Cryogenic Aerospace Polymer Phase Transition (Materials & Aerospace Engineering) Solution

Problem Overview

We evaluate the continuity of the thermal conductivity function $k(T)$ at the glass transition threshold $T = 77\text{ K}$ :

k(T) = \begin{cases} 0.02 T + 0.5 & : T < 77 \\ 2.04 & : T = 77 \\ 0.0003 T^2 + 0.3 & : T > 77 \end{cases}

Step-by-Step Mathematical Solution

Step 1: Condition 1 — Check $k(77)$

From the explicit point specification in the piecewise definition:

k(77) = 2.04\text{ W}/(\text{m}\cdot\text{K}) \quad \text{(Defined)}

Condition 1 is satisfied.

Step 2: Condition 2 — Check $\lim_{T \to 77} k(T)$

Evaluate the left-hand and right-hand limits as $T$ approaches $77$ :

Left-hand limit ( $T \to 77^-$ ): $\lim_{T \to 77^-} (0.02 T + 0.5) = 0.02(77) + 0.5 = 1.54 + 0.5 = 2.04\text{ W}/(\text{m}\cdot\text{K})$
Right-hand limit ( $T \to 77^+$ ): $\lim_{T \to 77^+} (0.0003 T^2 + 0.3) = 0.0003(77^2) + 0.3 = 0.0003(5929) + 0.3 = 1.7787 + 0.3 = 2.0787\text{ W}/(\text{m}\cdot\text{K})$

Compare the one-sided limits:

\lim_{T \to 77^-} k(T) = 2.04 \neq 2.0787 = \lim_{T \to 77^+} k(T)

Because the left-hand and right-hand limits disagree, the overall two-sided limit does not exist (DNE):

\lim_{T \to 77} k(T) = \text{DNE}

Condition 2 fails.

Step 3: Classify the discontinuity

Both one-sided limits exist as finite real numbers ( $2.04$ and $2.0787$ ), but they are unequal.

Classification: Jump Discontinuity
Jump Magnitude: $\Delta k = |2.0787 - 2.04| = 0.0387\text{ W}/(\text{m}\cdot\text{K})$ .

Engineering Conclusion

Because the thermal conductivity exhibits a jump discontinuity of $0.0387\text{ W}/(\text{m}\cdot\text{K})$ at $77\text{ K}$ , heat flux through the insulation changes discontinuously. In aerospace rocket engineering, this sudden change induces localized thermal stress gradients across the fuel line wall during cryogenic fueling, requiring engineers to install flexible expansion bellows to prevent brittle insulation spalling and tank wall rupture.