Elementary Principles 260 - . 7.30 (a) E p , E k , Ws = 0 Q...

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7- 15 7.30 (a) Δ ± E p , Δ ± E k , ±± ± . WQ H h A T T h T T ss o s o =⇒ = − = 0 300 18 300 Δ bg kJ h kJ h (b) Clothed: 8 =. C s h T T o =⇒= ° 34 2 134 . Nude, immersed: 64 C s h T T o =⇒ = ° 34 2 316 . (Assuming T s remains 34.2 ° C) (c) The wind raises the effective heat transfer coefficient . (Stagnant air acts as a thermal insulator —i.e., in the absence of wind, h is low.) For a given T o , the skin temperature must drop to satisfy the energy balance equation: when T s drops, you feel cold. 7.31 Basis: 1 kg of 30°C stream 1 kg H 2 O(l)@30 o C 3 kg H 2 O(l)@ T f ( o C) 2 kg H 2 O(l)@90 o C (a) T f =+ = 1 3 30 2 3 90 70 o CC C b g DD (b) Internal Energy of feeds: C, liq. kJ kg C, liq. kJ kg ² . ² . U U 30 1257 90 3769 °= U V | W | (Table B.5 - neglecting effect of P on ² H ) Energy Balance: -= + + == = QW U E E U pk QW E p E k ΔΔ Δ Δ = = 0 0 ⇒− = 3 1 2 376 9 0 ² (. ( . U f kg) kJ / kg kg) kJ / kg ⇒= = ° ² .. UT ff 2932
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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