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FAD flow rate
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1.- Normal flow rate to real flow rate 2.- Real flow rate to normal flow rate
Normal flow rate data Operation absolute pressure Real flow rate data
7.53 80 kPa V =
101,325 Pa 600 kPa (g)
273.15 K 680.00 kPa
680,000 PaLocal atmospheric pressure Local atmospheric pressure
80 kPa Real volumetric flow rate
V =
Operating conditions 101,325 Pa Normal conditions
600 kPa (g) 273.15 K
10 °C 680,000 Pa
283.15 °C
Operation absolute temperature 7.53 Operation absolute temperature
V = 1.16
10 °C
283.15 K
Rev. cjc 26.02.2013
Normal flow rate data Operation conditions Real flow rate data
7.53 680,000 Pa V =
10 °C 283 K
1949 m.a.s.l.
600.0 kPa (g) Normal conditions
101,325 Pa
273.15 KH = 1949 m.a.s.l. H =
80.00 kPa (abs)
Real volumetric flow rate
Operation pressure V = Operation pressure
101,325 Pa
600.00 kPa (g) 273 K
80.00 kPa 680,000 Pa
680.00 kPa 283.15 °C
680,000 Pa 7.53
V = 1.16Operation absolute temperature Operation absolute temperature
Pop = Patm_loc + Pop
Vn = Nm3/h Patm_loc =
Pn = Pop = Pop =
Tn = Pop = top =
Pop =
Patm_loc = Patm_loc =
(Pn/Pop) * (Top/Tn) * Vn
Pn =
Pop = Tn = Pn =
top = Pop = Tn =
Top =
Vn = Nm3/h
Top = top + 273.15 m3/h Top =
top = top =
Top = Top =
3.- Normal flow rate to real flow rate Patm_oc = f(H)
Vn = Nm3/h Pop =
top = Top = top =
Hloc = Hloc =
Pop_g = Pop_g =
Pn =
Patm_Loc = 101,325* (1 -2,25577E-5 * H)^5,25588 Tn = Patm_Loc = 101,325* (1 -2,25577E-5 * H)^5,25588
Patm_loc = Patm_loc =
(Pn/Pop) * (Top/Tn) * Vn
Pop = Pop_g + Patm_loc Pn = Pop =
Pop_g = Tn = Pop_g =
Patm_loc = Pop = Patm_loc =
Pop = Top = Pop =
Pop = Vn = Nm3/h Pop =
m3/h
10 °C
283.15 K
Top = top +273.15 Top =
top = top =
Top = Top =
Rev. cjc. 01.08.2013
2.- Real flow rate to normal flow rate
Real flow rate data Operation absolute pressure
1.16 80 kPa
600 kPa (g) 600 kPa (g)
10 °C 680 kPa
680,000 PaLocal atmospheric pressure
80 kPa Normal volumetric flow rate
Normal conditions 680,000 Pa
101,325 Pa 101,325 Pa
273.15 K 273.15 K
283.15 °C
Operation absolute temperature V = 1.2
7.53
10 °C
283.15 K
Real flow rate data Operation conditions
1.2 680,000 Pa
10 °C 283.15 K
1949 m.a.s.l.
600 kPa (g) Normal conditions
101,325 Pa
273.15 K1949 m.a.s.l.
80.00 kPa (abs) Normal volumetric flow rate
Operation pressure 679,999.9 Pa
101,325 Pa
600.00 kPa (g) 273.15 K
80.00 kPa 283.15 °C
680.00 kPa V = 1.2
680,000 Pa 7.53
Operation absolute temperature
Pop = Patm_loc + Pop
m3/h Patm_loc =
Pop =
Pop =
Pop =
Vn = (Pop/Pn) * (Tn/Top) * V
Pop =
Pn =
Tn =
Top =
m3/h
top + 273.15 Vn = Nm3/h
4.- Real flow rate to normal flow rate Ploc = f(H)
m3/h Pop =
Top =
Pn =
atm_Loc = 101,325* (1 -2,25577E-5 * H)^5,25588 Tn =
Vn = (Pop/Pn) * (Tn/Top) * V
Pop =
Pop_g + Patm_loc Pn =
Tn =
Top =
m3/h
Vn = Nm3/h
10 °C
283.15 K
top +273.15
FAD volume flow rate
Free air delivery (FAD) is the volume of air delivered under the conditions of temperature and pressure existing at the compressor's intake (state 2).
1.- Normal flow rate (state 1) to FAD flow rate (state 2)
Normal air conditions (State 1)
480
101,325 Pa
0 -
0 °C
273 K
609.6 Pa
FAD conditions (State 2))
73,400 Pa
0.42 -
22 °C
2645.1 Pa
295 K
726.99 Rev. cjc 26.02.2013
100000 * EXP(153.5411 + 0.066953 *(D36) - 0.0000505796 * (D36)^ 2 + 0.00000002183911 * (D36) ^ 3 - 8990.134 * (D36) ^ -1 - 25.07797 * LN((D17+273.15)))
0 °C
273 K
609.6 Pa
Free air delivered (FAD)
Free air delivery is the volume of air delivered under the conditions of
V2 = V1 * (P1 - RH1 * Psat.water_1) / (P2 - RH2 * Psat.water_2) * (T2 / T1)
V1 = Nm3/h
P1 =
RH1 =
t1=
T1 =
Psat.water_1 = f(T1)
Psat.water_1 =
P2 =
RH2 =
t2=
Psat.water_2 = f(T2)
Psat.water_2 =
T2 =
V2 = m3/h (FAD)
Psat.water1 =
t1=
T1 =
Psat.water1 =
temperature and pressure existing at the compressor`s intake (state 2).
To obtain the volume flow rate at the intake conditions (2), knowing the
volume flowrate at the standard conditions (1), one applicates Boyle-Mariot
law between both cases. Since we want a value of dry air, Boyle-Mariot is
to be applied to the dry air.
Thus, dry air partial pressures are to be used.
- Since normal air is a dry air (da), the total air pressure at this conditions is
the same as the partial dry air pressure
with an air without water vapor
0
thus
- The partial pressure of the dry air (da) in the ambient air (2) is
Total pressure of ambient air (state 2)
Partial pressure of water vapor in sate 2
where the water vapor pressure is calculated as
with
Relative humidity of ambient air (Also, indicated as RH)
Thus, Boyle-Mariot is applied as
480
Normal conditions (1) Intake or local conditions (2)
0 °C 22 °C
273.15 K 295 K
101,325 Pa 73,400 Pa
0 - 0.42 -
609.6 Pa 2645.1 Pa
Pda_1 = P1 - Pw_1
Pw_1 =
Pda_1 = P1
Pda_2 = P2 - Pw_2
P2 :
Pw_2 :
Pw_2 = f2 * Psat.water_2
f2 :
Psat.water_2 = Pressure of saturated water vapor at ambient temperatute "t2"
Determination of dry air flowrate (V2) that is to be sucked at the compressor's
intake to obtain the desired volume flowrate (V1)
V1 = Nm3/h
t1 = t2 =
T1 = T2 =
P1 = P2 =
RH1 = RH2 =Psat.water_1 = Psat.water_2 =
V2 = V1 * (P1 - RH1 * Psat.water_1) / (P2 - RH2 * Psat.water_2) * (T2 / T1)
480
101,325 Pa
0 -
0 °C
609.6 Pa
73,400 Pa
0.42 -
22 °C
2645.1 Pa
295 K
273 K
726.99
Saturation pressure of water100000 * Exp(153.5411 + 0.066953 *(C3+273.15) - 0.0000505796 * (C3+273.15)^ 2 + 0.00000002183911 * (C3+273.15) ^ 3 - 8990.134 * (C3+273.15) ^ -1 - 25.07797 * Ln((C3+273.15)))
Valid 0 ºC < t < 100 ºCt = 0 ºC
609.6 Pa
V1 = Nm3/h
P1 =
RH1 =
t1=
Psat.wate1 =
P2=
RH2 =
t2=
Psat.wate_2 =
T2 =
T1 =
V2 = m3/h (FAD)
Psat =
Psat=
2.- FAD flow rate (state 2) to Normal flow rate (state 1)
FAD conditions (State 2))
726.99
73,400 Pa
0.42 -
22 °C
2645.1 Pa
295 K
Normal air conditions (State 1)
101,325 Pa
0 -
0 °C
273 K
609.6 Pa
480
100000 * EXP(153.5411 + 0.066953 *(D36) - 0.0000505796 * (D36)^ 2 + 0.00000002183911 * (D36) ^ 3 - 8990.134 * (D36) ^ -1 - 25.07797 * LN((D17+273.15)))
Water vapor pressure
V1 = V2 * (P2 - RH2 * Psat.water_2) / (P1 - RH1 * Psat.water_1) * (T1 / T2)
V2 = m3/h (FAD)
P2 =
RH2 =
t2=
Psat.water_2 = f(T2)
Psat.water_2 =
T2 =
P1 =
RH1 =
t1=
T1 =
Psat.water_1 = f(T1)
Psat.water_1 =
V1 = m3/h
P1 y P2 son presiones absolutas totales
de la mezcla de aire y vapor de agua
La presión parcial del vapor de agua a
La humedad relativa del aire seco es RH1 = 0
Air at normal conditions (state "1")
f
0 -
0 °C
609.6 Pa
Air at ambient conditions (estado "2")
f
0.42 -
22 °C
2645.1 Pa
FAD volume flowrate
La presión parcial del aire seco es Pa_1
La temperatura normal es t1 = 0°C
la temperatura normal es Pw_1
Pw_1 = (Pw_2 / Psat.water_2) * Psat.water_2
(Pw_1 / Psat.water_1) =
with "f" : air Relative Humidity (RH)
Pw_1 = f1 * Psat.water_1
Pw_1 = RH1 * Psat_1
RH1 =
t1 =
Pw_1=
Pw_2 = (Pw_2 / Psat.water_2) * Psat.water_2
(Pw_2 / Psat.water_2) =
with "f" : air Relative Humidity (RH)
Pw_2 = f2 * Psat.water_2
Pw_2 = RH2 * Psat_2
RH2 =
t2 =
Pw_2=
V2 = V1 * (P1 - RH1 * Psat.water_1) / (P2 - RH2 * Psat.water_2) * (T2 / T1)
480
101,325 Pa State 1: Normal air conditions
0 - State 2: Local ambient air conditions
0 °C and also compressor intake conditions
609.6 Pa
73,400 Pa Free air delivery (FAD) is the volume of
0.42 - air delivered under the conditions of
22 °C temperature and pressure existing at
2645.1 Pa the compressor's intake (state 2).
295 K
273 K
726.99
100000 * Exp(153.5411 + 0.066953 *(C3+273.15) - 0.0000505796 * (C3+273.15)^ 2 + 0.00000002183911 * (C3+273.15) ^ 3 - 8990.134 * (C3+273.15) ^ -1 - 25.07797 * Ln((C3+273.15)))
V1 = Nm3/h
P1 =
RH1 =
t1=
Psat.wate1 =
P2=
RH2 =
t2=
Psat.wate_2 =
T2 =
T1 =
V2 = m3/h (FAD)
Air density and mass flow rates
Air constantR = 287.0 J/(kg*K)
1.-Normal flow rate to mass flow rate. SI 3.- Mass flow rate imperial to Normalflow rate. SI
1,000 m = 0.792m = 0.359
Mass flowrate m = 1293.2
m =
1000
1.29 m = 1,293.2
m = 1293 kg/h 1.29
m = 0.3592 kg/s 1,000
2.- Mass flow rate to Normal flow rate. SI 4.- Normal flow rate to mass flow
rate imperial
m = 0.3592 kg/s
1000m = 1293.2 kg/h Mass flowrate
m =
1000
m = 1,293.2 kg/h 1.29
1.29 m = 1293
1,000 m = 0.3592
m = 0.792
Normal density 1 kg = 2.20
p / ( R * T) m = 0.3592p = 101,325 Pa m = 0.7920R = 287.0 J/(kg*K)T = 273 K
1.29
Vn = Nm3/h
Vn * rn
Vn = Nm3/h Vn = m / rn
rn = kg/Nm3
rn =
Vn =
Vn =
Vn * rn
Vn = m / rn Vn =
rn =
rn = kg/Nm3
Vn = Nm3/h
rn =
rn = kg/Nm3
Air constant3.- Mass flow rate imperial to Normal 5.- Actual density R =
Rg =
10 ºC MM =
lb/s 450.0 kPa (g) R =kg/s H = 1730 m.a.s.l.kg/h
101,325* (1 -2,25577E-5 * H)^5,25588 Nitrogen constant
82.20 kPa R =
kg/h Rg =
T = MM =
10 ºC R =T = 283.15 K
Molecular masses from [1]
4.- Normal flow rate to mass flow P =
450.0 kPa (g)
82.20 kPa
P = 532.2 kPaP = 532,197 Pa
p / ( R * T)
p = 532,197 Pa
kg/h R = 287.0 J/(kg*K)
kg/s T = 283 K
lb/s 6.55
Rev. cjc. 30.05.2013
lb
kg/slb/s
tact =
Pact_g =
patm_loc =
patm_loc =
kg/Nm3 tact + 273.15
Nm3/h tact =
Pact_g + Patm_loc
Pact_g =
patm_loc =
Nm3/h
Nm3/h ract =
kg/Nm3
ract = kg/m3
Rg / MM8314.41 [ J / (kmol*K)]
28.97 kg/kmol
287.0 J/(kg*K)
Nitrogen constant
Rg / MM
8314.41 [ J / (kmol*K)]
28.0134 kg/kmol
296.8 J/(kg*K)
Molecular masses from [1]
Nitrogen density and mass flow rates
Nitrogen constantR = 296.8 J/(kg*K)
1.-Normal flow rate to mass flow rate. SI 3.- Mass flow rate imperial to Normalflow rate. SI
1,000 m = 0.766m = 0.347
Mass flowrate m = 1250.5
m =
1000
1.25 m = 1,250.5
m = 1251 kg/h 1.25
m = 0.3474 kg/s 1,000
2.- Mass flow rate to Normal flow rate. SI 4.- Normal flow rate to mass flow
rate imperial
m = 0.3474 kg/s
1000m = 1250.5 kg/h Mass flowrate
m =
1000
m = 1,250.5 kg/h 1.25
1.25 m = 1251
1,000 m = 0.3474
m = 0.766
Normal density 1 kg = 2.20
p / ( R * T) m = 0.3474p = 101,325 Pa m = 0.7658R = 296.8 J/(kg*K)T = 273 K
1.25
Vn = Nm3/h
Vn * rn
Vn = Nm3/h Vn = m / rn
rn = kg/Nm3
rn =
Vn =
Vn =
Vn * rn
Vn = m / rn Vn =
rn =
rn = kg/Nm3
Vn = Nm3/h
rn =
rn = kg/Nm3
Air constant3.- Mass flow rate imperial to Normal 5.- Actual density R =
Rg =
10 ºC MM =
lb/s 450.0 kPa (g) R =kg/s H = 1730 m.a.s.l.kg/h
101,325* (1 -2,25577E-5 * H)^5,25588 Nitrogen constant
82.20 kPa R =
kg/h Rg =
T = MM =
10 ºC R =T = 283.15 K
Molecular masses from [1]
4.- Normal flow rate to mass flow P =
450.0 kPa (g)
82.20 kPa
P = 532.2 kPaP = 532,197 Pa
p / ( R * T)
p = 532,197 Pa
kg/h R = 296.8 J/(kg*K)
kg/s T = 283 K
lb/s 6.33
Rev. cjc. 03.07.2013
lb
kg/slb/s
tact =
Pact_g =
patm_loc =
patm_loc =
kg/Nm3 tact + 273.15
Nm3/h tact =
Pact_g + Patm_loc
Pact_g =
patm_loc =
Nm3/h
Nm3/h ract =
kg/Nm3
ract = kg/m3
Rg / MM8314.41 [ J / (kmol*K)]
28.97 kg/kmol
287.0 J/(kg*K)
Nitrogen constant
Rg / MM
8314.41 [ J / (kmol*K)]
28.0134 kg/kmol
296.8 J/(kg*K)
Molecular masses from [1]
Imperial standard flow rate to Normal flow rate
Imperial standard flow rate data Normal absolute pressure
101,325 Pa
1.0 Normal absolute temperature
273.15 KImperial standard temperature
60 °F Standard volumetric flow rate to
15.56 °C Normal volumetric flow rate
Imperial standard pressure
101,325 kPa 101,325 Pa
101,325 Pa
273.15 K
Standard conditions (Imperial) 288.71 °C
101,325 Pa 1.0
288.71 K 0.9461
1 = 0.9461
1 = 1.057
PN =
VS = Sm3/h
TN =
tS =
tS =
Vn = VS * (PS/Pn) * (Tn/TS)
PS = PS =
Pn =
Tn =
TS =
PS = VS = m3/h
TS = Vn = Nm3/h
Sm3/h Nm3/h
Nm3/h Sm3/h
Approximate methodAproximate equation for calculating the atmosphericpressure as a function of the height above sea level
p = 101,325* (1 -2,25577E-5 * H)^5,25588H = 1730 mp = 82.20 kPa
The Engineering Toolbox
http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html
Esta ecuación es una simplificación de la fórmulahipsométrica [2], en la que la temperatura ambiente se toma con un valor aproximado de 15,2 °CEsta ecuación aproximada produce un error máximode 0.1% cuando se aplica en le rango de alturas 0 m.sn.m. <= H <= 6000 m.s.n.m.
http://www.engineeringtoolbox.com/molecular-weight-gas-vapor-d_1156.html
Molecular mass of common gases and vapors
The molecular weight of a substance, also called molecular mass, is the
mass of one molecule of that substance, relative to the unified atomic
mass unit u equal to 1/12 the mass of one atom of carbon-12.
Gas or Vapor Gas or Vapor
kg/kmol
26.04
Air 28.966 Hydrogen Chloride
Ammonia (R-717) 17.02 Hydrogen Sulfide
Argon, Ar 39.948 Hydroxyl, OH
Benzene 78.11 Krypton
58.12
Iso-Butane (2-Metyl propane) 58.12 Methyl Alcohol
Butadiene 54.09 Methyl Butane
1-Butene 56.108 Methyl Chloride
cis -2-Butene 56.108 Natural Gas
trans-2-Butene 56.108 Neon, Ne
Isobutene 56.108
44.01
Carbon Disulphide 76.13 Nitrous Oxide
Carbon Monoxide, CO 28.011 N-Octane
Chlorine 70.906
Cyclohexane 84.16 Ozone
Deuterium 2.014 N-Pentane
30.07 Iso-Pentane
Ethyl Alcohol 46.07
Ethyl Chloride 64.515 Propylene
28.054
Fluorine 37.996
Helium, He 4.02
N-Heptane 100.2
Hexane 86.17
Hydrochloric Acid 36.47
Molecular Weight - Gases and Vapors
Molecular Weight
Acetylene, C2H2 Hydrogen, H2
N-Butane, C4H10 Methane, CH4
Nitric Oxide, NO2
Carbon Dioxide, CO2 Nitrogen, N2
Oxygen, O2
Ethane, C2H6
Propane, C3H8
Ethylene, C2H4
Gas or Vapor
kg/kmol kg/kmol
2.016 R-11 137.37
36.461 R-12 120.92
34.076 R-22 86.48
17.01 R-114 170.93
83.8 R-123 152.93
16.044 R-134a 102.03
32.04 R-611 60.05
72.15 Sulfur 32.02
50.488 Sulfur Dioxide 64.06
19 Sulfuric Oxide 48.1
20.179 Toluene 92.13
30.006 Xenon 131.3
28.0134 18.02
44.012
114.22
31.9988
47.998
72.15
72.15
44.097
42.08
Molecular Weight
Molecular Weight
Water Vapor - Steam, H2O
[1] The Engineering Toolbox
http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html
also
Atmospheric_temperature_pressure_and_density_as_function_of_the_height_above_sea_level.xlsmwww.piping-tools.net
Atmospheric_temperature_pressure_and_density_as_function_of_the_height_above_sea_level.xlsm
