Chapter 13
Flow Measurements
Problem 13.1
Velocity in an air ow is to be measured with a stagnation tube that has a resolu-
tion of 0.1-in. H2O. Find the minimum 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 7th edition.
s
2
=

Convert the pressure change from a unit of in. H2O to a unit of psf.
µ Å› µ Å›
0 03609 psi 144 psf
= (0 1 in. H2O)
in. H2O psi
= 0 520 psf
Note that this pressure value could also have been found using the hydrostatic
equation: = ( H2O) .
The minimum velocity is
s
2
=
air
s
µ Å› µ Å› µ Å›
0 520 lbf ft3 slug · ft
= 2
0 00233 slug lbf · s2
ft2
= 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, nd the corresponding
value of
Solution
Viscous e ects are characterized in Fig. 13.1. From the vertical axis of this gure
s
2
actual =

When neglecting viscous e ects, the corresponding formula is
s
2
approx =

The error is given by
approx actual
=

pactual
1 1
= p (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
So
25
min =

Ą ó
25 × 15 1 × 10 6
=
0 5 × 10 3
= 0 775 m/s
Pressure change is related to uid speed by
s
2
=

So
2

=
Å‚2 ´ Å‚ ´
1 2 kg/m3 0 7752 m2/s2
=
2
= 0 360 Pa
Problem 13.3
The average velocity of gasoline (S = 0.68, = 4 6 × 10 6 ft2 s) is measured
with a 2-in. diameter ori 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 ce meter is given by
p
= 2 (2)
Before can be looked up, piezometric head ( ) is needed. This is de ned by
µ Å› µ Å›

= + + (3)
gasoline 1 gasoline 2
Applying the manometer equation (Eq. 3.17 in 8th edition) yields
1 + gasoline ( + ) Hg ( ) gasoline ( + ( 2 1)) = 2 (4)
Rearranging Eq. (4)
µ Å› µ Å› µ Å›
Hg
+ + = 1 (5)
gasoline 1 gasoline 2 gasoline
Combining Eqs. (3) and (5)
µ Å›
Hg
= 1
gasoline
µ Å›
13 55
= (4 12 ft) 1
0 68
= 6 31 ft
To nd the ow coe cient calculate the parameter on the top axis of Fig. 13.13.
Re p
= 2


2 12
= 2 × 32 2 × 6 31
4 6 × 10 6
= 730 000
On Fig. 13.3, tracing the dashed line to = 2 6 = 0 333 and interpolating gives
0 606
Combining Eqs. (1) and (2) and substituting values gives
p
1 = 2
1
Ą ó
r
Å‚ ´
22 in.2
Ą ó
= 0 606 2 × 32 2 ft/s2 × (6 31 ft)
62 in.2
= 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 = 100 cm. Find the ow rate
in the pipe. Kinematic viscosity of water is = 10 6 m2 s.
Solution
Flow rate through a venturi meter is given by
p
= 2 (1)
Before can be looked up, piezometric head ( ) is needed. This is de ned by
µ Å› µ Å›

= + + (2)
H2O 1 H2O 2
where locations 1 and 2 are de ned in the sketch. Applying the manometer equation
(Eq. 3.17 in 8th edition) yields
1 H2O ( + 2 1) H2O ( ) + air ( ) + H2O ( ) = 2 (3)
Rearranging Eq. (3) gives
µ Å› µ Å› µ Å›
air
+ + = 1 (4)
2 1 2 2 H2O
Combining Eqs. (2) and (4), and letting air H2O 0 gives
=
= 1 m
122 CHAPTER 13. FLOW MEASUREMENTS
To nd the ow coe cient calculate the parameter on the top axis of Fig. 13.13.
Re p
= 2


0 06
= 2 × 9 8 × 1
10 6
= 266 000
On Fig. 13.3, tracing the dashed line to = 6 12 = 0 5 and interpolating gives
1 01
Substituting values into Eq. (1) gives
p
= 2
Ą ó
r
´
0 062 m2 Å‚
= 1 01 2 × 9 8 m/s2 × (1 0 m)
4
= 0 0126 m3/s
Problem 13.5
Air of density = 1 2 m3 s and speed 1 = 20 m/s is metered with an ori ce.
The ori ce diameter is 2 cm, and the pipe diameter is 4 cm. A di erential pressure
gage records the pressure di erence between pressure taps 1 and 2, which are sep-
arated by a vertical distance of = 8 cm. Find the reading on the pressure gage.
The kinematic viscosity of air is 14 6 × 10 6 m2 s.
Solution
123
When pressure di erence is measured with a transducer as shown, the pressure
reading is piezometric pressure, and ow rate through the ori ce meter is
s
2
= (1)

Rearranging Eq. (1) gives the pressure change.
µ Å›2

= (2)
2
The ow rate is
= 1 1
µ Å›
× 0 042
= m2 (20 m/s)
4
= 0 0251 m3 s
To nd the ow coe cient calculate the Reynolds number as de ned on the
bottom axis of Fig. 13.13.
4
Re =

Ą ó
4 0 0251 m3 s
=
(0 02 m) (14 6 × 10 6 m2/s)
= 109 000
Interpolating in Fig. 13.3 with = 2 4 = 0 5 gives 0 63.
The area of the ori ce is
µ Å›
× 0 022
= m2
4
= 3 142 × 10 4 m2
Substituting values into Eq. (2)
µ Å›2

=
2
µ Å›2
1 2 kg/m3 0 0251 m3 s
=
2 (0 63) × (3 141 × 10 4 m2)
= 9 65 × 103 Pa
So
= 9 65 kPa
124 CHAPTER 13. FLOW MEASUREMENTS
Problem 13.6
An engineer is considering the feasibility of a small hydroelectric power plant, and
she wishes to design a rectangular weir to measure the discharge of a small creek.
The weir will span the creek, which is 1.5 m wide, and the engineer estimates that
the maximum discharge will be 0.5 m3 s. If the creek level cannot rise above 1.2
m, calculate the height of the weir.
Solution
Discharge is
p
= 2 3 2 (1)
where the head on the weir is given by
= 1 2 (2)
and the ow coe cient is

= 0 40 + 0 05 (3)

Combining Eqs. (1) to (3)
µ Å›
p
1 2
= 0 40 + 0 05 2 (1 2 )3 2

µ Å›

1 2
0 5 = 0 40 + 0 05 2 × 9 81 (1 2) (1 2 )3 2 (4)

One way to solve Eq. (4) is to program the right side of the equation and then
substitute values into the equation, until a value of 0.5 is achieved. This was
done the results are
= 0 422
= 0 368
= 0 832 m