Thermal Unit Operation (ChEg3113)

Transcription

1 Thermal Unit Operation (ChEg3113) Lecture 10- Shell and Tube Heat Exchanger Design Instructor: Mr. Tedla Yeshitila (M.Sc.)

2 Today Review Steps in Shell and tube heat exchanger Example

3 Review Shell and tube exchanger Baffle, tube pitch, clearance True temperature difference The first 3 steps in shell and tube exchanger design

4 Hot fluid: shell side 4. Flow area, a S : a S = ID C`B 144Pr, ft2 Where C` is clearance, B is baffle spacing and Pr is tube pitch 5. Mass velocity, G s = W a s, lb/hr.ft 2 6. Obtain D e from Figure 28 or using D e = For square pitch: d e = 4 (P r 2 Πd 0 2 /4) Πd 0 For triangular pitch: d e = 4 (1 2 P r 0.86P r 1 2 Πd 0 2 /4) Πd 0 4 free area wetted perimiter, ft or,in where d0 is tube outside diameter Obtain μ from Figure 14 at T c or t c where lb/ft.hr=cp*2.42 Reynolds number, Re s = D e G s μ, unit less,in Square pitch Triangular pitch

5 Hot fluid: shell side 7. Using Re s obtain j H from Figure At T c or t c obtain c (from Figure 2), μ (from Figure 14 ) and (from Table 4) and compute cμ 9. To obtain ho, h o = j H 1 3 D cμ => h o /ϕ s = j H D 1 3 μ cμ μ w 0.14 where ϕs = μ 10. Tube-wall temperature, t w t w = t c + h o/ϕ s h io (T + h o c t c ) ϕ i ϕs 11. Obtain μ w at t w (from Figure 14 ) then ϕ s = μ 0.14 μ w 12. Corrected coefficient, h o = h o ϕ s ϕ s 1 3 μ w 0.14

6 Cold fluid: tube side 4. Flow area, a t : a t = Number of tubes flow area/tube Number of passes = N ta t` 144 n, ft2 Flow area per tube a t` from Table 10, in Mass velocity, G t = w, lb/hr.ft 2 a t 6. Obtain D (internal diameter of tube) from Table 10, ft Obtain μ from Figure 14 at T c or t c where lb/ft.hr=cp*2.42 Reynolds number, Re t = DG t μ, unit less 7. Using Re t obtain j H from Figure 24.

7 Cold fluid: tube side 8. At T c or t c obtain c (from Figure 2), μ (from Figure 14 ) and (from Table 4) and compute 9. To obtain hi, h i = j H D cμ 1 3 μ 1 3 cμ 1 3 μ w 0.14 where ϕt = μ μ w 0.14 h => i cμ =j ϕ H t D 10. Convert or correct h i to the surface at OD (h io ) ; h io ϕ t = h i ϕ t A i A = h i ϕ i ID OD 11. Obtain μ w at t w (from Figure 14 ) then ϕ t = μ μ w Corrected coefficient, h o = h o ϕ t ϕ t

8 13. Compute clean overall heat transfer coefficient U c : U c = h ioh o h io + h o 14. Compute design or dirty overall heat transfer coefficient (U D ) from U D = Q AΔt Where heat transfer surface A= a``ln t You can obtain external surface/lin ft (a``) from Appendix Table Dirt factor R d : R d = 1 U c 1 U D = U c U D U c U D

9 If Rd equals or exceed the required dirt factor, proceed under the pressure drop. Hot fluid: Shell side 1. For Re s in step 6 obtain f, ft 2 /in 2 (from Figure 29 ) 2. Number of crosses, N+1=12L/B 3. ΔP s = fg s 2 D s (N+1) 5.22*10 10 D e Sϕ s, psi (S is from Figure 6)

10 Cold fluid: tube side 1. For Re t in step 6 obtain f, ft 2 /in 2 (from Figure 26 ) 2. ΔP t = f G t 2 Ln D S ϕ t,psi (S is from Figure 6) 3. Obtain V2 using G 4n g` t (from Figure 27 ) then ΔP r= s 4.. ΔP T = ΔP t +ΔP r V 2 g`,psi

11 Example Calculation of a erosene-crude oil Exchanger 43,000lb/hr of a 42 0 API erosene leaves the bottom of a distilling column at F and will be cooled to F by 149,000 lb.hr of 34 0 API. Mid-continent crude coming from storage at F and heated to F. A 10psi pressure drop is permissible on both streams, and in accordance with Table 12, a combined dirt factor of should be provided. Available for this service is a 21 ¼ in. ID exchanger having in. OD, 13 BWG tubes 16`0`` long and laid out on 1 ¼ in. square pitch. The bundle is arranged for four passes, and baffles are spaced 5in. Apart. Will the exchanger be suitable. i.e. what is the dirt factor?

12 Given: Shell and tube exchanger: Solution Shell side ID= 21 ¼ Baffle space=5in. Passes=1 Tube side Number and length= 158, 16`0`` OD, BWG, pitch= 1in, 13 BWG, 1 ¼ in square Passes=4

13 At the end of this class: You will be able to design shell and tube exchanger design

14 End of lecture -10