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Zn: P2 (3.50) + 0 = 0, 10(1.056) + R32 (4.35) Ni: P2 (2.19)+0 = 1.00(1.056) + R32 (2.36) Solve to get R,=2.75L/hr 32

Zn: P2 (3.50) + 0 = 0, 10(1.056) + R32 (4.35) Ni: P2 (2.19)+0 = 1.00(1.056) + R32 (2.36) Solve to get R,=2.75L/hr 32

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Show how the answer derived to 2.75
Pick Unit 3 as the system: Steps 6 and 7: Balances: 3 H20 ZN, Ni Unknowns: 3 W, R32, P2 (all in L) Steps 8 and 9. Zn: P2 (3.50) + 0=0.10(1.056) + R32 (4.35) Ni: P2 (2.19) +0 = 1.00(1.056)+ R32 (2.36) Solve to get R,=2.75L/hr 32
Transcribed Image Text:Pick Unit 3 as the system: Steps 6 and 7: Balances: 3 H20 ZN, Ni Unknowns: 3 W, R32, P2 (all in L) Steps 8 and 9. Zn: P2 (3.50) + 0=0.10(1.056) + R32 (4.35) Ni: P2 (2.19) +0 = 1.00(1.056)+ R32 (2.36) Solve to get R,=2.75L/hr 32
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