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Calculation of a 2-4 Oil Cooler

Calculation of a 2-4 Oil Cooler.

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Calculation of a 2-4 Oil Cooler

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  1. Calculation of a 2-4 Oil Cooler A 33.5˚API oil has viscosity of 1.0 centipoise at 180˚F and 2.0 centipoise at 100˚F.49,600lb/hr of oil leaving a distilling column at 358˚F and is to be used in an absorption process at 100˚F.Cooling will be achieved by water from 90˚F to 120˚F.Pressure drop allowances of 10psi may be used on both streams along with a combined dirt factor of 0.004. engineering-resource.com

  2. Available for this service from a discontinued operation is 35in.ID 2-4exchanger having 454 1in.OD ,11BWG tubes 12׳0״ long laid out on 1¼-in.squre pitch. The bundle is arranged for six tube passes and vertical cut baffles are spaced 7in. apart. The longitudinal baffle is welded to the shell. • Is it necessary to use a 2-4 exchanger? • Will the available exchanger fulfill the requirements? engineering-resource.com

  3. 2-6 Shell and tube heat exchanger:- engineering-resource.com

  4. Temperature profile: T1(358) t2(120) T2(100) t1(90) L engineering-resource.com

  5. Solution:- • Exchanger shell side Tube side ID=35in. Number=454 Baffle spacing=7in. Length=12׳0״ Passes=2 OD,BWG=1in.,11 Pitch=1¼in.squre Passes=6 engineering-resource.com

  6. Hot fluid Cold fluid difference T1=358˚F t2=120˚F ∆t1=238˚F T2=100˚F t1=90˚F ∆t2=10˚F Temperature range:- (T1-T2) (t2-t1) 258˚F 30˚F engineering-resource.com

  7. LMTD:- LMTD= ∆t1-∆t2 ln(∆t1/∆t2) LMTD = 238-10 ln(238/10) LMTD =72˚F engineering-resource.com

  8. Correction factor:- • R= (T1-T2)/(t2 - t1) R=238/30 R=8.6 • S=(t2-t1)/(T1-t1) S=30/(358-90) S=0.112 engineering-resource.com

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  10. True temperature difference:- • ∆t=FT×LMTD From table: FT=0.93 LMTD=72˚F ∆t=0.93×72 ∆t=66.96˚F engineering-resource.com

  11. Heat balance:- • Oil Q=W ×cp×(T1-T2) Q=49,600×0.545×(358-100) Q=6,980,000Btu/hr • Water Q=m×cp×(t2-t1) Q=23,2666.67×1.0×(120-90) Q=6,980,000.1Btu/hr engineering-resource.com

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  13. Caloric temperatures:- • ∆t2/∆t1=10/238=0.042 • For API=33.5˚ and temperature range(258˚F) Kc=0.47(from table) • For Kc=0.47 and ∆t2/∆t1=0.042 • Fc=0.267 engineering-resource.com

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  15. Caloric temperature of hot fluid: Tc=T2+Fc×(T1-T2) Tc=100+0.267×(258) Tc=165˚F • Caloric temperature of cold fluid: tc=t1+Fc×(t2-t1) tc=90+0.267×(30) tc=98˚F engineering-resource.com

  16. Hot fluid: shell side • Flow area as=1/2(ID×C׳×B)/144PT as=1/2(35×0.25×7)/144×1.25 as=0.17ft2 • Mass velocity Gs=W/as Gs=49,600/0.17 Gs=292000lb/(hr)(ft2) engineering-resource.com

  17. Viscosity: At Tc=165F (from table) µ=1.12cp µ=1.12×2.42 µ=2.71lb/(ft)(hr) • Equivalent diameter: De=0.99 in. (from table) De=0.99/12 De=0.0825ft engineering-resource.com

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  20. Reynolds number: Res=DeGs/µ Res=0.0825×292000/2.71 Res=8900 jH=52.5 (from table) • Prandtl number:- Pr=(cµ/k) • For API=33.5˚ and µ=2.71 (from table) k(Pr)⅓=0.20Btu/(hr)(ft2)(˚F) engineering-resource.com

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  24. Film coefficient: ho=jH× (k/De) × (Pr)⅓×Φs ho/Φs= 52.5 ×0.2/0.0825 ho/Φs=127 • Cold fluid: tube side • Flow area: a׳t=0.455 in. square at=(Nt×a׳t)/(144×n) at=(454×0.455)/(144×6) at=0.239ft2 engineering-resource.com

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  26. Mass velocity: Gt=m/at Gt=232666.67/0.239 Gt=973500lb/(hr)(ft2) • Fluid velocity: V=Gt/(3600×ρ) V=973500/(3600×62.37) V=4.33fps engineering-resource.com

  27. Diameter: D=0.76 in./12 (from table) D=0.0633ft • Viscosity: At tc=98˚F µ=0.73 cp (from table) µ=0.73×2.42 µ=1.77 lb/(hr)(ft) engineering-resource.com

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  29. Reynolds number: Ret=D× Gt/μ Ret=(0.0633 ×973500)/1.77 Ret=348156 • Film coefficient: For V=4.33fps (from table) hi=1010×0.96 hi=970 Btu/(hr)(ft2)(ºF) engineering-resource.com

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  31. hio=hi×(ID/OD) hio=970×(0.76/1.0) hio=737 Btu/(hr)(ft2)(ºF) • Tube-wall temperature: tw=tc+ ho × (Tc-tc) (ho+hio) tw=98+ 127 × (165-98) (127+737) tw=108ºF engineering-resource.com

  32. At tw: μw=1.95×2.42 μw=4.72 lb/(hr)(ft) Φs=(μ/μw)¼ Φs=(2.71/4.72)¼ Φs=0.92 ho=127×0.92 ho=117 Btu/(hr)(ft2)(ºF) engineering-resource.com

  33. Clean overall coefficient Uc: • Uc= (hio×ho)/ (hio+ho) Uc=(737×117)/(737+117) Uc=101 Btu/(hr)(ft2)(ºF) engineering-resource.com

  34. Design overall coefficient UD: • UD=Q/(A× ∆t) A(total)=454×12ft×(0.2618ft2/lin ft) A=1425ft2 UD=6980000/(1425×66.96) UD=73.15 Btu/(hr)(ft2)(ºF) engineering-resource.com

  35. Dirt factor Rd: • Rd=(Uc-UD)/UcUD Rd=(101-73.15)/(101×73.15) Rd=0.00377 (hr)(ft2)(ºF)/Btu Rd (required) 0.004 Rd(calculated) 0.00377 engineering-resource.com

  36. Pressure drop: (on shell side) • For Res=8900 (from table) f=0.00215ft2/in.2 • No of crosses, N+1=12L/B N+1=(12 × 12)/7 N+1=20.1 ( Say,21) • Ds=35 in./12 Ds=2.92ft engineering-resource.com

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  38. S( specific gravity)=0.82 (from fig.) engineering-resource.com

  39. ∆Ps = f×Gs2×Ds×(N+1) 5.22×1010×De×s×Φs ∆Ps =0.00215×2920002×2.92×42 5.22×1010×0.0825×0.82×0.92 ∆Ps =7.0psi (allowable=10psi) engineering-resource.com

  40. Pressure drop: (on tube side) Ret =34815.6 (from fig.) f=0.000195ft2/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

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  43. 2-4 Shell and tube heat exchanger:- engineering-resource.com

  44. Only replace value of n=6 to n=4 • At=0.3585 • Gt=649000 • V=2.89fps • Ret=23210 • hi=760 Btu/(hr)(ft2)(ºF) • hio=577 Btu/(hr)(ft2)(ºF) • tw=110ºF • ho=117 Btu/(hr)(ft2)(ºF) engineering-resource.com

  45. Uc=94 Btu/(hr)(ft2)(ºF) • Rd=0.003 (hr)(ft2)(ºF)/Btu • F=0.00025 • ∆Pt=1.53 psi ,v2/2g =0.065 • ∆Pr=1.04 psi • ∆PT=∆Pt+∆Pr=2.57psi (allowable=10psi)s So,2-6 STHE is more suitable as compare to 2-4 STHE. engineering-resource.com

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