Handbook of Chemical Engineering Calculations was published by Oya FX Trading & Investments on 2015-03-14.

However,that figure is based on Ad = 0, which assumes no interfacial shear. For equivalent solids suspension, an exponent of 3/4 will be used to scale up from the small-scalesolids suspension speed of 465 r/min: 4 3/4NL = 465 40 = 82.7 r/min5. Now, Q = 500,000× (0.5)(185.8) = 46,450,000 Btu/h (13,610,000 W).Related Calculations. 7.15. 2.

The thermal effectiveness of a 2–4 multipass heat exchanger can be deter-mined from Fig. Calculate the terminal Reynolds number 4Γ/µ. A conservative approach is to calculate a superficial massvelocity assuming the condensate fills the entire tube and use the equations presented above forsingle-phase heat transfer inside tubes. The mass velocity G will be doubled because the exchanger is to be converted to two passes on the tube side. Thus, GL = WL1 = 7 = 3338.8 lb/(h)(ft2) a (0.62/12)2(π/4) Gv = Wv1 = 119 = 56,759.2 a (0.62/12)2(π/4) G E = G L + Gv(ρL /ρv)1/2 = 3338.8 + (56,759.2)(50/1.0)1/2 = 3338.8 + 401,348.1 = 404,686.9 lb/(h)(ft2) Di G E = (0.62/12)404,686.9 = 83,635.3 µ 0.25. A 6000-gal batch in a 10-ft-diameter tank with a dished head will put 414 gal in the 16-in-deep dished head (see Table 12.2).

Consequently, this layer of liquid must not exceed values assumed without beingappropriately accounted for.

Because P = (t2 − t1)/(T1 − t1) = 0.39, t2 = t1 + P(T1 − t1) = 68 + (0.39)(500 − 68) = 236.5◦FNow, by definition, wc(t2 − t1) = W C(T1 − T2), and R = wc/(W C), so T2 = T1 − R(t2 − t1). Here are the calculationsfor three of the nine increments:First Incrementa. Since Rev is less than 20,000 and ReL greater than 5000, ha = 5.03(k/Di )(cµ/ k)1/3(Di G E /µ)1/3 = 5.03[0.08/(0.62/12)]1.1979(16,609.4)1/3 = 238.0 Btu/(h)(ft2)(◦F) [1353 W/(m2)(K)]5. Therefore, ρL2 g L 1/3 nµWc,av hc = 0.767k = 0.767(0.08) 502(4.18 × 108)(1.0) 1/3 . Check Pages 251 - 300 of Handbook of Chemical Engineering Calculations in the flip PDF version. For the stratified-flow assumption, the further assumptionis made that the rate of condensation on the stratified layer of liquid running along the bottom of thetube is negligible. I did not think that this would work, my best friend showed me this website, and it does! Determine Rev and ReL . Determine Rev and ReL . Physical Properties ρL = liquid density = 87.5 lb/ft3 (1401.6 kg/m3) ρv = vapor density = 1.03 lb/ft3 (16.5 kg/m3) µ = liquid viscosity = 0.7 lb/(ft)(h) (0.29 cP) c = liquid specific heat = 0.22 Btu/(lb)(◦F) [0.92 kJ/(kg)(K)] k = liquid thermal conductivity = 0.05 Btu/(h)(ft)(◦F) [0.086 W/(m)(K)], Handbook of Chemical Engineering Calculations. Determine Rev and ReL . Like this book?

Son: My Life, My City, My Food PDF eBook L.A. Please Note: There is a membership site you can get UNLIMITED BOOKS, ALL IN ONE … Heat-transfer coefficients for falling films can also be predicted from Fig. Thus, Rev = Di Gv(ρL /ρv)1/2 = (0.62/12)401,348.1 µ 0.25 = 82,945.3 ReL = Di G L = (0.62/12)3338.8 µ 0.25 = 690.0d. Since Rev is greater than 20,000 but ReL less than 5000, ha = 5.03(k/Di ) (cµ/k)1/3(Di G E /µ)1/3. Thus, Ad = 0.250µ1L.173µv0.16 g2/3 Di2ρL0.553ρv0.78 (0.250)(0.72)1.173(0.0313)0.16 = (4.18 × 108)2/3(0.62/12)2(60)0.553(0.0374)0.78 = 8.876 × 10−56. The remaining 6000 − 414 = 5586 gal will fill the vertical-wall portion of the tank at a rate of 48.9 gal/in, for a total liquid depth of (5586 gal)/(48.9 gal/in) + 16 in = 130 in (3.3 m).

Calculate hc, the condensing coefficient if stratified flow is assumed. Suppose that by visual observation and sample analysis, a pitched-blade im- peller with a diameter D of 4 in (with four blades, each 0.8 in wide) operating at a speed N of 465 r/min was found to produce the level of agitation necessary for uniform suspension to three-fourths the total liquid level. Thus, 1 lb of the suspension will have a volume of 0.7/7.42 + 0.3/31.68 = 0.1038 gal, or a density of (1 lb)/0.1038 gal = 9.63 lb/gal, which is the same as a specific gravity of 9.63/8.337 = 1.16 (1160 kg/m3). Now, G L = 119/[(0.62/12)2(π/4)] = 56,759.2 lb/(h)(ft2); Gv = 7/[(0.62/12)2(π/4)] = 3338.8 lb/(h)(ft2); G E = G L + Gv(ρL /ρv)1/2 = 56,759.2 + 3338.8(50/1.0)1/2 = 56,759.2 + 23,608.7 = 80,367.9 lb/(h)(ft2); and Di G E /µ = (0.62/12) 80,367.9/0.25 = 16,609.4.c. Calculate h. For Ad = 8.876 × 10−5 and 4 /µ = 1711.3, h/ h0 is approximately 1.65 fromstep 1. The following equations can be used to predict heat-transfer coefficients for condensation insidehorizontal tubes: For stratified flow, ρL2 g L 1/3 nµW hc = 0.767kFor annular flows,a. This approach can be used for condensing on the outside of vertical tubes. 12.2 need be applied tothe power number. Welcome to Basic Principles and Calculations in Chemical Engineering.

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