E1C10 09/14/2010
13:4:38 Page 438
KNOWN d 1 ¼ 6 cm p 1 ¼ 93:7 kPa abs
b ¼ 0:4
T 1 ¼ 20
C ¼ 293 K
H ¼ 250 cm H 2 O
Properties (found in Appendix B)
Air : v ¼ 1:0 Â 10
À5 m
2 /s
Water : r H 2 O ¼ 999 kg/m
3
ASSUMPTIONS Treat air behavior as an ideal gas ðp ¼ rRTÞ
FIND Volume flow rate, Q
SOLUTION The orifice flow rate is found from Equation 10.12, which requires information
about E, C, and Y. From the given information, we calculate both the orifice area,
A d 0 ¼ 4:52 Â 10
À4 m
2 , and, from Equation 10.11, the velocity of approach factor, E ¼ 1:013.
The air density is found from the ideal gas equation of state:
r 1 ¼
p 1
RT 1
¼
93;700 N/m
2
287 N-m/kg
ð
Þ293 K
ð
Þ
¼ 1:114 kg/m
3
or use air-property tables. The pressure drop is p 1 À p 2 ¼ r H 2 O gH ¼ 24; 500 N/m
2 .
The pressure ratio for this gas flow, ðp 1 À p 2 Þ=p 1 ¼ 0:26. For pressure ratios greater than 0.1 the
compressibility of the air should be considered. From Figure 10.6, Y ¼ 0.92 for k ¼ 1.4 (air) and a
pressure ratio of 0.26.
As in the previous example, the discharge coefficient cannot be found explicitly unless the
flow rate is known, since C ¼ f ðRe d 1 ; bÞ. So a trial-and-error iterative approach is used. From
Figure 10.5, we start with a guess of K 0 ¼ CE % 0:61 ðor C ¼ 0:60Þ. Then,
Q ¼ CEYA o
ffiffiffiffiffiffiffiffi ffi
2Dp
r 1
s
¼ ð0:60Þð1:013Þð0:92Þð4:52 Â 10
À4 m
2
Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
ð2Þð24; 500 N/m
2
Þ
1:114 kg/m 3
s
¼ 0:053 m
3 /s
Check: For this flow rate, Re d 1 ¼ 4 Q/pd 1 y ¼ 7:4 Â 10
4 and from Figure 10.5, K 0 % 0:61 as
assumed. The flow rate through the orifice is taken to be 0.053 m
3 /s.
Example 10.6
A pump can often impose pressure oscillations into a pipe flow that affect pressure measurements.
Figure 10.12 plots separate pressure measurements taken upstream, p 1 , and downstream, p 2 , of an
orifice meter, as well as pressure differential measurements, Dp ¼ p 1 À p 2 , taken across the meter.
The systematic effect of the oscillations on the internal pipe pressure is seen in comparing p 1 and p 2 ,
which rise and fall together with a correlation coefficient of r p 1 p 2 ¼ 0:998. Estimate the contribution
to random standard uncertainty in estimating pressure differential due to the data scatter. The
following information is known:
p 1 ¼ 9:25 kPa s p 1 ¼ 2:813 kPa N ¼ 20
p 2 ¼ 7:80 kPa s p 2 ¼ 2:870 kPa N ¼ 20
Dp ¼ 1:45 kPa s Dp ¼ 0:188 kPa N ¼ 20
438 Chapter 10 Flow Measurements
13:4:38 Page 438
KNOWN d 1 ¼ 6 cm p 1 ¼ 93:7 kPa abs
b ¼ 0:4
T 1 ¼ 20
C ¼ 293 K
H ¼ 250 cm H 2 O
Properties (found in Appendix B)
Air : v ¼ 1:0 Â 10
À5 m
2 /s
Water : r H 2 O ¼ 999 kg/m
3
ASSUMPTIONS Treat air behavior as an ideal gas ðp ¼ rRTÞ
FIND Volume flow rate, Q
SOLUTION The orifice flow rate is found from Equation 10.12, which requires information
about E, C, and Y. From the given information, we calculate both the orifice area,
A d 0 ¼ 4:52 Â 10
À4 m
2 , and, from Equation 10.11, the velocity of approach factor, E ¼ 1:013.
The air density is found from the ideal gas equation of state:
r 1 ¼
p 1
RT 1
¼
93;700 N/m
2
287 N-m/kg
ð
Þ293 K
ð
Þ
¼ 1:114 kg/m
3
or use air-property tables. The pressure drop is p 1 À p 2 ¼ r H 2 O gH ¼ 24; 500 N/m
2 .
The pressure ratio for this gas flow, ðp 1 À p 2 Þ=p 1 ¼ 0:26. For pressure ratios greater than 0.1 the
compressibility of the air should be considered. From Figure 10.6, Y ¼ 0.92 for k ¼ 1.4 (air) and a
pressure ratio of 0.26.
As in the previous example, the discharge coefficient cannot be found explicitly unless the
flow rate is known, since C ¼ f ðRe d 1 ; bÞ. So a trial-and-error iterative approach is used. From
Figure 10.5, we start with a guess of K 0 ¼ CE % 0:61 ðor C ¼ 0:60Þ. Then,
Q ¼ CEYA o
ffiffiffiffiffiffiffiffi ffi
2Dp
r 1
s
¼ ð0:60Þð1:013Þð0:92Þð4:52 Â 10
À4 m
2
Þ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
ð2Þð24; 500 N/m
2
Þ
1:114 kg/m 3
s
¼ 0:053 m
3 /s
Check: For this flow rate, Re d 1 ¼ 4 Q/pd 1 y ¼ 7:4 Â 10
4 and from Figure 10.5, K 0 % 0:61 as
assumed. The flow rate through the orifice is taken to be 0.053 m
3 /s.
Example 10.6
A pump can often impose pressure oscillations into a pipe flow that affect pressure measurements.
Figure 10.12 plots separate pressure measurements taken upstream, p 1 , and downstream, p 2 , of an
orifice meter, as well as pressure differential measurements, Dp ¼ p 1 À p 2 , taken across the meter.
The systematic effect of the oscillations on the internal pipe pressure is seen in comparing p 1 and p 2 ,
which rise and fall together with a correlation coefficient of r p 1 p 2 ¼ 0:998. Estimate the contribution
to random standard uncertainty in estimating pressure differential due to the data scatter. The
following information is known:
p 1 ¼ 9:25 kPa s p 1 ¼ 2:813 kPa N ¼ 20
p 2 ¼ 7:80 kPa s p 2 ¼ 2:870 kPa N ¼ 20
Dp ¼ 1:45 kPa s Dp ¼ 0:188 kPa N ¼ 20
438 Chapter 10 Flow Measurements