Elementary Principles 491 - 11.25 (cont'd) b. At...

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11- 20 b. At steady-state , dT dt m T m T ss =⇒ =⇒ = 04 0 3 0 5 5 9 0 0559 403 . ± . ± . . Tm s =°⇒ = 24 0333 C k g h r ± . c. Separate variables and integrate the balance equation: dT mT dt s Tt f 40 3 10 0 . ± . = z z dT T t 134 0559 10 23 .. E = z t =− L N M M O Q P P = 1 134 0559 23 134 0559 10 48 . ln . bg hr 11.26 a. Integral energy balance tt == 02 0 to min QU M C T v = ° ⋅° ΔΔ 250 kg 4.00 kJ 60 C kg C kJ 20 400 10 4 . Required power input: ± . Q = × = 4.00 10 kJ 1 min 1 kW 20 min s 1 kJ s kW 4 60 333 b. Differential energy balance: MC dT dt Q v = ± dT dt Qt = 0001 . ± tT = = ° 0 , Integrate: dT Q dT T Qdt t 20 00 20 o C o C z z z = + . ±± Evaluate the integral by Simpson's Rule (Appendix A.3) kJ s ± Qdt 0 600 30 3 3 34 3 33 53 94 45 05 86 67 58 59 5 2 3 43 74 14 75 46 27 08 09 0 1 0 0 3 4 8 3 0 z = + +++++++++ ++ + + + + + + + += ⇒= = ° T 600 s 0001 34830 548 ej oo C+ C/kJ kJ C c. Past 600 s,
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