Kaplan Chemical Engineering Problems & Solutions, 4 th Edition Errata

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Reynard Warren
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1 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 = All water vapor in the flue gas is from combustion because air used is moisture free and fuel is also moisture free Then moles of water vapor in flue gas = 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 = 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 = x kg mol = ( x) kg mol Hence N in air fed to combustion chamber = [3.76( x)] kg mol N is a key component and assumed inert. Therefore N + O in flue gas = 3.76( x) + x kg mol = x kg mol In 00 kg mol of flue gas, N + O Thus, equating the two x = From which x % excess air used = (4.56/5.35)(00) = 9.8 %

2 .3 Mol% of N in flue gas: Kaplan Chemical Engineering N in air = 3.76( ) = kg mol On the basis of 00 kg mol, the flue gas composition is Component kg mol mol % CO H O O N Total Mol % of N in flue gas 74.8%.4 Average molecular weight of the flue gas: Page 7: Average molecular of flue gas = = 8.7 Exhibit.5: Add F as shown below Page 7: Example.8 solution: Line : Add wet between Total and air t Change to 544. Line 6: Change to 544. Change 5.7 to Line 8: Change 5.7 to Line 9: Change 8. to.5 a e T ta wet air

3 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

4 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: a nd line: Insert = sign before 64 Item Phase Change from liquid to solid: sert betwee a ca K

5 Page 3: Item cooling of solid: nd 73 line of solution: Insert 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 by th line: Change to (Retain sign) Change 454 to th line of solution: change to Page 3: Example 3.4: Page 33: Page 35: 4 th line up from bottom of solution: Change.34 to.4 Change to rd line up from bottom of solution: Change to nd line up from bottom of solution: Change to.38 Last line: Change to Problem 4.3, Equation for A: Change to After equation for t, insert the missing lines as follows = s = 4.8 h

6 Page 36: Page 46: Page 59: Page 60: Page 6: Page 6: nd line up from bottom: Change to Bottom line: Change 58.3 to 57.9 Calculation of B: Insert brackets after division sign as shown below: G s = 50,000/( B) = 46,968 dv After equation for D.4V, in the equation, 50 0, C a e t 5 dv 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 to 333 Problem 7., line : Insert = sign before.0 C CV in V A CV in the denominator.

7 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 m s = 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

8 Page 34: Page 4: Problem 5., line 9, Equation for I H : Insert = sign before Problem 5., first expression for, nd line: Insert = sign before 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 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

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