Lecture_18 - Part VI Heat effects from property changes...

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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 Lee-Kessler 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 Lee-Kessler formulation. We could use an analytical cubic EOS for the entire calculation... Instead, we introduce a new state function, enthalpy, that is convenient for constant-pressure processes. Start with dU = dQ + dW = dQ - Pdv . Consider H = U + Pv ; this is also a state function, because U , P , v are. From this definition, we have dH = dU + Pdv + vdP = dQ + vdP So why is this convenient? For constant pressure problems,
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