NORD Projektierung GB
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Transcript of NORD Projektierung GB
NORD - InformationCalculation Methods and Examples
Getriebebau NORD, Schlicht + Küc henmeister GmbH & Co.Rudolf-Diesel-Str. 1, D- 22941 BargteheideTelefon: 04532 / 4010, Telefax: 04532 / 401 253
NORD - Information
Physical Formulae
Linear motion Rotating motion
Distance s = v ∗ t Angular ϕ = ω ∗ t
Velocity (speed) v = st
Angular velocity ω = ϕτ = 2 ∗ π ∗ n
Accelleration a = vt
Angular accelleration α = Wt
Force F = m ∗ a Torque M = J ∗ r = F ∗ r
Power P = F ∗ v Power P = M ∗ ωWork W = F ∗ s = P ∗ t Work W = M ∗ ϕ = P ∗ t
Kinetic energy Wkin = 12 ∗ m ∗ v2 Rotating energy Wrot =
12 ∗ J ∗ ω2
Formulae of drive engineering
Rolling resistance, -force FR = m ∗ g ∗ 2D
∗ (µL ∗ d2 + f) + c
: µL, f, c, s. tables_____ _1_,_ _2_,_ _3_._
or FR = We ∗ m We for wheel / rail steel
s. diagramme 1, 2.
Sliding resitance, -force FG = m ∗ g ∗ µ µ s. table 4
Static friction-force FH = m ∗ g ∗ µO µO s. table 4
Windload F = A ∗ PW
Moment of inertia with refernce Jred = 91,2 ∗ m ∗
vnM
2 Translation
to the motor shaft or Jred = J ∗
nnM
2 Rotation
Speed n = v ∗ 60π ∗ D
Torque M = P ∗ 9550
n
Friction power PR = F ∗ v
1000 ∗ ηTranslation
or PR = M ∗ n9550
Rotation
Acceleration power PB = m ∗ a ∗ v1000 ∗ η Translation
or PB = J ∗ n2
91,2 ∗ 1000 ∗ tB ∗ η Rotation
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Hubleistung PHub = m ∗ g ∗ v
1000 ∗ η
Beschleunigung aB = 9,55 ∗ v ∗ (MH ± ML)
(Jred + JM + JBre + JZ) ∗ n
Beschleunigungszeit tB = vaB
Beschleunigungsweg sB = v2
2 ∗ aB
Verzögerung av = 9,55 ∗ v ∗ (MB ± ML)
(Jred + JM + JBre + JZ) ∗ n
Verzögerungszeit tV = vaV
Verzögerungsweg sV = v2
2 ∗ aV
zulässige Schalthäufigkeit zzul = 1 − ML ⁄ MH
1 + (Jred + JBre + JZ) ⁄ JM ∗ zo
Positioniergenauigkeit Positioniergenauigkeit = ± 0,25 * sv
Bremsarbeit WB = (Jred + JM + JBre + JZ) ∗ n2
182,5 ∗
MB
MB ± ML
Lebensdauer der Bremsbeläge LN = Wzul
WB ∗ z
Übersetzung i = n1
n2 =
M2
M1 =
d2
d1 =
z2
z1
Wirkungsgrad η = Pab
Pzuη s. Tabelle 5
rücktreibender Wirkungsgrad η G’ = 2 − 1
η G
Querkraft FQ = 2 ∗ M2
D ∗ fZ ∗ ≤ FQzul fZ s. Tabelle 6
Betriebsfaktor fB = M2max
M2
Massenbeschleunigungsfaktor maf = Jred
JM + JZ + JBre
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a
aB
av
A
c
d
dO
d1
d2
D
f
fBfZ
F
FG
FH
FQ
FQvorh
FQzul
FR
FW
g
i
iV
J
JBre
JM
Jred
JZ
LN
m
maf
mG
mL
mO
Formulae symbols and unities
Acceleration
Acceleration (start up)
Deceleration (braking)
Area (wind)
Additional factor for secondary friction
Diameter (bearing spigot diameter)
Pinion or sprocket diameter
Pinion diameter
Chain sprocket diameter
Diameter of the travelling wheel or cable drum or of thesprocket
Lever arm of rolling friction
Service factor
Additional factor for overhung load
Force, rolling resistance
Sliding friction
Static friction
Overhung load
Existing overhung load
Permissible overhung load
Rolling resistance
Wind load
Gravity (constant: 9,81)
Reduction
Additional reduction (gear, chain, belt ...)
Moment of inertia
Moment of inertia of the brake
Moment of inertia of the motor
Moment of inertia with reference to the motorshaft
Moment of inertia of the z-fan
Brake service life until readjustment
Weight (mass)
Inertia mass acceleration factor
Mass of counter weight
Mass with full load
Mass without load
m/s2
m/s2
m/s2
m2
-
m
m
m
m
m
m
-
-
N
N
N
N
N
N
N
N
m/s2
-
-
kgm2
kgm2
kgm2
kgm2
kgm2
h
kg
-
kg
kg
kg
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M
MB
MH
ML
MN
M1
M2
M2max
n
nM
nN
n1
n2
PW
P
Pab
Pzu
PB
PHub
PN
PR
r
s
sB
sV
t
tB
tV
v
W
We
Wkin
Wrot
Wzul
WB
x
Torque
Braking torque
Run up torque
Torque with full load (with reference to the motor shaft)
Rated torque
Input torque
Output torque
Maximum permissible output torqe
Speed
Motor speed
Rated speed
Input speed
Output speed
Wind pressure
Power
Required power
Supplied power
Acceleration power
Lifting power
Rated power
Fricition power
Radius
Distance
Start up distance
Braking distance
Time
Start up time
Braking time
Velocity (speed)
Work
Standard rolling friction
Kinetic energy
Rotating energy
Braking work until readjustment
Braking work
Number of drives
Nm
Nm
Nm
Nm
Nm
Nm
Nm
Nm
1/min1/min1/min1/min1/min
N/m2
kW
kW
kW
kW
kW
kW
kW
m
m
m
m
s
s
s
m/s
J
N/t
J
J
J
J
-
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z
zzul
zO
z1
z2
αηηG
ηG’
µµL
µO
ϕω
Starting frequency
Permissible starting frequency
Starting frequency with no load
Number of gear teeth pinion
Number of gear teeth gear wheel
Angular acceleration
Efficiency
Efficiency of gear unit
Reverse operating efficiency
Coefficient of friction
Coefficient of friction for bearings
Coefficient of friction (static)
Angular
Angular velocity
s/h
s/h
s/h
-
-
1/s2
-
-
-
-
-
-
°1/s
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table 1: coefficient of friction for bearings µL
table 2: lever arm of rolling friction f
table 3: rim friction on the wheels c
table 4: static friction and sliding friction µ
table 5: efficiency η
table 6: additional factor inderming overhung loads fz
Friction bearing Sliding bearings
µL 0,005 0,1
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f
steel / steel 0,0005 m
wood / steel 0,0012 m
polymer / steel 0,002 m
hardrubber / steel 0,0077 m
hardrubber / concrete 0,01 - 0,02 m
rubber / concrete 0,015 - 0,035 m
Friction ofanti friction bearings
Friction ofsleeve bearings
Friction of guide rollers
c 0,003 0,005 0,002
Static friction µO Sliding friction µdry greased dry greased
steel / steel 0,11 - 0,40 0,10 0,10 - 0,30 0,01 - 0,10
steel / last iron 0,18 - 0,25 0,10 0,16 - 0,25 0,05 - 0,10
steel / wood 0,50 - 0,70 0,10 0,20 - 0,50 0,02 - 0,10
steel / polymer 0,20 - 0,50 0,10 - 0,35
steel / rubber 0,40 - 0,50
wood / wood 0,40 - 0,80 0,16 0,20 - 0,50 0,04 - 0,16
fZ
helical gears 1,1 z = 17 teeth
chain sprockets 1,4 z = 13 teeth
chain sprockets 1,2 z = 20 teeth
pulleys 1,7 by tensioning influence
pulleys 2,5 by tensioning influence
ηchain 0,90 - 0,96 per complete wrap of the rope around the drum
wire ropes 0,90 - 0,95 per complete wrap
flat polymer belts 0,93 - 0,98 per complete wrap of the rope depending on the material
V-belts 0,85 - 0,95 per complete wrap
rubber belts 0,80 - 0,85 per complete wrap
polymer belts 0,80 - 0,85 per complete wrap
helical inline gear 0,95 - 0,98 oil lubricated depending on the number of the stages
worm gear 0,30 - 0,93 oil lubricated depending on the number of starts of the worm
NORD - Information
Example I.1: Drive arrangement for crane
Mass without load of the crane mO 13800 kg
Mass without load of the mk 1800 kg
Load mL 15000 kg
Velocity v 0,17 / 0,66 m/s = 10/40 m/min
Diameter of the travelling wheel D 0,4 m
Number of drives x 2
Additional reduction iv 4,24
Mounting position B 3
Switching frequencies z 60 s/h
Efficiency η 0,85
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Motor arrangements
Standard rolling friction We
We = WO + 30 N/t W0 = 36 N/t s. diagram
We = 36 N/t + 30 N/t = 66 N/t 30 N/t additional for rim friction
Power P (at maximum velocity)
P = We ∗ m ∗ v1000 ∗ η
PO = 66 N ⁄ t ∗ (13,8 t + 1,8 t) ∗ 0,66 m⁄s
1000 ∗ 0,85 = 0,80 kW (without load)
PL = 66 N ⁄ t ∗ (13,8 t + 1,8 t + 15,0 t) ∗ 0,66 m⁄s
1000 ∗ 0,85 = 1,57 kW (with load)
Pmax = PL
2 ∗
mO + 2 ∗ (mK + mL )mO + mK + mL
= 1,57 kW
2 ∗
13,8 t + 2 ∗ (1,8 t + 15 t)13,8 t + 1,8 t + 15 t
Pmax = 1,22 kW (one-sided trolley)
Motor data
Type 100 L/80-20 WU Bre16 Z (2 pieces)
Rated output power PN 0,55 / 2,2 kW
Rated speed nN 670 / 2740 1/min
Rated torque MN 7,8 / 7,7 Nm
Permissible no-load starting frequency zo 4000 / 1400 s/h
Motor moment of inertia JM 0,0060 kgm2
Additonal moment of inertia Jz 0,0113 kgm2
Brake moment of inertia JBre 0,0001 kgm2
Braking torque MB (brake 16 adjusted to 8 Nm ) 8 Nm
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Gear arrangment
Wheel speed n L
nL = v ∗ 60π ∗ D
nL = 0,66 m⁄s ∗ 60
π ∗ 0,4 m = 32 1⁄min
Gear output speed n 2
n2 = nL ∗ iv
n2 = 32 1⁄min ∗ 4,24 = 136 1⁄min
Acceleration factor of mass m af
maf = Jred
JM + Jz + JBre
maf = 0,0810 kgm2
0,0060 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 4,7
Starting freqeuency per hour: 180 (60 times acceleration, switching, deceleration)
⇒ Type of load C, fB = 1,6
Output torque M a
Ma = PN ∗ 9550
n2 ∗ fB
Ma = 2,2 kW ∗ 9550
136 1⁄min ∗ 1,6 = 247 Nm
For service factor fB = 1,6 the output torque of the gear is 247 Nm.
Reduction i
i = nN
n2
i = 2740 1⁄min
136 1⁄min = 20
Complete type: SK 22-100 L/80-20 WU Bre 16 ZPN = 0,55 / 2,2 kW
i = 20,03
n2 = 33 / 137 1/min
Mounting position B 3
Shaft ø 30 x 60 mm
Brake 16 Nm adjusted to 8 Nm
Special provision: special rotor
high inertia fan
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Example I.2: Drive arrangement for a trolley
Mass without load mk 1800 kg
Load mL 15000 kg
Velocity v 0,08 / 0,33 m/s = 5/20 m/min
Wheel diameter D 0,3 m
Number of drives x 1
Additional reduction iv 4
Mounting position B 5
Switching frequency z 60 s/h
Efficiency n 0,85
Pairing of material steel / steel
Guiding rim friction
Type of bearings (4 wheels) antifriction bearings
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Drive resistance:
FR = m ∗ g
2D
∗ ( µ L ∗ d2
+ f ) + c
Fro = 1800 kg ∗ 9,81 m
s2
20,3 m
∗ (0,005 ∗ 0,06
2 + 0,0005 m ) + 0,003m
Fro = 129,5 N ( without load)
FRL = 16800 kg ∗ 9,81 ms2
∗
20,3 m
(0,005 ∗ 0,06
2 m + 0,0005 m ) + 0,003 m
FRL = 1208,6 N (with load)
Power P (calculation for 2-poles gearmotors)
P = F ∗ V
1000 ∗ η
Po = 129,5 N ∗ 0,33 m
1000 ∗ 0,85 s = 0,05 kW
PL = 1208,6 N ∗ 0,33
1000 ∗ 0,85 = 0,47 kW
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Motor arrangement
Motor data
Type 100 L/8-2 WU Bre10 Z
Rated output power PN 0,4 / 1,6 kW
Rated speed nN 670 / 2740 1/min
Rated torque MN 5,7 / 5,6 Nm
Hochlaufmoment MH 9,2 / 8,6 Nm
No-load switching frequency zo 4200 / 1500 s/h
Motor moment of inertia JM 0,0045 kgm2
Moment of high inertia fan Jz 0,0113 kgm2
Brake moment of inertia JBre 0,0001 kgm2
Braking torque MB (brake 10 adjusted on 6 Nm) 6 Nm
Load torque M
M = P ∗ 9550x ∗ nN
MO = 0,05 KW ∗ 9550
2740 1⁄min = 0,2 Nm (without load)
ML = 0,47 kW ∗ 9550
2740 1⁄min = 1,6 Nm (with load)
Reduced moment of inertia J red
Jred = 1x
∗ 91,2 ∗ m ∗
vnN
2
JredO = 91,2 ∗ 1800 kg ∗
0,33 m⁄s2740 1⁄min
2
= 0,0024 kgm2
JredL = 91,2 ∗ 16800 kg ∗
0,33 m⁄s2740 1⁄min
2
= 0,0222 kgm2
Acceleration aB
aB = 9,55 ∗ v ∗ (MH − ML)
(Jred ⁄ η + JM + JBre + JZ) ∗ nN
aB = 9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 0,2 Nm)
(0,0024 kgm2 ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min = 0,52 m⁄s2 (without load)
aB = 9,55 ∗ 0,33 m⁄s ∗ (8,6 Nm − 1,6 Nm)
(0,0222 kgm2 ⁄ 0,85 + 0,0045 kgm2 + 0,0001 kgm2) ∗ 2740 1⁄min = 0,19 m⁄s2 (with load)
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Decceleration a V
av = 9,55 ∗ v ∗ (MB + ML ∗ η2)
(Jred ∗ η + JM + JBre + JZ) ∗ nN
aVO = 9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 0,2 Nm ∗ 0,852)
(0,0024 kgm2 ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min = 0,39 m⁄s2 (without load)
aVL = 9,55 ∗ 0,08 m⁄s ∗ (6 Nm + 1,6 Nm ∗ 0,852)
(0,0222 kgm2 ∗ 0,85 + 0,0045 kgm2 + 0,0001 kgm2 + 0,0113 kgm2) ∗ 670 1⁄min = 0,24 m⁄s2 (with load)
Permissible switching fr equency z zul
zzul = 1 − ML ⁄ MH
1 + (Jred + JZ + JBre) ⁄ JM ∗ z0
zzul = 1 − 1,6 Nm ⁄ 8,6 Nm
1 +(0,0222 kgm2 + 0,0113 kgm2 + 0,0001 kgm2) ⁄ 0,0045 kgm2 ∗ 1500 s⁄h = 142 s⁄h
The perm. switching frequency is calculated for the acceptance: starting 2-pole with load (every time) is not correct
because of delay for switching and running on the 8-pole.
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Gear arrangements
Wheel speedl nL
nL = v ∗ 60π ∗ D
nL = 0,33 m⁄s ∗ 60
π ∗ 0,3 m = 21 1⁄min
Gear unit output speed n 2
n2 = nL ∗ iv
n2 = 21 1⁄min ∗ 4 = 84 1⁄min
Mass acceleration factor maf
maf = Jred
JM + Jz + JBre
maf = 0,022 kgm2
0,0045 kgm2 + 0,0113 kgm2 + 0,0001 kgm2 = 1,4
Circuit m per hour: 180 (each 60 accelerations, switching, decelerations)
⇒ type of load B
⇒ fB ≥ 1,3
Output torque M a
Ma = PN ∗ 9550
n2 ∗ fB
Ma = 1,6 kW ∗ 9550
84 1⁄min ∗ 1,3 = 236 Nm
For service factor fB = 1,3 the output torque of the gear is 236 Nm.
Reduction i
i = nN
n2
i = 2740 1⁄min
84 1⁄min = 33
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Complete type: SK 22 F - 100 L/8-2 WU Bre 10 Z
PN = 0,4 / 0,16 kW
i = 34,69
n2 = 19/79 1/min
Mounting position B 5
Shaft ø 30 x 60 mm
Flange ø 160 mm oder 200 mm
Brake 10 Nm adjusted on 6 Nm
Special provision: special rotor (WU-silumin rotor)
high inertia fan
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NORD - Information
Example II.1: Drive unit for vert ical motion
Mass without load mO 50 kg
Load mL 200 kg
midle drum diameter Dm 0,208 m
Max. lifting speed v 0,24 m/s = 14,4 m/min
Operation cycle 8 h/Tag, 40 % ED
Starting frequency z 360 Hubbewegungen/h
Efficiency η 0,8
Positioning accuracy ± 1 mm
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Motor arrangement
Power P
P = m ∗ g ∗ v1000 ∗ η
PL = (50 kg + 200 kg) ∗ 9,81 m⁄s2 ∗ 0,24 m⁄s
1000 ∗ 0,8 = 0,74 kW
To get the required accuracy of ± 1 mm we have to choose a polechanging motor.
Motor data
Typ 80 L/4-2 Bre8
Rated output power PN 0,60 / 0,75 kW
Rated speed nN 1400 / 2830 1/min
Synchronous speed nsyn 1500 / 3000 1/min
Rated torque MN 4,1 / 2,5 Nm
Run-up torque MH 7,4 / 5,7 Nm
No-load switching frequency zo 2500 / 1800 s/h
Motor moment of inertia JM 0,00165 kgm2
Brake moment of inertia JBre 0,00007 kgm2
max. braking work until readjustment Wzul. 7 * 107 J
Brake reaktion time t2 0,015 s
(DC-connection)
Braking torque MB 8 Nm
Load torque M
M = P ∗ 9550
nN
ML = 0,74 kW ∗ 9550
2830 1⁄min = 2,5 Nm
Switching torque M U
MU = 2 * MH4
MU = 2 * 7,4 Nm = 14,8 Nm
reduced moment of inertia J red
Jred = 91,2 ∗ m ∗
vnN
2
Jred = 91,2 ∗ (50 kg ∗ 200 kg) ∗
0,24 m⁄s2830 1⁄min
2
= 0,00016 kgm2
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z0 = 2320 / 1620 s/h = max. perm. switching frequency with no load
For this application A 4-2 polemotor (Dahlander-connection) is used. Therefor the half of the z o is criteria.
Permissible switching frequency zzul
up motion: zzul = 1 − ML ⁄ MH
1 + (Jred + JBre) ⁄ JM ∗
zO
2
zzul = 1 − 2,5 Nm ⁄ 5,7 Nm
1 + (0,00016 kgm2 + 0,00007 kgm2) ⁄ 0,00165 kgm2 ∗ 1620 s⁄h
2 = 399 s⁄h (2 poles)
down motion: zzul = 1 − ML ⁄ MU
1 + (Jred + JBre) ⁄ JM ∗
zO
2
zzul = 1 − 2,5 Nm ⁄ 14,8 Nm
1 + (0,00016 kgm2 + 0,00007 kgm2) ⁄ 0,00165 kgm2 ∗ 2320 s⁄h
2 = 846 s⁄h (4 poles)
The mechanical braking depends on the positioning speed. The max. braking distance depends on down motion.
Deceleration a v
av = 9,55 ∗ v ∗ nN4 ⁄ nN2 ∗ (MB − ML ∗ η2)
(Jred ∗ η + JM + JBre) ∗ nN4
av = 9,55 ∗ 0,24 m⁄s ∗ 1400 1⁄min ⁄ 2830 1⁄min ∗ (8 Nm − 2,5 Nm ∗ 0,82)
(0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ 1400 1⁄min = 2,80 m⁄s2
In case of calculation the deceleration time we have to use the increased speed for the down motion.
The cause is the delay for switching and the over-synchronous speed.
Load speed n L
nL = nsyn ± ML ⁄ MN ∗ (nsyn − nN) +: down motion, -: up motion
down motion: nL = nsyn + ML ∗ η2 ⁄ MN ∗ (nsyn − nN)
nL = 1500 1⁄min + 2,5 Nm ∗ 0,82
4,1 Nm ∗ (1500 1⁄min − 1400 1⁄min) = 1539 1⁄min
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Increased speed during braking time ∆∆n
∆n = ± 9,55 ∗ ML ∗ t2
Jred + JM + JBre+: down motion, -: up motion
down motion: ∆n = 9,55 ∗ ML ∗ η2 ∗ t2Jred ∗ η + JM + JBre
∆n = 9,55 ∗ 2,5 Nm ∗ 0,82 ∗ 0,015 s
0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2 = 124 1⁄min
Deceleration time t v (Braking time)
tv = v ∗ (nL + ∆n) ⁄ nN2
a
tv = 0,24 m⁄s ∗ (1539 1⁄min + 124 1⁄min) ⁄ 2830 1⁄min
2,80 m⁄s2 = 0,05 s
Deceleration distance sv (Braking distance)
sv = v ∗
nL + ∆nnN2
2
2 ∗ a
sv = 0,24 m⁄s ∗
1539 1⁄min + 124 1⁄min
2830 1⁄min
2
2 ∗ 2,80 m⁄s2 = 0,004 m
Positioning accuracy
The positioning accuracy is about ± 25 % from the deceleration distance s v.
Positioning accuracy = ± 25 % * sv = ± 0,25 * 0,004 m = ± 0,001 m
Braking work WB
WB = (Jred ∗ η + JM + JBre) ∗ n2
N4
182,5 ∗
MB
MB ± ML
WB = (0,00016 kgm2 ∗ 0,8 + 0,00165 kgm2 + 0,00007 kgm2) ∗ (1400 1⁄min)2
182,5 ∗ 8 Nm
8 Nm ± 0 = 20 J
Because of the same number of up- and down-motion the load torque = 0 Nm.
Brake service life until readjustment LN
LN = Wzul
WB ∗ z
LN = 7 ∗ 107J
20 J ∗ 360 1⁄h = 9720 h
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Gear arrangements
Gear output speed n2
n2 = v ∗ 60π ∗ Dm
n2 = 0,24 m⁄s ∗ 60π ∗ 0,208 m
= 22 1⁄min
^
Mass acceleration factor maf
maf = Jred
JM + JBre
maf = 0,00016 kgm2
0,00165 kgm2 + 0,00007 kgm2 = 0,09
Switching per hour: 1080 ( each 360 accelerations, change-over, decelerations)
⇒ kind of load A, f B = 1,2
Output torque Ma
Ma = PN ∗ 9550
n2 ∗ fB
Ma = 0,75 kW ∗ 9550
22 1⁄min ∗ 1,2 = 391 Nm
Reduction i
i = nN
n2
i = 2830 1⁄min
22 1⁄min = 129
Complete type: SK 2382 A - 80 L/4-2 Bre8PN = 0,60 / 0,75 kWi = 131,86n2 = 11 / 21 1/minMounting position H 1Hollow shaft ø 35 mmBrake 8 NmInsolating material class F
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Example III.1: Turntable drive for processing table
Determine the size of a cd-geared motor for a tuntable with 3 work stations (α = 120°)
Table weight without load m O 500 kg Positioning accuracy = ± 1 mm
Table diameter D 2 m Sprocket reduction i v 3,76
Positions of load α 120° Duty factor ED 60 %
Spaced at radius R 1 m Pulse number 360 Takte/h
Ball bearing ring diameter d 2 m Time of run 16 h/Tag
Cycle time for 120 ° turn t ges 6 s Efficiency η 0,8
Load m L (3 x 750 kg) m L 2250 kg Mounting position V 6
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Distance s ( at a rotation of 120 ° )
s = D ∗ π3
= 2 m ∗ π
3 = 2,094 m
Acceleration time t B or Deceleration time t V
tB = tV = 1 s (acceptance data)
Table speed n T
nT = sges ∗ 60
π ∗ D ∗ ( t − ( tB + tV ) ⁄ 2 ) =
2,094 m ∗ 60π ∗ 2 m ∗ ( 6 s − ( 1 s + 1 s ) ⁄ 2 )
= 4 1⁄ min
Table circumferential velocity v (Ball bearing ring)
v = π ∗ d ∗ nT
60 =
π ∗2 m ∗ 4 1⁄min
60 = 0,42 m⁄s
Moment of inertia J
J = 18
∗ mO ∗ D2 + 14
∗ mL ∗ d2 = 18
∗ 500 kg ∗ (2 m)2 + 14
∗ 2250 kg ∗ (2 m)2 = 2500 kgm 2
Friction power P R (static)
PR = (mO + mL) ∗ g ∗ µ L ∗ v
1000 ∗ η = (500 kg + 2250 kg ) ∗ 9,81 m⁄s2 ∗ 0,005 ∗ 0,42 m⁄s
1000 ∗ 0,8 = 0,07 kW
with µL = 0,005 for friction bearing
Acceleration power P B (dynamic)
PB = J ∗ nT
2
91,2 ∗ 1000 ∗ tB ∗ η = 2500 kgm 2 ∗ (4 1⁄min)2
91,2 ∗ 1000 ∗ 1 s ∗ 0,8 = 0,55 kW
Power P
P = PR + PB (friction + acceleration)
P = 0,07 kW + 0,55 kW = 0,62 kW
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NORD - Information
Motor data
Type 90 S/8-2 Bre 10
Rated power PN 0,25 / 1,1 kW
Rated speed nN 700 / 2810 1/min
Rated torque MN 3,4 / 3,7 Nm
Hochlaufmoment MH 4,0 / 5,7 Nm
No-load switching frequency zo 9000 / 1500 s/h
Motor moment of inertia JM 0,00235 kgm2
Brake moment of inerta JBre 0,00007 kgm2
Braking torque MB (adjusted at 8 Nm) 8 Nm
Load torque ML
M = PR ∗ 9550
nN =
0,07 kW ∗ 95502810 1⁄ min
= 0,2 Nm
Reduced moment of inertia Jred
Jred = J ∗
nT
nN
2 = 2500 kgm2 ∗
4 1⁄min
2810 1⁄min
2 = 0,00507 kgm2
Permissible switching frequence zzul
zzul = 1 − ML ⁄ MH
1 + (Jred + JBre) ⁄ JM ∗ zO = 1 − 0,2 Nm ⁄ 5,7 Nm
1 + (0,00507 kgm2 + 0,00007 kgm2) ⁄ 0,00235 kgm2 ∗ 1500 s⁄h = 453 s⁄h
Acceleration aB
aB = 9,55 ∗ v ∗ (MH − ML)
Jred ⁄ η + JM + JBre) ∗ nN = 9,55 ∗ 0,42 m⁄s ∗ (5,7 Nm − 0,2 Nm)
(0,00507 kgm2 ⁄ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ 2810 1⁄min = 0,90 m⁄s2
Acceleration time tB (start up time)
tB = vaB
= 0,42 m⁄s0,90 m⁄s2 = 0,47 s
Acceleration distance sB (start up distance)
sB = v2
2 ∗ aB = 0,42 m⁄s2
2 ∗ 0,90 m⁄s2 = 0,098 m
Change-over torque MU
MU = 2 ∗ MH8 = 2 ∗ 4,0 Nm = 8,0 Nm
At the change over the speed increased and the motor is decelet ad generator-style.
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NORD - Information
Change-over delay a U
aU = 9,55 ∗ v ∗ (1 − nN8 ⁄ nN2) ∗ (MU + ML ∗ η2)
(Jred ∗ η + JM + JBre) ∗ (nN2 − nN8) =
aU = 9,55 ∗ 0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min) ∗ (8,0 Nm + 0,2 Nm ∗ 0,82)
(0,00507 kgm2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ (2810 1⁄min − 700 1⁄min) = 1,79 m⁄s2
Change-over time t U
tU = v ∗ (1 − nN8 ⁄ nN2)
aU =
0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min)1,79 m⁄s2 = 0,18 s
Change-over distance s U
sU = (v ∗ (1 − nN8 ⁄ nN2))2
2 ∗ aU =
0,42 m⁄s ∗ (1 − 700 1⁄min ⁄ 2810 1⁄min))2
2 ∗ 1,79 m⁄s2 = 0,028 m
Deceleration a V
aV = 9,55 ∗ v ∗ nN8 ⁄ nN2 ∗ (MB + ML ∗ η2)
(Jred ∗ η + JM + JBre) ∗ nN8 =
9,55 ∗ 0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min ∗ (8,0 Nm + 0,2 Nm ∗ 0,82)(0,00507 kgm2 ∗ 0,8 + 0,00235 kgm2 + 0,00007 kgm2) ∗ 700 1⁄min
= 1,80 m⁄s2
Deceleration time tV
tv = v ∗ nN8 ⁄ nN2
aV =
0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min
1,80 m⁄s2 = 0,06 s
Deceleration distance s V
sV = (v ∗ nN8 ⁄ nN2 )2
2 ∗ aV =
(0,42 m⁄s ∗ 700 1⁄min ⁄ 2810 1⁄min)2
2 ∗ 1,80 m⁄s2 = 0,003 m
Distance s with velocity v
s = sges - sB - sU - sV = 2,094 m - 0,098 m - 0,028 m - 0,003 m = 1,965 m
Time t with velocity v
t = sv =
1,965 m0,42 m⁄s
= 4,68 s
Pulse duration t ges (total cycle time)
tges = tB + t + tU + tV = 0,47 s + 4,68 s + 0,18 s + 0,06 s = 5,39 s
The required cylce time of 6s is not reached. There is a possibility of driving for a longer period in creep speed.
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NORD - Information
Positioning accuracy
The positioning accuracy is about ± 25% of the deceleration way s V.
Positioning accuracy = ± 0,25 * sv = ± 0,25 * 0,003 m = ± 0,00075 m = ± 0,75 mm
Gear arrangements
Gear output speed n 2
n2 = nT * iV = 4 1/min * 3,76 = 15 1/min
Mass acceleration factor maf
maf = Jred
JM+JBre =
0,00507 kgm2
0,00235 kgm2 + 0,00007 kgm2 = 2,1
Switching per hour: 1080 (each 360 accelerations, change-over and decelerations)
→ kind of load B, f B = 1,5
Output torque M a
Ma = PN ∗ 9550
n2 ∗ fB =
1,1 kW ∗ 955015 1⁄min
∗ 1,5 = 1050 Nm
Reduction i
i = nN
n2 =
2810 1⁄min
15 1⁄min = 187
Complete type: SK 43 - 90 S/8-2 Bre 8PN = 0,25 / 1,1 kWi = 169,86n2 = 4/16 1/minMounting position V6Shaft ø 45 x 90 mmBrake 8 Nm
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