E1C10 09/14/2010
13:4:38 Page 437
Example 10.4
A 10-cm-diameter, square-edged orifice plate is used to meter the steady flow of 16
C water through
an 20-cm pipe. Flange taps are used and the pressure drop measured is 50 cm Hg. Determine the pipe
flow rate. The specific gravity of mercury is 13.5.
KNOWN d 1 ¼ 20 cm H ¼ 50 cm Hg d 0 ¼ 10 cm
Water properties (properties are from Appendix B)
m ¼ 1:08 Â 10
À3 N-s/m
2
r ¼ 999 kg/m
3
ASSUMPTIONS Incompressible flow of a liquid Y ¼ 1
ð
Þ
FIND Volume flow rate, Q
SOLUTION Equation 10.12 is used with an orifice plate, and it requires knowledge of the
product CE. The beta ratio is b ¼ d 0 =d 1 ¼ 0:5, so the velocity of approach factor is calculated from
Equation 10.11:
E ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À b
4
p
¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À 0:5
4
p
¼ 1:0328
We know that C ¼ f Re d 1 ; b
ð
Þ. The flow Reynolds number is estimated using Equation 10.3, which
can be modified with the relation, v ¼ m=r,
Re d 1 ¼
4Q
pd 1 v
¼
4rQ
pd 1 m
We see that without information concerning Q, we cannot determine the Reynolds number, and so C
cannot be determined explicitly.
Instead, a trial-and-error iteration is undertaken: Guess a value for K 0 (or for C) and iterate. A
good start is to guess a value at a high value of Re d 1 . This is the flat region of Figure 10.5, so choose a
value of K 0 ¼ CE ¼ 0:625.
Based on the manometer deflection and Equation 10.13 (see Ex. 10.2),
Q ¼ CEA 0 Y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2gH½ðS m =SÞ À 1
p
¼ K 0 A 0 Y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2gH½ðS m =SÞ À 1
p
¼ ð0:625Þðp=4Þð0:10 mÞ
2 ð1Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ð9:8 m/s 2 Þð0:50 mÞ½ð13:5=1Þ À 1
p
¼ 0:054 m
3 /s
Next we have to test the guessed value for K 0 to determine if it was correct. For this value of Q,
Re d 1 ¼
4rQ
pd 1 m
¼
4 999 kg/m
3
ð
Þ0:054 m
3 /s
ð
Þ
p 0:20 m 1:08 Â 10
À3 N-s/m 2
À
Á
À
Á ¼ 3:2 Â 10
5
From Figure 10.5, at this Reynolds number, K 0 % 0:625. The solution is converged, so we conclude
that Q ¼ 0:054 m
3 /s.
Example 10.5
Air flows at 20
C through a 6-cm pipe. A square-edged orifice plate with b ¼ 0:4 is chosen to meter
the flow rate. A pressure drop of 250 cm H 2 O is measured at the flange taps with an upstream
pressure of 93.7 kPa abs. Find the flow rate.
10.5 Pressure Differential Meters 437
13:4:38 Page 437
Example 10.4
A 10-cm-diameter, square-edged orifice plate is used to meter the steady flow of 16
C water through
an 20-cm pipe. Flange taps are used and the pressure drop measured is 50 cm Hg. Determine the pipe
flow rate. The specific gravity of mercury is 13.5.
KNOWN d 1 ¼ 20 cm H ¼ 50 cm Hg d 0 ¼ 10 cm
Water properties (properties are from Appendix B)
m ¼ 1:08 Â 10
À3 N-s/m
2
r ¼ 999 kg/m
3
ASSUMPTIONS Incompressible flow of a liquid Y ¼ 1
ð
Þ
FIND Volume flow rate, Q
SOLUTION Equation 10.12 is used with an orifice plate, and it requires knowledge of the
product CE. The beta ratio is b ¼ d 0 =d 1 ¼ 0:5, so the velocity of approach factor is calculated from
Equation 10.11:
E ¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À b
4
p
¼
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À 0:5
4
p
¼ 1:0328
We know that C ¼ f Re d 1 ; b
ð
Þ. The flow Reynolds number is estimated using Equation 10.3, which
can be modified with the relation, v ¼ m=r,
Re d 1 ¼
4Q
pd 1 v
¼
4rQ
pd 1 m
We see that without information concerning Q, we cannot determine the Reynolds number, and so C
cannot be determined explicitly.
Instead, a trial-and-error iteration is undertaken: Guess a value for K 0 (or for C) and iterate. A
good start is to guess a value at a high value of Re d 1 . This is the flat region of Figure 10.5, so choose a
value of K 0 ¼ CE ¼ 0:625.
Based on the manometer deflection and Equation 10.13 (see Ex. 10.2),
Q ¼ CEA 0 Y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2gH½ðS m =SÞ À 1
p
¼ K 0 A 0 Y
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2gH½ðS m =SÞ À 1
p
¼ ð0:625Þðp=4Þð0:10 mÞ
2 ð1Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2ð9:8 m/s 2 Þð0:50 mÞ½ð13:5=1Þ À 1
p
¼ 0:054 m
3 /s
Next we have to test the guessed value for K 0 to determine if it was correct. For this value of Q,
Re d 1 ¼
4rQ
pd 1 m
¼
4 999 kg/m
3
ð
Þ0:054 m
3 /s
ð
Þ
p 0:20 m 1:08 Â 10
À3 N-s/m 2
À
Á
À
Á ¼ 3:2 Â 10
5
From Figure 10.5, at this Reynolds number, K 0 % 0:625. The solution is converged, so we conclude
that Q ¼ 0:054 m
3 /s.
Example 10.5
Air flows at 20
C through a 6-cm pipe. A square-edged orifice plate with b ¼ 0:4 is chosen to meter
the flow rate. A pressure drop of 250 cm H 2 O is measured at the flange taps with an upstream
pressure of 93.7 kPa abs. Find the flow rate.
10.5 Pressure Differential Meters 437