Page : Page 6 and 7: Example., Line : Change 50 to 0 Replace present solutions of problems. to.4 on pages 6 and 7 by the following revised solutions:. Carbon/Hydrogen mass ratio: CO in the flue gas is 0 mol%, therefore CO in the flue gas = 0.(00) = 0 kg mol Mol fraction of water vapor in flue gas =.35/6 = 0.065 All water vapor in the flue gas is from combustion because air used is moisture free and fuel is also moisture free. 0.065 Then moles of water vapor in flue gas = 0 0. 65 kg mol H O 0.0 Amount of carbon in 0 kg mol of CO = 0() = 0 kg Amount of H in 0.65 kg mol of H O = 0.65() =. kg Hence carbon/hydrogen mass ratio = 0/.3 = 5.634. Percent excess air: O used up for formation of CO = 0 kg mol of O O consumed for formation of H O = 0.5(0.65) = 5.35 kg mol of O O remaining unused (not known). Assume x kg mol Therefore total oxygen used for combustion = 0 + 5.35 + x kg mol = (5.35 + x) kg mol Hence N in air fed to combustion chamber = [3.76(5.35 + x)] kg mol N is a key component and assumed inert. Therefore N + O in flue gas = 3.76(5.35 + x) + x kg mol = 57.6 + 4.76x kg mol In 00 kg mol of flue gas, N + O Thus, equating the two 57.6 + 4.76x = 79.35 From which x % excess air used = (4.56/5.35)(00) = 9.8 %
.3 Mol% of N in flue gas: Kaplan Chemical Engineering N in air = 3.76(5.35 + 4.565) = 74.79 kg mol On the basis of 00 kg mol, the flue gas composition is Component kg mol mol % CO 0.00 0.00 H O 0.65 0.65 O 4.56 4.56 N 74.79 74.79 Total 00.00 00.00 Mol % of N in flue gas 74.8%.4 Average molecular weight of the flue gas: Page 7: Average molecular of flue gas =0 44 + 0.65 8 + 4.56 3 + 74.79 8 = 8.7 Exhibit.5: Add F as shown below Page 7: Example.8 solution: Line : Add wet between Total and air t Change 505.5 to 544. Line 6: Change 505.5 to 544. Change 5.7 to 54.05 Line 8: Change 5.7 to 54.05 Line 9: Change 8. to.5 a e T ta wet air
Page 8: Page 3: Page 3: Problem. solution, 5 th line: Change 557. to 357. Example 3.: Line : Replace 000 by 900 Data table, row, column 3: Change 00 to 900 Data table, row 5, last column: Change.60 to.600 Data table, row 6, last column: Change.6 to.60 Examples 3.3 and 3.4: Data on SO : Insert decimal point in 755 to give 75.5 Add data above statement of example 3.3 as follows: Entropy of SO at 5 0 C ca K Example 3-3: Line : Change 538 to 537 Line : Change -0 to -00
Page 4: Example 3. solution: Exhibit 3. Page 5: Line 8, equation for Q: Make Add R as subscript to H in H Problem 3., last line on page: Change 06 to 0 6 (i.e. make 6 exponent of 0) Page : Item Gas Cooling: Insert sy b bef re S t ive S Item Gas Condensation: Insert = between Item Liquid Cooling: 5960 68 a nd line: Insert = sign before 64 Item Phase Change from liquid to solid: sert betwee 769 97.5 a ca K
Page 3: Item cooling of solid: nd 73 line of solution: Insert 4.656 between = and ln 97.5 Insert sign before.94 to give 3rd line of solution: change.34 to.4 4 th line of solution: Replace 9.778 by 5.574 5 th line: Change 9.778 to 5.574 (Retain sign) Change 454 to 454.6 6 th line of solution: change 0.4653 to 0.8076 Page 3: Example 3.4: Page 33: Page 35: 4 th line up from bottom of solution: Change.34 to.4 Change 38.85 to 38.86 3 rd line up from bottom of solution: Change 38.85 to 38.86 nd line up from bottom of solution: Change 0.8549 to.38 Last line: Change 0.358 to 0.3306 Problem 4.3, Equation for A: Change 78.45 to 78.54 After equation for t, insert the missing lines as follows = 7334.975 s = 4.8 h
Page 36: Page 46: Page 59: Page 60: Page 6: Page 6: nd line up from bottom: Change 8.398 to 8.338 Bottom line: Change 58.3 to 57.9 Calculation of B: Insert brackets after division sign as shown below: G s = 50,000/(0.06667B) = 46,968 dv After equation for D.4V, in the equation, 50 0, C a e t 5 dv 900 50 After Exhibit 7.4 title, 5 th C line equation for V C, insert = sign between V and A A 0.88( s) e In 6 th line, Change 4.4 to 4.46 Delete before C V to give C V = 4.46 Line 5, Calculation of : After = sign, insert before Line 9, calculation of dv : After = sign insert before d Problem 7.5, line 6: After.8, insert = sign Problem 7.7, line : Change 4873 to 4874 Problem 7.0, line : Change 7363.6 to 333 Problem 7., line : Insert = sign before.0 C CV in V A CV in the denominator.
Page 86: Problem 0.7, Equation for B: After = sign, Replace 0.47 by 0.59 Page 87: From top, 4 th line: Make i in yi subscript to y. Page 03: Page : Problem.3: t e i e be i i wit f x kg move Y to next line behind x / 60 to make Y / 60 Problem 4.: x Also insert and behind Y / 60 t ive a Y / 60 x x Line 3, Schmidt number: On RHS delete 00 and delete from 0.65 Reset ca cu ati f Sc i t s u ber as f ws: 3 0.0P g 0 kg 0.0 cp /P cp 0.0m g cm s Schmidt number, Sc = = cm D AB kg 5 m.85.6 0 3 m s 3 0.0 0 = 0. 65 5.85.6 0 h G In expression for, on RHS, delete from 0.65 k Y h G In expression for, Change 954 to 955 ky Equati f wi t e se te ce T eref re, by substituti i t e wet bu b equati : Change 954 to 955
Page 34: Page 4: Problem 5., line 9, Equation for I H : Insert = sign before 9.903 Problem 5., first expression for, nd line: Insert = sign before 6.8 0 G T Problem 5., equation for ln K: sert sign before G in 0 Problem 5.5, line 9: Change subscript r in H r to R to give H R Problem 5.5, line 9: sert betwee si a t ive 66000 and change.66 to.6658, also delete from 5.9 G RT 0 Page 45: Problem 6., Equation for cv dt dt : Insert = sign after cv dt dt t e se te ce starti wit F r stea y state e er y ba a ce, i sert dt after dt Equation for cv Page 47: d ( T T dt s ) : Insert = sign after d( T Ts ) cv dt Problem 6.3, second line: Insert before = sign the following: 3 0.5