Roller bearings must be oversized to live long, and here's an improved choice, with the
usual design first.
Most producers stick to the standard and indicate a dynamic load applicable for 1M turns, but a flywheel at 39Hz makes 25G turns in 20 years. Life shall then vary with the power 10/3 of the load. But SKF observes a faster improvement at light loads and tells "no wear at all under this limit" which I shall use through this applet:
http://www.skf.com/skf/productcatalogue/calculationsFilter?lang=en&newlink=&prodid=&action=Calc5
that computes losses if you choose a bearing and an oil viscosity in the process.
390kN per side is below the fatigue load of the NNCF 5048 CV (shaft d=240mm, bore D=360mm). Maximum 1000rpm impose a wheel of D=8.4m, undesired. With an oil viscosity of 6.6mm
2/s, loss is 27N*m or 2.8kW per bearing (screenshot), the pair summing
over 10h 3.3% of the stored energy.
Several smaller bearings at each shaft end accept more rmp and reduce the loss torque. This wins the 5m wheel back at identical power loss. It needs added hardware to align the bearings and share the load, like calottes working similarly to bogies.
3.3% energy loss isn't bad. Compare with transformers and high voltage lines working at full load instead of medium load: they must lose nearly that. But the big roller I suggested at saposjoint.net/Forum/viewtopic.php?f=66&t=1974 on Sat Jul 09, 2011 and on pictures 3 and 4 in the present thread shall reduce losses.
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How much do the
big rollers dissipate? Data exists for railways
http://www.engineeringtoolbox.com/rolling-friction-resistance-d_1303.html
but is pessimistic for hardened steel, and bearings are >10 times better.
First, how wide is the shaft? I keep the material and long-run contact pressure of existing bearings, with 20 rolls (uncritical), of which the 10 lower work. Because force is proportional to deformation in a cylindrical contact, and cos
2 has 0.5 mean value, 10 rolls bear 5 times the load of the lowest roll. Since the rolls have the shaft's diameter times pi/20, and bearing capacity is proportional to the diameter
http://en.wikipedia.org/wiki/Contact_(mechanics)#Contact_between_two_cylinders_with_parallel_axes
a shaft with a diameter multiplied by pi/4 running on a huge roller bears as much. Or rather,
keep the shaft's diameter, and have a roller 3.7 times larger, since 1/R add.
Ho do losses vary with the diameter? Unclear... With the identical stress, contact width increases as sqrt(R) or 2.5 over the previous 20 rolls, and the equivalent length of rolling resistance shall do the same at most. Then, as the big roller is supported by its smaller shaft, this loss occurs once, instead of twice at a roll bearing. The main improvement comes from the radius increased 6.4 times over the rolls. The combination
cuts loss by 5.
The roller's shaft must be supported, but at a smaller speed wasting less power - or if you prefer, this rolling force is slashed by the radius ratio. And the roller's shaft itself can run on a big roller.
This shall bring rolling losses
over 10h under 1% of the stored energy.
Roll bearings need a long contact line, so the big roller would be longer than a roll bearing, and the shaft accordingly narrower. Fine, as the shaft's diameter has margin, and this reduces loss. Elastic bending at the shaft may become a limit, but the roller can be inclined (already done at turbogenerators' hydrodynamic bearings) and grinded slightly non-cylindrical.
Marc Schaefer, aka Enthalpy