CHE222_solutions6

CHE222_solutions6 - ChE 222: Problem Set #6 Due: March...

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ChE 222: Problem Set #6 Due: March 7,2008 (8:OO am) , [g) 1 .) Koretsky, Problem 2.29 c i2\ 2.) Koretsky, Problem 2.30 15) 3.) Koretsky, Problem 2.40 (4) 4.) Koretsky, Problem 2.50 I Y ) 5.) Felder and Rousseau, Problem 9.1
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2.29 Mass balance dn . - -- dt nin - "out = nin Separating variables and integrating: Energy balance Neglecting ke andpe, the unsteady energy balance, in molar units, is written as: The terms associated with flow out and heat are zero. Integrating both sides with respect to time from the empty initial state to the final state gives: since the enthalpy of the inlet stream remains constant throughout the process. The work is given by:
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"2'2 = "2 [h,n - pLv2 I Rearranging, ~2 = hi, - Patv2 = ~i,, + envin - U2 - Uin = Pinvin - Pat v2
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Mass balance valve maintains pressure in system / constant v Separating variables and integrating: I TI = 200 OC XI = 0.4 V= 0.01 m3 Energy balance -b Integrating Substituting in the mass balance and solving for Q
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We can look up property data for state 1 and state 2 from the steam tables: So the mass in each state is: And for energy and enthalpy Solving for
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CHE222_solutions6 - ChE 222: Problem Set #6 Due: March...

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