Slagathor
Mechanical
- Jan 6, 2002
- 129
I have a customer that would like a flywheel added to pump system so as to prevent downsurge in a pipeline during power failure. Much cost and time has gone into developing this "solution". The problem, as I see it, is that no one bothered to see how challenging the development of a flywheel this large, at this speed, would be.
I designed a flywheel for a mixed load (combination of inertial and brake loading) dynamometer some years back...and it scared the heck out of me. That wheel peaked at 1500 RPM...and was about 20" in diameter. It was a plain disc (not ideal for RPM limit vs diameter, but most effective from a mfg cost perspective). At 3600 RPM...the stresses multiply nearly 6 fold from 1500...
I have a spreadsheet I created, which takes the general solution of a spinning disc problem from Timoshenko Part II (Advanced Theory and Problems) and makes a usable quick ideal world stress calculator. The question I have...is safety factor. Because the potential for catastrophic failure and loss of life is non-trivial with a flywheel...I have heard of huge safety margins being applied...ie max stresses of 10% of yield. This is to allow for the uncertainty of micro defects creating stress concentrations in the material. Is it really necessary to be this conservative if you are mfg from quality alloy plate...like 4140?
Example 1:
4142 High Carb...Heat Treated has a yield of 90,000 psi and a UTS of 140,000 psi. This is big bucks....probably $3000 for a 24"x24"x7" round raw material drop out. So...if we could manage to make a 6.75" thick x 23.5" OD Flywheel..the peak stress will be 11,950 psi at the ID (assuming a 3.5" rotor bore ID which is used to align a pair of heavy hubs, which actually bolt the flywheel). The radial stresses at the inner surface are of course zero. About 30-40 off center...I will have through holes for bolting up of hubs...so...to determine stresses there...I will have to look at the combined tensile (6200 psi) and radial (4300 psi) stresses...calculate from there...and THEN apply a stress concentration factor for the through hole. But the time I am done with this...I can see a real world stress potential in the range of 20,000 psig.... 200,000 yield on the material...?? Not going to happen
So this means go smaller on the diameter...which will drive up the thickness and weight.
Example 2:
Same material. Limit OD to 15.5". This will result in a peak stress...when conbined stresses..and concentration...is taken into account of about 9000 psig....or 10% of yield. The problem is this 15.5" OD wheel will need to be 35" thick..and it will weigh 2000#. Now..I have a shaft...not a wheel...and I start to get into rotor dynamics issues...mfg costs notwithstanding.
So...back to my question....how much safety factor for a wheel machined from high quality wrought alloy plate?
By the way...any PEs out there want to stamp this thing?
I designed a flywheel for a mixed load (combination of inertial and brake loading) dynamometer some years back...and it scared the heck out of me. That wheel peaked at 1500 RPM...and was about 20" in diameter. It was a plain disc (not ideal for RPM limit vs diameter, but most effective from a mfg cost perspective). At 3600 RPM...the stresses multiply nearly 6 fold from 1500...
I have a spreadsheet I created, which takes the general solution of a spinning disc problem from Timoshenko Part II (Advanced Theory and Problems) and makes a usable quick ideal world stress calculator. The question I have...is safety factor. Because the potential for catastrophic failure and loss of life is non-trivial with a flywheel...I have heard of huge safety margins being applied...ie max stresses of 10% of yield. This is to allow for the uncertainty of micro defects creating stress concentrations in the material. Is it really necessary to be this conservative if you are mfg from quality alloy plate...like 4140?
Example 1:
4142 High Carb...Heat Treated has a yield of 90,000 psi and a UTS of 140,000 psi. This is big bucks....probably $3000 for a 24"x24"x7" round raw material drop out. So...if we could manage to make a 6.75" thick x 23.5" OD Flywheel..the peak stress will be 11,950 psi at the ID (assuming a 3.5" rotor bore ID which is used to align a pair of heavy hubs, which actually bolt the flywheel). The radial stresses at the inner surface are of course zero. About 30-40 off center...I will have through holes for bolting up of hubs...so...to determine stresses there...I will have to look at the combined tensile (6200 psi) and radial (4300 psi) stresses...calculate from there...and THEN apply a stress concentration factor for the through hole. But the time I am done with this...I can see a real world stress potential in the range of 20,000 psig.... 200,000 yield on the material...?? Not going to happen
So this means go smaller on the diameter...which will drive up the thickness and weight.
Example 2:
Same material. Limit OD to 15.5". This will result in a peak stress...when conbined stresses..and concentration...is taken into account of about 9000 psig....or 10% of yield. The problem is this 15.5" OD wheel will need to be 35" thick..and it will weigh 2000#. Now..I have a shaft...not a wheel...and I start to get into rotor dynamics issues...mfg costs notwithstanding.
So...back to my question....how much safety factor for a wheel machined from high quality wrought alloy plate?
By the way...any PEs out there want to stamp this thing?