Question

Problem 3.30 Coal Components Wet coal being used in a power plant at a rate of 11000.0 lb/min has the following composition: Component Weight % (dry basis) Ash Sulfur 7.20 3.50 5.00 Hydrogen Carbon 15.2 Nitrogen 1.60 7.50 Oxygen The wet coal contains 5.08 Ibm H20 per Ibm of dry coal. Determine the molar flow rate of each element in the wet coal other than those in the ash. S: Ib-mol/min Ib-mol/min Ib-mol/min Ib-mol/min o b -mol/min

Answer 1

→ Mass Sourate of wer cool = 11000 llom i min Uan coal wer con contains het x ve I'm & H2O in wer 5.08 16m 42 per llum og dry of dry cool in wer cool cool = 5.08 x I'm 1 min Thus, Dry coal + 420 x + 5.08 x 6.08 x x= 11000 Goog = wer cool . - 11000 liooo : 16ml min = 1809,21 lbm 1 min Theis Mass Mass Lowrate flowrate of of wet Dry coal cool = x= 5.08x = 1809.21 16m 1 min 9190.79 lom/ min Sulfur in उस coal = (3.5) * Amount of dy cool C100 = = 0.035x 63. 32235 1809.21 lbm /min Similarly, Hydrogen in coal Carbon in alcool Nitrogen is cool Oxygen in coal = (5) x 1809.21 = 90. 4605 16mimin - (15.3.) x 1809-21 - 1360. 5259 lan mun ( 1 ) 1808.21 - 28.94736 lomimum = 7.5) *1909,21 138. 69075 lomimin

in wer coal / 2 x Atonie Mass 4 wer cool Mol. wh. of H2O I ^_9190079 18 1021. 1989 16mimin o in wer wal = x wer coal 1x Atomic mass of o Mol. wt. of H2O (16 x 9190. 79 18 8169.591 lbml min cool & wer coal) is :- Thus, s mass . = - 1111. 6594 Gml min of each element (Dry 63. 32235 16mimin 90.4605 + 1021, 1989 1360. S25a 16ml min 28.94736 Iom/min 135.69075 + 8169.sal1 = N o = 8305. 28185 16m/min On Dinding AEOMIC Mortar - weights, we ger Moiar Molar flowrares Component MW (lomol l min) 32 Molar Flowrates 63-32235/32 = 1111.65941 1360. S2SP1 28.94736/in 8305. 28185/16 = 1.9788 11. 6594 113. 3771 2.6677 $19.0801