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Unformatted text preview: Part VI — Heat e ff ects from property changes Lecture 18 — Enthalpy Example: cooling at constant pressure. Suppose we cool isobutylene at 80bar and 752.2K to 500K. a) Compute the work done. W = P Δ V = P ( v 2 v 1 ). To compute initial and final v, need EOS. This is not an ideal gas: T c = 417 . 9 K , P c = 40bar. So we need some EOS to compute the change in volume. We can use LeeKessler correlation, Haile p. 120 ff , given an acentric factor of ω = 0 . 194. b) Compute Q for the process. Δ U = Q + W ; we know W , so we just need Δ U . But, V is not constant, and the system is not an ideal gas, so dU has contributions from both dT and dV : dU = C v dT + ( T γ v P ) dv We cannot easily compute γ v = ( ∂ P/ ∂ T ) v using the LeeKessler formulation. We could use an analytical cubic EOS for the entire calculation... Instead, we introduce a new state function, enthalpy, that is convenient for constantpressure processes. Start with dU = dQ + dW = dQ Pdv . Consider H = U...
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 Spring '08
 PROF.MARANNAS
 Thermodynamics, Enthalpy, EoS, state function

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