Elementary Principles 182 - 6.27 Basis: 12500 L 1 mol 273 K...

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6- 16 6.27 Basis : 12500 1 273 K 103000 5285 L h mol 22.4 L(STP) 293 K Pa 101325 Pa mol / h = . n (mol/h) @ 35 o C, 103 KPa 528.5 (mol/h) @ 20 o C, 103 KPa Inlet : () 2 *o rH O o 2 35 C 0.90 42.175 mmHg 101325 Pa 0.04913 mol H O/mol 103000 Pa 760 mmHg hp y P × == = Outlet : y p P 1 20 17 535 101325 0 02270 = HO o 2 2 C mmHg 103000 Pa Pa 760 mmHg mol H O / mol * . . di Dry air balance : 1 0 04913 1 0 02270 5285 5432 −= = .. . . bg b g nn oo mol / h Inlet air : 22 4 308 101325 13500 mol h L(STP) mol K 273 K Pa 103000 Pa L/h = Total balance : 5432 5285 147 22 . =+ = mol / h Condensation rate : 14 7 18 02 1 0 265 . . . mol h g H O 1 mol H O kg 1000 g kg / h 2 2 = 6.28 Basis : 10000 1 492 29 8 24 82 ft lb - mol 359 R 550 R in Hg 29.92 in Hg lb - mol / min 3o o ft (STP) 3 min . . = n 1 lb-mole/min n 1 lb-mole/min 24.82 lb-mole/min 40 o F, 29.8 in.Hg 65 o F, 29.8 in.Hg 90 o F, 29.8 in.Hg y 1 [lb-mole H 2 O(v)/lb-mole] y 1 [lb-mole H 2 O(v)/lb-mole] y 0 [lb-mole H 2 O(v)/mol
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