Heat Exchanger Design Calculation - Shell and Tube Problem Solution

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

Outline • 2-4 Shell and tube heat exchanger • Why we use it ? • Problem 8.1 engineering-resource.com

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

SOLUTION engineering-resource.com

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

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

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

Data Available Hot Fluid (n-butyl alcohol) • Inlet temperature (T1) = 210 0F • Outlet temperature (T2) = 105 0F • Mass Flow rate (mh) = 33114 lb/hr Cold Fluid (Water) • Inlet temperature (t1) = 95 0F • Outlet temperature (t2) = 115 0F engineering-resource.com

T1 1 Tx t2 2 4 T2 3 2 t1 1 Diagram mh = 33114 lb/hr • (n-butyl alcohol) 210 0F105 0F • (Water) 115 0F 95 0F Temperature Profile engineering-resource.com L

Step #1 Heat Duty • Qh = mhCph(T1 - T2) (1) • mh = 33,114 lb/hr • Cph = 0.69 Btu/lboF (from fig.2) • Qh = 33114*(0.69)*(210-105) Btu/hr = 2399109.3 Btu/hr engineering-resource.com

engineering-resource.com

Step # 1 contd. • Mass flow rate of water • As Qh = Qc • mc = Qh / {Cpw*(t2 – t1)} = 2399109.3 / {1*(115 - 95)} = 119955.46 lb/hr engineering-resource.com

Step # 2 LMTD Calculation • (n-butyl alcohol) 210 0F105 0F • (Water) 115 0F 95 0F • LMTD = (T1-t2) – (T2-t1) ln(T1-t2)/(T2-t1) = (210 – 115 ) – (105 - 95) ln(210 – 115 )/(105 - 95) = 37.75 0F engineering-resource.com

True temperature Difference • Δt = FT * LMTD • R = T1 – T2 = 210 - 105 t2 – t1 115 – 95 = 5.25 • S = t2 – t1 = 115 - 95 T1 – t1 210 – 95 = 0.174 • FT= 0.95(from fig 19) • Δt = 0.95 * 37.75 = 35.860F engineering-resource.com

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

Step # 4a Shell Side Calculations • Hot Fluid (n-butyl alcohol) • Flow area (as) = I.D*C*B n*PT*144 • as = (19.25)*(.25)*(7) (2)*(1)*144 • = 0.117 ft2 engineering-resource.com

Step # 5a • Mass velocity • Gs = W/as = 33114 0.117 = 283025.6 lb/hr.ft2 engineering-resource.com

Step # 6a • Reynold Number Res • Res = De * Gs / μ • De = 4*(PT2 – (3.14/4)*do2) 3.14 * do = 4 * (12 – (3.14/4)*0.752) 3.14 * 0.75 = 0.95/12 = 0.0789ft from figure 14 • μ = 1cp * 2.42 = 2.42 • Re = 9356 engineering-resource.com

Step # 7a jHFactor from figure 28 • jH = 54 Step # 8a • ho = jH * (k / De) * (C μ / k)1/3 from Table 4 • k = 0.096 Btu/ft.0F • ho = 54*(0.096 / 0.0789)*(0.69*2.42/0.096)1/3 = 170 Btu / hr.ft2.0F engineering-resource.com

Step # 4b Tube Side Calculations • Tubes flow area from Table 10 • at = 0.302 in2 / tube = 204 * (0.302) / (144 * 4) = 0.1069 ft2 engineering-resource.com

Step # 5b • Mass velocity Gt • Gt = w/at = 119955.46 0.1069 = 1122127.78 lb / hr ft2 engineering-resource.com

Tube Side Velocity • V = Gt / p = 1122127.78 62.5 *3600 = 4.987 fps OR = 1.52 ms-1 engineering-resource.com

Step # 6b • Reynold Number Ret • Ret = di * Gt / μ from figure 17 • μ = 0.7 * 2.42 = 1.694 lb / ft hr from table 10 • di = 0.620 in = 0.0516 ft • Ret = 34180.5 engineering-resource.com

Step # 7b • Tube side heat transfer coefficient hi from Figure 25 • hi = 1240 Btu / hr ft2 0F • hio = 1240 * ID / OD = 1240 * 0.620 / 0.75 = 1025 Btu / hr ft2 0F engineering-resource.com

Step # 8 • Clean Overall Coefficient Uc • Uc = hio * ho hio + ho = 145.8 Btu / hr ft2 0F engineering-resource.com

Step # 9 • Design Overall Coefficient UD from Fourier Equation • UD = Q/A. Δt From Table 10 • a’’ = 0.1963 *ft2/ lin. Ft • A = 204 * 0.1963 * 16 = 640.72 ft2 • UD = 2399109.3 / 640.72 * 35.86 = 104.47 Btu / hr . Ft2 .0F engineering-resource.com

Step # 10 • Rd = Uc-Ud Uc*Ud = 145.8 - 104.47 145.8 * 104.47 = .0027 hr ft2 0F/Btu engineering-resource.com

Step # 11a • Pressure drop: (on shell side • For Res= 9356 • (from fig.29) f=0.0035 ft2/in.2 • No of crosses, N+1=12L/B N+1=(12 × 16)/7 N+1=27.42 ( Say,28) • Ds=19.25 in./12 Ds=1.604 ft s=? engineering-resource.com

Step # 11a engineering-resource.com

Step # 11a • ∆Ps = f×Gs2×Ds×(N+1) 5.22×1010×De×s×Φs ∆Ps =0.0035× 283025.62×1.604×28 5.22×1010× 0.0789ft ×?×1 ∆Ps =7.0psi (allowable=10psi engineering-resource.com

Step # 11b • Pressure drop: (on tube side) • Ret = 34180.5(from fig.26) f=0.0002ft2/in.2 ∆Pt=(f×Gt2×L×n)/(5.22×1010×Ds×Φt) ∆Pt= 4 psi Gt=973500,v2/2g=0.13 (from fig.) ∆Pr=(4×n×v2)/(2g×s) ∆Pr=3.2 psi ∆PT=∆Pt+∆Pr=7.2psi(allowable=10psi) engineering-resource.com

Step # 11b engineering-resource.com

Heat Exchanger Design Calculation - Shell and Tube Problem Solution