Elementary Principles 262 - 7.35 Basis: Given feed rate 200...

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7-17 7.35 Basis: Given feed rate 200 kg H 2 O(v)/h 10 bar, sat’d, ± . H = 2776 2 kJ / kg ² [ n 3 kg H O(v) / h] 2 1 0 b a r , 2 5 0 o C, ± H = 2943 kJ / kg ² [ ± n H 2 3052 kg H O(v) / h] 10 bar, 300 C, kJ / kg 2 o = ² Q (kJ / h) ± H from Table B.6 (saturated steam) or Table B.7 (superheated steam) Mass balance: 200 23 += ²² nn (1) ΔΔ Δ ² , ² , ² ² . ² , ² EEW Kp QH n n Q Energy balance: in kJ h = == 0 32 2943 200 2776 2 3052 bg b gbg (2) (a) ² n 3 300 = kg h () ² 1 100 2 n = ² . 2 225 10 4 Q kJ h (b) ² Q = 0 ,( ) ² 12 306 2 n = kg h , ² n 3 506 = 7.36 (a) T saturation @ 1.0 bar = 99.6 °C ⇒= T f 99 6 . D C H O (1.0 bar, sat'd) kJ / kg, kJ / kg H O (60 bar, 250 C) kJ / kg 2 2 = = ± . ± . . HH lv 417 5 26754 10858 D Mass balance: kg Energy balance: kg)(1085.8 kJ / kg) = 0 (2) , mm H mH vl EQEW vv ll = ⇒+−=+− = 100 1 0 100 0 11 ² ±±±±± ( ² , , ² Δ (,) .. 70 4 29 6 kg, kg y v 29 6 0296 . . kg vapor 100 kg kg vapor kg (b) is unchanged. T The temperature will still be the saturation temperature at the given final
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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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