2 4 Shell and Tube Heat exchanger engineeringresource

2 -4 Shell and Tube Heat exchanger engineering-resource. com

Outline 2 -4 Shell and tube heat exchanger l Why we use it ? l Problem 8. 1 l engineering-resource. com

Problem Statement l 33, 114 lb/hr of n-butyl alcohol at 210 0 F is to be cooled to 105 0 F using water from 95 to 115 0 F. Available for the purpose is a 19¼ in. ID, twopass shell exchanger with 204 tubes ¾. OD , 16 BWG, 16’ 0’’ long on 1 -in. square pitch arranged for four passes. Vertically cut baffles are spaced 7 in. apart. Pressure drops of 10 psi are allowable. l What is the Dirt factor ? engineering-resource. com

SOLUTION engineering-resource. com

Data Available Shell Side Data l l l Inside Shell Diameter Number of Passes Baffle spacing Baffle type Allowable Pressure Drop = = = engineering-resource. com 19¼ in 2 7 in Vertically Cut 10 psi

Data Available Tube Side Data l l l l Outside Diameter of Tubes BWG Length of tubes Tubes Pitch Number of tubes Number of tube passes Allowable Pressure Drop engineering-resource. com = = = = ¾ in 16 16’ 0’’ 1 in. Square 204 4 10 psi

Location of Fluids Tube Side Fluid l As water has more scaling tendency than n-butyl alcohol that is why it is taken in tube side Shell Side Fluid l n-butyl alcohol certainly engineering-resource. com

Data Available Hot Fluid (n-butyl alcohol) l l l Inlet temperature (T 1) Outlet temperature (T 2) Mass Flow rate (mh) = = = 210 0 F 105 0 F 33114 lb/hr Cold Fluid (Water) l l Inlet temperature (t 1) Outlet temperature (t 2) engineering-resource. com = = 95 0 F 115 0 F

Diagram mh = 33114 lb/hr l (n-butyl alcohol) 210 0 F 105 0 F l (Water) 115 0 F 95 0 F T 1 1 Temperature Profile t 2 Tx 2 4 T 2 3 2 t 1 1 engineering-resource. com L

Step #1 Heat Duty Qh l mh l Cph = = = mh. Cph(T 1 - T 2) (1) 33, 114 lb/hr 0. 69 Btu/lbo. F (from fig. 2) Qh = = 33114*(0. 69)*(210 -105) Btu/hr 2399109. 3 Btu/hr l l engineering-resource. com

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Step # 1 contd. Mass flow rate of water l As Qh = Qc l mc = Qh / {Cpw*(t 2 – t 1)} l = 2399109. 3 / {1*(115 - 95)} = 119955. 46 lb/hr engineering-resource. com

Step # 2 LMTD Calculation l (n-butyl alcohol) l (Water) LMTD = l 210 0 F 115 0 F (T 1 -t 2) – (T 2 -t 1) 105 0 F 95 0 F = ln(T 1 -t 2)/(T 2 -t 1) (210 – 115 ) – (105 - 95) ln(210 – 115 )/(105 - 95) = 37. 75 0 F engineering-resource. com

True temperature Difference l Δt = FT * LMTD l R = S = = FT l Δt = = = T 1 – T 2 = 210 - 105 t 2 – t 1 115 – 95 5. 25 t 2 – t 1 = 115 - 95 T 1 – t 1 210 – 95 0. 174 0. 95 (from fig 19) 0. 95 * 37. 75 = 35. 860 F l l engineering-resource. com

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Step # 3 Tc and tc l These liquids are not viscous and the viscosity correction will be negligible l (μ / μ w )s = ( μ / μ w )t = l Average temperatures can be used engineering-resource. com 1

Step # 4 a Shell Side Calculations l Hot Fluid (n-butyl alcohol) l Flow area (as) = I. D*C*B n*PT*144 l as = (19. 25)*(7) (2)*(1)*144 l = 0. 117 ft 2 engineering-resource. com

Step # 5 a Mass velocity l Gs = W/as = 33114 0. 117 l = 283025. 6 lb/hr. ft 2 engineering-resource. com

Step # 6 a Reynold Number Res l Res = De * G s / μ l De = 4*(PT 2 – (3. 14/4)*do 2) 3. 14 * do = 4 * (12 – (3. 14/4)*0. 752) 3. 14 * 0. 75 = 0. 95/12 = 0. 0789 ft from figure 14 lμ = 1 cp * 2. 42 = 2. 42 l engineering-resource. com

Step # 7 a l j. H = j. H Factor from figure 28 54 Step # 8 a ho = k l ho = = l l = j. H * (k / De) * (C μ / k)1/3 from Table 4 0. 096 Btu/ft. 0 F 54*(0. 096 / 0. 0789)*(0. 69*2. 42/0. 096)1/3 170 Btu / hr. ft 2. 0 F engineering-resource. com

Step # 4 b Tube Side Calculations l Tubes flow area from Table 10 l at = 0. 302 in 2 / tube = 204 * (0. 302) / (144 * 4) = 0. 1069 ft 2 engineering-resource. com

Step # 5 b l Mass velocity Gt l Gt = = = w/at 119955. 46 0. 1069 1122127. 78 lb / hr ft 2 engineering-resource. com

Tube Side Velocity l V = = Gt / p 1122127. 78 62. 5 *3600 4. 987 fps = OR = 1. 52 ms-1 engineering-resource. com

Step # 6 b Reynold Number Ret l Ret = di * Gt / μ from figure 17 lμ = 0. 7 * 2. 42 = from table 10 l di = 0. 620 in = l Ret = 34180. 5 l engineering-resource. com 1. 694 lb / ft hr 0. 0516 ft

Step # 7 b Tube side heat transfer coefficient hi from Figure 25 l hi = 1240 Btu / hr ft 2 0 F l hio = 1240 * ID / OD = 1240 * 0. 620 / 0. 75 = 1025 Btu / hr ft 2 0 F l engineering-resource. com

Step # 8 Clean Overall Coefficient Uc l Uc = hio * ho hio + ho l = 145. 8 Btu / hr ft 2 0 F engineering-resource. com

Step # 9 Design Overall Coefficient UD from Fourier Equation l UD = Q/A. Δt From Table 10 l a’’ = 0. 1963 *ft 2/ lin. Ft l. A = 204 * 0. 1963 * 16 = 640. 72 ft 2 l UD = 2399109. 3 / 640. 72 * 35. 86 = 104. 47 Btu / hr. Ft 2. 0 F l engineering-resource. com

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Step # 10 l Rd = Uc-Ud Uc*Ud = 145. 8 - 104. 47 145. 8 * 104. 47 =. 0027 hr ft 2 0 F/Btu engineering-resource. com

Step # 11 a l Pressure drop: (on shell side For Res= 9356 l (from fig. 29) f=0. 0035 ft 2/in. 2 l No of crosses, N+1=12 L/B N+1=(12 × 16)/7 N+1=27. 42 ( Say, 28) l Ds=19. 25 in. /12 Ds=1. 604 ft s=? engineering-resource. com l

Step # 11 a engineering-resource. com

Step # 11 a engineering-resource. com

Step # 11 a l ∆Ps = f×Gs 2×Ds×(N+1) 5. 22× 1010×De×s×Φs ∆Ps =0. 0035× 283025. 6 2× 1. 604× 28 5. 22× 1010× 0. 0789 ft ×? × 1 ∆Ps =7. 0 psi (allowable=10 psi engineering-resource. com

Step # 11 b Pressure drop: (on tube side) l Ret = 34180. 5(from fig. 26) f=0. 0002 ft 2/in. 2 ∆Pt=(f×Gt 2×L×n)/(5. 22× 1010×Ds×Φt) ∆Pt= 4 psi Gt=973500, v 2/2 g=0. 13 (from fig. ) ∆Pr=(4×n×v 2)/(2 g×s) ∆Pr=3. 2 psi ∆PT=∆Pt+∆Pr=7. 2 psi(allowable=10 psi) l engineering-resource. com

Step # 11 b engineering-resource. com