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- 1. Design of shelland tube heat exchanger Submitted By: Mahmoud Mohammed Supervision by: Dr. Mohammed El dalli Dr. Asma Mustafa Chemical & Petroleum Engineering Department Faculty of Engineering- Al-khoms Elmerghib University
- 2. Gas oil at200oC is to be cooled to 40oC. The oil flow-rate is 22,500 kg/h. Cooling water is available at 30oC and the temperature rise is to be limited to 20oC. The pressure drop allowance for each stream is 100 kN/m2. Design a suitable exchanger for this duty. Description Symb ol Valu e Unit Hot stream mass flow rate mh 22500 kg/h Inlet Temperature Ti,h 200 oC Outlet Temperature To,h 40 oC Cold stream Inlet Temperature Ti,c 30 oC Outlet Temperature To,c 50 oC Pressure drop ΔP 100 kN/m 2 𝑄𝑐 = 𝑚𝑐𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐) 𝑄ℎ = 𝑚ℎ𝐶ℎ (𝑇𝑖,ℎ − 𝑇𝑜,ℎ) 𝑄ℎ = 𝑄𝑐 𝑚ℎ𝐶ℎ 𝑇𝑖,ℎ − 𝑇𝑜,ℎ = 𝑚𝑐𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐) 𝑚𝑐 = 𝑚ℎ𝐶ℎ(𝑇𝑖,ℎ − 𝑇𝑜,ℎ) 𝐶𝑐(𝑇𝑜,𝑐 − 𝑇𝑖,𝑐) mc= 98181.82 kg/h Step#1
- 3. Physical properties ofwater at 40 oC Gas oil at oC 120 Heat Capacity Cc= 4.18 kJ/kg.oC Ch= 2.28 kJ/kg.oC Thermal conductivity kc= 6.31E-04 kW/m.oC kh= 0.125 W/m.oC Viscosity μc= 6.71E-01 mN/m2 s μh= 0.17 mN/m2 s Density ρc= 992.8 kg/m3 ρh= 850 kg/m3 Prandtl number Prc= 4.444976 Prh= 3.1008 ℃ Number of material Material name Thermal conduct. 7 Stainless steel(18/8) 16 W/m.oC Select Shell Type Type of shell One shell passes Two shell passes Divided-flow shell Split- flow shell Step#2 Step#3
- 4. To,h Ti,c ΔT1 Ti,h To,c ΔT2 ∆𝑇𝑙𝑚 = ∆𝑇1 −∆𝑇2 ln( ∆𝑇1 ∆𝑇2 ) ΔTlm= 51.69771 oC 𝑅 = 𝑇𝑖,ℎ − 𝑇𝑜,𝑐 𝑇𝑜,𝑐 − 𝑇𝑖,𝑐 𝑆 = 𝑇𝑜,𝑐 − 𝑇𝑖,𝑐 𝑇𝑖,ℎ − 𝑇𝑖,𝑐 ΔTm= 48.59585 oC F= 0.94 ∆𝑇𝑚= 𝐹 ∗ ∆𝑇𝑙𝑚 Step#4
- 5. 𝐴𝑜 = 𝑞/(𝑈.∆𝑇𝑙𝑚) 𝑞 = 𝑚ℎ𝐶ℎ(𝑇𝑖,ℎ − 𝑇𝑜,ℎ) q= 2280 KW Ao= 75.06814 m2 𝐴𝑠 = 𝜋 𝐷𝑜 𝐿 𝑁𝑡 = 𝐴 𝐴𝑠 𝐴𝑐 = 𝜋 𝐷𝑖 2 4 𝐴𝑡 = 𝑁𝑡 ∗ 𝐴𝑐 Surface area of one tube Number of Tubes Cross-sectional area " one tube " Total tube area Do= 16 mm t= 1.6 mm Di= 12.8 mm L= 2.44 m As =0.122648 m2 Nt=613 AC =0.000129 m2 At =0.078881 m2 Step#5 Step#6 Step#7
- 6. No. of passes=4 Geometry of tube arrangement: Triangular Pitch K1= 0.175 n1= 2.285 Db= 0.569232 m (Shell inside-bundle)diameter = 91.32309 mm Shell inside diameter = 660.5554 mm Pt= 20 mm Step#8
- 7. 𝑢𝑡 = 𝑚 𝜌𝑤 ∗𝐴𝑡 Tube Water velocity, ut= 0.34825431 m/s Tube velocity for Passes= 1.39301726 m/s Heat transfer coefficient hi= 7069.86228 W/m2.oC 𝑑𝑒 = 𝐴 𝑑𝑜 (𝑃𝑡 2 − 𝐵𝑑𝑜 2 ) de= 11.3608 mm Baffle spacing lB= 0.132111m Tube pitch Pt= 20mm Total area As= 0.017453m2 Mass velocity Gs= 358.0977kg/m2.s Linear velocity us= 0.421291m/s A 1.1 B 0.917 Re= 23931.04 Pr= 3.1008 Step#9 Step#10
- 8. 1.00E-04 1.00E-03 1.00E-02 1.00E-01 1.00E+00 1.00E+01 1.00E+02 1.00E+031.00E+04 1.00E+05 1.00E+06 Heat transfer factor, j h Reynolds number, Re Buffle cut, percent 15 25 35 45 𝑗ℎ = 𝑐𝑅𝑒𝑚 jh= 0.0038935 Heat transfer coefficient in shell hs= 1494.9521 W/m2.oC c= 0.4275 m= -0.466 do 16 mm Di 12.8 mm Kw 16 W/m .oC Ho 1494.952 W/m2.oC hi 7069.862 W/m2.oC hod= 5000 W/m2.oC hid= 4255.319 W/m2.oC Uo= 689.1578 W/m2.oC Error%= 10.2653 % Step#11
- 9. jf = 8.0722Re-1.004 R²= 1 jf = 0.0169Re-0.09 R² = 1 jf = 0.0725Re-0.296 R² = 1 jf = 0.0438Re-0.241 R² = 1 jf = 0.0303Re-0.206 R² = 1 0.001 0.01 0.1 1 10 100 1000 10000 100000 1000000 Heat transfer factor, j f Reynold,Re Reynold vs. Heat transfer factor 𝑗𝑓 = 𝑧𝑅𝑒𝑤 z= 0.0438 w= -0.241 ΔPt= 31.7643 kPa Step#12
- 10. 𝑗𝑓 = 𝑥𝑅𝑒𝑦 0.01 0.1 1 10 110 100 1000 10000 100000 1000000 Heat transfer factor, j f Reynold number, Re 15 25 35 45 x= 0.1939 y= -0.148 ΔPs= 28.2544 kPa jf= 0.043601
- 11. 𝐸𝑥𝑐ℎ𝑎𝑛𝑔𝑒𝑟 𝐶𝑜𝑠𝑡, $1000= 𝑒 ∗ 𝐴𝑟𝑒𝑎 𝑟 Total area = 75.06814 m2 Exchanger Cost= 78874.27 $ Pressure factor= 1.1 Type factor= 1 Purchased cost = (bare cost from figure) * Type factor * Pressure factor Purchased cost= 86761.7 $ "Time base 2004" 𝐶𝑜𝑠𝑡 𝑎𝑡 2019 = 𝐶𝑜𝑠𝑡 𝑎𝑡 2004 ∗ 𝐼𝑛𝑑𝑒𝑥 𝑣𝑎𝑙𝑢𝑒(𝑎𝑡 2019) 𝐼𝑛𝑑𝑒𝑥 𝑣𝑎𝑙𝑢𝑒(𝑎𝑡 2004) Purchased cost= 131398.2 $ "Time base 2019" Step#13 e= 2.2083 r= 0.828