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curriculum-project-hub/examples/TH-141/examples/41届复赛三-1-混注石油/solution.typ
T
sjfhsjfh c73a2c903f feat(examples): migrate TH-141 sample to declarative layout (WU-6)
Real-content fixture for the end-to-end pipeline. TH-141 (39 parts: 22 segment,
15 lemma, 2 example) migrated from the prototype's typst `#let parts` manifest
to ADR-0008: declarative manifest.toml (project+info+ordered parts+targets) +
per-element element.toml (kind + scalars; examples carry `source`). Content .typ
files copied byte-identical (no math corruption); per-element main.typ + meta.toml
dropped (wiring is now generated). Part order matches source exactly; 5 lemmas
have no proof.typ (optional); no cross-file imports / paralearn refs / figs.

Co-Authored-By: Claude Opus 4.8 (1M context) <noreply@anthropic.com>
2026-06-22 01:33:40 +08:00

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参考态满足 $p = beta T$。混合过程总体积不变,仍属等容过程,故混合后亦满足 $p = beta T$。即整个过程始终有 $p = beta T$
代入 $sigma(T, p)$,令 $tau equiv T \/ "K"$,得 $sigma$ 退化为 $tau$ 的单变量函数
$ sigma(tau) \/ (10^(-3) "N/m") = 60 + 0.065 tau - 24.0 thin (beta tau) + 3.15 thin (beta tau)^2 . $
代入 $beta = 7.28 times 10^(-3)$,注意 $24.0 times 7.28 times 10^(-3) = 0.1747$$3.15 times (7.28 times 10^(-3))^2 = 1.670 times 10^(-4)$
$ sigma(tau) \/ (10^(-3) "N/m") = 60 - 0.1097 tau + 1.670 times 10^(-4) tau^2 . $
$tau$ 求导取零:
$ tau_("mix") = 0.1097 / (2 times 1.670 times 10^(-4)) approx 328.4 , $
$T_("mix") approx 328.4 thin "K" approx 55.3 thin "℃"$
混合的热平衡。设参考态石油温度 $T_0 = 373.15 thin "K"$、密度 $rho_0 = 0.8 rho_w$、比热容 $c_0 = 0.6 c_w$;注入水温度 $T_w$、密度 $rho_w$、比热容 $c_w$。等体积混合且绝热给出
$ rho_0 c_0 (T_0 - T_("mix")) = rho_w c_w (T_("mix") - T_w) , $
代入 $rho_0 c_0 = 0.48 thin rho_w c_w$
$ 0.48 (T_0 - T_("mix")) = T_("mix") - T_w , $
解出
$ T_w = T_("mix") - 0.48 (T_0 - T_("mix")) approx 306.9 thin "K" approx 33.7 thin "℃" . $