Application Notes for Valve Sizing
Sizing examples
Inhalt
Ventilberechnung bei Flüssigkeiten
Ventilberechnung bei Wasserdampf
Ventilberechnung bei Gas und Dampf
Ventilberechnung bei Luft
Durchfluss durch Rohrleitungen
Seite
2
5
7
9
11
General information
These application notes provide a simplified procedure for
sizing control valves in standard service conditions.
The data required to size a valve, such as the nominal size,
nominal pressures and KVS coefficient, is contained in the
SAMSON data sheets for self-operated regulators and control
valves.
Self-operated regulators and control valves can be sized
accurately according to the IEC 60534 procedure. For most
applications, however, the following sizing equations
formulated in the VDI/VDE Guideline 2173 provide a sufficient
degree of accuracy.
In order to calculate the valve flow coefficient KV, the operating
data specified in the diagram on the right are required.
p1
p2
H100
H0
t1
.
V W
Typical sizing coefficients
Explanatory notes
Control valves and self-operated regulators
Rated travel · The rated travel H100 is the amount of movement of
the valve closure member from the closed position to the designated full open position published by the manufacturer for each
control valve series.
KV · The flow coefficient KV is defined as the number of cubic me.
ters per hour (volume flow rate V) of water at 5 to 30 °C that will
flow through a control valve at a specified travel H with a differential pressure (Dp = p1 – p2) of 1 bar.
KVS · The KVS coefficient is the expected flow coefficient KV of the
valve at rated travel H100 indicated and published by the manufacturer for each valve type (series).
KV100 · This value is the effective (actual) flow coefficient KV of the
valve at rated travel H100. It must not deviate by more than ±10
% from the specified KVS coefficient.
p1
p2
Dp
H
Upstream pressure in bar
Downstream pressure in bar
Differential pressure (p1 – p2) across the valve in bar
Travel in mm
V
W
r
r1
t1
Volume flow rate in m³/h
Mass flow rate in kg/h
Density (general) in kg/m³
Upstream density (for gases and steam) in kg/m³
Upstream temperature in ° C
.
Self-operated regulators
KVS
Safety factor S =
KV
KVS
KV
Indicated KVS
Calculated KV
For self-operated regulators: S » 1.3 to 5
To ensure the proper operation of a self-operated regulator, the
kinematic viscosity n of the medium to be controlled must not
exceed 1 · 10–4 m²/s = 100 cSt.
All pressures stated are absolute pressures in bar, if not specified
otherwise
Fig. 1 · Operating data for determining the KV coefficient
Edition August 2012
Application Notes
AB 04 EN
Liquids
Calculation of the maximum volume flow rate of liquids with different
densities
.
.
æ rA
÷
VB = VA × çç
÷
è rB ø
1000 × Dp
r × Dp0
.
V = KV ·
Equation (1) generally applies to liquids:
(1)
.
Diagram 1 shows the relationship between V, KV and Dp for liquids that have
a density of r = 1000 kg/m³ at a temperature of t = 20 °C.
Example 1
Flow rate calculation for a valve
Determine:
Volume flow rate of acetone (m³/h) in a fully open
valve
Type 3241-1 Pneumatic Control Valve · DN 40
Pressure drop Dp = p1 – p2 · Density in kg/m³ for acetone
Calculate according to equation (2):
Known:
Solution:
Symbols and units
p1
Absolute presssure
p2
in bar
Pressure drop (differential pressure) in
Dp
bar
Reference differential pressure of 1 bar
Dp0
according to the definition of K V. It is
omitted in the other calculations to
simplify them.
Density in kg/m³
r
.
Flow rate in m³/h
V
KV coefficent in m³/h
KV
Minimum differential pressure across the
Dpmin
valve in bar
Differential pressure in bar created
DpWirk
at the restriction in the flow path for flow
measurement purposes
KVS
Valve flow coefficient in m³/h
KVS =
Dp =
r =
.
1000 × Dp
V = KVS ·
r
.
25 m³/h 1)
p1 – p2 = 0.5 bar
800 kg/m³
1000 × 0.5
= 19.76 m³/h
800
V
=
25 ·
1)
The KVS coefficient has a permissible tolerance of
±10 % and, as result, the calculated volume
(2)
.
flow rate V , too .
Example 2
Pressure reducing valve (self-operated regulator) for water
Determine:
Type ... Pressure Reducing Valve, KVS and nominal size
of the valve
Volume flow rate of water · Pressure drop Dp
Density r in kg/m³ for water
Calculate KV coefficient according to equation (3)
derived from equation (1):
Known:
Solution:
.
KV = V ·
r
1000 × Dp
(3)
After calculating the KV coefficient, select the KVS
coefficient of the valve. As a general rule: KVS » 1.3 · KV
Solution according to Diagram 1:
KV » 8.2 m³/h can be read from Diagram 1 for
.
Dp = 2.1 bar and V = 12 m³/h.
Flow velocity in the pipe:
The following can be read from Diagram 4 (page 11) for
.
V =
Dp =
r =
12 m³/h
p1 – p2 = 2.1 bar
1000 kg/m³
KV =
12 ·
1000
= 8.2 m³/h
1000 × 2.1
KVS = 1.3 · KV = 1.3 · 8.2 = 10.7 m³/h
Selected: Type 41-23, DN 40, KVS = 20
Determining the safety factor:
KVS 20
=
S =
» 2.4
KV 8.2
.
V = 12 m³/h and DN 40: wpipe » 2.8 m/s
Note:
The flow velocity wpipe must not exceed 2 m/s for HVAC
and district heating applications!
2
AB 04 EN
m3
h
1000
800
Kv
600
500
400
00
20
00
12 00
10
0
80
0
60 0
50
0
40
300
200
0
30 0
25
0
20
0
16
0
12
0
10
80
100
80
60
50
40
60
50
40
30
25
20
16
30
20
10
8
12
10
8
6
5
4
6
5
3
4
3.0
2.5
2.0
1.6
1.2
1.0
0.8
0.6
0.5
0.4
0
0.3 5
0.2 0
0.2
6
0.1
2
0.1 0
0.1
8
0.0
6
0.0 5
0.0 4
0.0
30
0.0 25
0.0 20
0.0 6
1
0.0 2
1
.
0 0 10
0.0
2
1
0.8
0.6
0.5
0.4
0.3
0.2
0.1
0.08
0.06
0.05
0.04
0.03
0.02
p
0.01
10
20
30 40
60 80 100
0.1
200 300 400 600 800 1000
mbar
0.2
2
0.3 0.4
0.6 0.8 1
3
4
6
8 10
20
30 40
bar
.
V in m³/h
KV in m³/h
Dp in bar
.
V = KV ·
1000 × Dp
r
Diagram 1 · Volume flow rate diagram for water r = 1000 kg/m³, t = 20 °C
AB 04 EN
3
Liquids
Example 3
Self-operated flow regulator for water
Determine:
Required minimum differential pressure Dp for flow
rate control
Type 42-36 Flow Regulator
Differential pressure at restriction of 0.2 bar
DN 40 · KVS 20 · Flow rate of water
Calculate according to equation (4) for flow regulators
with incompressible media 1):
Known:
Solution:
æ . 2
V ÷
Dpmin = Dprestriction + ç
çç KVS ÷÷
è
ø
KVS
Dprestriction
=
=
20 m³/h
0.2 bar
V
=
10 m³/h water 1)
Dpmin
=
æ 10 2
0.2 + ç ÷ = 0.45 bar
è 20 ø
.
(4)
1)
Contact SAMSON for flow rate control of air,
steam or gases
Dprestriction = 0.2 or 0.5 bar; depending on regulator version
Example 4
Determining the pressure loss with water
Determine:
Known:
Pressure loss Dp = p1 – p2 in a fully open valve
Type 4 Temperature Regulator (self-operated) · DN 50
.
Volume flow rate V of water
Density r in kg/m³ for water
Calculate pressure loss according to equation (5) derived
from equation (1):
Solution:
æ . 2
V ÷
r
·
Dp = ç
çç KVS ÷÷
1000
è
ø
KVS =
.
V
r
=
=
Dp =
32 m³/h
10 m³/h
1000 kg/m³
æ 10 2 1000
= 0.097 bar » 0.1 bar
ç ÷ ·
1000
è 32 ø
(5)
Solution according to Diagram 1:
The pressure loss Dp » 0.1 bar can be read from Diagram
.
1 for V = 10 m³/h and KVS 32
4
AB 04 EN
Steam
A modified version of equation (9) applies to steam:
W = KV · m · Z
Z
m
(6)
The dimensionless compressibility factor Z is defined
as follows: Z = 14.2 · p1 × r1. Determine Z from Table 2
using the upstream pressure p1 and differentiating
between saturated steam and superheated steam.
Determine the dimensionless head loss coefficient m from
Table 1 or, interpolate values from Diagram 2 using
c = 1.135.
Symbols and units
p1
Absolute pressure
p2
in bar
Pressure loss (differential pressure) in bar
Dp
W
Mass flow rate in kg/h
KV
KV coefficient in m³/h
m
Pressure loss coefficient, dimensionless
Z
14.2 · p1 × r1 , dimensionless
Example 5
Determining the flow rate in a valve
Determine:
Known:
Steam flow rate in kg/h in a fully open valve
Type 3241-2 Electric Control Valve
Steam temperature · Upstream and downstream pressures
p2
Calculate
p1
KVS =
t
=
p1 =
p2
=
p1
40 m³/h 1)
200 °C
4 bar p2 = 3 bar
3
= 0.75
4
Determine m from Table 1 or Diagram 2
Determine Z from Table 2 using the upstream pressure and
temperature
m =
Z =
W =
0.92
38.5
40 · 0.92 · 38.5 = 1417 kg/h
Solution:
W = KVS · m · Z
(6)
Example 6
Steam pressure reducing valve (self-operated)
Determine:
Type ... Pressure Reducing Valve, KVS and nominal size
of the valve
Steam flow rate · Steam temperatures · Upstream and
downstream pressures
p2
Calculate
p1
Determine m from Table 1 · Determine Z from Table 2
using the upstream pressure and temperature
Calculate KV from equation (7):
Known:
Solution:
KV =
W
Z ×m
Example 7
Determining pressure loss with steam
Determine:
Known:
Pressure loss Dp = p1 – p2 in a fully open valve
Type 4 Temperature Regulator
Steam flow rate · Steam temperature · Upstream pressure
Select compressibility factor Z from Table 2 depending on
pressure and temperature upstream of the valve
m=
W
Z × KVS
p2
from diagram 2 for m = 0.701
p1
p2 = 0.89 · p1 -> Pressure loss Dp = p1 – p2
Find the ratio
AB 04 EN
The KVS coefficient has a tolerance of ±10 %,
and, as a result, the calculated steam flow rate W,
too.
W
t
p1
p2
p1
=
=
=
m
Z
=
=
=
1
71.3
1000
= 14 m³/h
713
. ×1
KVS =
1.3 · KV = 1.3 · 14 = 18.2 m³/h
Selected: Type 39-2, DN 40, KVS = 20
KVS =
W =
t
=
p1 =
Z =
m
(8)
1000 kg/h
corresponds to saturated steam
7 bar p2 = 2 bar
2
= 0.286
7
KV =
(7)
Determine the KVS value of the valve from the following
equation using the calculated KV value. In general: KVS »
1.3 · KV
Solution:
1)
=
p2
=
p1
p2 =
Dp =
20 m³/h
1000 kg/h
Corresponding to saturated steam
7 bar
71.3
1000
= 0.701
713
. × 20
0.89
0.89 · 7 = 6.23 bar
p1 – p2 = 7 – 6.23 = 0.77 bar
5
Table 1 · Pressure loss coefficient m as a function of p2/p1
Pressure ratio p2/p1
Pressure loss coefficient m
0 to 0.6
0.70
0.75
0.80
0.85
0.90
0.95
0.99
1.0
0.96
0.92
0.86
0.77
0.66
0.48
0.22
Table 2 · Compressibility factor Z for steam · All pressures stated as absolute pressures in bar
Compressibility factor Z for ...
p1
in bar
Superheated steam at the following temperatures ...
60 °C
1.13
2.27
80 °C
1.1
2.21
100 °C
1.07
2.15
120 °C
1.04
2.09
140 °C
1.02
2.04
160 °C
0.99
1.99
180 °C
0.97
1.95
200 °C
0.95
1.90
250 °C
0.90
1.81
300 °C
0.86
1.73
350 °C
0.83
1.66
400 °C
0.80
1.59
3.31
4.42
3.22
4.29
3.14
4.18
3.06
4.08
2.99
3.98
2.92
3.89
2.86
3.81
2.71
3.62
2.59
3.46
2.49
3.32
2.39
3.19
0.1
0.2
1.16
2.27
0.3
0.4
3.37
4.45
0.5
0.6
5.53
6.58
5.37
6.45
5.23
6.28
5.10
6.12
4.98
5.97
4.86
5.84
4.76
5.72
4.52
5.43
4.33
5.19
4.15
4.98
3.99
4.78
0.7
0.8
7.65
8.71
7.53
8.62
7.33
8.39
7.15
8.17
6.97
7.97
6.82
7.79
6.67
7.63
6.34
7.25
6.06
6.91
5.80
6.64
5.59
6.37
0.9
1.0
9.76
10.8
9.70
10.8
9.44
10.5
9.19
10.2
8.98
9.98
8.77
9.76
8.58
9.53
8.16
9.07
7.90
8.66
7.37
8.30
7.18
7.98
1.1
1.2
11.9
12.9
11.5
12.6
11.3
12.3
11.0
12.0
10.8
11.8
10.5
11.4
10.0
10.9
9.50
10.4
9.10
10.0
8.70
9.60
1.3
1.4
13.9
15.0
13.7
14.7
13.3
14.3
13.0
14.0
12.7
13.7
12.3
13.4
11.8
12.7
11.2
12.1
10.8
11.6
10.4
11.2
1.5
1.6
16.0
17.0
15.8
16.9
15.4
16.4
15.0
16.0
14.7
15.6
14.3
15.3
13.6
14.5
13.0
13.9
12.4
13.3
12.0
12.8
1.7
1.8
18.0
19.1
17.9
19.0
17.5
18.5
17.0
18.0
16.6
17.6
16.3
17.2
15.4
16.4
14.7
15.6
14.1
14.9
13.6
14.4
1.9
2.0
20.1
21.1
20.1
21.1
19.5
20.6
19.0
20.0
18.6
19.6
18.1
19.1
17.3
18.2
16.5
17.3
15.8
16.6
15.2
16.1
2.2
2.4
23.2
25.2
22.6
24.7
22.1
24.1
21.5
23.5
21.0
23.1
20.0
21.8
19.1
20.8
18.3
20.0
17.6
19.2
2.6
2.8
27.2
29.3
26.8
28.9
26.0
28.1
25.5
27.5
24.9
26.8
23.6
25.5
22.6
24.3
21.5
23.2
20.8
22.4
3.0
3.2
31.0
33.4
31.0
33.1
30.2
32.2
29.4
31.4
28.8
30.7
27.3
29.1
26.0
27.8
24.9
26.6
24.0
25.6
3.4
3.6
35.4
37.4
35.2
37.3
34.3
36.3
33.4
35.4
32.6
34.6
31.0
32.8
29.6
31.3
28.2
29.9
27.2
28.9
3.8
4.0
39.4
41.4
38.3
40.4
37.4
39.4
36.5
38.5
34.7
36.5
33.0
35.1
31.6
33.3
30.4
32.0
4.5
5.0
46.4
51.4
45.6
50.8
44.4
49.4
42.8
48.2
41.1
45.7
39.1
43.6
37.3
41.8
36.1
40.0
5.5
6.0
56.4
61.4
56.0
61.2
54.4
59.5
53.0
57.9
50.2
54.9
47.8
52.3
46.7
50.2
44.2
48.2
6.5
7.0
66.3
71.3
64.6
69.7
62.9
67.8
59.4
64.2
56.6
61.1
54.2
58.3
52.2
56.2
8.0
9.0
81.2
91.0
79.9
90.2
77.6
87.7
73.4
82.6
69.8
78.7
67.0
75.0
64.3
72.4
10.0
11.0
101
101
97.9
92.2
87.4
83.2
80.4
111
108
102
96.5
92.1
88.5
12.0
121
130
118
128
111
121
105
114
99.7
109
96.7
105
140
150
139
150
130
139
123
132
118
125
113
121
13.0
14.0
15.0
16.0
17.0
160
149
141
134
129
170
159
150
143
137
18.0
19.0
180
189
169
178
159
168
151
161
146
154
20.0
21.0
199
209
188
198
177
187
168
178
162
170
23.0
229
248
218
238
205
224
195
213
187
203
27.0
29.0
268
258
242
230
216
288
279
261
248
236
31.0
33.0
308
328
300
322
280
299
264
282
253
270
35.0
37.0
348
368
343
365
318
338
301
319
286
304
39.0
41.0
388
408
387
356
376
337
354
320
338
25.0
6
Saturated
steam
AB 04 EN
Gases and vapors
Symbols and units
p1
Absolute pressure
p2
in bar
Pressure loss (differential pressure) in bar
Dp
W
Mass flow rate in kg/h
KV
KV coefficient in m³/h
Density in kg/m³
r1
p2
p1
1
c = 1.135
0.9
c = 1.3
c = 1.4
The calculation procedure provides approximate values
meeting practical needs
c = 1.66
0.8
c = 1.66
For monatomic gases such as
helium, argon, krypton
c = 1.4
For diatomic gases such as
hydrogen, nitrogen, air,
chlorine gas, city gas
c = 1.3
For triatomic and multiatomic gases
such as carbon dioxide, propane,
butane, methane, acetylene, ammonia
c = 1.135 For steam
0.7
W = 14.2 · KV · m · p1 × r1
The values for m and r1can be found in Diagrams 2
and 3 respectively.
Diagram 2 · Pressure loss coefficient m in
p2
as a function of
p1
0.6
0.2
For
0.3
p2
< 0.6 ---> m = 1
p1
0.4
0.5
0.6
0.7
0.8
0.9
1 m
Example 8
Mass flow rate with gas
Determine:
Known:
Propane gas mass flow rate W in kg/h in a fully open valve
Type 3241-1 Pneumatic Control Valve, DN 50 ·
Upstream and downstream pressures
Determine the upstream density r1 from Diagram 3
p2
p2
Calculate
· Find m from Diagram 2 for the given
p1
p1
c
=
and
1.3
Solution:
W = 14.2 · KVS · m · p1 × r1
Reducing pressure valve (self-operated) for nitrogen
Determine:
Type ... Reducing Pressure Valve, KVS and nominal size
of valve
Nitrogen network · Flow rate· Upstream and downstream
pressures
p2
Calculate · Find m from diagram 2 (c = 1.4) · Find r1
p1
from diagram 3 for p1 = 5 bar
Solution:
KV =
W
14.2 × m × p1 × r1
(10)
Example 10
Determining the pressure loss with nitrogen
Determine:
Known:
Pressure loss Dp = p1 – p2 in a fully open valve
Type 3241-2 Electric Control Valve, DN 20 · Nitrogen
network · Nitrogen flow rate · Upstream pressure
Find r1 from Diagram 3 for p1 = 5 bar
m=
W
14.2 × KVS × p1 × r1
p
Find the ratio 2 from Diagram 2 with m = 0.791
p1
p2 = 0.815 · p1 -> Pressure loss Dp = p1 – p2
AB 04 EN
40 m³/h 1)
2.7 bar
p2 = 2.2 bar
5 kg/m³
2.2
= 0.815
2.7
m = 0.805
W = 14.2 · 40 · 0.805 · 2.7 × 5 = 1680 kg/h
The KVS coefficient has a tolerance of ±10 %,
and, as a result, the calculated mass flow rate W,
too.
p1 =
W =
p2
=
p1
5 bar
230 kg/h
3
= 0.6
5
m =
r1 =
0.97
6.2 kg/m³
230
= 3.00 m³/h
14.2 × 0.99 × 5 × 6.2
KV =
KVS =
p2 = 3 bar
1.3 · KV = 1.3 · 3.00 = 3.90 m³/h
Selected: Type 44-1, G ¾, KVS = 4
Determine KVS coefficient from the calculated KV.
As a rule: KVS » 1.3 · KV
Solution:
KVS =
p1 =
r1 =
p2
=
p1
1)
(9)
Example 9
Known:
(9)
(11)
KVS =
p1 =
W =
r1 =
m
=
p2
=
p1
p2 =
Dp =
4 m³/h
5 bar
250 kg/h
6.2 kg/m³
250
= 0.791
14.2 × 4 × 5 × 6.2
0.815
0.815 · 5 = 4.08 bar
5 – 4.08 = 0.92 bar
7

100
80
kg
m3
4
60
50
40
7
30
20
1
10
8
6
5
4
9
10
3
11
8
3
5
6
2
2
1
0.8
0.6
0.5
0.4
0.3
0.2
0.1
0.08
0.06
0.05
0.04
0.03
0.02
0.01
0.008
0.006
0.005
0.004
0.003
0.002
p
0.001
0,01
1
0.02 0.03
0.06
0.1
0.04
0.08
0.2
0.3 0.4
0.6 0.8 1
If the operating temperature t deviates considerably from 0 °C,
correct r1 using the equation rt = r1 × 273
273 + t
1
2
3
4
2
3
Chlorine gas
Butane
Propane
Carbon dioxide
4
6
5
6
7
8
8 10
Air
Nitrogen
Acetylene
Ammonia
20
30 40
bar
9 Methane
10 City gas
11 Hydrogen
Diagram 3 · Density of gases r or r1 as a function of pressure at 0 °C
8
AB 04 EN
Air
The calculation procedure provides approximate values meeting practical
needs. The equation derived for dry air is as follows:
W = 15.3 · m · KV · p1 × r1
(12)
Equation (6) can be written as follows:
W = KV · m · Z
(13)
The value for Z can be obtained from Table 4 using p1. The
value for m can be found in Table 3 (interpolated values
can be determined from Diagram 2 for c = 1.4).
Example 11
Air flow rate
Determine:
Given:
Mass flow rate W in kg/h in a fully open valve
Type 41-23 Self-operated Pressure Regulator, DN 50
Upstream and downstream pressures · Temperature
p2
Determine
p1
Determine the value for m from Table 3 or Diagram 2
Determine the value for Z from Table 4 using the upstream
pressure and temperature
Solution:
W = KVS · m · Z
(13)
Example 12
Pressure reducing valve (self-operated) for air
Determine:
Type ... Pressure Reducing Valve, KVS and nominal size of
valve
Upstream and downstream pressures · Temperature
Compressed air as medium
p2
Determine
p1
Determine the value for m from Table 3 or Diagram 2
Determine the value for Z from Table 4 using the upstream
pressure and temperature
Given:
Solution:
KV =
W
Z ×m
(14)
Determine KVS coefficient from the calculated KV.
As a rule: KVS » 1.3 · KV
Example 13
Determining the pressure loss with air
Determine:
Given:
Pressure loss Dp = p1 – p2 in a fully open valve
Type 42-24 Differential Pressure Regulator, DN 50
Flow rate of compressed air · Upstream pressure
Temperature
Determine the value for Z from Table 4 using the upstream
pressure and temperature
Solution:
m=
W
Z × KVS
Determine the ratio
(15)
p2
for m = 0.884 from Diagram 2 or
p1
Table 3
p2 = 0.75 · p1 -> Pressure loss Dp = p1 – p2
AB 04 EN
Symbols and units
p1
Absolute pressure
p2
in bar
Pressure drop (differential pressure) in bar
Dp
W
Mass flow rate in kg/h
KV
KV coefficient in m³/h
Density in kg/m³
r
m
Pressure loss coefficient, dimensionless
Z
Compressibility factor, dimensionless
Note:
Please contact SAMSON AG to obtain more detailed information on sizing flow regulators for air.
KVS =
t
=
p1 =
p2
=
p1
32 m³/h 1)
20 °C
4 bar p2 = 3 bar
3
= 0.75
4
m =
Z =
W =
0.884
66
32 · 0.884 · 66 = 1867 kg/h
1)
The KVS coefficient has a tolerance of ±10 %,
and, as a result, the calculated mass flow rate W,
too.
p1
t
W
p2
p1
=
=
=
m
Z
=
=
=
5 bar
20 °C
190 kg/h
3
= 0.6
5
p2 = 3 bar
0.982
82.60
KV =
190
= 2.34 m³/h
0.982 × 82.60
KVS =
1.3 · KV = 1.3 · 2.34 = 3.04 m³/h
Selected: Type 44-1, G ½, KVS = 3.2
KVS =
W =
t
=
p1 =
Z =
m
=
p2
=
p1
32 m³/h
3270 kg/h
20 °C
7 bar
115.6
3270
= 0.884
115.6 × 32
0.75
p2 =
0.75 · 7 = 5.25 bar
Dp =
p1 – p2 = 7 – 5.25 = 1.75 bar
9
Table 3 · Pressure loss coefficient m as a function of p2/p1
Pressure ratio p2/p1
Pressure loss coefficient m
0.527
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
0.99
1
0.982
0.978
0.935
0.884
0.818
0.730
0.623
0.448
0.207
Table 4 · Compressibility factor Z for air · All pressures stated as absolute pressures in bar
Compressibility factor Z for …
Dry air for following temperatures …
100° C
150° C
p1
in bar
0° C
20° C
50° C
200° C
250° C
300° C
0.1
0.2
1.71
3.42
1.65
3.30
1.57
3.15
1.47
2.93
1.38
2.77
1.30
2.60
1.24
2.47
1.18
2.31
0.3
0.4
5.13
6.84
4.96
6.61
4.74
6.29
4.39
5.85
4.13
5.50
3.89
5.20
3.71
4.95
3.55
4.72
0.5
0.6
8.55
10.26
8.26
9.90
7.87
9.42
7.32
8.79
6.88
8.24
6.50
7.79
6.18
7.42
5.92
7.09
0.7
0.8
11.97
13.68
11.56
13.22
11.00
12.58
10.22
11.72
9.61
11.00
9.09
10.40
8.64
9.79
8.27
9.45
0.9
1.0
15.40
17.10
14.86
16.50
14.15
15.72
13.18
14.65
12.36
13.75
11.69
13.00
11.12
12.35
10.62
11.81
1.1
1.2
18.83
20.50
18.15
19.80
17.30
18.20
16.07
17.52
15.10
16.50
14.32
15.60
13.60
14.70
13.00
14.12
1.3
1.4
22.10
24.00
21.42
23.10
20.45
22.00
18.25
10.70
17.85
19.25
16.90
18.19
16.09
17.30
15.35
16.55
1.5
1.6
25.65
27.30
24.75
26.40
23.60
25.15
21.68
23.35
20.06
21.95
19.46
20.80
18.55
19.78
17.70
18.88
1.7
1.8
29.10
30.80
28.10
29.70
26.70
28.30
24.80
26.35
23.40
24.75
22.05
23.35
20.90
22.25
20.10
21.20
1.9
2.0
32.25
34.20
30.25
33.00
29.60
31.45
27.55
29.27
26.15
27.70
24.65
25.95
23.50
24.70
22.50
23.10
2.2
2.4
37.65
41.10
36.40
39.60
34.70
37.75
32.20
35.15
30.20
33.30
28.50
31.20
27.20
29.70
25.90
28.35
2.5
42.70
41.40
39.30
36.60
34.40
32.45
30.90
29.50
2.6
44.50
42.60
40.90
38.05
35.75
33.80
32.15
30.70
2.8
47.80
46.20
44.00
41.70
38.45
36.35
34.55
33.10
3.0
51.30
49.55
47.40
43.95
41.25
38.90
37.10
35.45
3.2
54.30
52.40
49.90
46.60
43.90
41.60
39.80
37.70
3.4
58.25
56.20
53.50
49.80
46.70
44.20
42.00
40.10
3.6
62.20
59.60
56.60
52.70
49.40
46.80
44.50
42.60
3.8
4.0
65.00
68.20
62.70
66.00
59.75
62.90
55.60
58.55
51.50
55.00
49.40
52.00
46.90
49.40
44.80
47.20
4.5
77.00
74.40
70.70
65.80
61.80
58.50
55.60
48.50
5.0
86.90
82.60
78.75
73.20
68.75
65.00
61.75
59.20
5.5
94.00
90.90
87.40
80.60
75.60
71.60
68.00
64.90
6.0
102.06
98.90
94.30
87.90
82.40
77.90
74.15
70.90
6.5
111.0
107.2
101.10
95.20
88.40
84.50
80.40
76.80
7.0
119.6
115.6
110.0
102.2
96.90
90.90
86.40
82.70
8.0
136.8
132.2
125.7
117.1
110.0
104.0
97.9
94.5
9.0
162.2
148.6
141.6
131.8
123.6
116.9
111.2
106.2
10.0
171.0
165.0
157.2
146.5
137.5
130.0
123.5
118.1
11.0
188.3
181.5
173.0
160.7
151.0
143.2
136.0
130.0
12.0
205.0
198.0
182.0
175.2
165.0
156.0
147.0
141.0
13.0
221.0
214.2
204.5
182.5
178.5
169.0
160.9
153.5
14.0
240.0
231.0
220.0
197.0
192.5
181.9
173.0
165.5
15.0
256.5
247.5
236.0
216.8
200.6
194.6
185.5
177.0
16.0
273.0
264.0
251.0
253.5
219.5
208.0
197.8
188.8
17.0
291.0
281.0
267.0
248.0
234.0
220.5
209.0
201.0
18.0
308.0
297.0
283.0
263.5
247.5
233.5
222.5
212.0
19.0
20.0
322.5
342.0
302.5
330.0
296.0
314.5
275.6
292.7
261.6
277.0
246.5
259.6
235.0
247.0
225.0
231.0
22.0
376.5
364.0
347.0
322.0
302.0
285.0
272.0
259.0
24.0
411.0
395.0
377.5
351.5
333.0
312.0
297.0
283.5
26.0
445.0
428.0
409.0
380.5
357.5
338.5
321.5
307.0
28.0
478.0
462.0
440.0
417.0
384.5
363.5
345.5
331.0
30.0
513.0
495.5
474.0
439.5
412.5
389.0
371.0
354.5
10
AB 04 EN
Flow rate through pipelines
wmax of self-operated
regulators 1)
wmax of district
heating 1)
Nominal size DN in mm/inch
40
0
3
25 00
20 0
0
m3
h
10000
8000
6000
5000
4000
3000
2000
50
65
80
10
1
12 50
5
0
1000
800
600
500
400
300
20
4
3 0
25 2
200
¼
"
"
15
100
80
60
50
40
30
20
10
8
6
5
4
3
2
1
0.8
0.6
0.5
0.4
0.3
1)
0.2
Only for water
.
Diagram 4 · V -w diagram for gases, vapors and liquids
Diagram 4 shows the following relationship:
.
Vpipe = Apipe · wpipe
.
(16)
DN
æ DN 2
Determining velocity and nominal size with Apipe = ç
÷
è 18.8 ø
Velocity w of the medium
æ 18.8 2
wpipe = Vpipe · ç
÷
è DN ø
.
(17)
Nominal size DN
.
AB 04 EN
V
wpipe
Volume flow rate in m³/h
Flow velocity in m/s
Cross-section of nominal size, registered as
straight
Nominal size
The flow rate of gases in m³/h determined for standard
temperature and pressure using Diagram 4 can be converted into kg/h (operating conditions). See Example
14.
VG
.
DN = 18.8 ·
V pipe
wpipe
Apipe
Gas flow rate, based on standard temperature
and pressure
(18)
11
Example 14
Compressed air flow rate
Determine:
Flow rate of compressed air for operating conditions and
standard pressure and temperature
Pipe nominal size · Pressure p in the pipe
Flow velocity
Read volume flow rate in m³/h based on the operating
conditions from Diagram 4
How high ist the mass flow rate of the compressed air in
kg/h?
Given:
Solution:
.
W=V ·r
Nominal size = DN 32
p
= 5 bar
wpipe = 7 m/s
.
= 20 m³/h
V
Diagram 3 shows:
r = 6.3 kg/m³
In the operating state:
W = 20 · 6.3 = 126 kg/h
(19)
.
VG is the volume flow rate of gas in m³/h at a standard
temperature of 0 °C and at a standard pressure of
1013 mbar.
At standard temperature and pressure:
1 m³ air -> 1.293 kg
.
VG =
W
126
=
= 97.5 m³/h
1.293 1293
.
SAMSON AG · MESS- UND REGELTECHNIK
Weismüllerstraße 3 · 60314 Frankfurt am Main · Germany
Phone: +49 69 4009-0 · Fax: +49 69 4009-1507
Internet: http://www.samson.de
AB 04 EN
2012-12
Specifications subject to change without notice