Flow Measurement - ROYAL MECHANICAL | Specially designed

Report
The orifice meter consists of an accurately machined and
drilled plate concentrically mounted between two flanges.
The position of the pressure taps is somewhat arbitrary.
The orifice meter has several practical advantages
when compared to venturi meters.
• Lower cost
• Smaller physical size
• Flexibility to change throat to pipe diameter
ratio to measure a larger range of flow rates
Disadvantage:
• Large power consumption in the form of
irrecoverable pressure loss
The development of the orifice meter equation is similar to
that of the venturi meter and gives:
V 
2 pa  pb 
C0
1 
4

q  V S0
where:
–= ratio of orifice diameter to pipe diameter ≈ 0.5 usually
S0 = cross sectional area of orifice
V = bulk velocity through the orifice
C0 = orifice coefficient ≈ 0.61 for Re > 30,000
There is a large pressure drop much of which is not
recoverable. This can be a severe limitation when
considering use of an orifice meter.
Fluid Meters: Their Theory and Applications,
6th ed., American Society of Mechanical
Engineers, New York, 1971 pp. 58-65.
Rotameters fall into the category of flow
measurement devices called variable area
meters. These devices have nearly constant
pressure and depend on changing cross
sectional area to indicate flow rate. Rotameters
are extremely simple, robust devices that can
measure flow rates of both liquids and gasses.
Fluid flows up through the tapered tube and
suspends a ‘float’ in the column of fluid. The
position of the float indicates the flow rate on a
marked scale.
Three types of forces must be
accounted for when analyzing
rotameter performance:
• Flow
• Gravity
• Buoyancy
Buoyancy
Gravity
For our analysis neglect drag effect
Flow
Mass Balance
Assume Gradual Taper
V1S  V2 S
Q
V1  V2 
S
Flow Between Float and Tube
Q
S
V3 
 V1
S  S f  S3
S3 is annular flow area at plane 3
Momentum Balance
Note:
• p3 = p2
• Must account for force due to float
QV3 V1    p1  p2 S   gzS Vf   f gVf
p
 Q
 gz   

S
2

S  gV f   b   
1   
S

 S3 
Mechanical Energy Balance
1 2
p
2
ˆ
W0  V3  V1  gz 
 hf
2



2
V
Assume: h f  K R 3
2
(Base velocity head on
smallest flow area)
2
2






p
1 2
S
S
2
2


 gz  V1  V1    K RV1   

2
S3 
S3  




Combining Momentum and Mechanical Energy Balance
2

S 
S  gVf b    1  Q 
 Q  
   1  1  K R   
  1   

2 S  
S3  
 S   S3  S



2
2
After Some Manipulation
Q  S3
S Sf
1  K R S S f 
2
2 gVf  f  
Sf

Assuming Sf ≈ S a discharge coefficient can be
defined
1 2
CR  1  K R 
Q  S3C R
2 gV f  f  
Sf

CR must be determined experimentally. As Q increases the float rides
higher, the assumption that Sf = S is poorer, and the previous expression
is more nearly correct.
Measure by determining RPM of turbine (3) via sensor (6).
Turbine meters are accurate but fragile.
When fluid is passed through a U-bend, it imposes a force on the tube
wall perpendicular to the flow direction (Coriolis force). The
deformation of the U-tube is proportional to the flow rate. Coriolis
meters are expensive but highly accurate.
A 2 in. Schedule 40 pipe carries 35º API distillate at 50° F (SG=0.85). The flow
rate is measured by an orifice meter which has a diameter of 1.5 in. The
pressure drop across the orifice plate is measured by a water manometer
connected to the flange taps. If the manometer reading is 20 in. of H2O, what
is the flow rate of the oil in GPM ?
Uo 

Co
2( Pa  Pb )
1  4

do
1.5 in

 0.726
d p 2.067 in
P   g h  ( Pa  Pb )
Assum e: Co  0.61  N Re  30,000
P  (1  0.85)  62.4
Uo 
0.61
1  (0.726) 4
lbm 
lbm
ft   20 

32
.
2

ft

502
.
3




ft 3 
s 2   12 
ft s 2

lb 
2   502.3 m 2 
ft s 
ft

 3.120
lb
s
53.04 m3
ft
.
2
 1.5 

ft 
2
 d o 
ft 7.48 gal 60s
12 

Q
U o 
 3.120 

 17.2 GPM
4
4
s
ft 3
min
.
N RE
 1.5   3.120 ft   53.04 lbm 

ft   


s  
ft 3 
d o U o    12  


 6840 30,000


 4 lbm 
 6.719710

ft
s

4.5 cP  


cP




Now what ???

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