Lecture_18 - Part VI Heat e ff ects from property changes...

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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 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 = 0 . 194. b) Compute Q for the process. U = Q + W ; we know W , so we just need 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 v- P ) dv We cannot easily compute 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...
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This note was uploaded on 04/01/2008 for the course CHE 220 taught by Professor Prof.marannas during the Spring '08 term at Pennsylvania State University, University Park.

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Lecture_18 - Part VI Heat e ff ects from property changes...

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