Problem 8.127
Air at 200 °F and 60 psia flows in a 4-in.-diameter pipe at a rate of lb
0.52 s. Determine the
pressure at the 2-in.-diameter throat of a Venturi meter placed in the pipe.
Solution 8.127
βγ
ρβ
−
=====
−
12
4
2( ) 2in. lb
, where 0.5 and 0.52
4in. s
(1 )
VT
pp d
QCA Q
D (1)
Physical Properties of Air at Standard Atmospheric Pressure (BG/EE Units)a
Temperature
(°F)
Density,
ρ
(slugs/ft3)b
Specific
Weightc,
γ
(lb/ft3)
Dynamic
Viscosity,
μ (lb ·
s/ft2)
Kinematic
Viscosity,
ν
(ft2/s)
Specific
Heat
Ratio, k
(—)
Speed
of
Sound,
c (ft/s)
−40 2.939 E – 3 9.456 E – 2 3.29 E – 7 1.12 E – 4 1.401 1004
−20 2.805 E – 3 9.026 E – 2 3.34 E – 7 1.19 E – 4 1.401 1028
0 2.683 E – 3 8.633 E – 2 3.38 E – 7 1.26 E – 4 1.401 1051
160 1.990 E – 3 6.404 E – 2 4.22 E – 7 2.12 E – 4 1.399 1220
180 1.928 E – 3 6.204 E – 2 4.34 E – 7 2.25 E – 4 1.399 1239
200 1.870 E – 3 6.016 E – 2 4.49 E – 7 2.40 E – 4 1.398 1258
aBased on data from R. D. Blevins, Applied Fluid Dynamics Handbook, Van Nostrand Reinhold Co., Inc.,
New York, 1984.
bTo obtain EE units (lbm/ft3) multiply by 32.174.
cDensity and specific weight are related through the equation
γ
=
ρ
g. For this table g = 32.174 ft/s2.
Also, from the table above,
µ
−⋅
=×
7
2
lb s
4.49 10 ft so that
From Section 8.6.1,
5
6.88 6.88
1.0054 1.0054 0.99
Re 2.74 10
V
d
C
=−=− =
×
From Eq. (1):
2
312
34
3
2( )
ft 2
2
.11 (0.99) ft slug
s412
7.63 10 (1 0.5 )
ft
pp
π
−
−
=
×−
or
Problem 8.128
A 2.5-in.-diameter flow nozzle meter is installed in a 3.8-in.-diameter pipe that carries water
at 160°F. If the air–water manometer used to measure the pressure difference across the
meter indicates a reading of 3.1 ft, determine the flowrate.
Solution 8.128
From Table B.1 Physical Properties of Water (BG/EE Units): 3
slugs
1.896 ft
ρ
=,
or
6.78 n
V
C= (3)
h
= 3.1 ft
air
Trial and error solution using the figure above for Cn = Cn (Re,
β
= 0.658):
Assume Cn = 0.99. From Eq. (3), ==
ft
6.78(0.99) 6.71 s
V
1.00
0.98
Re =
ρ
VD/
μ
0.6
0.4
0.2
Problem 8.129
A 0.064-m-diameter nozzle meter is installed in a 0.097-m-diameter pipe that carries water
at 60 °C. If the inverted air–water U-tube manometer used to measure the pressure
difference across the meter indicates a reading of 1 m, determine the flowrate.
Solution 8.129
From Table B.2 Physical Properties of Water (SI Units): 3
kg
983.2 m
ρ
=,
or
=×
5m
Re 2.04 10 , where ~ s
VV
(2)
or
2.14 n
V
C= (3)
h
= 1 m
air
γ
Trial and error solution using the figure above for Cn = Cn (Re,
β
= 0.660):
Assume Cn = 0.99. From Eq. (3), ==
m
2.14(0.99) 2.12 s
V
1.00
0.98
0.6
0.4
0.2
Problem 8.130
A 50-mm-diameter nozzle meter is installed at the end of a 80-mm-diameter pipe through
which air flows. A manometer attached to the static pressure tap just upstream from the
nozzle indicates a pressure of 7.3 mm of water. Determine the flowrate.
Solution 8.130
Thus, with 2
4
n
A
d
π
=,
()
2
2
4
3
N
2 71.5 m
0.050 m kg
41.23 (1 0.625 )
m
n
QC
π
=
−
or
Q = 0.0230Cn. Assume Cn = 0.97 so that =
3
m
0.0223 s
Q
h
= 7.3 mm H2O
Thus, assume Cn = 0.963 so that ==
3
m
0.0230(0.963) 0.0221 s
Q
and
π
==
3
2
m
0.0221 m
s4.40 s
(0.08 m)
4
V
1.00
0.98
0.6
0.4
0.2
Problem 8.131
Water flows through the Venturi meter shown in the figure below. The specific gravity of
the manometer fluid is 1.52. Determine the flowrate.
Solution 8.131
Hence,
4
2( 1)
(1 )
TV
gSG h
QCA
ρ
ρβ
−
=− or
1
2
22
4
ft 2
2 32.2 (1.52 1) ft
312
s
ft
412 (1 0.5 )
V
QC
π
−
=
−
Thus,
3 in.6 in.
2 in.
SG = 1.52
Q
3 in.6 in.
Q
V
(1)
•
(2)
•
so
6
6.88
0.1198 1.0054
2.38 10
QQ
−
=−
×
Using a root solving method, =
3
ft
0.117 s
Q
checking Re,
Problem 8.132
Air flows through the Venturi meter shown in the figure below. The specific gravity of the
manometer fluid is 1.52. Determine the flowrate. Would compressibility effects be
important? Explain.
Solution 8.132
Thus, from Eq. (1),
1
2
22
34
3
lb
2 15.8
3ft
ft slug
412 2.38 10 (1 0.5 )
ft
V
QC
π
−
=
×−
or
Q = 5.84CV
3 in.6 in.
2 in.
SG
= 1.52
Q
3 in.6 in.
Q
V
(1)
•
(2)
•
so
5
6.88
5.84 1.0054
3.08 10
QQ
−
=−
×
.
Using a root finding method, =
3
ft
5.78 s
Q
Problem 8.133
The scale reading on the rotameter shown in the figure below is directly proportional to the
volumetric flowrate. With a scale reading of 2.6, the water bubbles up approximately 3 in.
How far will it bubble up if the scale reading is 5.0?
Solution 8.133
For the rotameter,
Q
KSR=× where
S
R= scale reading and
K
is a constant.
Thus,
By dividing these two equations,
2
2
(5.0)
3
(2.6) ft
12
h
=
or 0.925 ft 11.1 in.
h
==
Rotameter
3 in.
0
6
5
4
3
2
1
6
5
(2)
•