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Physics / Chapter 18: Temperature, Heat, and the First Law of Thermodynamics / Section 18-5

Thermal Expansion — Question 10

Section 18-5 · Physics pages 2910–2911

Question 10 A differential support can cancel first-order thermal drift A mechanism has output separation D = LA −LB. Both rods expand freely through the same temperature change, with αA = 12.0 × 10−6 K−1 and αB = 20.0 × 10−6 K−1. At the reference temperature, require D0 = .200 m and zero first-order temperature dependence. The linkage transmits the difference of lengths without loading the rods. D A B The ideal linkage outputs a length difference, not a sum. (a) Find the two required reference lengths. (b) If rod B is actually 1.0 mm longer than designed while A is correct, find the reference separation error and the additional drift during a 50.0 K rise.
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Question 10 - Solution A differential support can cancel first-order thermal drift D A B (a) Method and calculation LA −LB = .200, αALA −αBLB = 0. LA = 20 12LB, LB = .300 m, LA = .500 m. Equal absolute expansions cancel in the output even though each rod grows. (b) Interpretation and further calculation δD0 = −.0010 m = −1.0 mm. dD dT = αA(.500) −αB(.301) = −2.00 × 10−8 m/K. ∆Dthermal = −2.00 × 10−8(50) = −1.00 µm. The static length error and residual thermal drift are different effects. Physical check Cancellation requires both rods to share the same temperature change and their coefficients to remain valid.

Question 10

A differential support can cancel first-order thermal drift

A mechanism has output separation D=LA−LBD=L_A-L_B. Both rods expand freely through the same temperature change, with αA=12.0×10−6K−1\alpha_A=12.0\times10^{-6}\,\mathrm{K^{-1}} and αB=20.0×10−6K−1\alpha_B=20.0\times10^{-6}\,\mathrm{K^{-1}}. At the reference temperature, require D0=.200mD_0=.200\,\mathrm m and zero first-order temperature dependence. The linkage transmits the difference of lengths without loading the rods.

Diagram shown in the original SVG worksheet.

  1. Find the two required reference lengths.

  2. If rod B is actually 1.0 mm longer than designed while A is correct, find the reference separation error and the additional drift during a 50.0K50.0\,\mathrm K rise.

Question 10 - Solution

A differential support can cancel first-order thermal drift

Diagram shown in the original SVG worksheet.

(a) Method and calculation

LA−LB=.200,αALA−αBLB=0.L_A-L_B=.200,\qquad\alpha_AL_A-\alpha_BL_B=0.LA=2012LB,LB=.300m,LA=.500m.L_A=\frac{20}{12}L_B,\quad L_B=.300\,\mathrm m,\quad L_A=.500\,\mathrm m.Equal absolute expansions cancel in the output even though each rod grows.

(b) Interpretation and further calculation

δD0=−.0010m=−1.0mm.\delta D_0=-.0010\,\mathrm m=-1.0\,\mathrm{mm}.dDdT=αA(.500)−αB(.301)=−2.00×10−8m/K.\frac{dD}{dT}=\alpha_A(.500)-\alpha_B(.301)=-2.00\times10^{-8}\,\mathrm{m/K}.ΔDthermal=−2.00×10−8(50)=−1.00μm.\Delta D_{\mathrm{thermal}}=-2.00\times10^{-8}(50)=-1.00\,\mu\mathrm m.The static length error and residual thermal drift are different effects.

Physical check

Cancellation requires both rods to share the same temperature change and their coefficients to remain valid.

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