13-Practice Problems.pdf

Chapter 13

Flow Measurements

Problem 13.1

Velocity in an air ß ow is to be measured with a stagnation tube that has a resolu-

tion of 0.1-in. H 2 O. Find the minimum ß uid speed in ft/s that can be measured.

Neglect viscous e ! ects and assume that the air is at room condition.

Solution

Fluid speed for a stagnation tube is given by Eq. (5.19) in the 7 th edition.

! =

2!"

Convert the pressure change from a unit of in. H 2 O to a unit of psf.

0$03609 psi

in. H 2 O

¶Μ

144 psf

psi

!" = (0$1 in. H 2 O )

= 0$520 psf

Note that this pressure value could also have been found using the hydrostatic

equation: !" = (# H 2 O )%& .

The minimum velocity is

! =

2!"

# air

¶Μ

0$520 lbf

ft 2

ft 3

0$00233 slug

slug · ft

lbf · s 2

= 21$1 ft/s

117

118

CHAPTER 13. FLOW MEASUREMENTS

Problem 13.2

Air velocity is measured with a stagnation tube of diameter ' = 0$5 mm. Pressure

in the stagnation tube causes water in a U-tube to rise to a height &$ Find the

minimum velocity ! that can be measured with the stagnation tube if the aim is

that viscous e ! ects contribute an error less than 5 % . Also, Þ nd the corresponding

value of !"$

Solution

Viscous e ! ects are characterized in Fig. 13.1. From the vertical axis of this Þ gure

! actual =

2!"

#( !

When neglecting viscous e ! ects, the corresponding formula is

! approx =

2!"

The error ) is given by

) = ! approx ! ! actual

! actual

= 1!

1*( !

(1)

1*( !

Algebraic manipulation of Eq. (1) gives

( ! = (1+)) 2

So, a 5% error is associated with

( ! = (1+0$05) 2

= 1$103

From Fig. 13.1 in the textbook, this occurs at a Reynolds number of about 25.

Thus

RE = 25 = !'

119

! MIN = 25+

= 25×

15$1×10 !6

0$5×10 !3

= 0$775 m/s

Pressure change is related to ß uid speed by

! =

2!"

!" = #! 2

´³

1$2 kg/m 3

0$775 2 m 2 /s 2

= 0$360 Pa

Problem 13.3

The average velocity of gasoline (S = 0.68, + = 4$6 × 10 !6 ft 2 * s) is measured

with a 2-in. diameter ori Þ ce meter in a 6-in. diameter pipe. The manometer uses

mercury with dimensions of & = 4 in. and , = 3 in. Find ! 1 $

Solution

120

CHAPTER 13. FLOW MEASUREMENTS

Discharge and velocity are related by

- = . 1 ! 1

(1)

and discharge for an ori Þ ce meter is given by

- = /. "

2%!&

(2)

Before / can be looked up, piezometric head ( !&) is needed. This is de Þ ned by

0 gasoline +1

!& =

(3)

Applying the manometer equation (Eq. 3.17 in 8th edition) yields

" 1 +0 gasoline (,+&)!0 Hg (&)!0 gasoline (,+(1 2

!1 1 )) = " 2

(4)

Rearranging Eq. (4)

0 gasoline +1

0 Hg

0 gasoline

= &

(5)

Combining Eqs. (3) and (5)

0 Hg

0 gasoline

!& = &

13$55

0$68

= (4*12 ft )

= 6$31 ft

To Þ nd the ß ow coe " cient /2 calculate the parameter on the top axis of Fig. 13.13.

RE #

/ =

2%!& '

2×32$2×6$31

2*12

4$6×10 !6

= 7302000

On Fig. 13.3, tracing the dashed line to '*3 = 2*6 = 0$333 and interpolating gives

/ # 0$606$

Combining Eqs. (1) and (2) and substituting values gives

! 1 = / . "

. 1

2%!&

2 2 in. 2

32$2 ft/s 2

= 0$606

6 2 in. 2

2×

×(6$31 ft )

= 1$36 ft/s

121

Problem 13.4

Water speed is measured with a venturi meter. Throat diameter is 6 cm, pipe

diameter is 12 cm, and height on the manometer is 4 = 100CM . Find the ß ow rate

in the pipe. Kinematic viscosity of water is + = 10 !6 m 2 * s.

Solution

Flow rate through a venturi meter is given by

- = /. "

2%!&

(1)

Before / can be looked up, piezometric head (!&) is needed. This is de Þ ned by

0 H 2 O +1

!& =

(2)

where locations 1 and 2 are de Þ ned in the sketch. Applying the manometer equation

(Eq. 3.17 in 8th edition) yields

" 1 !0 H 2 O (,+1 2 !1 1 )!0 H 2 O (4)+0 air (4)+0 H 2 O (,) = " 2

(3)

Rearranging Eq. (3) gives

0 $ 2 % +1

1! 0 air

0 H 2 O

= 4

(4)

Combining Eqs. (2) and (4), and letting 0 air *0 H 2 O # 0 gives

!& = 4

= 1 m

Plik z chomika:

Inne pliki z tego folderu:

Inne foldery tego chomika: