6.5.1 Potentially explosive atmospheres
Reference should be made to IEC 60079-0:2007 for general requirements.
NOTE It is possible that reference should also be made to other parts of the IEC 60079 series.
6.5.2 Cabling
The specification for the cabling of electrical instrumentation will be defined by the instrumentation design engineer and will be influenced by the type of instrument in question. Nevertheless, the following simple rules can be given:
a) signal cables should be as short as possible;
b) shielded cables which are earthed only at one point should be used;
c) weak signals should be amplified before they are transmitted through the cables;
d) power cables should be separated from instrument cables and should only cross instrumentation lines at right angles;
e) signal lines should be shielded from electrical lines.
6.5.3 Electronic equipment
The installation of electronic equipment should be carried out in accordance with the Code of Practice appropriate for the intended use.
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Annex A (informative)
Principles of measurement and computation
A.1 Formulae A.1.1 General
In all formulae, d, D and β refer to actual flowing conditions. In particular, when the flowing temperature differs from the temperature at which these dimensions were measured (usually 20 °C), the values shall be corrected for thermal expansion (see ISO 5167-1:2003, 5.4.4.1).
An explanation of the symbols used can be found in ISO 5167-1:2003, Clause 4.
A.1.2 Formulae common to all devices
Mass flowrate: [1 4] 0,5 2[2 1]0,5
m 4
q = −β − Cεπd ∆pρ (A.1)
Volume flowrate: 1
1
or
m m
V VR
R
q q
q q
ρ ρ
= = (A.2)
where 1 1
1 1 R R R
R
p T Z p T Z
ρ =ρ (A.3)
Subscript “1” refers to the flow condition at the upstream pressure tapping cross-section.
Subscript “R” refers to given conditions of pressure and temperature.
Reynolds number: 1 1
1 1 1
4 m 4 V
D V D q q
Re = ν = πDà = πDν (A.4)
A.1.3 Limits of use of primary devices
The formulae given for C and ε in all parts of ISO 5167 for the various primary devices can be applied only when certain quantities lie within given limits.
These limits of use are recalled in Table A.1.
A.2 Example of computation
A.2.1 General
Four detailed examples are shown below which deal with a compressible fluid and the discharge coefficient depending on β and ReD.
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As will be seen later, it may be convenient to consider the discharge coefficient C as the sum of two terms, C = C∞ + CRe, where C∞ is the discharge coefficient obtained for an infinite Reynolds number. Table A.2 shows the formulae giving C∞ and CRe for each type of device.
Reference should be made to the table of iterative computations in ISO 5167-1:2003, Annex A.
Depending on the quantity which is to be calculated, additional equations derived from Equation (A.1) may be useful. Table A.3 shows the equation needed in the four types of problem usually encountered together with the quantities which have to be known to perform the calculations.
In all examples, 10-digit numbers are listed which is much more accurate than can be justified for practical purposes, but which can be helpful when checking the accuracy of the computer programs.
In each case, the aim is to solve an equation f(X) = X; so if Xi is the ith approximation to the true answer δI can be defined as f(Xi) − Xi and the iterative algorithm inAnnex A of ISO 5167-1:2003 becomes
1 1
1 1
( ) )
( ) ( )
n n n n
n n
n n n n
f X X X X
X X
f X X f X X
+ −
− −
− −
⎡ ⎤
⎣ ⎦
= −
− − +
( (A.5)
An initial value, X1, is required; then X2 = f(X1); then the above equation can be used for n = 2, … Equation (A.5) can be rewritten as
(1 ) ( )
n n n
X +1= −E f X (A.6)
where
1 1
( ( ) ) ) ( )
( ) ( ) ( )
n n n n
n n n n n n
f X X f X f X
E f X X f X f X X
−
− −
− ⎡⎣ − ⎤⎦
= ⎡⎣ + − − ⎤⎦
( 1
(A.7) Then, given an initial value, X1, Equation (A.6) can be used for subsequent iterations with E1= 0 and
Equation (A.7) for n = 2, …
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© ISO 2008 – All rights reserved 43 Table A.1 — Limits of use d DβReD Type of device mm mm Roughness Criteria Corner tappings orifice plate D and D/2 tappings orifice plate W 12,5 50 uDu 1 000 0,10 uβ u 0,75 ReDW 5 000 for 0,10uβ u 0,56 ReDW 16 000β2 forβ > 0,56 See ISO 5167-2:2003, Tables 1 and 2 Flange tappings orifice plate W 12,5 50 uDu 1 000 0,10 uβ u 0,75 ReDW 5 000 and ReD W 170β2D* See ISO 5167-2:2003, Tables 1 and 2 ISA 1932 nozzle — 50 uD u 500 0,30 uβ u 0,80 70 000 uReDu 107 for 0,30 uβ u 0,44 20 000 uReDu 107 for 0,44 uβ u 0,80See ISO 5167-3:2003, Table 1 Long-radius nozzle — 50 uD u 630 0,20 uβ u 0,80 104uReDu 107Ra/Du 3,2 × 10−4 Rough-cast convergent Venturi tube— 100 uDu 800 0,30 uβ u 0,75 2 × 105 uReDu 2 × 106Ra < 10−4d** Ra < 10−4D*** Rough-welded convergent Venturi tube— 200 uD u 1 200 0,40 uβ u 0,70 2 × 105 uReDu 2 × 106Ra < 10−4d** Ra≈ 5 × 10−4D*** Machined convergent Venturi tube— 50 uD u 250 0,40 uβ u 0,75 2 × 105 uReDu 1 × 106Ra < 10−4d** *** Venturi nozzle W 50 65 uD u 500 0,316 uβ u 0,775 1,5 × 105uReD u 2 × 106See ISO 5167-3:2003, Table 2 Key * Where D is in millimetres. ** Throat roughness criterion. *** Convergent section roughness criterion. NOTE For all devices,∆p/p1 u 0,25 when used with compressible fluids.
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Table A.2 — C∞ and CRe for orifice plates for D > 71,12 mm: C = C∞ + CRe
Type of device Equations Equation
number
,corner ,
2 8
,corner 0,5961 0,0261 0,216
C C C L
C β β
∞ ∞ ∞
∞
= +
= + −
1 1
10 7 4
, (0,043 0,080e 0,123e ) 4 1
L L
C L β
β
− −
∞ = + −
− 1,1 1,3
2 2
0,031(M' 0,8M' )β
− −
(A.8.1)
Orifice plates
0,7 0,3
6 6
10 3,5 10
0,000 521 (0,018 8 0,006 3 )
Re D D
C A
Re Re
β β
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
= ⎜⎝ ⎟⎠ + + ⎜⎝ ⎟⎠
1 1
10 7 4
0,11 (0,043 0,080e 0,123e ) 4 1
L L
A β
β
− −
− + −
−
(A.8.2)
L1, M2′ and A are as defined in ISO 5167-2:2003, 5.3.2.1.
Table A.3 — Iteration equations
Known parameters Quantities to be computed Equations Equation number
f(qm) = CKq (A.9.1)
d D ∆p qm
(1 4) 0,5 4 2 2 1
Kq= −β − επd ∆pρ (A.9.2)
2 2 0,25
( ) (1 )
f β = +C ε Kβ − (A.9.3)
qm ∆p D β 2 2
1
8 m
p D
Kβ q
ρ ⎛ ⎞
∆ ⎜π ⎟
= ⎜⎝ ⎟⎠
(A.9.4)
f(∆p) = K∆pε−2 (A.9.5)
d D qm ∆p ( 4) 2
1 2
8 1 m
p q
K Cd
β
∆ ρ
− ⎛ ⎞
= ⎜⎝π ⎟⎠ (A.9.6)
f(D) = KDC−0,5 (A.9.7)
qm ∆p β D ( 4) 2 0,25
1 4
8 1 m
D q
K p
β ρ β ε
⎡ − ⎛ ⎞ ⎤
⎢ ⎥
=⎢⎢⎣∆ ⎜⎝π ⎟⎠ ⎥⎥⎦
(A.9.8)
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© ISO 2008 – All rights reserved 45 A.2.2 Determination of D — Example
See Figure A.1 for an example of a flowchart.
Assume an orifice plate metering facility using flange taps has to be designed for the following conditions:
⎯ fluid: steam
⎯ maximum flowrate: 1 kgãs−1
⎯ maximum diameter ratio: 0,65
⎯ maximum pressure differential: 0,5 × 105 Pa (500 mbar)
⎯ pressure: 10 × 105 Pa (10 bar)
⎯ temperature: 773,15 K (500 °C)
⎯ λd = 16 × 10–6 K−1
⎯ λD = 11 × 10–6 K−1 Use the following typical data:
⎯ ρ1 = 2,825 1 kgãm−3
⎯ à1 = 28,5 ì 10−6 Paãs
⎯ κ = 1,276
The exit criterion chosen is 10−6 (0,000 1 %). The calculation procedure is then:
*** Assessing starting values
1) ε, applying equation for expansibility [ISO 5167-2:2003, 5.3.2.2]:
( 4 8) 21 1/
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,983 201 997 0.
For manual calculations using a calculator with a memory, it is useful to store the value of β4, since it is required in a number of the subsequent equations
2) KD, applying Equation (A.9.8): ( 4) 2 0,25
1 4
8 1 m
D q
K p
β ρ β ε
⎡ − ⎛ ⎞ ⎤
⎢ ⎥
=⎢⎢⎣ ∆ ⎜⎝π ⎟⎠ ⎥⎥⎦
KD = 0,072 295 778 11
In each case, p1, T1, ρ1, U1, K shall also be known.
3) Applying Equation (A.8.1) for corner tappings:
2 8
,corner 0,596 1 0,026 1 0,216
C∞ = + β − β
C = C∞, corner = 0,600 244 522 0
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For all the other devices, C∞ could be readily calculated at this stage.
4) The starting value of D is obtained from Equation (A.9.7): f(D) = KD C−0,5 D1 = f(D) = 0,093 314 435 6
5) Reynolds number from Equation (A.4):
ReD = 478 758,419 9
NOTE For most practical purposes, it is possible to stop the calculation here, since the final result will not be significantly different from D1 and will be eventually rounded up to the next commercially available pipe diameter.
The final result obtained by the complete computation would be:
D = 0,092 707 108 61
From the previous calculation of D, the nearest commercially available pipe diameter, D = 0,102 m, would be selected by the designer of the metering station.
A.2.3 Computation of β — Example Refer to Figure A.2 for an example of a flowchart.
It is now necessary to calculate the orifice diameter d for the same conditions as in A.2.2, i.e.
⎯ fluid: steam
⎯ maximum flowrate: 1 kgãs−1
⎯ maximum pressure differential: 0,5 × 105 Pa (500 mbar)
⎯ pressure: 10 × 105 Pa (10 bar)
⎯ temperature: 773,15 K (500 °C)
⎯ pipe diameter at ambient: D0 = 0,102 m
⎯ λd = 16 × 10−6 K−1
⎯ λD = 11 × 10−6 K−1
⎯ ρ1 = 2,825 1 kg ⋅ m−3
⎯ à1 = 28,5 ì 10−6 Paãs
⎯ K = 1,276
The exit criterion being still 10−6, the calculation would be:
*** Assessing starting values:
1) D is obtained from Equation (7): D = D0 [1 + λD (T − T0)]
D = 0,102 538 560 0
2) ReD, from Equation (A.4):
1
4 m
D q
Re = πDà ReD = 435 690,453 9
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© ISO 2008 – All rights reserved 47 3) β being unknown, it is convenient and reasonable to use ε = 0,97 as the starting value, except in the
case of incompressible fluids, for which ε = 1.
4) β being unknown, it is convenient either
⎯ to use a fixed starting value of C, e.g. 0,60 for orifice plates and 0,99 for all types of nozzles, or
⎯ to use as a starting value C = C∞ (in the case of classical Venturi tubes C is a constant).
The second method is preferable when the diameter ratio β (and D for orifice plates using flange tappings) is a known parameter; in such a case, C is calculated from C = C∞ + CRe in the iteration steps, where C∞ has already been calculated.
In the case of orifice plates using flange tappings where D is not known and β is known, the starting value of C can be taken as equal to C∞,corner, i.e. the value of C∞ that would be obtained for corner tappings. In the iteration steps, C has to be computed as:
C = C∞,corner + C∞,L + CRe
where the last two terms have to be recalculated at each step.
In most practical cases however, it will be sufficient to assume C = C∞ and make no iteration.
5) Kβ, from Equation (A.9.4):
2 2 1
8 m
p D
Kβ q
ρ ⎛ ⎞
∆ π
= ⎜⎜ ⎟⎟
⎝ ⎠
Kβ = 19,264 708 61
6) Starting value of β from Equation (A.9.3): f(β) = (1 + C2ε2Kβ) −0,25 β1 = f(β) = 0,603 764 155 8
*** First iteration step
7) ( 4 8) 2 1/
1
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,984 182 761 4
8) C, from Equations (A.8.1) and (A.8.2):
C = C∞ + CRe C = 0,607 261 036 6
9) Next value of β from Equation (A.9.3): f(β) = (1 + C2ε2Kβ)−0,25 β = f(β1) = 0,596 831 560 9
No correction being made at the first step (E1 = 0), the starting value for the second step is:
β2 = f(β1) = 0,596 831 560 9
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*** Second iteration step
10) ( 4 8) 2 1/
1
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,984 300 372 0
11) C, from Equations (A.8.1) and (A.8.2).
C = 0,607 076 664 5
12) Next value of β from Equation (A.9.3): f(β) = (1 + C2ε2Kβ)−0,25 β = f(β2) = 0,596 879 546 2
13) Deviation in f(β2) is given by: 2 2 2 2
2 2 2 1
[ ( ) ] 1
( ) [2 ( ) ] f X X
E f X X f X X
= − −
− −
E2 = 5,526 344 567 × 10−7
which is less than the exit criterion. The iteration is then stopped.
d = βD d = 0,061 203 169 16 m
d0 from Equation (4): d = d0 [1 + λd(T − T0)] d0 = 0,060 736 711 22 m A.2.4 Computation of qm — Example
Refer to Figure A.3 for an example of a flowchart.
Assume the metering station is now used to measure a flowrate with a plate of diameter d0 = 0,061 m in the following conditions:
⎯ fluid: steam
⎯ pressure differential: 0,481 × 105 Pa (481 mbar)
⎯ pressure: 10 × 105 Pa (10 bar)
⎯ temperature: 773,15 K (500 °C)
⎯ ρ1 = 2,825 1 kgãm−3
⎯ d0 = 0,061 m
⎯ D0 = 0,102 m
⎯ à1 = 28,5 ì 10−6 Paãs
⎯ λd = 16 ×10−6 K−1
⎯ λD = 11 ×10−6 K−1
⎯ κ = 1,276
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© ISO 2008 – All rights reserved 49 The exit criterion being 10−6, the calculation would be:
*** Assessing starting values:
1) d is obtained from Equation (4): d = d0 [1 + λd (T – T0)]
d = 0,061 468 480 00
2) D, from Equation (7): D = D0 [1 + λD (T – T0)]
D = 0,102 538 560 0
3) β, from β = d/D β = 0,599 466 971 3
4) ( 4 8) 2 1/
1
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,984 857 929 9
5) Kq, from Equation (A.9.2): (1 4) 0,5 2(2 1)0,5
q 4
K = −β − ε πd ∆pρ
Kq = 1,632 671 123 6) C, from Equation (A.8.1):
C = C∞ = 0,602 425 043 2
7) The starting value for qm from Equation (A.9.1): f(qm) = CKq qm,1 = f (qm) = 0,983 561 971 8
*** First iteration step
8) Reynolds number from Equation (A.4):
1
4 m
D
Re q
Dà
= π
ReD = 428 528,561 9
9) New estimate of C, from C = C∞ + CRe C = 0,607 176 725 2
10) New estimate of qm, from Equation (A.9.1): f (qm) = CKq qm = f (qm,1) = 0,991 319 905 8
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No correction being made at the first step (E1 = 0), the starting value for the second step is:
qm,2 = f (qm,1) = 0,991 319 905 8
*** Second iteration step
11) New Reynolds number from Equation (A.4):
1
4 m
D
Re q
Dà
= π
ReD = 431 908,619 7
12) New value of C, from C = C∞ + CRe C = 0,607 163 108 8
13) New value of qm, from Equation (A.9.1): f (qm) = CKq qm = f (qm,2) = 0,991 297 674 7
14) Deviation of f (qm,2):
2 2 2
2 2 2 2 1
[ ( ) ]
1
( ) [2 ( ) ]
f X X
E f X X f X X
= − −
− −
E2 = − 6,408 057 577 ×10−8
which is less than the exit criterion. The iteration is then stopped and the result is:
qm = 0,991 297 674 7 kg.s−1
A.2.5 Determination of ∆p — Example Refer to Figure A.4 for an example of a flowchart.
Assume the pressure differential is required for the maximum flowrate of the same facility if the plate has a diameter of d0 = 0,050 m.
⎯ fluid: steam
⎯ flowrate: 1 kgãs−1
⎯ pressure: 10 × 105 Pa (10 bar)
⎯ temperature: 773,15 K (500 °C)
⎯ density: 2,825 1 kgãm−3
⎯ d0 = 0,050 m
⎯ D0 = 0,102 m
⎯ à1 = 28,5 ì 10−6 Paãs
⎯ λd = 16 × 10−6 K−1
⎯ λD = 11 × 10−6 K−1
⎯ κ = 1,276
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© ISO 2008 – All rights reserved 51 10−6 being the exit criterion, the calculation would be:
*** Assessing starting values
1) d is obtained from Equation (4): d = d0 [1 + λd (T − T0)]
d = 0,050 384 000 00
2) D, from Equation (7): D = D0 [1 + λD (T − T0)]
D = 0,102 538 560 0
3) β, from β = d/D β = 0,491 366 369 9 4) ReD from Equation (A.4):
1
4 m
D q
Re = πDà
ReD = 435 690,453 9
5) C, from Equations (A.8.1) and (A.8.2): C = C∞ + CRe C = 0,603 572 933 9
6) K∆p, from Equation (A.9.6):
4 2 1 2
8(1 ) m
p q
K Cd
β
∆ ρ− ⎛ ⎞
= ⎜ ⎟
⎝π ⎠ K∆p = 115 091,115 8
7) ε is taken as equal to 0,97.
8) Starting value for ∆p, from Equation (A.9.5): f(∆p) = K∆pε−2
∆p1 = f (∆p) = 122 320,242 1
*** First iteration step
9) 1 (0,351 0,256 4 0,93 8) 1 p21 1/
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,964 125 846 1
10) Next value of ∆p, from Equation (A.9.5): f(∆p) = K∆pε−2
∆p = f (∆p1) = 123 815,310 0
No correction being made at the first step (E1 = 0), the starting value for the second step is:
∆p2 = f (∆p1) = 123 815,310 0
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*** Second iteration step
11) New value of ε: ( 4 8) 2 1/
1
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,963 680 936 9
12) Next value of ∆p, from Equation (A.9.5): f(∆p) = K∆pε–2
∆p = f(∆p2) = 123 929,661 7
13) Deviation of f(∆p2):
2 2 2
2 2 2 2 1
[ ( ) ]
1
( ) [2 ( ) ]
f X X
E f X X f X X
= − −
− −
E2 = − 7,641 976 350 × 10−5
For manual computation, one would stop here. The calculation is carried on to show the effect of the rapid scheme.
*** Third iteration step
14) The starting value for step 3 is obtained from: ∆pn+1 = (1 − En) f(∆pn)
∆p3 = 123 939,132 4
NOTE If substitution iteration was continued [∆pn+1 = f(∆pn) for all steps], a total of 5 iteration steps would be necessary to conform to the exit criterion.
15) New value of ε: ( 4 8) 2 1/
1
1 0,351 0,256 0,93 1 p
p κ
ε = − + β + β ⎡⎢⎢⎣ − ⎜⎛⎝ ⎞⎟⎠ ⎤⎥⎥⎦
ε = 0,963 644 081 9
16) Next value of ∆p, from Equation (A.9.5): f(∆p) = K∆pε−2
∆p = f(∆p3) = 123 939,141 4
17) Deviation on f (∆p3): 1
1 1
[ ( ) ][ ( ) ( )]
1
( ) [ ( ) ( ) ]
n n n n
n n n n n n
f X X f X f X
E f X X f X f X X
−
− −
− −
= + − −
E3 = − 6,017 524 711 × 10−9
The iteration is then stopped, the exit criterion being met. The result is ∆p = 123 939,141 4 Pa.
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© ISO 2008 – All rights reserved 53
Key
a ε and p1 have to be known for compressible fluids only.
b For incompressible fluids, ε = 1.
c For classical Venturi tubes and Venturi nozzles, C = C∞, and no loop is necessary.
d For the first step, E1 = 0 but proceed to “NO”, except for classical Venturi tubes and Venturi nozzles.
e Except for flange tapping orifice plates, only CRe has to be computed here, then it is added to previously computed C∞.
Figure A.1 — Flowchart example — Computation of pipe diameter D
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54 © ISO 2008 – All rights reserved Key
a ε and p1 have to be known for compressible fluids only.
b For incompressible fluids, ε = 1 and need not be computed further again.
c For classical Venturi tubes, C is a constant and need not be computed further again. If in addition the fluid is incompressible, no iteration is necessary.
d For the first step, E1 = 0 but proceed to "NO", except for Venturi tubes operated with incompressible fluids.
Figure A.2 — Flowchart example — Computation of diameter d and diameter ratio β
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© ISO 2008 – All rights reserved 55
Key
a ε and p1 have to be known for compressible fluids only.
b For incompressible fluids, ε = 1 and need not be computed further again.
c For classical Venturi tubes and Venturi nozzles, C = C∞ and no iteration is necessary.
d For the first step, E1 = 0 but proceed to “NO”, except for classical Venturi tubes and Venturi nozzles.
e Only CRe has to be computed here, then added to already computed C∞.
Figure A.3 — Flowchart example — Computation of flowrate qm
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a ε and p1 have to be known for compressible fluids only.
b For incompressible fluids, ε = 1 and no loop is necessary.
c For the first step, E1 = 0 but proceed to “NO”, except for classical Venturi tubes and Venturi nozzles.
Figure A.4 — Flowchart example — Computation of pressure ∆p
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Annex B (informative)
Computation of compressibility factor for natural gases
B.1 Calculation of density, ρ
The density of a gas may be calculated by means of either of the following equations:
u
pM
ρ = R TZ (B.1)
0 0 0
0
pT Z p TZ
ρ = ρ (B.2)
where
Z is the compressibility factor;
ρ is the density;
M is the molecular weight of the gas;
Ru is the universal gas constant.
The subscript 0 refers to a reference state of temperature and pressure.
Z is a function of the composition of the gas.
B.2 Calculation of compressibility factor, Z
Modern methods for the computation of Z aim to cover the entire range of transmission metering conditions and gas compositions. These are described in ISO 12213 [2].
ISO 12213 has three parts: 1) introduction and guidelines; 2) calculation using molar composition analysis;
3) calculation using physical properties. All of these parts were published in 2006 and are based on AGA Report Number 8 [7].
The calculation of Z using molar composition analysis, also known as the “detailed method”, uses up to 21 components and has been thoroughly evaluated for a broad range of typical natural gas pipeline temperatures, pressures and gas compositions (see Reference [9]). High accuracy measurements on five gravimetrically prepared reference natural gas mixtures were made by four leading laboratories in Europe and North America. The gas compositions were selected by European and North American pipeline company representatives, and are characteristic of a wide range of commercial natural gases found world-wide.
The calculation of Z using physical properties, known as the “gross method”, uses a simplified input data set comprising any three from superior (gross) calorific value (heating value), relative density, carbon dioxide content and nitrogen content, together with pressure and temperature. With this limited information, the equation predicts Z within the respective pressure and temperature ranges of 0 MPa to 12 MPa (0 bar to 120 bar) and 265 K to 335 K (−8 °C to 62 °C) with an accuracy of about 0,1 %, about the same as the detailed method.
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The detailed and gross calculation methods have also been published in Technical Monographs by the Groupe Européen de Recherches Gazières (GERG) and are known as the Master (or Molar) GERG-88 Virial Equation and Standard (or Simplified) GERG-88 Virial Equation (SGERG).
Consistent thermophysical property calculations over a range of pipeline operating conditions are also needed for general orifice meter calibrations using sonic nozzles and cross-meter checking. The GRI/AGA8 detail equation provides highly accurate, internally consistent derived thermophysical properties at standard pipeline operating conditions. These properties include the speed of sound, heat capacity, enthalpy, and entropy required for sonic nozzle and other metering calculations used to evaluate and calibrate orifice, turbine and ultrasonic meters. It is not recommended that the SGERG equation be used for calculating derived thermophysical properties.
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Annex C (informative)
Orifice plate assembly
Recommended orifice plate assemblies are illustrated below.
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1 raised face (RF) flange 2 gasket
3 orifice plate
Figure C.1 — Standard RF orifice flange assembly
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Key
1 raised face (RF) flange 2 ‘O’ rings
3 orifice plate
4 dowel pins (for location)
Figure C.2 — Dowelled orifice flange assembly
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1 raised face (RF) flange 2 gasket
3 orifice plate (locates on flange RF outside diameter)
Figure C.3 — Orifice flange assembly
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1 raised face (RF) flange 2 gasket
3 orifice plates (locates in flange face recess)
Figure C.4 — Tongued faced orifice flange assembly
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1 raised face (RF) flange 2 gasket
3 integral carrier
Figure C.5 — Integral carrier orifice flange assembly
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1 single chamber orifice body 4 orifice plate 7 locking nut
2 upstream sealring 5 sealing bar 8 downstream sealring
3 ‘O’ ring 6 locking bar 9 winding mechanism to remove orifice plate
Figure C.6 — Single chamber orifice assembly
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1 ring type joint (RTJ) flange 2 orifice plate (integral male RTJ)
Figure C.7 — Standard RTJ orifice flange assembly
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1 ring type joint (RTJ) flange 2 ring type joint gasket
3 orifice plate (integral female RTJ)
Figure C.8 — Standard RTJ orifice flange assembly
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Bibliography
[1] ISO 5168:2005, Measurement of fluid flow — Procedures for the evaluation of uncertainties [2] ISO 12213 (all parts), Natural gas — Calculation of compression factor
[3] ISO/IEC Guide 98-31) , Uncertainty of measurement — Part 3: Guide to the expression of uncertainty in measurement (GUM:1995)
[4] IEC 60079-0:2007, Explosive atmospheres — Part 0: Equipment — General requirements [5] IEC 60584 (all parts), Thermocouples
[6] IEC 60751, Industrial platinum resistance thermometer sensors
[7] AGA Report Number 8, Compressibility factor of natural gas and related hydrocarbon gases. American Gas Association, January 1994
[8] BRAIN, T.J.S. and REID, J. Measurement of orifice plate edge sharpness. Measurement and Control, 6, 1973, pp. 377-383
[9] GERG Technical Monograph, High Accuracy Compressibility Factor Calculation for Natural Gases and Similar Mixtures by Use of a Truncated Virial Equation. GERG TM2 1988
[10] Matheson Gas Products, The Matheson unabridged gas data book. East Rutherford, New Jersey, 1974
[11] MILLER, R.W. Flow measurement engineering handbook. Third edition, 1996
[12] NORMAN, R., RAWAT, M.S. and JEPSON, P. Buckling and eccentricity effects on orifice metering accuracy. International Gas Research Conference, 1983
[13] READER-HARRIS, M.J., GIBSON, J., HODGES, D., NICHOLSON, I. and RUSHWORTH, R. The performance of flow nozzles at high Reynolds number. In: Proc. 14th International Flow Measurement Conference, FLOMEKO 14, Sandton, South Africa, September 2007
[14] SAVIDGE, J.L. and BEYERLEIN, S.W. GRI Report No. 93/D181, Technical Reference Document for 2nd Edition of AGA Report No. 8. Reprinted March 1995. Gas Research Institute, Chicago, Illinois
[15] STARLING, K.E. and SAVIDGE, J.L. Compressibility factors of natural gas and other related hydrocarbon gases. American Gas Association, Transmission Measurement Committee Report No. 8 and American Petroleum Institute, MPMS chapter 14.2, second edition
[16] ZEDAN, H.F. and TEYSSANDIER, R.G. The effect of recesses on the discharge coefficient of a flange tapped orifice plate, ASME Symposium on Mass Flow Measurement, 1984
1) ISO/IEC Guide 98-3 will be published as a reissue of the Guide to the Expression of Uncertainty in Measurement (GUM),1995.
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