Heat Transfer Equipment Design [PDF]

Citation preview

Examp1e 12.1. Ca1cu1ation of a Horizontal n-propanol Condenser. A horizontal 1-2 condenser is required for the condensation of 60,000 lb/hr of substantially pure n-propanol (propyl alcohol) coming from the top of a distilling column operating. At ;lì psig, at which pressure it boils at 244°F. Water at 85°F will be used as the cooling medium. A dirt factor of 0.003 is required with allowable pressure drops of 2.0 psi for the vapor and 10.0 psi for the water,Because of the location of the condenser, assume that 8’0” tubes are used. Tubes are to be % in. OD, 16 BWG on 1-f-ìn. triangular pitch. Solution: (1) Heat balance: n-Propanol, Q = 60,000 X 285 =



17,100,000 Btu/hr



Water, Q = 488,000 X 1(120 — 85) = (2) Δt



244 244 0 Δt



17,100,000 Btu/hr



Higher Temp Lower Temp Differences



120



124



85



159



35



35



= LMTD = 141°F



The exchanger is in true counterflow, since the sheìl-.side fluid is isothermal. (3) Tc and tc: The influence of the tube-wall temperature is included in the condensing film coefficient. The mean ta = 102.5°F can be used for tc. (a) Assume UD = 100: Condensing film coefficients will generally range from 150 to 300 Assuming a film coefficient of 1000 for water Uc will range from 130 to 230.



EQ Number of tubes = EQ (b) Assume four tube pass: The quantity of water is large, but the condenser will have a large number of tubes, making a two-pass assumption inadvisable. From the tube counts (Table 9): 773 tubes, four passes, 3/4 in. OD on 15/6 in. triangular pitch . Nearest count: 766 tubes in a 31 in. ID shell (c) Corrected coefficient UD : EQ EQ Hot fluid: shell side, n-propanol



Cold fluid: tube side, water



(4’) Assume maximum baffle space. This will be 321/2,31,and 321/2 in. equal to 96 in.or 2 baffles and 3 crosses for side-to-side flow. As = ID X C’B/144Pt (Eq. (7.I) = 31 X 0.1875 X 31/144 X o.937



(4) Flow area, at= 0.302 in2 [Table 1O] = Ntat/144n [Eq. (7.48)] = 766 X 0302/144 X 4 = 0.402 ft2



= 1.34 ft2 (5’) Gs = W/as (for pressure drop only)



(5) Gt = w/at = 488,000/0.402 = 1,210,000 lb/(hr)(ft2)



[Eq. (7.2)] Vel, V = Gt/3600p = 60,000/1.34 = 1,210,000/3600 X 62.5 = 5.42 fps = 44,700 lb/(hr)(ft2) (6) At ta = 102.5°F, Loading, G’’=



W/LNt2/3 [Eq. (12.43)] μ = 0.72 X 2.42 = 1.74 lb/(ft)(hr)



= 60,000/8 X 7662/3 D = O.62/12 = 0.0517 ft [Table 10) =



89.3 (lb)/(hr)(lin ft) Ret = DGt/μ (for pressure drop only)



Assume h- = h0 = 200 = 0.0517 X 1,210,000/1.74 = 36,200 From (10) hio = 1075 hi = 1300 [Fig. 2.5] EQ hio =



hi X ID/OD [Eq. (6.5)]



= 102.5 + 200/1275(244 — 102.5) =1300 X O.62/0.75 =125°F = 1075 Btu/(hr) (ft2) (°F) tf = (Tv + tw)/2 [Eq. (12.19)) = (244 + 125)/2 = 184.5°F kf = 0.094 Btu/(hr (ft2)(°F/ft) [Table 4] sf= 0.80 [Table 6] μf = 0.62 cp [Fig. 14] From Fig. 12.9 or Eq. (12.42) h- = h0 = 172 Btu/(hr)(ft2)(°F)



Based on 172 instead of the assumed 200 a new va1ue of tw and tf could be obtained to give a more exact value of h based on fluid properties at a value of tf, more nearly correct. It is not necessary in this example because the condensate properties will not change material1y. Pressure Drop (1`) At Tv = 244°F



(1) For Ret = 36,200,



μvapor = 0.010 X 2.42



F=



= 0.0242 lb/(ft)(hr) [Fig. 15] De = 0.55/12 Res = DeGs/μ = 0.0458 X f = 0.00141



= 0.0458 ft [Fig. 28] [Eq. (7.3)] 44,700/0.0242 = 84,600 ft2/in.2 [Fig. 29]



0.00019 ft2/in.2 [Fig. 26]



(2’) No. of crosses, N +, 1 = 3



(2) ΔPt = EQ



MoL wt. = 60.1



=



[Eq. (7.45)]



EQ = 3.3 psi



Density. EQ = 0.238 lb/ft3 s = 0.238/62.5 = 0.00382 Ds =



31/



12



= 2.58 ft (3)ΔPr = (4n/s)(V2/2g’) [Eq. (7.46)]



(3’) ΔPs = EQ



[Eq. (12.47)] =



(4 X 4/1)0.20 = 3.2 psi



EQ =



1.2 psi



(4) ΔPT = ΔPt + ΔPr [Eq. (7.47)] = 3.3 + 3.2 = 6.5 psi



(13) Clean overall ccefficient Uc: EQ (14) Dirt factor Rd: UD from (c) = 101 EQ



Summary 172



h outside



UC



148.5



UD



101



1075



Rd Calculated 0.0032 Rd Required



0.003



1.2



Calculated ΔP



6.5



2.0



Allowable ΔP



10.0



The first trial exchanger is satisfactory and will be Shell side ID = 31 in. Baffle space = Passes = 1



Tube .side 31 in.(approx.)



Number and length = 766,8’0” OD, BWG, pitch = 3/4 in.,16 BWG,15/16-in Passes = 4



It is interesting at this point to compare a vertical condenser with the horizontal condenser which fulfilled the process conditions of Example 12. The horizontal and vertical condensing film coefficients are both affected by W and Nt, and the best basis for comparison is obtained when



the number of tubes in both condensers is the same. To this end a vertical condenser will be assumed which employs the same tube count as that of the preceding example except that the tube length may be 12 Or 16 ft as needed to account for the lower coefficients obtained in vertical condensation.



Example 12.2. Design of a Vertical n-Propanol Condenser. Process conditions will be taken from Example 12.1. To prevent corrosion of the shell the water will flow in the tubes. Solution: (1) Heat balance: same as Example 12.1, Q = 17,100,000 Btu/hr (2) Δt: same as Example 12.1, Δt = 141°F (3) Tc ànd tc: same as Example 12.1 I In condensation calculations the omission of the resistance of the tube metal may introduce a significant error and should be checked. Trial: (a) Assume UD = 70. The equation for the condensing film coefficient gives greater values for horizontal tubes than for vertical tubes. It will consequently be necessary to reduce the value of UD. EQ Nearest common tube length: EQ (b) The layout of Example triangular pitch and four (c) Corrected coefficient A = 766 X 12’0” X 0.1963 EQ



12.1, using 3/4 in. OD tubes on 15/16.in. passes will be retained for comparison. UD: = 1805 ft2



Hot fluid: shell side, n-propanol



Cold fluid: tube side, water



(4’) See (1’) under Pressure Drop



(4)—(1O) Same as Example 12.1



De = 0.75/12 = 0.0625 ft



hio = 1075



Loading, G’ = W/3.14NtDo (Eq. (12.36)] = 60,000/3.14 X 766 X 0.0625 = 399 lb/(hr) (lin ft) Assume h- = h0 = 100 EQ



[Eq. (5.31)]



= 1O2.5 + (100/1175(244 — 102.5) = 114.5°F tf = (Tv + tw)/2 [Eq. (12.19)) = (244 + 114.5)/2 =



179°F



kf = 0.0945 Btu/(hr) (ft2) (°F/ft) [Table 4] sf = 0.76 μf = 0.65 cp (4G’/μ = 1025)(Fig. 14) From Fig. 12.9 or Eq. (12.41), h- = h0 = 102 Btu /(hr)(ft2)



Pressure Drop (1’) It arrange minimum N + 1 =



will be necessary to the 12-ft bundle into a number of bundle crosses or 5. The spacing will be .



(1) Same as Example 12.1 except for the tube length.



B =144/5 =29in. as = ID X c’B/144PT [Eq. (7.1)] = 31 X 0.1875 X 29/144 X 0.937 = 1.25. sq ft G = W/as [Eq. (7.2)] = 60,000/1.25 = 48,000 lb/(hr(ft2) At Tv = 244°F, [Fig. 15] μvapor =0.010X2.42.=0.0242lb/(ft)(hr) De = 0.55/12 = 0.0458 ft [Fig. 28] Res = DeGs/μ [Eq. (7.3)] = 0.0458 X 48,000/0.0242 91,000 f = 0.0014 ft2/in.2 (2’) No. of crosses, N + 1 = 5 [Eq. (7.43)] s = 0.00381 [Example 12.1) Ds = 31/12 = 2.58 ft (3’) ΔPs = EQ



[Eq. (12.47)]



EQ =



(2)ΔPs = EQ EQ =



(3) ΔPr as in Examp1e 12.1 = 3.2 psi (4) ΔPT = ΔPs + ΔPT = 5.0 + 3.2 = 8.2 psi



2.3 psi



The pressure drop is high, and if it cannot be compensated for by elevating the condenser, it will be necessary to use the half-circle support baffles as in Exmp1e 7.8 (13) Clean overall coefficient Uc: EQ (14) Dirt factor Rd: UD from (c) = 67.2 EQ Summary 102



h outside



UC



93.2



UD



67.2



1075



Rd Calculated 0.00415 Rd Required



5.0 psi



0.0030



2.3



Calculated ΔP



8.2



2.0



Allowable ΔP



10.0



Discussion. The condenser is somewhat safe from a heat-transfer standpoint but exceeds the allowable pressure drop, although not seriously. The advantage of horizontal condensation may be observed from the Uc of 148.5 in the horizontal condenser as compared with 92 in a vertical condenser in an identical service. The vertical condenser has advantages, however, when the condensate is to be subcooled. The final vertical condenser is Shell side ID = 31 in. Baffle space = Passes = 1



Tube .side 29 in.



Number and length = 766,12’0” OD, BWG, pitch = 3/4 in.,16,15/16-in Passes = 4