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4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient?

4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient?

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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4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient? Solution: a. For zero-order reaction, C = Co - kt 30 mg/L = 100 mg/L - k x (5 hr) k = 14 mg/(L.hr) b. For first-order reaction, C = Coe-kt 30 mg/L = 100 mg/L x e-kx (5hr) k = 0.24 hr¹
Transcribed Image Text:4. A treatment pond with an influent waste concentration of 100 mg/L produces an effluent concentration of 30 mg/L. The detention time inside the treatment pond is 5 hr. (5 pts) If the treatment reaction is zero-order, what is the decay rate? b. If the treatment reaction is first-order, what is the decay coefficient? Solution: a. For zero-order reaction, C = Co - kt 30 mg/L = 100 mg/L - k x (5 hr) k = 14 mg/(L.hr) b. For first-order reaction, C = Coe-kt 30 mg/L = 100 mg/L x e-kx (5hr) k = 0.24 hr¹
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