In my post “How Does Calculating Your Own Alignment Targets Work?”, I discussed using a hot check to determine proper alignment targets. Thermal growth is not rocket science. However, it is often misunderstood. I see people causing harm by compensating for thermal growth incorrectly.
Let’s look at a simple blower illustration:
If we assume that the blower expands evenly and therefore set the motor high, we could be going in exactly the wrong direction.
Let’s take another look:
The temperature change across the blower is not uniform. The discharge end of the blower will expand more than the suction end. In my next post I will plot out the actual movement and the right compensation.
We really want to create a collaborative blog. Feel free to offer your opinion/question.
11 Comments
In over-hung pump as we have one housing for both bearings and provided that the inboard and outboard bearings have same temperature ,angular misalignment stay the same in hot and cold condition.
I’m not so sure about that, Hamid. In an over-hung pump, if it is pumping a hot process fluid, the bowl is hotter than the bearing end. Since it is pinned at the bottom, the bowl end will grow up. If there is no foot under the power end, maybe it will grow straight up (neglecting pipe strain for the moment), maybe not. If there is a foot and it is being constrained at the bowl and the foot, there is definitely a good chance for it to grow at an angle. Throw the piping issue into the equation and who knows where it will grow–that why it needs to be measured.
I really do not like setting pre-determined alignment targets due to the ever changing thermal conditions. At my company I recommend that critical equipment to be aligned be scanned with a thermal imager prior to alignment and appropiate thermal growth calculations be made based on the results. It would be nice if alignment equipment had this capability.
Matt,
Thank you for your contribution. I am so pleased to see your two entries. We welcome all input, debate, & ideas.
In the spirit of debate, I pose this question: a) When we use temperature delta’s and expansion coefficients to calcualte thermal growth… b) are we not assuming that the machine will grow from a fixed plane in space (the base)… c) in a perpendicular direction from the base? So….. Given that the metal will indeed expand, what effect does pipe strain have on thermal growth?
Yes pipe strain, loading, looseness and other issues affect any alignment but those items usually are unwanted variables that we want to try to minimize or eliminate through engineering controls. Piping strain we try to get rid of with constant or variable pipe hangers or by other means. Looseness, loading are more difficult to engineer out but possible. Temperature changes also can be lessened through but since they are also related to the environment it becomes relative to the current conditions. For example a large pump my heat up to a consistant temp, say 300f due to the fluid it pumps but the motor is cooled by ambient air. This means in the middle of summer when cooling is less effective than cooling in the winter the alignment will vary greatly. So basically the alignment is relative to the time of year. This doesn’t always apply but when it does it needs to be accounted for. As for growing perpendicular the larger machines I am referring to usually are located on massive foundations and grow uniformly. The whole issue is that thelarger the machine the more difficult it is to come up with set values.
Hi,
We have a case on which environmental temperature is 25Degc and pedestal temperature is 65Degc for Main Motor and 50Degc for compressor .Also shaft height for motor is 315mm and for compressor is 250mm.Using delta and above mentioned values we found shaft center rise for motor is 0.14mm and compressor shaft center rise is 0.07mm.Thus we consider 0.14mm for round vertical.
Br
Hamidreza
One thing I like to do when I suspect temperature is an issue is to use a thermal imager on like equipment prior to the alignment. This gives me a good idea of what the operating conditions may be once the machine is back in service. I take temps of each support and get an average temp for each between the shim plane and shaft centerline. I do this for both the “sister” machine and the one I am aligning. The operating machine is T2 (T hot) and the one I am aligning is T1(T cold). This way the delta T accounts for the current conditions.
I am very pleased and grateful for your responses and comments! I love that this site is relatively non-commercial (we ARE proud of our company and the things we sell so a little of that does creep into our posts). I love that we are both teaching and learning. Matt, I am going to dust off and old report from a thermal gowth measurement. The machine train had a GE Frame 6 gas turbine driving a gearbox, driving a compressor and lastly another compressor. The whole train is stretched 35 meters from end to end. And…..the movement was UNBELIEVABLE. I measured it comming down and going back up. Might be a week or so before I have the time.
Anyone have any particular alignment subjects that they would like us to dive into?…. or if anyone would like to author a post, let me know. Have a great day all.
I made several OL2R measurents with a Fixturlaser XA, between a big generator with 3.000RPM and rotor mass of 10.000Kg. It was coupled to a big FLENDER gear box, which was reducing the turbine speed of 7.500RPM.
It is really difficult to get good hot values due to the environement and vibration conditions.
In cold conditions with the machines stopped it is easy to measure.
Most important is to correct the calculate the difference from the point where we place the OL2R supports and the real center line of the coupling. This is something which is the key for our success… Kind regards. Luis
Dear Dave,
Good day,
Thermal growth calculation : all these while I’ve been using the formula T X L X C for compensation on the hot alignment, I hope that I’m doing the right thing.
Kindly advise,
Mr. Mel Bunsu.
Hello,
I have a motor having thermal expansion 0.19 mm.
SO what should be my TIR on radial ?
12 O clock- 0
6 O Clock- +0.38 ??
Can anyone help ?