4343.2log1.20.078260.8357209,9453.2log

ff

10

10

b

a=

+

+=

Therefore trial value is okay.

13.1=b

a

f

f

13.1=bf

0.006125


6/9


Page 6 of 9

fb

= 0.004797

a. Velocity of the smaler pipe.

Ga

= Va

Ga

= 889.4qa

Ga

= Va = 889.4q

a

Va(78.624) =(889.4)(0.8357)

Va

= 9.4535 ft/sec . . . Ans.

b. Pressure drop thru line a orb.

2aa21 G0.235465fpp =

( )2aa21 889.4q0.235465fpp = ( ) ( )( )[ ]221 0.8357889.40.0061250.235465pp =

p1

- p2

= 796.76 lb/sq ft . . . Ans.

c. Head developed by pump.

Head = (796.76 lb/sq ft) / (78.624 lb/cu ft)Head = 10.134 ft . . . Ans.

6_005 It is desired to heat 27 lb/min of dry air from 40 to 1040 F (moving-stream temperatures) by passing itthrough a heated horizontal section of smooth pipe, having an i.d. of 2 inches. The heat is to be supplied byelectrical wires wrapped around the outside of the pipe, and these wires will supply 4.0 kw of electricalpower/ft of heated pipe, distributed uniformly over the heated length. The air is to leave the heated section ata pressure of 1 atm. abs. What must be the length of the heated section and the pressure of the air enteringthe section?

Solution:

wa

= 27 lb/min

= 1,620 lb/hr= 0.45 lb/sec

cp

= 0.24

t1

= 40 F

t2

= 1040 F

a. Heated Length

Q = wac

p(t

1- t

2)

Q = (1,620)(0.24)(1040 - 40)Q = 388,000 Btu/hr = 113.95 kw

L = heated length = 113.95 kw / 4 kw/ftL = 28.5 ft . . . Ans.

b. Pressure entering the section, p1.

p2

= 1 atm abs. = 2116.8 lb/sq ft abs.


7/9


Page 7 of 9

D = 2 inches = 0.166667 ftrh

= D/4 = 0.041667 ft

gc

= 32.2 ft x pound matter / (sec)(sec)(pound force)

( )

( )22a

0.041667

0.454

D

4wG ==

G = 330 lb/(sec)(sq ft)

( )( )

11

1G1

p

4604053.34

p

TR +==

11

p

26,670=

( )( )

2116.8

460104053.34

p

TR

2

2G2

+==

37.82 = ( )

2

21m

+=

2

37.8p

26,670

1

+

=m

18.9p

13,335

1

+=m

Viscosity of air at average temperature.tave

= (1/2)(40 + 1040) = 540 F

= 0.06978 lb/hr-ft = 1.93833 x 10-5

lb/sec-ft

( )( )2,837,496

101.93833

3300.166667

DGN

5Re =

==

= 1Smooth pipe:Equation 6-8.

0.32NRe

0.1250.00140f +=

( )0.002476

2,837,496

0.1250.00140f

0.32=+=

Use Equation 6-9d, T

m= 540 F + 460 F = 1000 R

hc

2m

1

2

c

2

mG

22

21

r2g

LGfln

g

G

T2R

pp+

=

( )( )( )

( )( )( )

( )( )( )( )( ).04166732.22

33028.50.002476

p26,670

37.8ln

32.21

330

100053.342

2116.8p2

1

2221

+

=

( )2864

705.56

p3382ln

106,680

2116.8p 122

1+

=

By trial and error:


8/9


Page 8 of 9

p1

= 42,269 lb/sq ft abs.

p1

= 20 atm abs. . . . Ans

6_006 Water is pumped over a high pass in mountain country from a lake in the valley. The pump is located 10 ftabove the lake surface. From the pump, the pipe line extends 2000 ft to the top of the pass, which has an

altitude of 1010 ft above the lake surface. The pipe line then runs 1000 ft down the other side of the pass,losing 500 ft in altitude. The pipe line next travels another 1000 ft horizontally and empties into a reservoir.

Neglecting end losses and assuming that the lengths of pipe are reported as equivalent lengths(thus containing allowance for bends, etc.), compute the shaft horsepower which must be delivered to thepump to move water through this system at a rate of 120 gal/sec. The pipe line has an i.d. of 1.5 ft and ismade of steel. the over-all pump efficiency is 75 per cent. The discharge end of the pipe is several feet abovethe surface of the reservoir.

Solution:

D = 1.5 ft steelq = 120 gal/secPimp-Efficiency = 0.75

Equation 6-9a.

m

12

chc

m2

m21

zz

g

g

r2g

LGfpp

+

=

z

2- z

1= 1010 ft - 500 ft - 10 ft = 500 ft

L = 2000 ft + 1000 ft + 1000 ft = 4000 ftq = 120 gal/sec x 231 cu n/gal x 1 cu ft / 1728 cu inq = 16.042 cu ft/sec

w = q = (62.4)(16.042) = 1001 lb/sec

( )

( )22 1.5

10014

D

4w

A

wG ===

G = 566.45 lb/sec-ftrh

= D/4 = 0.375 ft

g = 32.17g

c= 32.2

m

= 1/62.4 = 0.01603 cu ft/lb

viscosity at 68 F, = 2.43 lb/hr-ft = 6.75 x 10-4

lb/sec-ft

( )( )1,258,778

106.75

566.451.5N

4Re=

=

Equation 6-8


9/9


Page 9 of 9

( ) 1.2fN3.2logf

1Re10 +=

( ) 1.2f1,258,7783.2logf

110 +=

f = 0.00354

( )( ) ( )( )( )( )

( )( )

( )( )0.01603

500

32.2

32.17

0.37532.22

40000.01603566.450.00354pp

2

21 +=

p

1- p

2= 34,178.3 lb/sq ft

( ) ( )( )550

16.04234,178.3

550

qpphp 21 =

=

hp = 996.9 hp

Checking if height 1000 ft is reached.

If L = 2000 ftz

2- z

1= 1010 ft - 10 ft = 1000 ft

( )( ) ( )( )( )( )

( )( )

( )( )0.01603

1000

32.2

32.17

0.37532.22

20000.01603566.450.00354pp

2

2'1 +=

p

1- p

2= 63,832.9 lb/sq ft

( ) ( )( )550

16.04263,832.9

550

qpphp 2'1 =

=

hp = 1,861.8 hp > 996.9 jp

Therefore:

Shaft hp = 1,861.8 hp / 0.75 = 2,482 hp Ans.

- end -

6 flow of fluids

Documents