Elementary Principles 345 - 8.90 (cont'd) Product solution:...

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8-66 8.90 (cont’d) Product solution: nH n H m C dT As p ±± + z Δ 25 25 50 C bg = −× + −° ⋅° =− 0.154 .095 kmol kJ hk m o l 30 kg 3.5 kJ 50 C g C kJ h 11 7 1 1 0 25 259 4 A A . Crystals: nH n H m C dT Ap =+ z Δ hydration 25 50 (hydrate at 25 ° C , heat to 50 ° C) = ⋅− × + ⋅° 1.13 kmol 3H O s kJ m o l 154 kg 1.2 kJ C g C kJ h 2 A 366 10 50 25 36700 4 . HO , 5 0C 2 vn H n H C d T vp °= + L N M O Q P z : ± ΔΔ 25 50 (vaporize at 25 ° C , heat to 50 ° = ×+ = 0.877 kmol H O kJ h kJ h 2 4 39 10 32 4 50 25 39200 4 .. b g neglect out in Energy balance: kJ h 60 kJ h (Transfer heat from unit) Δ Δ E ii R QH n H n H b g == = −− + 259 36700 39200 2300 8.91 50 mL H SO g mL g HSO mo l 84.2 mL H O g g H O mol H O mol H O mol H SO 24 2 22 4 1834 917 0935 100 84 2 4 678 500 . . . =⇒ U V | | W | | ⇒= l ll r Ref: HO 2 , H SO @ 25 ° C ± (( ) , [ . Hl H O 15 C) kJ / (mol C)](15 25) C = 0.754 kJ / mol 2 oo o =⋅ 0 0754 ± ,. . (. . ) ( ) Hr T T kJ mol (91.7 + 84.2) g 2.43 J C 1 kJ 0.935 mol H SO
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