10-Practice Problems.pdf

Chapter 10

Flow in Conduits

Problem 10.1

Water at 20 o C ( ! = 10 !3 N · s/m 2 , " = 1000 kg/m 3 ) ß ows through a 0.5-mm tube

connected to the bottom of a reservoir. The length of the tube is 1.0 m, and the

depth of water in the reservoir is 20 cm. Find the ß ow rate in the tube. Neglect

the entrance loss at the junction of the tube and reservoir.

Solution

Applying the energy equation between the top of the water in the reservoir (1) and

the end of the tube (2) gives

# 1

$ +% 1

& 2' +( 1 +) ! = # 2

$ +% 2

& 2' +( 2 +) " +) #

Thepressureatpoints1and2isthesame(atmospheric),thevelocityinthereservoir

is zero, and there is no pump or turbine in the system. Also, the only losses are

CHAPTER 10. FLOW IN CONDUITS

friction losses in the tube. The energy equation simpli Þ es to

( 1 = % 2

& 2' +( 2 +) $

We will assume the ß ow is laminar, so % 2 = 2* The head loss due to friction in a

laminar ß ow is

) $ = 32 !+&

$, 2

Substituting into the energy equation gives

( 1 = % 2

& 2' +( 2 +32 !+& 2

$, 2

and replacing the variables with values

1*2 = & 2

9*81 +32 10 !3 ×1×& 2

9810×0*0005 2

0*102& 2 +13*05& 2 !1*2 = 0

Solving

& 2 = 0*092 m/s

The volume ß ow rate is

4 ×0*0005 2 ×0*092 = 1*8×10 !8 m 3 /s

= 1*8×10 !2 ml/s

To determine whether the ß ow is laminar, calculate the Reynolds number.

RE = "&,

= 10 3 ×0*092×0*0005

10 !3

= 46

Since the Reynolds number is less than 2000, the laminar ß ow assumption is justi-

Þ ed.

- = .& = /

Problem 10.2

An oil supply line for a bearing is being designed. The supply line is a tube

with an internal diameter of 16 in. and 10 feet long. It is to transport SAE 30W

oil ( ! = 2×10 !3 lbf · s/ft 2 0 " = 1*71 slugs/ft 3 ) at the rate of 0.01 gpm. Find the

pressure drop across the line.

Solution

First determine the Reynolds number to establish if the ß ow is laminar or turbulent.

The velocity in the line is

& = -

= 0*01 gpm ×0*00223 cfs/gpm

4 ×

16 × 12

2 ft 2

= 1*05 ft/s

The Reynolds number is

RE = "&,

= 1*71×1*05× 16 × 12

2×10 !3

= 4*68

The ß ow is laminar, so the head loss is

) $ = 32 !+&

$, 2

= 32

2×10 !3 ×10×1*05

1*71×32*2×

16 × 12

2 = 450 ft

The pressure drop is

!# = $) $

= 1*71×32*2×450

= 24*8×10 3 psf = 172 psi

CHAPTER 10. FLOW IN CONDUITS

Problem 10.3

Kerosene ( ! = 1*9 × 10 !3 N · s/m 2 , 1 =0.81) ß ows in a 2-cm diameter commer-

cial steel pipe ( 2 & = 0*046 mm). A mercury manometer ( 1 = 13*6) is connected

between a 2-m section of pipe as shown, and there is a 5-cm de ß ection in the

manometer. The elevation di ! erence between the two taps is 0.5 mm. Find the

direction and velocity of the ß ow in the pipe.

Solution

First Þ nd the di ! erence in piezometric pressure between the twopressure taps inthe

pipe. Flow is always in the direction of decreasing piezometric head. Take station

1 on the left and station 2 on the right. De Þ ne the distance 3 as the distance from

the center of the pipe at station 2 and the top of the mercury in the manometer.

Using the manometer equation from 1 to 2 gives

# 2 = # 1 +$ k (( 1 !( 2 )+$ k 3+$ Hg !)!$ k !)!$ k 3

Thus we can write

# 2 +$ k ( 2 !(# 1 +$ k ( 1 ) = ($ Hg !$ k )!)

# '(2 !# '(1 = ($ Hg !$ k )!)

Since $ Hg 4 $ k 0 # '(2 4 # '(1 0 the ß ow must be from right to left (uphill).

The energy equation from 2 to 1 is

# '(2 +% 2 " k & 2 +$ k ) ! = # '(1 +% 1 " k & 2 +$ k ) " +$ k ) #

Since the pipe has a constant area, & 1 = & 2 0 and there are no turbines or pumps in

the system, the equation reduces to

# '(2 !# '(1 = $ k ) # = ($ Hg !$ k )!)

The head loss can be expressed using the Darcy-Weisbach relation

) # = 5 +

& 2

5 +

& 2

2' =

$ Hg

$ k !1

Substituting in the values

13*6

0*81 !1

2×9*81

5& 2 = 0*155 m 2 6 s 2

= 5& 2 2

0*02

0*05×

Since 5 depends on the Reynolds number (and velocity), this equation has to be

solved by iteration. The relative roughness for the pipe is

, = 0*046

= 0*0023

From the Moody diagram (Fig. 10.8), the friction factor for a fully rough pipe

would be about 0.025. This would give a velocity of

0*155

0*025 = 2*49 m/s

& =

The corresponding Reynolds number is

RE = "&,

= 1000×0*81×2*49×0*02

1*9×10 !3

= 2*12×10 4

From the Moody diagram, the friction factor for this Reynolds number is 0.0305.

The velocity is corrected to

0*155

0*0305 = 2*25 m/s

& =

ThenewReynoldsnumberis 1*92×10 4 * Thefrictionfactoris0.030, givingavelocity

of 2.27 m/s. Further iterations would not signi Þ cantly change the value so

& = 2*27 m/s

and the ß ow is from right to left.

2 &

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