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ME405_Lecture_19

# ME405_Lecture_19 - ME 405 Fall 2006 Professor John M...

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ME 405 Fall 2006 Professor John M. Cimbala Lecture 19 10/20/2006 Today, we will : Continue our discussion of Unsteady Dilution Ventilation in Section 5.4 Do several example problems – unsteady dilution ventilation Discuss Removal by Solid Surfaces (Adsorption) in Section 5.5 Discuss how to measure the wall loss coefficient Do C a n d y Q u e s t i o n s f o r C a n d y F r i d a y

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Example (Example 5.5 in text – the Clever Outdoorsman) Given : A man sleeps overnight in a cabin with volume V = 32.65 m 3 . The rate of “fresh” air entering the cabin is 0.30 air changes per hour. The concentration of CO in the outside “fresh” air is c a = 10 PPM (11.4 mg/m 3 ). A kerosene space heater emits CO according to () 1500 1 sin 0.80 St t π ⎡⎤ ⎣⎦ =+ mg CO per hour, where t is in hours. To do : Calculate the concentration of CO in the cabin as a function of time. Solution : Death Coma Vomit, collapse

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ME405_Lecture_19 - ME 405 Fall 2006 Professor John M...

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