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Important Note:
The equations used are mainly derived from the technical information provided in the SDP/SI technical library see links below.
I
recommend that for serious work the linked information is more suitable.
The efficiency of a gear system is simple calculated as the
[output shaft power /Input shaft power ].100 %.
The output power is the (input power - the power losses). Power losses
in gear systems are associated primarily with tooth friction and lubrication churning losses.
Churning losses are relatively independent of the nature of the gears and the gear ratios - they are primarily realated
to the peripheral speed of the gears passing through the fluid. Churning losses are difficult to calculate and
estimates based on experience are often used in initial gear design.
The frictional losses are related to the gear design,the reduction ratio,the pressure
angle, gear size, and the coefficient of friction.
The notes below relate primarily to estimating /calculating the the part of the
efficiency of gear trains which is associated with the tooth friction.
A simple table is provided below showing the efficiencies of various gear types.
These efficiencies related to tooth friction losses only for single tooth meshes. For drive
trains the efficiencies or each mesh in the line is multiplied together ( 2 gears pairs of efficiency result in
a gear train efficiency of 81%.
α
n = normal pressure angle
vs = gear surface sliding velocity (m/s)
α
= pressure angle
β
= helix angle (deg)γ
= worm lead angle (deg)γ
= worm lead angle (deg)μ
= coefficient of frictionConsidering a spur gear a good first approximation for average operating conditions is that the power loss at each mesh can be approximated as P% of the potential power transmitted through the mesh. The efficiency is therefore..
Clarification of formula terms.
Gear TypeΦ
sign of Rg term External spur cosα
(Rg+1) Internal Spur
α
(Rg-1) Single Helical cosφ
n / cos 2β
φ
n = tanα
.cos
β
β
= helix angle (Rg+1) Worm GearsThe theoretical efficiency of a worm gear is provided on page worm gears and is shown again below as.
Chart of Worm Gear Efficiency
The graph below shows a worm gear efficiencies plotted against the lead angle for different coefficient of friction
Notes:
This chart agrees with the equation for worm gears having a normal pressure angle of 20 degrees...
An approximation for the friction coefficient for worm gears (Bronze -steel) is
μ = 0,04 vs-0,25
vs = sliding velocity (m/s)
For more information, please visit Best Cylindrical Gear.
Consider the two meshing gears below which are part of an epicyclic gear train. This is provided as and example of a typical component in an epicyclic gear. The two gears and the arms are rotating as shown .
Now it the arm was stationary the contact point P would have a instantaneous velocity = -ω2 R2. (velocities to the right are positive ). Now if the whole system was rotating as a rigid assembly with the velocity of the arm the instantaneous velocity -ω3 R2. Combining these two motions together, the linear velocity of the tooth engagement (gear 2 -> gear 1)is therefore
v12 = - ( ω2 R2 - ω3 R2) = - (ω2 - ω 3)R 2
Now the magnitude of the transmitted tangential force F12 x the tooth engagement velocity v12 is called the potential power and the power loss due to tooth friction is proportional to this power. Generally for spur gears ( and helical gears) it is sufficient to estimate the power loss as 1% of the potential power. For more accurate estimations the equations above can be used.
P = 0,01 F12v12
The potential power is not the actual power but is the but is the power
transmitted by the same gears operating on fixed centres at angular velocities of
( ω2 - ω3 ) for gear 2 and
( ω1 - ω3 ) for gear 2.
The actual pitch line velocity of the gear mesh is -(ω2) and
therefore the ration of the potential power to the actual power is
Now in cases where the arm is rotating faster than the gear the potential power can be greater than the actual power and the losses proportionally greater.
Example Epicyclic Gear efficiency calculation.Consider the epicyclic gear chain shown below. The input speed = 250 RPM (ACW) and the input torque = 2.5 Nm
ω2 = (250.2.π)/60 = 26,18rads/s. and
ω1 = 0
The power into the gear = Tω2 = 2,5.26,18 = 65 Watts.
Calculating forces
The torque on the arm 2 = M2 = 2.5Nm
The force (F2) at Radius R2 (= R1 + R3) : = -(M2/R2) : F2 = -2.5 /(0,1m+.025m) = -20 N (forces to right are positive)
Now all forces and torques on the link between Arm(2) and gears (3) & (4) are in equilibrium therefore
For the gear mesh between gear 3 and the fixed sun 2 the velocity of tooth engagement is calculated
by the product of the angular velocity of the arm 2 at pitch radius of R2. = ω2R1 = 26,18.0,075 = 1,96m/s
The potential power relates to the gear pair operated in isolation -
in this case for gear 3 to be in equilibrium (With forces and moments)
F2.R3 + F3.R3 = 0 and F2.R3 + F3R3 = 0 therefore F3 = -F2 = 20N
The potential power of the mesh between the sun gear 1 and the planet gear 3
= the gear engagement velocity(1,96m/s) x the tangential force (F3= 20N) = 1,96.20 =3,92W
Allowing 1% loss on a typical spur gear engagement the power loss at this engagement = 0,39W.
For the gear engagement between the gears 4 and 5 the angular velocity of gear five =
The velocity of tooth engagement between gear (4) and gear (5) =
R5 (ω5 - ω2) = 0,068(-5,95- 26,18)=-2.18 m/s
The potential power at this tooth engagement =
the gear engagement velocity (2,18 m/s). x the tangential force (F4 = -167N) = 2,18.167 = 364 W
Assuming 1% power loss then the loss at this tooth engagement = 3,64 W
The power input to the epicyclic gear = 65 W therefore the Gear efficiency = 100(1 - (3,64 + 0,39) /65 ) = 94%
Once the required demand torque in a particular application has been determined, the tendency is to determine motor power using the standard equation: kW = Nm x r/min / . However, gearbox efficiency must also be considered in calculations, and is often overlooked.
Gear type, bearing type, seal type and lubrication all contribute to the inefficiency of gearboxes. As gearboxes lose efficiency, so the input power required also increases to compensate, to deliver the required torque at the output.
The type of gear primarily determines how efficient a particular gearbox is going to be. Inefficiency per stage can amount to:
Spur gears &#; 0.5% to 3%
Helical gears &#; 0.5% to 3%
Bevel gears &#; 0.5% to 3%
Hypoid gears &#; 2% to 10%
Planet gears &#; 2% to 10%
Crossed helical gears &#; 5% to 50%
Cylindrical worm gears &#; 10% to 50%
Spiroid gears &#; 3% to 50%
Worm gearboxes are renowned for having the possibility of a high ratio in a small form factor, but also having a relatively low efficiency. As a result, consideration must always be given to application demand torque.
With multiple stage units, inefficiency per stage is compounded.
With gearbox type being the key determining factor in finding efficiency, bearings and seals have a slightly smaller contribution and usually are not able to be changed for each gearbox type.
In bigger design applications, losses for bearings and seals are obviously less of a concern when compared with smaller applications.
Lubrication type (grease or oil), viscosity, and amount also contributes to gearbox inefficiency, although, like bearings and seals, are only a secondary consideration to gearbox type.
The main thing to think about when looking at gearbox application design is the overall power loss from design inefficiency, and to compensate accordingly. Planetary, inline helical and bevel helical gearboxes are the most efficient, whereas worm gearboxes are usually the least efficient.
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