[XLS]piping-tools.netpiping-tools.net/Downloads/1.Listado/Dimensioning... · Web viewCompressed_Air_Manual_tcm46-1249312 The actual working pressure is approx Pact_work = 2. Working

16 2.2 1

20 1.9 2.8 225 2.3 3.5 3

32 1.8 2.9 4.4 4

40 1.8 2.3 3.7 5.5 5

50 1.8 2.0 2.9 4.6 6.9 6

63 1.8 2.0 2.5 3.6 5.8 8.6 7

75 1.9 2.3 2.9 4.3 6.8 10.3 8

90 2.2 2.8 3.5 5.1 8.2 12.3 9

110 2.7 3.4 4.2 6.3 10.0 15.1 10125 3.1 3.9 4.8 7.1 11.4 17.1 11

140 3.5 4.3 5.4 8.0 12.7 19.2 12

160 4.0 4.9 6.2 9.1 14.6 21.9 13

180 4.4 5.5 6.9 10.2 16.4 24.6 14

200 4.9 6.2 7.7 11.4 18.2 27.4 15

225 5.5 6.9 8.6 12.8 20.5 30.8 16250 6.2 7.7 9.6 14.2 22.7 34.2 17280 6.9 8.6 10.7 15.9 25.4 38.3 18

315 7.7 9.7 12.1 17.9 28.6 43.1 19

355 8.7 10.9 13.6 20.1 32.2 48.5 20400 9.8 12.3 15.3 22.7 36.3 54.7 21

450 11.0 13.8 17.2 25.5 40.9 61.5 22

500 12.3 15.3 19.1 28.4 45.4 68.3 23

560 13.7 17.2 21.4 31.7 50.8 24

630 15.4 19.3 24.1 35.7 57.2 25710 17.4 21.8 27.2 40.2 64.5 26800 19.6 24.5 30.6 45.3 27900 22.0 27.6 34.4 51.0 28

1000 24.5 30.6 38.2 56.7 291100 26.9 33.7 42.0 62.4 301200 29.4 36.7 45.9 68.0 311400 34.4 42.9 53.5 32

1600 39.2 49.0 61.2 33

Dn [mm] 2 1/2 3.2 4 6 10 16Presiones nominales PN [bar]

HDPE PE80 DIN 8074 / ISO 4427Espesor [mm]

1 2 3 4 5 6 7

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

2

3 Pipe Thickness [mm], according ASME B36.10M4

5 ASME B36.10M SCHEDULE / IDENTIFICATION6 Size 5 10 20 30 40 60 80 100 120 140 160 STD XS XXS

7 1/2 21.3 1.65 2.11 - 2.41 2.77 - 3.73 - 0 - 4.78 2.77 3.73 7.478 3/4 26.7 1.65 2.11 - 2.41 2.87 - 3.91 - 0 - 5.56 2.87 3.91 7.829 1 33.4 1.65 2.77 - 2.9 3.38 - 4.55 - 0 - 6.35 3.38 4.55 9.0910 1 1/2 48.3 1.65 2.77 - 3.18 3.68 - 5.08 - 0 - 7.14 3.68 5.08 10.1511 2 60.3 1.65 2.77 - 3.18 3.91 - 5.54 - 0 - 8.74 3.91 5.54 11.0712 3 88.9 2.11 3.05 - 4.78 5.49 - 7.62 - 0 - 11.13 5.49 7.62 15.2413 4 114.3 2.11 3.05 - 4.78 6.02 - 8.56 - 11.13 - 13.49 6.02 8.56 17.1214 5 141.3 2.77 3.4 - - 6.55 - 9.53 - 12.7 - 15.88 6.55 9.53 19.0515 6 168.3 2.77 3.4 - - 7.11 - 10.97 - 14.27 - 18.26 7.11 10.97 21.9516 8 219.1 2.77 3.76 6.35 7.04 8.18 10.31 12.7 15.09 18.26 20.62 23.01 8.18 12.7 22.2317 10 273 3.4 4.19 6.35 7.8 9.27 12.7 15.09 18.26 21.44 25.4 28.58 9.27 12.7 25.418 12 323.8 3.96 4.57 6.35 8.38 10.31 14.27 17.48 21.44 25.4 28.58 33.32 9.53 12.7 25.419 14 355.6 3.96 6.35 7.92 9.53 11.13 15.09 19.05 23.83 27.79 31.75 35.71 9.53 12.7 - 20 16 406.4 4.19 6.35 7.92 9.53 12.7 16.66 21.44 26.19 30.96 36.53 40.49 9.53 12.7 - 21 18 457 4.19 6.35 7.92 11.13 14.27 19.05 23.83 29.36 34.93 39.67 45.24 9.53 12.7 - 22 20 508 4.78 6.35 9.53 12.7 15.09 20.62 26.19 32.54 38.1 44.45 50.01 9.53 12.7 - 23 22 559 4.78 6.35 9.53 12.7 - 22.23 28.58 34.93 41.28 47.63 53.98 9.53 12.7 - 24 24 610 5.54 6.35 9.53 14.27 17.48 24.61 30.96 38.89 46.02 52.37 59.54 9.53 12.7 - 25 26 660 - 7.92 12.7 15.88 - - - - - - - 9.53 12.7 - 26 28 711 - 7.92 12.7 - - - - - - - - 9.53 12.7 - 27 30 762 6.35 7.92 12.7 15.88 - - - - - - - 9.53 12.7 - 28 32 813 - 7.92 12.7 15.88 17.48 - - - - - - 9.53 12.7 - 29 34 864 - 7.92 12.7 15.88 17.48 - - - - - - 9.53 12.7 - 30 36 914 - 7.92 12.7 15.88 19.05 - - - - - - 9.53 12.7 - 31 38 965 - - - - - - - - - - - 9.53 12.7 - 32 40 1016 - - - - - - - - - - - 9.53 12.7 - 33 42 1067 - - - - - - - - - - - 9.53 12.7 - 34 44 1118 - - - - - - - - - - - 9.53 12.7 - 35 46 1168 - - - - - - - - - - - 9.53 12.7 -

dext

36 48 1219 - - - - - - - - - - - 9.53 12.7 -

Index

1. Compressor system design 5. MotorsDesign according [1] Minimum nominal efficiency standard at 100% of nominal full load 1.- Ambient data 2.- Compressor 6. FAD 3.- After cooler Free Air Delivery 4.- Air receiver volume 5.- Dryer References 6.- Condensed water vapor 7.- Pressure drop

2. Equation's derivation Receiver volume Compressor flow rate

3. Atlas Copco Chapter 5. pdf

4. Pressure drop According example [5]

Rev. cjc. 15.04.2016

Minimum nominal efficiency standard at 100% of nominal full load

Dimensioning _compressed_air_installations_Atlas_Copco

cjcruz[at]piping-tools.net

www.piping-tools.net

http://www.piping-tools.net/

Compressor system design [1]

1.- Ambient data This example is based in an Atlas Copco publication (See Note 1)

Height above sea level H = 0 m.a.s.l.Normal ambient temperature t = 20 °CMaximum amb. temperature 30 °CRelative humidity 60 %

Local atmospheric pressure above sea level [11]101,325* (1 -2,25577E-5 * H)^5,25588

H = 0 m.a.s.l.101.325 kPa

1.013 bar(a)

2.- Compressor

Compressed Air RequirementThe nedd consists of three compressed air consumers, with following requirements

Table 1. Consumers' requirementsConsumer Air flow Pressure Dew Point

1 12 6 bar(g) 5 °C2 67 l/s (FAD) 7 bar 5 °C

3 95 l/s (FAD) 4 bar(g) 5 °C

Air flow rate of case 1. Units change from Nm³/min to l/s (FAD) Flow rate requirements

12101,325 Pa State 1: Normal air conditions

0 - State 2: Local amb. air conditions Total requirement to be supplied

1.- Ambient data2.- Compressor3.- After cooler4.- Air receiver volume5.- Dryer6.- Condensed water vapor7.- Pressure drop

tmax =f =

Pamb =

Pamb =Pamb =

Nm3/min

V2 = V1 * (P1 - RH1 * Psat.water_1) / (P2 - RH2 * Psat.water_2) * (T2 / T1) Eq. (21) V1 =

V1 = Nm3/min V2 =P1 = V3 =

RH1 =

0 °C and also compressor intake

273.15 K conditions Considering a safety margin of#VALUE! Pa SM =101,325 Pa Free air delivery (FAD) is the volume The Safety Factor is

0.6 - of air delivered under the conditions SF =30 °C of temperature and pressure existing The required flow rate is

303 K at the compressor's intake (state 2).

#VALUE! Pa SF =#VALUE!#VALUE! l/s (FAD)

Selected compresor FAD conditions (State 2, intake)450.0 l/s FAD 1620.00

1620.0 m³/h FAD 101,325 Pa7.00 bar(g) 0.6 -

30 °CInititial receiver's pressure H2O_SaturationPressure_t

#VALUE! Pa7.00 bar (g) 303 K7.00 bar (g)

Normal air conditions (State 1)Final receiver's pressure 101,325 Pa

0 -7 bar (g) 0 °C

0.5 bar (g) 273 K6.50 bar (g) H2O_SaturationPressure_t

#VALUE! Pa

Compressor´s discharge temperatureHelp variables

t1= Vreq_ =T1 =

Psat.water_1 =P2=

RH2 =t2=

T2 = Vreq =Psat.water_2 =

V2 = m3/min (FAD) Vreq =V2 = Vreq =

Vreq = V2 = m3/h (FAD)

Vreq = P2 =Pcomp = RH2 =

t2=Psat.water_2 =

Prcv_initial = Pcomp Psat.water_2 =Pcomp = T2 =

Prcv_initial =

P1 =Prcv_final= Prcv_initial - DPop RH1 =Prcv_initial = t1=DPop = T1 =

Prcv_final= Psat.water_1 =Psat.water_1 =

DataAir specific heat ratio Kelvin units constant

k = 1.4 - Kelv = 273.15 KCompressor´s efficiency Inlet conditions

80 %Local atmospheric pressure 30 ºC

101.33 kPa 303.15 KLocal atmospheric temperature

30 °C 101.325 kPaCompressor manometric discharge pressure 101.325 kPa

7.00 bar (g) Outlet conditions

700.0 kPa101.33 kPa801.33 kPa

Compressor powerNormal density

Mass flow rate of dry air p / ( R * T)#VALUE! p = 101,325 Pa

With Safety Factor R = 286.9 J/(kg*K)FS = 1.15 T = 273 K

1.29Ambient conditions

101.33 kPa Air mass flowrate30 °C m =

RH = 60 % Q = #VALUE!1.29

Air isentropic exponent #VALUE! kg_da/h1.4 #VALUE! kg_da/s

Compressor Inlet temperature7.0 bar (g) 30 °C700 kPa(g) 303.15 K

hc = Tin = tin + Kelvtin =

Patm = Tin =Pin = Patm

tatm = Patm =Pin =

Pout(g) =Pout = Pout(g) + Patm

Pout(g) =Patm =Pout =

rn =Vreq_N = Nm3/h

rn = kg/Nm3

Patm =tatm = Q * r

Nm3/hrn = kg/Nm3 mda =

g = mda =

Pout(g) = t1 =Pout(g) = T1 =

Eficiency0.8 - Specifice heat

Cp = AirSpecificHeat_tCp = #VALUE! kJ/kg

CompressorCompressor inlet conditions

Height above sea level H = 0 m.a.s.l.Normal ambient temperature t = 20 °CMaximum ambient temperature 30 °CRelative humidity 60 %

1.01 bar(a)

Design FAD flow rate with a 450.0 l/s FADDesign normal flow rate (dry air) #VALUE!Safety factor on flow rate SF = 1.15 -

CompressorCompressor eficiency 0.8 -Compressor power Wc = #VALUE! kWCompressor motor Wm = 185 kW

3.- After cooler Approach temperature After cooler air outlet presure

bar(g)to an approach temperature of 2.7°C 7 bar(g)to 11°C of ambient air temperature, or 7 bar(g)

hc =

tmax =f =

Pamb =

Vreq =Vreq_N = Nm3/h

hc =

Most aftercoolers are sized to cool the air PAC_air_in = Pc_out Pc_out =

PAC_air_in =

water available temperature

0.140 barLet 6.860 bar

10 KAvailable water

25 °CPressure drop 2 barFor the best results, size the aftercooler for a 1 to 2 psi pressure drop. Let After cooler air outlet temperature

Let 25 °C14.0 kPa 10 K

0.140 bar 35 ºC

Compressor discharge temperature After cooler air inlet temperature335.2 ºC

335.2 ºC335.2 °C

Enthalpy of inlet water Enthalpy of outlet waterH2O_Enthalpy_t_p H2O_Enthalpy_t_p

25 °C 55 °C2 bar 1.5 bar

#VALUE! KJ/kg #VALUE! KJ/kg

Water outlet temperature Water change of specific enthalpy

35 °C #VALUE! KJ/kgApproach #VALUE! KJ/kg

App = 20 K #VALUE! KJ/kg

55 °C

Water heat flow rateWater outlet pressure

2 bar

0.5 bar

1.5 bar

PAC_air_out = PAC_air_in - DPAC_air

2.7 K < DtApproach < 11 K DPAC_air =PAC_air_out =

Dtapp =

tw_in =pw =

7 kPa < DPAC < 14 kPa tAC_air_out = tw +Dtapp

tw =DPAC_air = Dtapp =DPAC_air = tAC_air_out =

tcomp_disch = tAC_air_in = tcomp_disch tcomp_disch =tAC_air_in =

hw_in = hw_out =tw_in = tw_out =Pw_in = Pw_out =hw_in = hw_out =

tw_out = tair_out + App Dhw = hw_out - hw_in

tair_out = hw_out =hw_in = Dhw =

tw_out =

DHw = mw * Dhw

Pw_out = Pw_in - DPw_AC

Pw_in =DPw_AC =Pw_out =

AftercoolerAfter cooler inlet conditions

Air side inletAir inlet pressure 7 bar(g)Air inlet temperature 335.2 °C

Air mass flow rateAir heat flow rateAir pressure drop

Water mass flow rate Water heat flow rate

Water pressure drop

4.- Air receiver volume

Air receiver volume

Frequencyf = 1 cycle/30 s

t = 30 s/cyclePressure change in receiver

PAC_air_in =tAC_air_in =

ATlas Copco formula [1] (Note 2)

Buffer time (Note 1)Vrec =0.25*(Q /( fmax * DPL_U ) )*( Tin_receiver/Tin_comp)

Vrec = 0.25*(Q*t / DPL_U )*(Tin_receive /Tin_comp)

.

during a cycle0.5 bar V = 450 l/s (FAD)

Compressor FAD flow rate t = 30 s/cycle450.0 l/s FAD 0.5 bar

Compressor intake temperature Temperature at the air receiver inlet

20 °C 308.15 K293.15 K Compresors maximum intake temperature

Receiver inlet air temperature 293.15 K7,095 l

35.0 °C 7.1035.0 °C

308.15 KReciver inlet pressure

For notes, see next page6.860 bar (g)6.860 bar (g) For a general case, see:

Air receiver pressure drop.To estimate de pressure drop in Receiver outlet pressure

an air reciever, one asssumes6.860

1 #VALUE!0.5 #VALUE!1.5

Assuming an air velocity Receiver outlet temperaturev = 10 m/s Since it is assumed an isothermal process

For an air temperaturet = 10 ºC 35.0

the density is 35.0AirDensity_t

#VALUE! kg/m³The kinematic pressure is

#VALUE! kg/m³v = 20 m/s

#VALUE! PaThe pressure drop is

1.5#VALUE!#VALUE! Pa

DPop =

QC = DPL_U =

tcomp_in = Tin_receiver =Tcomp_in =

Tin_Comp =trcv_in = tAC_air_out Vrec =

tAC_air_out = Vrec = m3 trcv_in =Trcv_in =

Prcv_in = PAC_air_out

PAC_air_out =Prcv_in = www.piping-tools.net

Prcv_out = Prcv_in - DPrcv

Krcv = Kin + Kout Prcv_in =Kin = DPrcv =Kout = Prcv_out =Krcv =

trcv_out = trcv_in trcv_in =trcv_out =

r =r =

PK = (r/2) * v^2r =

PK =

DPrcv = Krcv * PK Krcv =PK =DPrcv =

.


#VALUE! bar

4. Air receiver volume

Air receiver inlet conditions Air inlet pressure 6.860 bar(g)(Initial state)Air intlet temperature 35.0 °C

Air mass flow rateAir pressure drop

5.- Dryer

Ambient conditions Inlet dryer temperature30 °C60 % 35.0 °C

H = 0 m.s.n.m. 35.0 °C1.013 bar

Inlet dryer pressure (initial)Dryer inlet temperature

ºC #VALUE! bar (g)35.0 ºC #VALUE! bar (g)

DPrcv =

Prcv_in =

trcv_in =

tcomp_in = tdryer_in = trc-out

fcomp_in = trcv_out =tdryer_in =

Pcomp_in =

Pdryer_in_ini = Prcv_out_ini tdryer_in = trcv_out Prcv_out_ini =trcv_out = Pdryer_in_ini =

35.0 ºCCompresor normal flow rate

1620.0 m³/h FADAbsolute humidity of entering air 0.45 m³/s FADSince until this point there is not water extraction, the absolute humidity is the Air mass flowratesame as at the ambient conditions.

#VALUE! kg_da/s#VALUE! kg_w/kg_da #VALUE! kg_da/s#VALUE! kg_w/kg_da #VALUE! kg_da/h

Inlet dryer data Oulet dryer temperature35.0 °C

#VALUE! kg_w/kg_da 5 ºCH = 0 m.a.s.l. 5 ºC

Dryer inlet pressure Dryer outlet pressure

1.013 bar 1.013 bar1.013 bar 1.013 bar

Mass flow rate Outlet dryer relative humidityThe air shall leave the dryer at a dew

#VALUE!saturated condition at a dry bulb

Volume flow rateV = 0.45 m³/s 100 %

Inlet dryer relative humidity

#VALUE! %

tdryer_in =

Vreq =Vreq =

mdryer = mda xdryer_in = xcomp_in mda =xcomp_in = mdryer =xdryer_in = mdryer =

tdryer_in = tdryer_out = tDP

xdryer_in = tDP =tdryer_out =

pdryer_in = Pamb pdryer_out = pdryer_in

Pamb = pdryer_in =pdryer_in = pdryer_out =

mrcv = mda mda = kgda/s point temperature of tDP. That is in a

temperature tdb = tDP

fdryer_out =

fdryer_in = Sicro_Relative_Humidity_tdb_x_H

fdryer_in =

Heta flow rate to be extracted

Inlet data35.0

#VALUE!#VALUE!

q =Outlet data

#VALUE! 5#VALUE! 100#VALUE! #VALUE!#VALUE!#VALUE! Change in water content#VALUE!#VALUE! #VALUE!

q = #VALUE! kW #VALUE!#VALUE!

t1 =f1 =x1 =

mda * ( (h1 - h2) +x1*hW1 - x2*hW2 )

mda = kgda/s t2 =hda,1 = kJ/kgda f2 =hda,2 = kJ/kgda x2 =x1 = kgw/kgda

x2 = kgw/kgda

hw1 = kJ/kgw Dx = x1 - x2

hw2 = kJ/kgw x1 =x2 =Dx =

q=mda⋅[ (hda,1−hda,2 )+x1⋅hw,1− x2⋅hw,2 ] ( g)

x1

x2

f1

5. Dryer

Dryer inlet conditions Air inlet pressure 1.013 bar Air intlet temperature 35.0 °C

#VALUE!#VALUE! %

Air mass flow rateVolume flow rateAir pressure drop

Condensed water vapor Extracted heat rate

6.- Condensed water vapor

Pdryer_in =

tdryer_in =

xdryer_in = kgw / kgda

fdryer_in =

t1t2

Cooling and liquefaction

If the afterccoler is a stand alone aftercooler, the water will not be contaminated.

Compressor manufacturers may include aftercoolers within the compressor package. In general these compressors are referred to as integral aftercoolers. In these case, the water will be contaminated

The water comming out of the refrigerant dryer wil be contaminated, since this water has been carried out by an oil contaminated air

7.- Pressure drop

Aftercooler

Aftercooler pressure drop [3]Pressure drop in th system [1]Estimation example Let

Air receiver pressure drop.To estimate de pressure drop inan air reciever, one asssumes

Assuming an air velocity

v =

For an air temperaturet =

the density is

7 [kPa] < DPAC < 14 [kPa]

DPAC =DPAC =

Karcv =Kin =Kout =

Karcv =

r =r =

Dry air gas constantAmbient air absolute humidity 286.9

Sicro_Dew_Point_tdb_f_H Wate vapor gas constanttdb = 20 ºC 461.5

60 %H = 0 m.a.s..l. Dry air standard conditions

#VALUE! 101,3250

The required dew point of the dryed air is 05 ºC 273.15

Note 1. This example could not consider exactly the publication [1], Kelvinsince different equipments do not use allways the outputs from Kelv = 273.15previous equipment.

Table 2. Consumer's requirements, with all air flows in l/s(FAD) and pressures in bar (g)Consumer Air flow Pressure Dew Point

1 #VALUE! l/s (FAD) 6 bar(g) 52 67 l/s (FAD) 5.99 bar(g) 53 95 l/s (FAD) 4 bar(g) 5

#VALUE! l/s (FAD) Selected required flow rate

Flow rate requirements Minimum pressure required at

#VALUE! l/s FAD consumption points (cp) Selected compressor' pressure

67 l/s FAD 6 bar (g)95 l/s FAD Total pressure loss between

Total requirement to be supplied compressor and the consumption points

Rda =xcomp_in =

Rw =f =

xcomp_in = kgw/kgda P1 =f1 =t1=

tDP = T1 =

Sum Vi =

Vreq =

Pcp = Pcomp =

#VALUE! l/s FAD (Estimated in #7)Considering a safety margin of #VALUE! bar

15 % Defining the operational pressure changeThe Safety Factor is in the receiver as

1.15 0.5 bar The required flow rate is

Compresor's pressure1.15

#VALUE! l/s FAD #VALUE! bar (g)#VALUE! l/s FAD

See sheet 6. FAD, equation (21)

State 2: FAD1620.00101,325 Pa

0.6 - #VALUE! Pa

303 K

State 1: Normal conditions101,325 Pa

0.0 - #VALUE! Pa

273 K

Normal value of the required flow rate#VALUE! Nm³/h#VALUE! Nm³/h

Discharge temperature

DPloss =

DPop =

SF * Vreq_ Pcomp = Pcp + DPloss + DPop

Pcomp =

V1 = V2 * (P2-RH2*Psat.water_2) / (P1-RH1*Psat.water_1 ) * (T1 / T2)

Vreq = V2 = m3/h (FAD)

P2 =RH2 =

Psat.water_2 =T2 =

P1 =RH1 =

Psat.water_1 =T1 =

V1 =Vreq_N =

Microsoft Equation 3.0

(21) q. ETT

PP

VV2

1

1_.11

2_.2221

=watersat

watersat

PRHPRH

303.15 K101.3 kPa801.3 kPa

k = 1.4 -0.8 -608 K

335.2 ºC

FromCompressor_power_and_air_discharge_ temperature

Compressor outlet pressure Compressor's power

70.4 kPa #VALUE!700 kPa(g) Cp = #VALUE!

770.4 kPa 303.150.8

Compression ratio r = 7.60r = 1.4

770.4 kPa W= #VALUE!101.33 kPa From sheet "Motors", selected motor:

r = 7.60 W = 185

Note. Selection based on [5]

Tdisch = Tin * ( 1 + ( ((Pdisch/Pin)^((k-1)/k) -1) / hc ) )Tin =Pin =Pout =

hc =Tdisch =tdisch =


Pout= Patm + Pout(g) W =m * Cp * T1 / hc * ( r^( (g-1) / (g) ) - 1 )

Patm = mda =Pout(g) =Pout= T1 =

hc =

Pout / Pin g =Pout=Pin=

(e) W

1PP cm

P

k1-k

1

21p

c

T

h

=

=

c

kk

in

desc

indesc

PP

TTh

11

1


FromCompressor_power_and_air_discharge_ temperature

Compressor outlet conditions

Design pressure 7 bar(g)Outlet temperature 335.2 °C

Compressed air quality from an oil lubricated compressor is regarded as sufficient.

If the after cooler is part of the compressor, the cooling water will be a contaminatedwater.If the cooling occur in an heat exchanger, after the compresion, the cooling waterwill be a clean water.

Specific heat capacity of air at Air heat flow rateatmospheric pressure (at sea level)Inlet #VALUE!

335.2 °C #VALUE!


Pcomp_out =tcomp_out =

DHair = mair * Dhair mda =

tAC_air_in = DHair =

(e) W

1PP cm

P

k1-k

1

21p

c

T

h

=


#VALUE! kJ/(kg*K) #VALUE!

Outlet35.0 °C

#VALUE! kJ/(kg*K)

Average air specific heat

#VALUE! kJ/(kg*K)#VALUE! kJ/(kg*K)

#VALUE! kJ/(kg*K)

Air specific enthalpy change

#VALUE! kJ/(kg*K)335.2 °C35.0 °C

#VALUE! KJ/kg

Heat balance

#VALUE! KW#VALUE! KJ/kg#VALUE! kg/s

Mass flow rate of water required in the after cooler

#VALUE! kg/s

cp_AC_in = DHair =

tAC_air_out =cp_AC_out =

cp_AC_air_ave = (cp_AC_air_in + cp_AC_air_out) / 2cp_AC_air_in =cp_AC_air_out =

cp_AC_air_ave =

DhAC_air = h_AC_air_in - h_AC_air_out

DhAC_air = cp_AC_air_ave * (t_AC_air_in -t_AC_air_out)cp_AC_ave =tAC_air_in =tAC_air_out =DhAC_air =

DHair = DHw DHair = mw * Dhw

DHair / Dhw = mw mw = DHair / Dhw

DHair = Dhw =mw =

mw =

After cooler outlet conditions

Air side outletAir outlet pressure 6.860 bar(g)Air outlet temperature 35.0 °C

S kg/s#VALUE! KW

14.0 kPa

S kg/s This is clean water, because in the heat exchanger there is not#VALUE! KW a point were water and air with oil could come in contact

0.5 bar

Refill compressor flow rate

Normal flow rate (Rem. is dry air) 6.860#VALUE! #VALUE!

PAC_air_out =tAC_air_out =

mair = DHair =DPAC_air =

mw = DHw =DPw_AC =

Air receiver volume (Note 3)

Prcv_out_ini = Prcv_in - DPrcv_loss

Prcv_in =Vreq_N = Nm3/h DPrcv_loss =.

.

)13( op

rcv

N

NNbuffer P

TTPVV

D

= t(18) 1V _

arg_ Nreq

rcv

N

N

op

echNcomp V

TT

PP

V D

=t

#VALUE! #VALUE!

Application #VALUE!0.5

30 s/cycle #VALUE!#VALUE! Nm³/s1.01325 bar

0.5 bar308.15 K #VALUE!

273.15 K 60#VALUE! m³ 0.5

1.01325273.15308.15

#VALUE!Air receivers volume calculation #VALUE!

Note 1

Buffer time is the time interval in which the the receiver is supplying the required air flow

bar (g)bar and is able to deliver the required air flow rate until its pressure reach its minimum

bar(g)

Since it is assumed an isothermal process

°C°C For a comparison of the Atlas Copco equation with other

equations published in the web and with the deducted

equation, see

Air receivers volume calculation

Note 3

The receiver volume, or buffer volume, is the tank volume needed to deliver the required

Note 4

Refill time is the time required by the compressor to increase the

Vreq_N = Nm3/s Prcv_out_ini =Prcv_out_final = Prcv_out_ini - Dpop Prcv_out_ini =

Vrcv = tbuffer * Vn * PN/ ( DPop) * (Trcv/TN) DPop =tbuffer = Prcv_out_final =

Vn =Pn =DPop = Vcomp_N = Vrcv*(1/tcharge)*(DPop/PN)*(TN/Trcv) + Vreq_N

Trcv = Vrcv =

TN = tcharge = Vrcv = DPop =

PN =TN =Trcv =

Vreq_N =www.piping-tools.net Vcomp_N =

rate whithout receiving any compressor flow. The receiver is initially at a pressure P initial

value Pfinal. A minimum of 15 minutes is recommended as a buffer time [10].

A large buffer time requires a large DPop


flow rate with a supply pressure pfinal_g , without compressor contribution, during

a time tbuffer.

Note 2This equation does not agree with the derived equation nor with the in the web proposed equations. The Atlas Copco equation replaces the value of the ambient pressure, in this case 1.013 bar, by a value 0.25

.

.

.

.

.

.

.



receiver pressure to its highest value.

4. Air receiver volume

Air receiver outlet conditions Air outlet pressure #VALUE! #VALUE! bar(g)Air outlet temperature 35.0 °C

S kg/s#VALUE! kPa

Dryer pressure drop(from #7)

0.09 bar

Dryer outlet pressure (initial)

#VALUE! bar(g)0.09 bar

#VALUE! bar(g)

Prcv_out_ini =Prcv_out_final =

trcv_out =

mair =DParc =

DPdryer =

Pdryer_out_ini = Pdryer_in:ini - DPdryer Pdryer_in_ini =DPdryer =

Pdryer_out_ini =

Dryer outlet pressure (final)

#VALUE! bar (g)0.5 bar (g)

#VALUE! bar (g)

Dryer intlet: State 1 Enthalpy of water in state 1H2O_Enthalpy_t_p

Density in state 1 35.0Sicro_Density_tdb_x_H p = 1.013

35.0 ºC #VALUE!#VALUE! %

H = 0 m.a.s.l. Enthalpy of water in state 2#VALUE! H2O_Enthalpy_t_p

5Dry air mass flow rate p = 1.013

#VALUE! #VALUE!

Absolute humidity in state 1 Enthalpy of air in state 1#VALUE! Sicro_Enthalpy_tdb_x_H

35.0Dryer outlet: State 2 #VALUE!

#VALUE!Absolute humidity in state 2

Sicro_Absolute_Humidity_tdb_f_H Enthalpy of air in state 25 ºC Sicro_Enthalpy_tdb_f_H

100 % 5H = 0 m.a.s.l. 100

#VALUE! H = 0.000#VALUE!

Pdryer_out_final = Pdryer_out_ini - DPop Pdryer_out_ini =DPop =

Pdryer_out_final =

hw1 =t1 =

r1 =t1 = hw1 =x1 =

r1 = kgda/m³ hw2 =t2 =

m1 = kgda/s hw2 =

x1 = kgw / kgda ha1 =t1 =x1 =ha1 =

x2 =t2 = ha2 =f2 = t2 =

f2 =x2 = kgw / kgda

ha2 =

Condensed water vapor flow rateºC% #VALUE!

#VALUE!#VALUE!#VALUE!

ºC%

mw_cond = mda * Dxmda = kgda/s

kgw/kgda Dx = kgw/kgda

mw_cond = kgw / smw_cond = kgw / h

kgw/kgda

kgw/kgda

kgw/kgda

kgw/kgda

Dryer outlet conditions Air outlet pressure #VALUE! bar (g) #VALUE! bar (g)Air outlet temperature 5.0 °C

#VALUE!100 %

#VALUE! kg/s The dryer exit presure, when the receiver isV = 0.45 m³/s supplying at its minimum pressure is

0.090 bar #VALUE! bar (g)The maximum pressure drop between the dryer

#VALUE! exit and the las consumption point has been q = #VALUE! kW estimated as

0.15 barThus, the supplyed pressure is

#VALUE! bar (g)0.15 bar

#VALUE! bar#VALUE!

Pdryer_out_ini =Pdryer_out_final =

tdryer _out =

xdryer_out = kgw / kgda

fdryer_out =

mair =

DPdryer = Pdryer_out_final is

mw_cond = kgw / h

DPpipes =

Psupplyed = Pdryer_out_final - DPpipes Pdryer_out_final =DPpipes =Psupplyed =

In general these compressors are referred to as integral aftercoolers. In these case, the water will be contaminated

The water comming out of the refrigerant dryer wil be contaminated, since this water has been carried out by an oil contaminated air

Aftercooler pressure drop [3]

Pressure drops on compresor`s equipment

14 kPa0.14 bar Aftercooler

Air receiverAir receiver pressure drop. The kinematic pressure is Oil filterTo estimate de pressure drop in Refrig.dryeran air reciever, one asssumes #VALUE! kg/m³ Dust filter

v = 20 m/s Total pressure drop in equipments1 #VALUE! Pa

0.51.5 The pressure drop is Pressure drop in pipes between

compressor exit and consumptionAssuming an air velocity 1.5 points (Assumed)

10 m/s #VALUE!#VALUE! Pa

For an air temperature #VALUE! bar Total pressure drop10 ºC

AirDensity_t #VALUE! kg/m³


PK = (r/2) * v^2r =

Kin + Kout

PK = DPequip =

DParcv = Karcv * PK Karcv =PK = DPpipes =

DParcv =DParcv =

DPloss =DPequip =DPpipes =DPloss =

Rev. cjc. 15.04.2016

page 1 of 20

J/(kg*K)

J/(kg*K)

Pa -°CK

K

page 2 of 20

°C°C°C

Selected required flow rate

450.0 l/s FAD

Selected compressor' pressure

7 bar (g)

page 3 of 20

page 4 of 20

page 5 of 20

kg_da/skJ/kgK - - -kW

From sheet "Motors", selected motor:kW

g-1) / (g) ) - 1 )

Return

page 6 of 20

Compressed air quality from an oil lubricated compressor is regarded as sufficient.

If the after cooler is part of the compressor, the cooling water will be a contaminated

If the cooling occur in an heat exchanger, after the compresion, the cooling water

page 7 of 20

kg_da/sKJ/kg

Return

KW

page 8 of 20

page 9 of 20

This is clean water, because in the heat exchanger there is not

page 10 of20

bar (g)bar

Return

(18) 1V _arg

_ Nreqrcv

N

N

op

echNcomp V

TT

PP

V D

=t

bar (g)

bar (g)barbar (g)

m³

sbarbarKKNm³/sNm³/s

page 11 of 20

Buffer time is the time interval in which the the receiver is supplying the required air flow

and is able to deliver the required air flow rate until its pressure reach its minimum

The receiver volume, or buffer volume, is the tank volume needed to deliver the required

)*(DPop/PN)*(TN/Trcv) + Vreq_N

rate whithout receiving any compressor flow. The receiver is initially at a pressure P initial

, without compressor contribution, during

.

page 12 of 20

page 13 of 20

page 14 of 20

Density in state 1

H2O_Enthalpy_t_p r1 = Sicro_Density_tdb_f_HºC t1 = 30 ºCbar f1 = 50 %

H = 0 m.a.s.l.r1 = 1.155154124 kgda/m³

Dry air mass flow rate

H2O_Enthalpy_t_p m1 = V1 * r1ºC V1 = 5 m³/sbar r1 = 1.155154124 kgda/m³

m1 = 5.77577062 kgda/sAbsolute humidity in state 1

x1 = Sicro_Absolute_Humidity_td .Sicro_Enthalpy_tdb_x_H t1 = 30 ºC

ºC f1 = 50 %H = 0 m.a.s.l.x1 = 0.013311014 kgw / kgda

Absolute humidity in state 2

x2 = Sicro_Absolute_Humidity_td .Sicro_Enthalpy_tdb_f_H t2 = 10 ºC

ºC f2 = 100 %% H = 0 m.a.s.l.m.a.s.l. x2 = 0.007630516 kgw / kgda

kJ/kgw

kJ/kgw

kgw / kgda

kJ/kgda

kJ/kgda

page 15 of 20

page 16 of 20

page 17 of 20

page 18 of 20

page 19 of 20

Return

page 20 of 20


0.14 bar

#VALUE! bar0.08 bar0.09 bar0.08 bar

Total pressure drop in equipments#VALUE! bar

Pressure drop in pipes betweencompressor exit and consumptionpoints (Assumed)

0.15 bar

Total pressure drop

#VALUE! bar0.15 bar

#VALUE! bar

DPequip + DPpipes

Return

Enthalpy of water in state 1hw1 = H2O_Enthalpy_t_pt1 = 30 ºCp = 1.01325 barhw1 = 125.7520702 kJ/kgwEnthalpy of water in state 2hw2 = H2O_Enthalpy_t_pt2 = 10 ºCp = 1.01325 barhw2 = 42.091837 kJ/kgw

Enthalpy of air in state 1ha1 = Sicro_Enthalpy_tdb_f_Ht1 = 30 ºCf1 = 50 %H = 0 m.a.s.l.ha1 = 64.01163777 kJ/kgdaEnthalpy of air in state 2ha2 = Sicro_Enthalpy_tdb_f_Ht2 = 10 ºCf2 = 100 %H = 1.01325 m.a.s.l.

29.22398957 kJ/kgda

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1f

as definedfrequency buffer a of use make literature Some

r

r

t

t

t

=D

=

=D

=D

=

=

Vf

m

Vm

cyclekgmm

cyclekgmm

f

buff

buffer

buffer



can T

product theThus,


(e) TR

obtainsona

(c) TR


(d) ³v³V


=

=

=

=

pv

RT

pvmV

pVV

skgm

kgm

sm

r

r





can T

product theThus,


(e) TR

obtainsona

(c) TR


(d) ³v³V


=

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=

=

pv

RT

pvmV

pVV

skgm

kgm

sm

r

r

(9) R1

T1

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R1

T1

gets one

8)( 1 equation into

(e) R1

T

equation Replacing

_

__

=D

=D

=D

=

incomp

incompincomp pvm

fm

pvmf

m

Vf

m

pvmV

r

r

)10(³][

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1

1

obtains one

(f) 1

equation into

Replacing

_

3

_

_

_

__

_

__

_

__

_

__

__

KTKT

barPs

cyclesf

smV

barPmV

R

TT

PfV

PV

TT

PP

Vf

V

PT

TP

Vf

V

PT

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vmf

V

Vvm

incomp

rcv

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rcv

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incompincomp

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rcv

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rcv

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op

rcv

incomp

incompincomp

incompincomp

D

=

D

=

D

=

D=

D

=

=




)12(

)11( ³][

gets one

)10( ³][

equation into

1f

Replacing

_

__

_

__

_

3

_

_

_

3

_

_

op

rcv

incomp

incompincompbuffer

incomp

rcv

op

incompbufferincomp

incomp

rcv

op

incompbuffer

incomp

incomp

rcv

op

incomp

incomp

buffer

PT

TP

VV

TT

PV

PV

KTKT

barPsmV

cycles

barPmV

KTKT

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smV

barPmV

D

=

D

=

D

=

D

=

=

t

t

t

t

)15(

)14(

(13) and (12) equations arranging-re and

)13(

obtains one

)12(

equation in relation thisreplacing and

has onelaw gas ideal theusing state, Normal the withstateinlet s'compressor theRelating

_

__

_

__

_

__

N

rcv

op

NNbuffer

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op

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TT

PP

VV

PT

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PT

TP

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TNPV

TP

V

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=

D

=

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t

t

t

t

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gets one

by density normal de replacing and

rateflow mass Required

rateflow mass Compressor

__arg

arg__

__

arg__

_

_

RTVVPm

TRPVVm

TRP

VVm

VVm

Thus

Vm

Vm

NNreqNcompNech

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NNreqNcomp

N

NN

NNreqNcomp

echNNreqNNcomp

NNreqreq

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=

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=

t

t

r

r

trr

r

r

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1

1

obtains one

(f) 1

equation into

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_

3

_

_

_

__

_

__

_

__

_

__

__

KTKT

barPs

cyclesf

smV

barPmV

R

TT

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PV

TT

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Vf

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PT

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PT

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rcv

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D

=

D

=

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=

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=

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gets one

by density normal de replacing and

rateflow mass Required

rateflow mass Compressor

__arg

arg__

__

arg__

_

_

RTVVPm

TRPVVm

TRP

VVm

VVm

Thus

Vm

Vm

NNreqNcompNech

echN

NNreqNcomp

N

NN

NNreqNcomp

echNNreqNNcomp

NNreqreq

NNcompcomp

=D

=D

=

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t

t

r

r

trr

r

r

(18) 1V

1V

V 1

(17) 1 equation From

interval timegivena in

"" system thesupply toair ssimultaneu

ithreceiver w therefill ocapacity t s'Compressor

_arg

_

arg__

__arg

__arg

arg

_

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N

N

op

echNcomp

oprcv

N

NechNreqNcomp

opN

rcvNreqNcompNech

opN

rcvNreqNcompNech

ech

Nreq

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PP

V

PTT

PVV

PTTVVP

PTTVVPV

V

D

=

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=

=D

D=

t

t

t

t

t

Saturation pressure of water

From [4], page 6.2, equation (6). Valid for a range of 0 to 200ºCexp( -5800.2206/(t+273.15) + 1.3914993 + -0.048640239*(t+273.15) + 0.000041764768*(t+273.15 )^2 + -0.000000014452093*(t+273.15 )^3 + 6.5459673*ln(t+273.15 ) )

0 ºC611.2 Pa

Using the VBA function Eq. (22)Sicro_Saturated_vapor_pressure_t(t)

0 ºC#VALUE! kPa#VALUE! Pa

Using the Steamdat VBA functionsH2O_SaturationPressure_t

0 °C#VALUE! bar

#VALUE! Pa

For derivation of equation (21), see sheet FAD


(18) 1V

1V

V 1

(17) 1 equation From

interval timegivena in

"" system thesupply toair ssimultaneu

ithreceiver w therefill ocapacity t s'Compressor

_arg

_

arg__

__arg

__arg

arg

_

Nreqrcv

N

N

op

echNcomp

oprcv

N

NechNreqNcomp

opN

rcvNreqNcompNech

opN

rcvNreqNcompNech

ech

Nreq

VTT

PP

V

PTT

PVV

PTTVVP

PTTVVPV

V

D

=

D

=

=D

D=

t

t

t

t

t

(21) q. ETT

PP

VV2

1

1_.11

2_.2221

=watersat

watersat

PRHPRH

Energy balanceEda,1+ Ew,1=Eda,1+ Ew,1−qmda⋅hda,1+ mw,1⋅hw,1=mda⋅hda,2+ mw,2⋅hw,2+q (a ) Water mass balancemw1=mw1+mw_cond

with


frommda⋅x1=mda⋅x2+mw_cond (d )mw_cond=mda⋅x1−mda⋅x2

mw_cond=mda⋅( x1−x2 ) (e )and frommda⋅hda,1+mw,1⋅hw,1=mda⋅hda,2+mw,2⋅hw,2+q (a )q=mda⋅hda,1+mw,1⋅hw,1−mda⋅hda,2−mw,2⋅hw,2q=mda⋅hda,1−mda⋅hda,2+mw,1⋅hw,1−mw,2⋅hw,2

q=mda⋅(hda,1−hda,2)+mw,1⋅hw,1−mw,2⋅hw,2 ( f )Replacing in ( f )mw1=mda⋅x1 (b )mw2=mda⋅x2 (c )q=mda⋅(hda,1−hda,2)+mw,1⋅hw,1−mw,2⋅hw,2 ( f )q=mda⋅(hda,1−hda,2)+mda⋅x1⋅hw,1−mda⋅x2⋅hw,2


Energy balanceEda,1+ Ew,1=Eda,1+ Ew,1−qmda⋅hda,1+ mw,1⋅hw,1=mda⋅hda,2+ mw,2⋅hw,2+q (a ) Water mass balancemw1=mw1+mw_cond

with




mw_cond=mda⋅( x1−x2 ) (e )and frommda⋅hda,1+mw,1⋅hw,1=mda⋅hda,2+mw,2⋅hw,2+q (a )q=mda⋅hda,1+mw,1⋅hw,1−mda⋅hda,2−mw,2⋅hw,2q=mda⋅hda,1−mda⋅hda,2+mw,1⋅hw,1−mw,2⋅hw,2

q=mda⋅(hda,1−hda,2)+mw,1⋅hw,1−mw,2⋅hw,2 ( f )Replacing in ( f )mw1=mda⋅x1 (b )mw2=mda⋅x2 (c )q=mda⋅(hda,1−hda,2)+mw,1⋅hw,1−mw,2⋅hw,2 ( f )q=mda⋅(hda,1−hda,2)+mda⋅x1⋅hw,1−mda⋅x2⋅hw,2


Rev. cjc. 15.04.2016

Page 1 of 10

Page 2 of 10


(18) 1V

(17) 1

)16( 11

)15(

)14(

)13(

_arg

_

__arg

__arg

_

__

Nreqrcv

N

N

op

echNcomp

opN

rcvNreqNcompNech

NNreqNcompNech

N

rcv

op

NNbuffer

incomp

rcv

op

incompincompbuffer

op

rcv

N

NNbuffer

VTT

PP

V

PTTVVPV

RTVVPm

TT

PPVV

TT

PP

VV

PT

TPVV

D

=

D=

=D

D=

D=

D

=

t

t

t

t

t

t

Page 3 of 10

Page 4 of 10

Page 5 of 10

Page 6 of 10

Page 7 of 10

Page 8 of 10

Page 9 of 10

Eq. (22)exp( -5800.2206/(t+273.15) + 1.3914993 + -0.048640239*(t+273.15) + 0.000041764768*(t+273.15 )^2 + -0.000000014452093*(t+273.15 )^3 + 6.5459673*ln(t+273.15 ) )

Page 10 of 10

(21) q. ETT

PP

VV2

1

1_.11

2_.2221

=watersat

watersat

PRHPRH


mw_cond=mda⋅( x1−x2 ) (e )and frommda⋅hda,1+ mw,1⋅hw,1=mda⋅hda,2+mw,2⋅hw,2+q (a )q=mda⋅hda,1+mw,1⋅hw,1−mda⋅hda,2−mw,2⋅hw,2q=mda⋅hda,1−mda⋅hda,2+mw,1⋅hw,1−mw,2⋅hw,2

q=mda⋅(hda,1−hda,2)+ mw,1⋅hw,1−mw,2⋅hw,2 ( f )Replacing in ( f )mw1=mda⋅x1 (b )mw2=mda⋅x2 (c )q=mda⋅(hda,1−hda,2)+ mw,1⋅hw,1−mw,2⋅hw,2 ( f )q=mda⋅(hda,1−hda,2)+ mda⋅x1⋅hw,1−mda⋅x2⋅hw,2




mw_cond=mda⋅( x1−x2 ) (e )and frommda⋅hda,1+ mw,1⋅hw,1=mda⋅hda,2+mw,2⋅hw,2+q (a )q=mda⋅hda,1+mw,1⋅hw,1−mda⋅hda,2−mw,2⋅hw,2q=mda⋅hda,1−mda⋅hda,2+mw,1⋅hw,1−mw,2⋅hw,2

q=mda⋅(hda,1−hda,2)+ mw,1⋅hw,1−mw,2⋅hw,2 ( f )Replacing in ( f )mw1=mda⋅x1 (b )mw2=mda⋅x2 (c )q=mda⋅(hda,1−hda,2)+ mw,1⋅hw,1−mw,2⋅hw,2 ( f )q=mda⋅(hda,1−hda,2)+ mda⋅x1⋅hw,1−mda⋅x2⋅hw,2



5.1 Example of dimensioning compressed air installations

A result of 225 l/s (FAD) comes from a pressure P = 100 kPa and a temperature of t = 30 ºC.

Result is valid for relative air humidity RH2 = 0 % (instead of 60%) , t2 = 35 °C (instead of 30 °C)and ambient local pressure Pi = 0.74 bar (instead of 1 bar)

For a pressure P = 74 kPa, a temperature of t = 35 ºC and a relative humidity of RH = 0 %, the flow rate is 308.9 l/s (FAD)

This equation does not agree with the derived equation nor with the in the web proposed equations. In all other equations, the constant 0.25 is replaced for a pressure. In this case, the value 0.25 should be replaced by the value of the compressor's inlet pressure P = 1 bar.So the result would be 6.972 * ( 1/0.25) = 27.89m³. See sheet Receiver.

according Atlas Copco, 5.1 kW

[1]

Q = 309 l/s(FAD)

A result of 225 l/s (FAD) comes from a pressure P = 100 kPa and a temperature of t = 30 ºC.

= 35 °C (instead of 30 °C)

For a pressure P = 74 kPa, a temperature of t = 35 ºC and a relative humidity of RH = 0 %, the flow rate is 308.9 l/s (FAD)

This equation does not agree with the derived equation nor with the in the web proposed equations. In all other equations, the constant 0.25 is replaced for a pressure. In this case, the value 0.25 should be replaced by the value of the compressor's inlet pressure P = 1 bar.So the result would be 6.972 * ( 1/0.25) = 27.89m³. See sheet Receiver.

Pressure drop in th system [1]Estimation example

Aftercooler pressure drop [3]

Let14 kPa

0.14 bar

Air receiver pressure drop. The kinematic pressure isTo estimate de pressure drop inan air reciever, one asssumes #VALUE! kg/m³

v = 20 m/s1 #VALUE! Pa

0.51.5 The pressure drop is

Assuming an air velocity 1.5v = 10 m/s #VALUE!

#VALUE! PaFor an air temperature #VALUE! bar

t = 10 ºCthe density is

AirDensity_t #VALUE! kg/m³


DPAC =DPAC =

PK = (r/2) * v^2r =

Karcv = Kin + Kout

Kin = PK =Kout =

Karcv =DParcv = Karcv * PK Karcv =PK =

DParcv =DParcv =

r =r =


Aftercooler 0.14 barAir receiver #VALUE! barOil filter 0.08 barRefrig.dryer 0.09 barDust filter 0.08 barTotal pressure drop in equipments

#VALUE! bar

Pressure drop in pipes betweencompressor exit and consumptionpoints (Assumed)

0.15 bar

Total pressure drop

#VALUE! bar0.15 bar

#VALUE! bar

DPequip =

DPpipes =

DPloss = DPequip + DPpipes

DPequip =DPpipes =DPloss =

[5] http://www.nrcan.gc.ca/energy/regulations-codes-standards/products/6885

Minimum nominal efficiency standard at 100% of nominal full load(Premium - most stringent level)

Corresponds to Table 3 in the CAN/CSA 390-10

Item Power (HP) Power (kW)Energy Efficiency Standard (Percentage)

Open Enclosed2 Pole 4 Pole 6 Pole 2 Pole

1 1 0.75 77 85.5 82.5 772 1.5 1.1 84 86.5 86.5 843 2 1.5 85.5 86.5 87.5 85.54 3 2.2 85.5 89.5 88.5 86.55 4 3 86.5 89.5 89.5 88.56 5 3.7 86.5 89.5 89.5 88.57 5.5 4 86.5 89.5 89.5 88.58 7.5 5.5 88.5 91 90.2 89.59 10 7.5 89.5 91.7 91.7 90.2

10 15 11 90.2 93 91.7 9111 20 15 91 93 92.4 9112 25 19 91.7 93.6 93 91.713 30 22 91.7 94.1 93.6 91.714 40 30 92.4 94.1 94.1 92.415 50 37 93 94.5 94.1 9316 60 45 93.6 95 94.5 93.617 75 55 93.6 95 94.5 93.618 100 75 93.6 95.4 95 94.119 125 90 94.1 95.4 95 9520 150 110 94.1 95.8 95.4 9521 175 132 95 95.8 95.4 95.422 200 150 95 95.8 95.4 95.4

See CSA C390-10 for interpolating minimum performance requirements.

http://www.nrcan.gc.ca/energy/regulations-codes-standards/products/6885

Minimum nominal efficiency standard at 100% of nominal full load(Energy Efficient - least stringent level)

Corresponds to Table 2 in the CAN/CSA C390-1


Open Enclosed2 Pole 4 Pole 6 Pole 8 Pole 2 Pole

1 1 0.75 75.5 82.5 80 74 75.52 1.5 1.1 82.5 84 84 75.5 82.53 2 1.5 84 84 85.5 85.5 844 3 2.2 84 86.5 86.5 86.5 85.55 4 3 84 86.5 86.5 86.5 85.56 5 3.7 85.5 87.5 87.5 87.5 87.57 5.5 4 85.5 87.5 87.5 87.5 87.58 7.5 5.5 87.5 88.5 88.5 88.5 88.59 10 7.5 88.5 89.5 90.2 89.5 89.5

10 15 11 89.5 91 90.2 89.5 90.211 20 15 90.2 91 91 90.2 90.212 25 19 91 91.7 91.7 90.2 9113 30 22 91 92.4 92.4 91 9114 40 30 91.7 93 93 91 91.715 50 37 92.4 93 93 91.7 92.416 60 45 93 93.6 93.6 92.4 9317 75 55 93 94.1 93.6 93.6 9318 100 75 93 94.1 94.1 93.6 93.619 125 90 93.6 94.5 94.1 93.6 94.520 150 110 93.6 95 94.5 93.6 94.521 175 132 94.5 95 94.5 93.6 9522 200 150 94.5 95 94.5 93.6 95

file:///E:/Motores/Electric%20Motors%20%281%20to%20500%20HP_0.746%20to%20375%20kW%29%20_%20Office%20of%20Energy%20Efficiency.htm

23 250 185 94.5 95.4 95.4 94.5 95.424 300 225 95 95.4 95.4 95.425 350 260 95 95.4 95.4 95.426 400 300 95.4 95.4 95.427 450 335 95.8 95.8 95.428 500 375 95.8 95.8 95.4


- -

- - - - - -


Minimum nominal efficiency standard at 100% of nominal full load Minimum nominal efficiency standard at 100% of nominal full load (Premium - most stringent level) (Energy Efficient - least stringent level)

Corresponds to Table 3 in the CAN/CSA 390-10 Corresponds to Table 2 in the CAN/CSA C390-1Energy Efficiency Standard (Percentage)


Enclosed Open4 Pole 6 Pole 2 Pole

85.5 82.5 1 1 0.75 75.586.5 87.5 2 1.5 1.1 82.586.5 88.5 3 2 1.5 8489.5 89.5 4 3 2.2 8489.5 89.5 5 4 3 8489.5 89.5 6 5 3.7 85.589.5 89.5 7 5.5 4 85.591.7 91 8 7.5 5.5 87.591.7 91 9 10 7.5 88.592.4 91.7 10 15 11 89.5

93 91.7 11 20 15 90.293.6 93 12 25 19 9193.6 93 13 30 22 9194.1 94.1 14 40 30 91.794.5 94.1 15 50 37 92.4

95 94.5 16 60 45 9395.4 94.5 17 75 55 9395.4 95 18 100 75 9395.4 95 19 125 90 93.695.8 95.8 20 150 110 93.696.2 95.8 21 175 132 94.596.2 95.8 22 200 150 94.5

23 250 185 94.524 300 225 9525 350 260 9526 400 300 95.427 450 335 95.828 500 375 95.8

Minimum nominal efficiency standard at 100% of nominal full load(Energy Efficient - least stringent level)

Corresponds to Table 2 in the CAN/CSA C390-1Energy Efficiency Standard (Percentage)

Enclosed4 Pole 6 Pole 8 Pole

82.5 80 7484 85.5 7784 86.5 82.5

87.5 87.5 8487.5 87.5 8487.5 87.5 85.587.5 87.5 85.589.5 89.5 85.589.5 89.5 88.5

91 90.2 88.591 90.2 89.5

92.4 91.7 89.592.4 91.7 91

93 93 9193 93 91.7

93.6 93.6 91.794.1 93.6 9394.5 94.1 9394.5 94.1 93.6

95 95 93.695 95 94.195 95 94.1


95 95 94.595.4 9595.4 9595.495.495.8


- -

- - - - - -

Minimum nominal efficiency standard at 100% of nominal full load (Energy Efficient - least stringent level)

Corresponds to Table 2 in the CAN/CSA C390-1Energy Efficiency Standard (Percentage)

Open Enclosed4 Pole 6 Pole 8 Pole 2 Pole 4 Pole 6 Pole 8 Pole

82.5 80 74 75.5 82.5 80 7484 84 75.5 82.5 84 85.5 7784 85.5 85.5 84 84 86.5 82.5

86.5 86.5 86.5 85.5 87.5 87.5 8486.5 86.5 86.5 85.5 87.5 87.5 8487.5 87.5 87.5 87.5 87.5 87.5 85.587.5 87.5 87.5 87.5 87.5 87.5 85.588.5 88.5 88.5 88.5 89.5 89.5 85.589.5 90.2 89.5 89.5 89.5 89.5 88.5

91 90.2 89.5 90.2 91 90.2 88.591 91 90.2 90.2 91 90.2 89.5

91.7 91.7 90.2 91 92.4 91.7 89.592.4 92.4 91 91 92.4 91.7 91

93 93 91 91.7 93 93 9193 93 91.7 92.4 93 93 91.7

93.6 93.6 92.4 93 93.6 93.6 91.794.1 93.6 93.6 93 94.1 93.6 9394.1 94.1 93.6 93.6 94.5 94.1 9394.5 94.1 93.6 94.5 94.5 94.1 93.6

95 94.5 93.6 94.5 95 95 93.695 94.5 93.6 95 95 95 94.195 94.5 93.6 95 95 95 94.1

95.4 95.4 94.5 95.4 95 95 94.595.4 95.4 95.4 95.4 9595.4 95.4 95.4 95.4 9595.4 95.4 95.495.8 95.4 95.495.8 95.4 95.8

- - - -

- - - - - - - - - - - -

Free Air Delivery (FAD)

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)1,450

101,325 Pa0 -0 °C

273 K

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)



p

p (-) humidity relativeair :

(Pa) pressure water saturated:p

(c) pppressure partialr Water vapo

w_sat

v

w_sat

w_satv

=

=

where

(b) - p ppressure partialair Dry

pressureor water vappartial:pairdry of pressure partial:p

air humid of pressure total:pp p

ha)(index air humida For

(a) TT

pp v v

TRvp

TRvp

2 and 1 statesfor andTRvp(ha)air humid

theof fraction (a)air dry the toappliedlaw gas Ideal

haa

v

a

ha

aha

1

2

a2

a112

1

1a1

2

2a2

a

v

v

p

p

=

=

=

=

=


(d) TT

- p- p vv

gets on

(a) TT

pp v v

equation into themreplacingandair humid theof pressures

total theare Pand Pwhere- p p

2 state for the and- p p

as 1 state of b) (Eq. presurepartialair dry theDenoting

1

2

22

1112

1

2

a2

a112

21

22a2

11a1

=

=

=

=

v

v

v

v

pp

p

p

(21) TT

p- pp- p

vv

obtains one

(d) TT

- p- p vv

equation into

(c) pp equation Replacing

1

2

2w_sat_22

1w_sat_1112

1

2

22

1112

w_satv

=

=

=

v

v

pp

(21) TT

p- pp- p

vv1

2

2w_sat_22

1w_sat_1112

=

#VALUE! Pa

FAD conditions (State 2))98,000 Pa

0.4 - 22 °C

#VALUE! Pa295 K

#VALUE!

Psat.water_1 =

P2 =RH2 =

t2=Psat.water_2 = f(t2)Psat.water_2 =

T2 =

V2 = m3/h (FAD)

FAD volume flow rateFree air delivery (FAD) is the volume of air delivered under the conditionsof 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 Pa0 -0 °C

273 K

#VALUE! Pa

FAD conditions (State 2))73,400 Pa

0.42 - 22 °C

#VALUE! Pa295 K

#VALUE!

Normal densityp / ( R * T)

2.- FAD flow rate (state 2) to Normal flow rate (state 1) p = 101,325R = 286.9

FAD conditions (State 2)) T = 273#VALUE! 1.2998,000 Pa

0.4 - 22 °C

#VALUE! Pa


V1 = Nm3/hP1 =

f1 = RH1 =t1=

T1 =Psat.water_1 = f(t1)Psat.water_1 =

P2 =f2 = RH2 =

t2=Psat.water_2 = f(t2)Psat.water_2 =

T2 =V2 = m3/h (FAD)


rn =

V2 = m3/h (FAD) rn =P2 =

RH2 =t2=

Psat.water_2 = f(t2)Psat.water_2 =


(21) TT

p- pp- p

vv

obtains one

(d) TT

- p- p vv

equation into

(c) pp equation Replacing

1

2

2w_sat_22

1w_sat_1112

1

2

22

1112

w_satv

=

=

=

v

v

pp

(21) TT

p- pp- p

vv1

2

2w_sat_22

1w_sat_1112

=

295 K

Normal air conditions (State 1)101,325 Pa

0 -0 °C

273 K

#VALUE! Pa

#VALUE!

T2 =

P1 =RH1 =

t1=T1 =

Psat.water_1 = f(t1)Psat.water_1 =

V1 = Nm3/h

Rev. cjc. 15.04.2016

Free air delivery (FAD) is the volume of air delivered under the conditionsof temperature and pressure existing at the compressor's intake (state 2).

(21)

p / ( R * T)PaJ/(kg*K)K

kg/Nm3

[1] http://www.atlascopco.dk/Images/CAM_05_CALCULATION_tcm48-705084.pdf

[1a]

Compressed_Air_Manual_tcm46-1249312

[1b] http://pdf.directindustry.fr/pdf/atlas-copco-compresseurs/manuel-air-comprime-atlas-copco/8358-185247-_60.html

[2]

Calculating receivers in compressed air systemshttp://www.engineeringtoolbox.com/compressed-air-receivers-d_846.html

Compressed Air Receivers

http://www.engineeringtoolbox.com/compressed-air-receivers-d_846.html

[3]

[4]

[5]


[10] Instruments Plant Systemshttp://www.chagalesh.com/snportal/uploads/chagalesh/karafarinan%20farda/jozveh/process/8.pdf

[11]

http://www.ecompressedair.com/library-pages/aftercoolers.aspx

http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html

http://www.ecompressedair.com/library-pages/aftercoolers.aspx

http://www.engineeringtoolbox.com/air-altitude-pressure-d_462.html

http://pdf.directindustry.fr/pdf/atlas-copco-compresseurs/manuel-air-comprime-atlas-copco/8358-185247-_60.html



Documents

[XLS]piping-tools.netpiping-tools.net/Downloads/1.Listado/Dimensioning... · Web viewCompressed_Air_Manual_tcm46-1249312 The actual working pressure is approx Pact_work = 2. Working