Max temp range allowed that's still in range of tolerance?

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Discussion Overview

The discussion revolves around determining the maximum temperature range for measuring a titanium jet engine disk that has been machined to specific tolerances. Participants explore the implications of thermal expansion on the measurements of the disk, considering the initial temperature of 80°F and the tolerances provided by the customer.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the maximum temperature using the formula for thermal expansion but expresses uncertainty about the simplicity of the approach.
  • Another participant points out that the disk's diameter is subject to manufacturing tolerances, suggesting the need for multiple calculations.
  • Further contributions propose that there are four potential calculations based on the disk's size and customer acceptance limits, although some participants suggest that not all calculations are necessary.
  • Disagreement arises regarding the validity of the calculated temperature range, with one participant asserting that the results cannot be correct if the customer measures at the same temperature as the supplier.
  • Participants explore the concept of negative thermal expansion and its implications for temperature calculations, leading to further calculations that yield different temperature results.
  • Some participants express that their calculations appear reasonable, while others question the relevance of certain results based on the context of the measurements.

Areas of Agreement / Disagreement

There is no consensus on the correct approach to calculating the maximum temperature range, as participants present competing views and calculations. The discussion remains unresolved regarding the implications of thermal expansion and the appropriate temperature range for the measurements.

Contextual Notes

Participants highlight limitations related to the assumptions made about the disk's diameter and the effects of thermal expansion, as well as the dependence on the definitions of tolerances and measurements.

nchin
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Homework Statement



A 24" diameter titanium jet engine disk is machined to tolerance or +/- 0.002". Customer will accept +/- 0.003". The part is machined and measured in you plant at 80F. What is the max temp range for the customer to measure the part which you should specify in the contract?

Homework Equations



Delta L = L initial (alpha)(Tf - Ti) [alpha = coefficient of thermal expansion]

The Attempt at a Solution



0.003 = 24" (4.8 x 10^-6)(Tf - 80)--->Tf = 106 F??
I don't think its this simple. I am pretty sure I am wrong. help please!
 
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Your calculation is correct if the disk was exactly 24" diameter. But it was manufactured with a tolerance, so its diameter might not be exactly 24".
 
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AlephZero said:
Your calculation is correct if the disk was exactly 24" diameter. But it was manufactured with a tolerance, so its diameter might not be exactly 24".

so do i make two calculations? One with initial length as 24 + 0.002 and the other 24 - 0.002?
 
You could say there are four calculations. The disk size might be 24 + 0.002 or 24 - 0.002, and the customer might measure 24 + .003 or 24 - .003.

But if you think about it a bit, you don't need to do all four. (If you can't see why not, just do all four and see what you get).
 
AlephZero said:
You could say there are four calculations. The disk size might be 24 + 0.002 or 24 - 0.002, and the customer might measure 24 + .003 or 24 - .003.

But if you think about it a bit, you don't need to do all four. (If you can't see why not, just do all four and see what you get).

0.003 = 23.997" (4.8 x 10^-6)(Tf - 80)
Tf = 106.04

0.003 = 24.003" (4.8 x 10^-6)(Tf - 80)
Tf = 106.038

0.002 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 97.3597

0.002 = 23.998" (4.8 x 10^-6)(Tf - 80)
Tf = 97.3625

so the range would be from 97 F - 106 F?
 
The answer can't be right, because if the customer measured at 80 F they would measure the same size as the supplier.

If the disk size is 24. + .002 and the maximum the customer will accept is 24 + .003, what is the thermal expansion between those measurements? (It's not .002, or .003).

Also, for some of the four calculations, the expansion is negative and the temperature will be below 80 F.
 
AlephZero said:
The answer can't be right, because if the customer measured at 80 F they would measure the same size as the supplier.

If the disk size is 24. + .002 and the maximum the customer will accept is 24 + .003, what is the thermal expansion between those measurements? (It's not .002, or .003).

Also, for some of the four calculations, the expansion is negative and the temperature will be below 80 F.

thermal expansion = .001?

0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 88.679?

0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
Tf = 88.33?
 
nchin said:
0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 88.679?
That looks reasonable.
0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
Tf = 88.33?
If it were machined to a diameter of 23.998" at 80F but measured at 88.33F, what would it be measured to be?
 
nchin said:
thermal expansion = .001?

0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 88.679?

Those look right.

0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
Tf = 88.33?

If the size changes from 23.998 to 23.997, the "expansion" is negative and the temperature will be less than 80.
 
  • #10
AlephZero said:
Those look right.
If the size changes from 23.998 to 23.997, the "expansion" is negative and the temperature will be less than 80.
-0.001 = 23.998" (4.8 x 10^-6)(Tf - 80)
Tf = 71.32 F

thanks!
 
Last edited:
  • #11
haruspex said:
That looks reasonable.
If it were machined to a diameter of 23.998" at 80F but measured at 88.33F, what would it be measured to be?

-0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 71.32 F
 
  • #12
nchin said:
-0.001 = 24.002" (4.8 x 10^-6)(Tf - 80)
Tf = 71.32 F

That isn't what I asked.