By Richard M. Felder, Ronald W. Rousseau
Achieve a greater figuring out of chemical processes.
This textual content will give you a practical, informative advent to chemical approaches. This third version has been thoroughly revised to supply you with elevated readability, including:
- Hundreds of latest and revised difficulties and new case experiences disguise a broader spectrum of chemical engineering applications.
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Extra info for Solutions manual for elementary principles of chemical processes
Example text
00 dt dt s s s 100 . 00 kg a. Continuous, Steady State b. k = 0 ⇒ C A = C A0 c. 3 b v kg / h m a. 850m . 3 kg h Solve (1) & (2) simultaneously ⇒ m b. The flow chart is identical to that of (a), except that mass flow rates (kg/h) are replaced by masses (kg). The balance equations are also identical (initial input = final output). c. Possible explanations ⇒ a chemical reaction is taking place, the process is not at steady state, the feed composition is incorrect, the flow rates are not what they are supposed to be, other species are in the feed stream, measurement errors.
0791 . 5 0 b g . ln ΔP + 52068 . From the plot above, ln V = 04979 ml s ⇒ n = 04979 . ≈ 05 . 5 = 132 mL s = 180 . mol s . = 544° R / 18 . 48 (a) T = 85° F + 4597 . = 474° R − 460 = 14° F (b) T = −10° C + 273 = 263 K × 18 (c) ΔT = (d) 85° C 10 . °K 85° C 18 . 8° R = 85° K; = 153° F; = 153° R . °C . 8D F ΔT (D F) = . D F ⇒ ΔT (D R) = 169 . 0311FG D C IJ ; H D LK b1000 - 100gD L . 4D F ⇒ −9851 . 6D FB ⇒ 156 . 2539 b g b g ⇓ T ° C = 1810 . 596 b g b g . mV→136 . mV ⇒1856 . °C →2508 . 51 (a) ln T = ln K + n ln R n= b b g .
D. 70 The results of trials 6 and 12 are impossible since the flow rates are negative. You cannot blend a 10% H2 mixture with a 50% H2 mixture and obtain a 60% H2 mixture. Results are the same as in part c. 90 mg urea / ml Dialysate b b Dialyzing fluid 1500 ml / min a. 0 ml / min b g b g . 0 − 175 . 8 mg urea / min Urea removal rate: 190 b. c. 7 − 11. 24 a. 6 kmol / min, n3 = 561 . kmol / min n1 = b. 33 m / s 18 s 1 561 . kmol m3 1 min s A = πD 2 = ⇒ D = 108 . m 4 min 0123 . 9 C =3 b gb g Dye concentration: A = 018 .