Heat transfer Q--equilibrium temperature of shrink fit onto shaft

[tfry200]

Hello,

I have a problem I'm trying to solve. I have a shaft that need to shrink fit a rotor onto. Due to the presence of seals inside the shaft, I have a maximum allowable temperature of 175F. I am trying to determine how close I will come to this temperature during assembly based of the cooled shaft and heated rotor. The shaft is solid. I have all the geometries, part masses, materials, etc.

I can do the basic equilibrium temperature calculation, but due to cooling via convection, the numbers come out high. I was just wondering if someone could get me pointed in the right direction to find a closer equilibrium temperature (maximum temperature the shaft will attain before it starts cooling). It's been a while since my heat transfer courses!

Here is my assumption:
One dimensional conduction
Composite Cylinder
Both materials steel (rotor is actually aluminum and steel, but steel makes up the majority of the exposed material and the rotor is fully heated to temperature)

Thanks for any help you guys can give.

Tim

 

[desertfox]

Hi tfry

It may help if you upload a picture of your rotor and shaft but I would say from what you describe you will not get uniform heating and differential heating within the rotor will occur.
What is the intereference between rotor and shaft? is it possible you could just cool the shaft.

desertfox

 

[tfry200]

Here's a print of the situation. Interference at MMC is listed on there. 0.028mm max interference.

I'm planning to cool the shaft as well, but my minimum shaft temperature is -4F.

The reason I was thinking I could say the heating in the rotor is uniform is because we can leave it in the furnace long enough that it reaches the same temperature throughout. Or I guess you meant that cooling will not be uniform?

 

[tfry200]

Hopefully with attachment this time!

 

[desertfox]

hi tfry200

Have you got seals at both ends of the rotor it appears to me if you only have seals at the left hand end could they be put in after.
Not sure that if you have different materials assembled together you can avoid thermal stresses particularly if you have steel and aluminium together, your diagram doesn't really show much of the rotor construction so if your rotor gets to 250 degrees centigrade all over do you know what stresses may occur?
For sure cooling won't be uniform either, which was why I enquired about just cooling the shaft it would appear to be simpler to me.

desertfox

desertfox

 

[tfry200]

Desertfox,

The seals on this shaft are internal and are installed by the manufacturer. We can't change that, so are bound to their temperature range. The drawing I posted was of the shaft that we used to test the mechanical fit.

I know it would be imperfect, but was hoping I could let the shaft be solid steel and use some general heat transfer equations to get an approximation of the system. Looking back at my Heat xfer book, I see two equations that look like they could be used, I just don't quite remember if they are fully applicable. They are:

qr=(Ti-Tair)/(R_total)
and
qr=2(pi)L*k(T1-T3)/(ln(r2/r1))

Does it seems like I'm on the right track here, or am I way off in left field?

 

[tfry200]

My apologies. In above posting, I meant that I was hoping I could get an approximation of the system by assuming the ROTOR was solid steel, not the shaft. The shaft already is solid steel.

 

[desertfox]

hi tfry200

I am a little confused as to what your trying to achieve however if you are assuming that the rotor is all steel and you wish to determine how long to put it in the oven for so that it reaches the required temperature then this would be my approach

Q = m * Cp * (T2-T1) where m = mass of rotor

Cp = specific heat capacity of
steel

T2-T1= temperature change

Q= heat energy req

Now look at the power of your heat source XX Kw= XX * 10^3 W

divide Q by the power of your heat source in Watts and that will give you the time in seconds to heat the rotor.
Obviously I can't account for the efficiency of your furnace. So turn your furnace on let it reach the required temperature and put your rotor in for at least the time calculated by the formula above you may need to leave it slightly longer because obviously I can't account for the efficiency of your furnace.
I hope this is what you are trying to calculate.
My concern is that when you heat the rotor because the assembly is made up of steel and aluminium then the stresses induced between these materials might be very significant, however if your company as been doing this process for years without a problem then its probably okay.

desertfox

desertfox

 

[tfry200]

Hmm..well that is good information also, so thank you for that! But what I am really after is: 'what is the hottest temperature the chilled shaft will rise to once the heated rotor is dropped on?'

i.e.--after the rotor is dropped on, the rotor will begin to cool and the shaft will begin to get hotter. I need to know if the 440F rotor will raise my -4F shaft up past 175 degrees.

If you have any idea, it would be much appreciated.

Thanks!
Tim

 

[desertfox]

hi tfry200

You can use the same principle as above like this:-

Heat gained by shaft= Heat lost by rotor

M(s) * Cp * (T?-Ts)= M(r) * Cp * (Tr-T?)
Shaft Rotor

Where M(s) and M(r) are the masses of shaft and rotor

Cp = specific heat capacity of steel which will cancel out
in the above because both components are steel

T(s)= temperature of shaft on assembly

T(r)= temperature of rotor on assembly

T(?)= final temperature after assembly which you have to
solve for.

This should be very conservative as it assumes that all the heat from the rotor goes into the shaft which in practice will not happen as cooling of the rotor will occur by convection to the surrounding atmosphere as well as by conduction through the shaft.

solving for T(?)= (Mr * T(r)) + (Ms * T(s))/ (M(s) + M(r))

this is the most simple way I can think of getting a rough figure for the final temperature without getting into a complicated heat transfer process which frankly your situation is.

 

[desertfox]

Hi tfry200

I tried estimating masses of rotor and shaft from your file, I get the mass of rotor roughly to 1.38kg and shaft to 1.452kg, sadly based on the temps you give I get an equilibrum temp of around 100 degrees centigrade your looking for a max of about 80 degrees centigrade.
Bearing in mind that I have made no allowance for convection
I wonder if you could put some kind of heat sink under the exposed parts of the shaft after assembly which would take more heat away by conduction. (in effect it icreases the mass of the shaft in the above calculation).
So if the mass of the shaft was 2.1kg instead of 1.452kg the equilibrium temperature would be 78 degrees centigrade.

desertfox

 

[tfry200]

desertfox,

Thanks for the input. I tried to nail down a little more accuracy using your formula as:

Qr_alum+Qr_steel=-Qshaft

To account for the two materials in the rotor. I pulled the mass numbers (in kg) off CAD, and got:

Mr_alum=.199
Mr_steel=.955
Mshaft=1.924 (the actual shaft we will use is larger than the above test shaft)

And got my final temperature at 84 celsius.

So I'm cutting it close, but I will do as you suggest and try to pull as much heat as possible away from the shaft using a heat sink. Between that and losses due to convection hopefully I will be okay.

Thanks again for the help!

 

[btrueblood]

Once the rotor is seated on the shaft, blow on the exterior with an air nozzle (i.e. increase the convection term).

Having made shrink fits with those kinds of clearances before...good luck. We averaged about 2/3 not jamming halfway down the shaft. Then we redesigned the shaft to be well undersized for most of the length, only widening to the shrink fit in the area where it was needed.

 

[desertfox]

hi tfry200

Yes you may be okay with some heat sinks and if you have chance to do a test why not put some thermocouples in the area where the seals would be in the shaft and take readings.
I was looking through my heat transfer books at heat loss with convection and came across this formula:-


q"= h*(T(s)-T([∞]))

where q"= W/m^2 = heat flux

h = convection heat transfer coefficient

T(s)& T([∞])= Temp of metal surface & Temp of
surrounding

h in gases for free convection = 2 to 25 (W/m^2 * K)

where K temp in
Kelvin

If you know the surface area of the rotor from the above formula we can calculate the heat loss from the rotor in the first second ie W= (joules/second) if this heat loss is significant in the first second then it should give a further degree of comfort that the shaft won't get to hot.
I would use 2 W/m^2 * K as that would be conservative unless you want to verify what (h) actually is.
I think what were doing is pretty crude but it might get you in the right ball park.

desertfox

 

[tfry200]

Hmm... only about 7 watts with h=2. Hardly seems significant, does it?

I do see some stuff in my book about transient response...maybe I should look there? At that point I guess I would be kind of just going through it for the exercise, but it might not be a bad one to do. I'll let you know if I get a reasonable answer out of it.

 

[desertfox]

Hi tfry200

I agree not much going with convection and of course, less and less heat will be transfered as the rotor cools to both the air and shaft.
It would appear most of the heat would go by conduction into the shaft. If you take the 7 joules away from the heat lost in the shaft and apply the same logic as before how much does that reduce the equilibrium temperature by bearing in mind you were only 4 or 5 degrees off the target figure, also we have taken the worst case figure for (h)in practice that figure might be higher.
Might be best to do a test run and get some results practically.

desertfox

 

[chicopee]

Use dry ice to shrink the shaft and let it warm up under ambient condition when assembled with the rotor. Forget the calculations too much time wasted.