By Richard M. Felder, Ronald W. Rousseau

Achieve a greater knowing of chemical processes.

This textual content provides you with a pragmatic, informative advent to chemical techniques. This third version has been thoroughly revised to supply you with elevated readability, including:

  • Hundreds of latest and revised difficulties and new case reviews disguise a broader spectrum of chemical engineering applications.
  • Guidance for fixing difficulties that require unfold sheeting and equation-solving software.
  • A CD-ROM that offers an energetic studying setting. With this software program, scholars reply to questions and obtain quick suggestions, discover adaptations in approach parameters and spot the impact in their alterations on method operations, and more.
  • 2005 version icons within the textual content margin tell you while it really is such a lot useful to exploit the ICPP CD-ROM and the scholar Workbook.

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06243C A′ = 1334 . 419t ′ 60 drop primes ⇒ b g b C A mol / L = 214 . 00 atm 101325 . 4 N cm2 2 2 1 atm 100 cm (d) 280 cm Hg 10 mm 101325 . 696 psi b (h) 325 mm Hg − 760 mm Hg = −435 mm Hg gauge (i) Eq. 81 m / s2 m3 h (m) 1N 1 kPa 2 1 kg ⋅ m / s 103 N / m2 . Pg (kPa) ⇒ h (m) = 0111 h Pg Pg = 68 kPa ⇒ h = 0111 . 92 × 1000 IJ FG K H IJ K kg 16 2 3 × 7 . 55 × π × m = 14 . 36 . 8066 m 150 m 12 m2 1N s2 1002 cm2 1 kg ⋅ m / s2 FG H IJ K lb f 022481 . 154 N 65 cm2 4 N = 100 . × 10 × = 2250 lb f 1N cm2 14 .

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 . − 1856 . °C dT 2508 = = 326 . °C / s dt 20 s T = KR n g = 1184 . 0 . ln K = ln 1100 . − 1184 . (ln200 . ) = 1154 . ⇒ K = 3169 . ⇒T = 3169 .

95 + 27316 . 016034gb9119. 93 + 27316 . 33 kmol min (e) V3 = (f) b g b gb g n3 T2 + 27316 . 33 2251 . + 27316 . 707 110. 61 127. 1 a. Continuous, Transient b. 00 c. 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.

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