As a chemical engineer, you are employed in a power plant industry and is assigned in the supervision oF the boiler operation. The boiler uses coal as its Fuel, supplied by a third—party supplier. The stack gas produced upon combustion of the coal analyzes 11.2% C02, 8.7% CO and 8.3% 02 at QBBDF. The air supplied in the boiler is saturated with water vapor and enters at 73°F. Assume a barometric pressure of T58 mmHg. a. To be able to complete your report around the boiler, you need to know the ratio of net H to carbon in the coal, as well as the ultimate analysis of the coal. Initial data From the coal says it contains 76.1% carbon, 6.9% moisture and 8% ash. Assume negligible nitrogen and sulfur content. b. How much excess air is supplied in the boiler? Also, a value is used to control the air entering the boiler, how much air is needed in Ft3flb of coal? c. In one day, the boiler makes use of 56 tons of coal, how many Ft3 of stack gas is produced? Establish basis as 168 lbmols of dry stack gas.

a. net H / C = 0.0382

ultimate analysis :

% C = 70.1 %

%H = 4.81 %

%O = 17.09 %

% ash = 8%

%N,%S = 0

b.   % excess air = 59.8207 %

ft³ air / lbm coal = 201.08 ft³ / lbm

c. volume of stack gas = 51.8835 x 10⁶ ft³ stack gas

Step-by-step explanation

GIVEN : 73 " F = 22.78 C & 750 mmig REQD AIR 100 % KH Stack Gas A . Net H / C RATIO & ULTIMATE ANALYSIS COAL FURNACE 900 F = 482.22 C 70 . 1 % C 11. 2 / CO2 6.9 % MoisturE 0.7% CO 8 % ACH 8.3 1 02 SOVN: @ Stack Gas Basis: 100 lbmoles DRY stack gas ncoz = 11. 2 lbmole n CO = 0-7 lbmode n 02 = 8.3 lbmole n N 2 = 100 - 11. 2 - 0.7 - 8.3 = 79. 81bmole total nc @ stack gas : 11.2 + 0.7 = 11. 9 lbmole total n 02 @ stack gas = 8:3+ 11:2+ 0.7 = 19:85 lbmole C Air inlet nN2 : nN 2 @ stack gas ( since N 2 @ coal zo ) n N2 - 79. 8 lbmole 1 02= 79.8 lbmole N2 x 0-21 1bmole 02 = 21. 213 1bmole 0 79 lbmote NZ n Air = noz+nN2: 79.8+ 21. 213 = 101. 013 lbmole @ 22 78 C , PK PHIO = 20 . 72 mmlg HO = PPHZO At120 INAIR = 101. 013 lbmole Air & _ 20 . 72 - 2.87 lbmole 710 - 20-72 102 used for Net H = 02 air - 02 stackgow = 21.213 - 19:85 = 1:343 1bmole Net H = 1.343 /bmode 02x 2atomox mole H20 2atom! = 5.452 1bmole # 16mole 02 latomo * Imole Hzo

1. excess air = 02 air - Oz theo * 100% On theo 02 theo : C + 02 CO2 2 H 2 + 02 2 H 20 FOR Carbon 02 = 142. 8 1bmc x /molec 1 mole 0 2 = 11 9/ mole Oz 12 1buc x Imole c For netht 02 = 5. 452 /bm netHy atom |mole #2 Imole Oz = 1. 34 3 /bmole On 2 atom I 2mole # 2 Oz theo = 11.91 + 1. 343 = 13. 273 16mole . excess air = 21.213 - 13-273 59- 8207 / 13. 273 ANSWER :. flair lbcoal nair = noz+ nN2+ wl/20 = 101.013+ 2:87 = 103. 88 lbmole Stair nkT 103. 88 bmotel [0. 7302 51 3 - atm ] [533x] Homolele J = 0 . 987 atom lbcoal = 203. 31 201. 08 ft air 203. 71 1bm /16m coal ANSWER C - fis of stack gas in so tons n Stackgow = 10olbmole Dry stack gas + n H20 = loot tho air + CW + Moisture = 100+ 25-10 + 14.04 + 2.87 = 105 . or lbmoles 18 18 Vstack gow = lov. of Hemoles x 0.7302 ft zatin Ibmotek 6.98 Fatm + SOtony 2080/box coat - 51 .8835x 10 f1 3 203- 71 1bmcoat ANSWER

Please see attached file