Descriptive Comparison between a Centrifugal & Reciprocating Compressor
Centrifugal Compressor Design and Performance .Full description
Control engineers, mechanical engineers and mechanical technicians will learn how to select the proper control systems for axial and centrifugal compressors for proper throughput and surge c…Description complète
Compressor Fundamentals and Performance Curves
Descriptive Comparison between a Centrifugal & Reciprocating Compressor
(Syn - Gas Conditions) Centrifugal Compressor Performance Calculations The Centrifugal Compressor Performance Calculations below shall be utilized to monitor the centrifugal compress or stage efficiencies to compare to the the original original performance data. In addition: a separate power calculation is included to determine the power on a change in process conditions (i.e. Moleweight, pressure, flow, temperature, etc...). (Reference: Compres Compressor sor Performance by M. Theodore Gres Gresh) h) The Following Input Data is required to perform a compressor performance analysis: (1) Inlet Conditions: FR = Inle Inlett Gas Flowrat Flowrate e SCFM T1 = Inlet Gas Temperature (F) P1 = Inlet Gas Pressure Press ure (psig) T2 = Discharge D ischarge Temp (F) P2 = Disc harge Press (psi) K = Specific Heat Ratio (c p /c v) and dimensionless MW = Moleweight of gas ( mole ) Rc = Universal Gas Constant (1545 lb-ft/R-mole) v = Specific Volume of Gas (ft3 /lb) Pc = Critical C ritical gas gas Pressure Press ure Tc = Critical Gas Temperature Temperature g = Gravitational Constant (32.174 ft/sec 2)
Applicable Equations are as follows: (1) Conversion of SCFM (flowrate) to ACFM (flowrate) SCFM = Standard Cubic Feet per Minute at operting condition ACFM = Actual Cubic Feet Feet per Minute at ambient conditions Ps = Standard pressure at ambient conditions (14.7 psi) Ts = Standard Temperature at ambient conditions (60 F) Zs = Standard Compressibility Compres sibility at ambient conditions (1.0)
ACFM CFM = SCFM SCFM⋅
Ps T1 Zs ⋅ ⋅ P1 Ts Z1
(2) Calculate the Volumetric Volumetric Flow coefficient coeffic ient for for the inlet inlet conditions v1 = Volumetric flow value value for inlet conditions (ft 3 /lb) Z1 = Compres Compressibility sibility value value for for inlet conditions R = Gas Constant (1545 mole-ft/R/Moleweight) T1 = Temperature at inlet conditions (F) p1 = Pressure at inlet conditions (psi)
v1 =
R Z1⋅ ⋅ T1 MW p1
(3) Calculate the Weight (mass) (mass ) Flow Flow (lbs./min) MF = Mass Mass Flowrate (lbs/min) v1 = Volumetric flow Conditions at inlet (ft 3 /lb) ACFM = Actual Cubic Feet per Minute (ft 3 /min) ACFM = MF⋅ v1
where;
MF =
ACFM v1
(4) Polytropic Exponent (n)
P2 P1 v1 ln v2 ln
n=
5) Determine the press pressure ure ratio ratio (rp) rp =
P2 P1
6) Calculate the generated generated head of of the compress com pressor or (polytropic) (polytropic) Rc
n ⋅ r Head poly = Z1⋅ ⋅ T1⋅ p MW n − 1 7) Polytropic Efficiency
T2 T1 P2 ln P1 ln
n=
n− 1 n
−
1
ηpoly =
k− 1 k
⋅
n n − 1
8) Calculate Gas Horsepower Hppoly =
MF⋅ Head poly
ηpoly⋅ 33000⋅
lb⋅ ft min⋅ hp
9) Adiabatic Conditions a) Calculate Adiabatic Head Headadia =
Rc MW
⋅
⋅ r T1⋅ p k 1 − k
k− 1 k
−
1
b) Adiabatic Efficiency
ηadia =
k− 1
T1⋅ rp
k
−
1
T2 − T1
c) Calculate the Adiabatic Horsepower
Hpadia =
MF⋅ Head adia
ηadia ⋅ 33000⋅
lb⋅ ft min⋅ hp
10) Horsepower Losses Bearing Horsepower Losses are generated by the following Formula only if the system uses Light Turbine Oil (32 SSU or ISO Grade 15). The following nomenclature shall be used in the equation for both the plain and thrust end bearings: a) Journal Bearing Calculations (Reference Shigley 5th Ed.) 1) Sommerfield Number (S)
r S= c
2 ⋅
⋅ N P
2) Determine the L/D Ratio L
Ld =
d
3) Determine the Torque Requirements T = f ⋅ W⋅
d 2
4) Determine the Power Lost in Hp T⋅ N
Hbearing =
1050
b) Thrust Bearing Calculations 1) Calculate Radial and Tangential slope parameters (Note: The Radial tilt is defined as Zero and the standand angle for the tangential tilt is 0.03333 degrees) mr =
R1 hc
⋅
and
γr
mθ =
R1 hc
⋅
γθ
where;
m = Slope parameter R1 = Thrust Pad ID (inches) hc = Thrust clearance at load point (inches) g = Angular movement (degrees) 2) Bearing Pad Loading Equation: 2
W⋅ hc 6 ⋅ ⋅ ω ⋅ ( R2
−
R1)
4
= Bearing_Loading
3) Minmum Film Thickness hmin hc
= film_thickness
4) Horsepower Calculation H⋅ hc
⋅ ω
2
⋅
( R2 − R1)
4
= Horsepower
Note: Need to reference Nomograph in Volume II of CRC Lubrication manual (pages (427-429)
where; R2 = Outside Radius of Bearing Pad(inches) R1 = Inside Radius of Bearing Pad (inches) m = Oil viscosity (mreyn) hc = Axial clearance of the rotor (inches) w = Angular Velocity (rad/sec) N = Rotational Speed (rpm) W = Unit Load of Bearing (lbf) hmin = Minimum film thickness of bearing (inches)
c) Gear Losses (addition of Horsepower) BHP = 0.05⋅ ( H ppoly_case1 + Hp.poly_case2 + Hbearing_losses + Hseal)
12) Power required for the motor to operate BHP
Powermotor =
Amps =
and
ηmotor
Powermotor 1.73⋅ E⋅ ηmotor⋅ PF
Problem: Analze C3 Compressor Performance to determine if start up on Natural Gas is acceptable and will not overload the 21,000Hp motor during start up. In addition; compare the performance data to the original performance data running on Syn Gas.
Solution #1: (makeup)
Note: Using Mollier diagram Methane as substitute for Natural Gas.
(1) Given input data: (Make-up Compressor Only) Calculating for Syn Gas First for Reference) a) Inlet Conditions
Discharge Conditions
T1 := 560⋅ R P1 := 343 ⋅
lb in
P1
=
T2 := 869⋅ R Patm
+
2
lb
P2 := 1205⋅
in −
357.7⋅ lb⋅ in
2
Gage
P2
=
+
2
Patm
3 lb
1.22⋅ 10
⋅
in
2
Makeup Compressor Characteristics −
Nspeed := 10800⋅ min N := 6 d := 19.0⋅ in
1 3
SCFM
:=
111965⋅
ft
min
(Note: Calculate the moleweight,k and critical T & P)
Gas Composition H2O := 47.36⋅ lb⋅ CO
:=
3727⋅ lb⋅
mol
C1 := 306.94⋅ lb⋅
hr mol
hr
H2 := 11988.18⋅ lb⋅
hr
CO2 := 1424⋅ lb⋅
mol
mol
N2 := 28.06⋅ lb⋅
hr
AR := 17.54⋅ lb⋅
mol hr
mol hr
mol hr
(1) Mole Weight of Each Gas MW h2o := 18.02⋅ mol
MW c1
MW co := 28.01⋅ mol
MW h2 := 2.02⋅ mol
MW co2 := 44.01⋅ mol
MW n2 := 28.02⋅ mol
:=
16.04⋅ mol
(2) Total Gas Mass in Mixture Mtotal := H2O + CO + CO2 + C1 + H2 + N2 + AR and Mtotal
=
4
1.754⋅ 10 ⋅ lb ⋅
mol hr
(3) Mcp values for each gas in the composition Mcp_h2o := 7.98
Mcp_h2 := 6.86
Mcp_co := 6.96
Mcp_n2 := 6.96
Mcp_co2 := 8.71
Mcp_ar := 4.97
Mcp_c1 := 8.38
(4) Percentages of each gas in the mixture Ph2o :=
Pco :=
Pco2 :=
Pc1
:=
H2O Mtotal CO Mtotal CO2 Mtotal C1 Mtotal
Ph2o
=
−
2.7⋅ 10
Pco = 0.212
Pco2 = 0.081
Pc1 = 0.018
3
MW ar := 39.94⋅ mol
Ph2 :=
Pn2 :=
Par :=
H2
Ph2 = 0.684
Mtotal N2
Pn2
Mtotal AR
Par
Mtotal
=
=
−
1.6⋅ 10
−
1 ⋅ 10
3
3
(5) Computing Gas Moleweight (Total) Mh2o := MWh2o⋅ Ph2o
Mc1
Mco := MWco⋅ Pco
Mh2 := MW h2⋅ Ph2
Mco2 := MW co2⋅ P co2
Mn2 := MW n2⋅ Pn2
MW
:=
MW
=
Mh2o + Mco + Mco2 + Mc1
:=
+
MW c1 ⋅ Pc1
Mar := MWar⋅ P ar
Mh2 + Mn2 + Mar
11.32
(6) Critical Pressures and Temperatures of each gas Critical Pressures Pcr_h2o
:=
Critical Temperatures
3208⋅ lb⋅ in
Pcr_co := 510⋅ lb⋅ in
2
−
2
−
2
:=
188⋅ lb⋅ in
Pcr_n2
:=
492⋅ lb⋅ in
Pcr_ar := 705⋅ lb⋅ in
−
−
Pcr_h2
−
2
2
−
Pcr_co2 := 1073⋅ lb⋅ in Pcr_c1 := 673⋅ lb⋅ in
−
2
Tcr_h2o Tcr_co
2
:=
1166⋅ R
242⋅ R
:=
Tcr_co2 := 548⋅ R Tcr_c1
:=
344⋅ R
Tcr_h2
:=
60⋅ R
Tcr_n2
:=
228⋅ R
Tcr_ar
272⋅ R
:=
DETERMINE the CRITICAL Pressure and Temp of Mixture
(1) Percentile of P cr_xx x Pxx Cp_h2o := Pcr_h2o⋅ Ph2o Cp_h2o
(3) Critical Temperature Tcritical_mix := Ct_h2o Tcritical_mix
=
+
Ct_co
+
Ct_co2 + Ct_c1 + Ct_h2
+
Ct_n2
+
Ct_ar
146.733⋅ R
(4) Determine the Specific Heat Ratio Mc_h2o Mc_co
:=
:=
Mcp_h2o⋅ Ph2o
Mcp_co⋅ Pco
Mc_co2 := Mcp_co2⋅ Pco2 Mc_c1
:=
Mcp_c1⋅ Pc1
Mc_h2
:=
Mcp_h2 ⋅ Ph2
Mc_n2
:=
Mcp_n2 ⋅ Pn2
Mc_ar := Mcp_ar⋅ Par
Mcp
:=
Mc_h2o
+
Mc_co + Mc_co2 + Mc_c1 + Mc_h2
+
Mc_n2
+
Mc_ar
Mcp = 7.059
k :=
Mcp
where
Mcp − 1.985
k
=
1.391
b) Determine Compress ibility Value by determining Tr1 , Pr1, Tr2 & Pr2. Compressibility values (Z) are determined from Compressibility Chart in above Reference (Pg. 112)
TR1 :=
PR1 :=
TR2 :=
PR2 :=
T1
where
Tcritical_mix P1 Pcritical_mix T2 Tcritical_mix
TR1 = 3.816 Z1 := 1.024
where
PR1 = 1.034
where
TR2 = 5.922 Z2 := 1.027
P2
PR2 = 3.526
where
Pcritical_mix
b) Conversion of SCFM to ACFM
ACFM :=
Ps T1 Zs SCFM⋅ ⋅ ⋅ P T 1 s Z1
ACFM
where;
=
4.839⋅ 10
c) Calculate the volumetric conditions at Inlet and Discharge Conditions Inlet Conditions Z1⋅ v1
:=
Dis charge Conditions
Rc ⋅ T1 MW P1
Z2⋅ v2
:=
v2
=
3
v1 = 1.519⋅
ft
lb
Rc ⋅ T2 MW P2 3
0.694⋅ ft ⋅ lb
−
1
3 3 ft ⋅
min
d) Calculate the Weight (mass) Flow (lbs./min) where;
ACFM = MF⋅ v1 and
MF
ACFM
MF
:=
n
1.564
3 lb
3.185⋅ 10
=
⋅
v1
min
e) Calculate the Polytropic Head Coefficent
P2 P1 v1 ln v2 ln
n :=
where
=
f) Determine the pressure ratio (r p) rp :=
P2
where
P1
rp = 3.41
g) Calculate the generated head of the compressor (polytropic)
Head poly := Z1⋅
Rc MW
⋅
⋅ r T1⋅ p n 1 − n
n− 1 n
−
1
5
Headpoly = 1.207 × 10 ⋅ ft
h) Polytropic Efficiency
ηpoly :=
k − 1 ⋅ n k n − 1
where
ηpoly = 0.78
i) Calculate Gas Horsepower MF⋅ Headpoly
Hppoly_makeup :=
ηpoly⋅ 33000⋅
wher e
lb⋅ ft min⋅ hp
Hppoly_makeup
=
j) Calculated Discharge Temperature n− 1
T2d_makeup
:=
T1⋅ rp
n
where
T2d_makeup
=
4
1.494⋅ 10 ⋅ hp
871.546⋅ R
Adiabatic Conditions 1) Calculate Adiabatic Head Head adia
Rc
:=
⋅
MW
Headadia
T1⋅
⋅ r p k − 1 k
k− 1 k
−
1
5
1.12 × 10 ⋅ ft
=
2) Adiabatic Efficiency
ηadia
:=
k− 1 k
T1⋅ rp
−
1
where
T2 − T1
ηadia = 0.746
3) Calculate the Adiabatic Horsepower MF⋅ Headadia
Hpadia
:=
ηadia ⋅ 33000⋅
lb⋅ ft
where
Hpadia
Note: Using Mollier diagram Methane as substitute for Natural Gas.
(1) Given input data: (Recycle Compressor Only) Calculating for Syn Gas First for Reference) Discharge Conditions
T1 := 560⋅ R P1 := 1133⋅
T2 := 570⋅ R lb
in P1
=
3
2
+
Patm
1.148⋅ 10 ⋅ lb⋅ in
P2 := 1220.79⋅ P2
2
−
=
lb 2
+
Patm
in 3 lb 1.235⋅ 10 ⋅ 2 in
Recycle Compressor Characteristics −
Nspeed := 10800⋅ min N := 1 d := 19.0⋅ in
4
1.447⋅ 10 ⋅ hp
min⋅ hp
Solution #2: (Recycle Case)
a) Inlet Conditions
=
1 3
SCFM
:=
487738⋅
ft
min
(Note: Calculate the moleweight,k and critical T & P) Gas Composition H2O := 206⋅ lb⋅ CO
:=
mol
C1 := 184519⋅ lb⋅
hr
16843⋅ lb⋅
mol
H2 := 421218⋅ lb⋅
hr
CO2 := 32517⋅ lb⋅
mol
N2 := 18768⋅ lb⋅
hr
mol hr mol hr
AR := 10999.7⋅ lb⋅ METH := 2474⋅ lb⋅
mol hr mol hr
mol hr
(1) Mole Weight of Each Gas MW h2o := 18.02⋅ mol
MW c1
MW co := 28.01⋅ mol
MW h2 := 2.02⋅ mol
MW co2 := 44.01⋅ mol
MW n2 := 28.02⋅ mol
:=
16.04⋅ mol
(2) Total Gas Mass in Mixture Mtotal := H2O + CO + CO2 + C1 + H2 + N2 + AR and Mtotal
=
5
6.851⋅ 10 ⋅ lb ⋅
mol hr
(3) Mcp values for each gas in the composition Mcp_h2o := 7.98
Mcp_h2 := 6.86
Mcp_co := 6.96
Mcp_n2 := 6.96
Mcp_co2 := 8.71
Mcp_ar := 4.97
Mcp_c1 := 8.38
Mcp_meth := 10.5
(4) Percentages of each gas in the mixture Ph2o :=
Pco :=
Pco2 :=
H2O Mtotal CO Mtotal CO2 Mtotal
Ph2o
=
−
3.007⋅ 10
Pco = 0.025
Pco2 = 0.047
4
MW ar := 39.94⋅ mol MW meth := 32.04⋅ mol
Pc1
C1
:=
Pc1 = 0.269
Mtotal
Ph2 :=
Pn2 :=
Par :=
H2
Ph2 = 0.615
Mtotal N2
Pn2 = 0.027
Mtotal AR
Par = 0.016
Mtotal
Pmeth :=
METH
Pmeth
Mtotal
=
−
3.611⋅ 10
3
(5) Computing Gas Moleweight (Total) Mh2o := MWh2o⋅ Ph2o
Mc1
Mco := MWco⋅ Pco
Mh2 := MW h2⋅ Ph2
Mco2 := MW co2⋅ P co2
Mn2 := MW n2⋅ Pn2
MW
:=
MW
=
Mh2o + Mco + Mco2 + Mc1
:=
+
MW c1 ⋅ Pc1
Mar := MWar⋅ P ar Mmeth := MW meth⋅ Pmeth
Mh2 + Mn2 + Mar + Mmeth
9.87
(6) Critical Pressures and Temperatures of each gas Critical Pressures Pcr_h2o
:=
Critical Temperatures
3208⋅ lb⋅ in
Pcr_co := 510⋅ lb⋅ in
2
−
2
−
2
:=
188⋅ lb⋅ in
Pcr_n2
:=
492⋅ lb⋅ in
Pcr_ar := 705⋅ lb⋅ in
−
2
1166⋅ R
242⋅ R
:=
Tcr_c1
:=
344⋅ R
Tcr_h2
:=
60⋅ R
Tcr_n2
:=
228⋅ R
Tcr_ar −
:=
Tcr_co2 := 548⋅ R
2
Pcr_meth := 1157⋅ lb⋅ in
Tcr_h2o Tcr_co
−
−
Pcr_h2
2
2
−
Pcr_co2 := 1073⋅ lb⋅ in Pcr_c1 := 673⋅ lb⋅ in
−
2
:=
Tcr_meth
272⋅ R :=
924⋅ R
DETERMINE the CRITICAL Pressure and Temp of Mixture
(3) Critical Temperature Tcritical_mix := Ct_h2o Tcritical_mix
+
Ct_co
+
Ct_co2 + Ct_c1 + Ct_h2
+
Ct_n2
+
Ct_ar
+
Ct_meth
175.807⋅ R
=
(4) Determine the Specific Heat Ratio Mc_h2o Mc_co
:=
:=
Mcp_h2o⋅ Ph2o
Mcp_co⋅ Pco
Mc_co2 := Mcp_co2⋅ Pco2 Mc_c1
Mcp
:=
:=
+
Mc_co
:=
Mcp_h2 ⋅ Ph2
Mc_n2
:=
Mcp_n2 ⋅ Pn2
Mc_ar := Mcp_ar⋅ Par
Mcp_c1⋅ Pc1
Mc_h2o
Mc_h2
Mc_meth
+
Mc_co2
+
Mc_c1
+
Mc_h2
+
:=
Mcp_meth⋅ Pmeth
Mc_n2
+
Mc_ar + Mc_meth
Mcp = 7.37
k :=
Mcp
where
Mcp − 1.985
k
=
1.369
b) Determine Compress ibility Value by determining Tr1 , Pr1, Tr2 & Pr2. Compressibility values (Z) are determined from Compressibility Chart in above Reference (Pg. 112)
TR1 :=
PR1 :=
TR2 :=
PR2 :=
T1
where
TR1 = 3.816 Z1 := 1.013
Tcritical_mix P1
where
PR1 = 1.034
where
TR2 = 5.922
Pcritical_mix T2 Tcritical_mix P2
Z2 := 1.013
where
PR2 = 3.526
Pcritical_mix
b) Conversion of SCFM to ACFM
ACFM :=
Ps T1 Zs SCFM⋅ ⋅ ⋅ P T 1 s Z1
where;
ACFM
=
3 3 ft
6.641⋅ 10
⋅
min
c) Calculate the volumetric conditions at Inlet and Discharge Conditions Inlet Conditions Z1⋅ v1
:=
Dis charge Conditions
Rc ⋅ T1 MW
Z2⋅
P1
v2
:=
v2
=
Rc ⋅ T2 MW P2
3
v1 = 0.537⋅
ft
lb
3
0.508⋅ ft ⋅ lb
−
1
d) Calculate the Weight (mass) Flow (lbs./min) where;
ACFM = MF⋅ v1 and
MF
=
ACFM
MF
:=
n
1.316
v1
4 lb 1.236⋅ 10 ⋅ min
e) Calculate the Polytropic Head Coefficent
P2 P1 v1 ln v2 ln
n :=
where
=
f) Determine the pressure ratio (r p) rp :=
P2
where
P1
rp = 1.076
g) Calculate the generated head of the compressor (polytropic)
Head poly :=
Rc
⋅ r Z1⋅ ⋅ T1⋅ p MW n − 1 n
n− 1 n
−
1
3
Headpoly = 6.604 × 10 ⋅ ft
h) Polytropic Efficiency
ηpoly :=
k − 1 ⋅ n k n − 1
where
ηpoly = 1.122
i) Calculate Gas Horsepower Hppoly_recycle
MF⋅ Head poly
:=
ηpoly⋅ 33000⋅
where
lb⋅ ft min⋅ hp
Hppoly_recycle
=
3
2.205⋅ 10 ⋅ hp
j) Calculated Discharge Temperature n− 1
T2d_recycle := T1⋅ rp
n
where
T2d_recycle
=
570⋅ R
Adiabatic Conditions 1) Calculate Adiabatic Head Head adia
:=
Headadia
Rc
k ⋅ r ⋅ T1⋅ p MW k − 1 =
k− 1 k
−
1
3
6.526 × 10 ⋅ ft
2) Adiabatic Efficiency
ηadia
:=
k− 1
T1⋅ rp
k
−
1
where
T2 − T1
ηadia = 1.123
3) Calculate the Adiabatic Horsepower
Hpadia
:=
MF⋅ Headadia
ηadia ⋅ 33000⋅
lb⋅ ft min⋅ hp
where
Hpadia
=
3
2.177⋅ 10 ⋅ hp
Bearing Losses
Given : Compressor Speed is 10800 rpm, ID of pad 4.125", OD of pad 11.18 (pad effective areas is 27 square inches), Radial clearance is 0.006 - 0.008", viscosity is 50EE-5 reyn, Length of journal is 1.625", . −
:= 7⋅ 10
D1 := 4.125⋅ in
N
6
⋅
10800⋅
:=
reyn
c
:=
rev P := 50⋅
min
R1
=
D1 2
PL := 50⋅
in
0.172⋅ ft
ω
lbf 2
:=
L := 1.625⋅ in
lb in
therefore;
R1 :=
0.006⋅ in
2
2⋅ π⋅ N 3
−
ω = 1.131 × 10 ⋅ s
1
=
−
0.032⋅ ft
1
⋅
lb⋅ s
−
1
a) Compressor and High Speed Gear Journal Bearing Calculations 1) Sommerfield Number (S)
R1 S := c
2 ⋅
⋅ N
S
PL
=
2.978
2) Determine the L/D Ratio Ld :=
L
Ld = 0.394
D1
3) Determine the Torque Requirements W = Rotor Weights (800 lb)
W1
:=
800⋅ lbf
Reference Figure 12-17, using the Sommerfield number & L/D ratio in Shigley 5th R1 c
⋅
f := 75
T := f ⋅ W1⋅
f :=
75⋅ c R1
f
=
0.218
D1 2
T
=
360⋅ lbf ⋅ in
4) Determine the Power Lost in Hp Hbearing := T⋅ N Htotal_4
:=
4⋅ Hbearing
Hbearing
=
9.818⋅ hp
Htotal_4
=
39.273⋅ hp
Note :Bearing count is 2 journals with this approximate size for the 2 compressors and additional size increase to 6" Dia & 6 " wide for the compressor and gear HSP). There are three (3) sets of bearings for the Motor and the Low speed gear shaft not accounted for. Use the same formula listed above for these calculations. H journal_total := Htotal_4 H journal_total
=
+
Htotal_6
+
Htotal_10
3
1.235⋅ 10 ⋅ hp
b) Thrust Bearing Calculations 1) Calculate Radial and Tangential slope parameters (Note: The Radial tilt is defined as Zero and the standand angle for the tangential tilt is 0.03333 degrees) γr := 0
0.03333⋅ 2⋅ π
γθ :=
γθ
=
hc := 0.012⋅ in
360 −
5.817⋅ 10
4
therefore;
mr :=
R1 hc
⋅
γr
and
mθ
:=
R1
⋅
hc
γθ
where;
m = Slope parameter R1 = Thrust Pad ID (inches) hc = Thrust clearance at load point (inches) g = Angular movement (degrees)
mr
=
0
mθ
=
0.1
2) Bearing Pad Loading Equation: DT1 := 4.125⋅ in + 1⋅ in
RT1 :=
DT2 := 11.18⋅ in
RT2 :=
DT1 2 DT2 2
Note: Need to reference Nomograph in Volume II of CRC Lubrication manual (pages (427-429)
2
W⋅ hc
6 ⋅ ⋅ ω ⋅ ( R2 − R1)
4
= 0.006
where; W := 0.006⋅ 6⋅ ⋅ ω⋅ ( RT2 − RT1) W
=
−
0.333⋅ lb⋅ s
2
3) Minmum Film Thickness hmin hc
= 0.95
hmin
:=
0.95⋅ hc
hmin
=
0.011⋅ in
hc
=
0.012⋅ in
4) Horsepower Calculation H⋅ hc
⋅ ω
2
⋅
( RT2 − RT1)
4
= 2.5
ω
=
3
−
1.131 × 10 ⋅ s
1
where; 2.5⋅ ( ) ⋅ ω
Hthrust
:=
Hthrust
=
2
⋅
( RT2 − RT1)
4
hc
23.744⋅ hp
Hthrust_total := 4 ⋅ Hthrust
Hthrust_total
94.976⋅ hp
=
5) The total accounted for Horsepower for bearings is: Htotal := H journal_total
+
Hthrust_total
Htotal
=
3
1.33⋅ 10 ⋅ hp
c) Seal losses are accounted for by using nomograph for Seal leakage. Reference Elliott sizing criteria charts. Hseal
:=
6 ⋅ 60⋅ hp
Hseal
360⋅ hp
=
d) Gear losses are accounted for by taking a 5% increase in the total Horsepower requirements of all the driven equipment. GearBHP := 0.05⋅ ( Hppoly_makeup + Hppoly_recycle GearBHP
=
+
Htotal + Hseal)
941.98⋅ hp
Conclusion: Total Horsepower required by the motor is as follows: BHP := GearBHP + ( Hppoly_makeup + Hppoly_recycle BHP
