Split_00008 - 8.25 a. 5500 L(ST P)/ min CH3 OH (v) 65o C n...

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8- 8 8.25 a. 2 3 5500 L(STP) 1 mol 245.5 mol CH OH(v)/min min 22.4 L(STP) n == An energy balance on the unit is then written, using Tables B.5 and B.6 for the specific enthalpies of the outlet and inlet water, respectively, and Table B.2 for the heat capacity of methanol vapor. The only unknown is the flow rate of water, which is calculated to be 2 1.13 kg H O/min. b. kg kJ 1 min 1 kW 1.13 2373.9 44.7 kW min 60 sec 1 kJ/s Q ⎛⎞ ⎜⎟ ⎝⎠ ± 8.26 a. 100 mol/s (30 o C) 0.100 mol H 2 O(v)/mol 0.100 mol CO/mol 0.800 mol CO 2 /mol n 2 mol/s (30 o C) 0.020 mol H O(v)/mol y 2 (0.980-y 2 m 4 y 4 kg H 2 O(v)/kg humid air (1-y 4 ) kg dry air/kg humid air m 3 kg humid air/s (50 o ( 0.002 / 1.002 ) kg H 2 O(v)/kg humid air ( 1.000 / 1.002 ) kg dry air/kg humid air H 2 O(v) only Basis : 100 mol gas mixture/s 5 unknowns: n 2 , m 3 , m 4 , y 2 , y 4 – 4 independent material balances, H 2 O(v), CO, CO 2 , dry air – 1 energy balance equation 0 degrees of freedom all unknowns may be determined) ( b. (1) CO balance: (100)(0.100) = (2) CO balance: (100)(0.800) =
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