Feb 1, 2009 - methanol condenser from 95 °C to 40 °C. Flow-rate of methanol 100,000 kg/h. Brackish water (seawater) will be used as the coolant, with a.
Designing Shell-and-Tube Heat Exchangers Using Softwares
Lecture 1: Kern’s Method
By: Majid Hayati University of Kashan, Kashan, I.R. IRAN 2014
Schedule – Kern’s Method Kern’s method
Introduction to Kern’s method
Algorithm of design procedure for shell-and-tube heat exchangers
Design procedure steps along with an example
3
Objectives This lecture on designing shell-and-tube HEs serves as an introduction lecture to the subject, and covers:
Introduction to “Kern’s method” definition along with its advantages and disadvantages Developing an algorithm for the design of shell-and-tube exchangers Finally, following up the procedure set out in the algorithm in an example
4
Introduction to Kern’s method Kern’s was based on experimental work on commercial exchanger Advantages: Giving reasonably satisfactory prediction of the heat-transfer coefficient for standard design Simple to apply Accurate enough for preliminary design calculations Accurate enough for designs when uncertainty in other design parameter is such that the use of more elaborate method is not justified
Disadvantage: The prediction of pressure drop is less satisfactory, as pressure drop is more affected by leakage and bypassing than heat transfer The method does not take account of the bypass and leakage streams 5
Design procedure for shell-and-tube heat exchangers (Kern’s method) End
Start from step 3
Start Step 1 Collect physical properties and HE specifications Step 2 Define duty Make energy balance if needed
No
Yes
Compare to estimated overall heat transfer coefficient
Accept all design parameters
Step 8 Determine overall heat transfer coefficient Yes
Estimate tube- and shell-side heat transfer coefficient-go to step 3
Question: Are pressure drops within specification?
No
Step 7 Calculate unspecified flow rates Calculate ΔTLMTD and ΔTM Step 3 Assume value of overall coefficient Uo,ass Step 4 Calculate tube number Calculate shell diameter
Estimate tube- and shell-side pressure drop Step 6 Estimate tube- and shell-side heat transfer coefficient Step 5 Determine fouling factors
Fig. 1: Algorithm of design procedure 6
Kern’s Method Design Example Design an exchanger to sub-cool condensate from a methanol condenser from 95 °C to 40 °C
Flow-rate of methanol 100,000 kg/h Brackish water (seawater) will be used as the coolant, with a temperature rise from 25° to 40 °C
7
Solution: Step 1 Collect physical properties and HE specifications: Physical properties Table 1 Physical properties at Methanol Water fluid mean temperature Cp (Kj/Kg °C)
2.84
4.2
μ (mNs/m2)
0.34
0.8
kf (W/m °C)
0.19
0.59
ρ (Kg/m3)
750
995
HE specifications: Coolant (brackish water) is corrosive, so assign to tube-side. Use one shell pass and two tube passes. At shell side, fluid (methanol) is relatively clean. So, use 1.25 triangular pitch (pitch: distance between tube centers). 8
Tube Arrangements The tubes in an exchanger are usually arranged in an equilateral triangular, square, or rotated square pattern (Fig. 2)
Fig. 2: Tube patterns 9
Tube Pattern Applications The triangular and rotated square patterns give higher heattransfer rates, but at the expense of a higher pressure drop than the square pattern. A square, or rotated square arrangement, is used for heavily fouling fluids, where it is necessary to mechanically clean the outside of the tubes.
The recommended tube pitch is 1.25 times the tube outside diameter; and this will normally be used unless process requirements dictate otherwise. 10
Step 2 Define duty, Make energy balance if needed To start step 2, the duty (heat transfer rate) of methanol (the hot stream or water, the cold stream) needed to be calculated.
Fig. 3: Streams definitions. •
Heat load = Q = m
h
100000 Cph (T1 - T2 ) = × 2.84 (95 - 40) = 4340 kW 3600 11
Step 2 (Cont’d) The cold and the hot stream heat loads are equal. So, cooling water flow rate is calculated as follow: .
Cooling water flow = mc =
Q CP c (t 2 _ t1)
=
4340 _
4.2 (40 25)
= 68.9 kg/s
The well-known “logarithmic mean” temperature difference (LMTD or lm) is calculated by:
ΔTLMTD
(T1 t 2 ) (T2 t1 ) (95 40) (40 25) 31 C T t (95 40) ln 1 2 ln T2 t1 (40 25) 12
Mean Temperature Difference The usual practice in the design of shell and tube exchangers is to estimate the “true temperature difference” from the logarithmic mean temperature by applying a correction factor to allow for the departure from true counter-current flow:
ΔTm Ft ΔTLMTD Where:
ΔTm = true temperature difference, Ft = the temperature correction factor. 13
Temperature Correction Factor The correction factor (Ft) is a function of the shell and tube fluid temperatures, and the number of tube and shell passes. It is normally correlated as a function of two dimensionless temperature ratios:
T1 T2 R t 2 t1 t 2 t1 S T1 t1 14
Step 2 For a 1 shell : 2 tube pass exchanger, the correction factor is plotted in Fig. 4.
Fig. 4: Temperature correction factor: one shell pass; two or more even tube passes (available in TEMA) 15
Step 2 (Cont’d) T1 T2 95 40 R 3.67 t 2 t1 40 25 (t 2 t1 ) 40 25 S 0.21 (T1 t1 ) 95 25 From Fig. 4, the correction factor (Ft) is 0.85.
ΔTm Ft ΔTLMTD 0.85 31 26 C 16
Step 3 Assume value of overall coefficient Uo,ass Typical values of the overall heat-transfer coefficient for various types of heat exchanger are given in Table 1. Fig. 5 can be used to estimate the overall coefficient for tubular exchangers (shell and tube). The film coefficients given in Fig. 5 include an allowance for fouling. The values given in Table 1 and Fig. 5 can be used for the preliminary sizing of equipment for process evaluation, and as trial values for starting a detailed thermal design.
From Table 2 or Fig. 5:
U=600 W/m2°C 17
Step 3 (Cont’d) Table 2: Typical overall coefficients
18
Step 3 (Cont’d)
Fig. 5: Overall coefficients (join process side duty to service side and read U from centre scale) 19
Step 4 Calculate tube number, Calculate shell diameter Provisional area:
Q 4340 × 10 3 A= = = 278 m2 U ΔTM 600 × 26
So, the total outside surface area of tubes is 278 m2 Choose 20 mm o.d. (outside diameter), 16 mm i.d. (inside diameter), 3 4.88-m-long tubes ( in. 16 ft ), cupro-nickel. 4 Allowing for tube-sheet thickness, take tube length: L= 4.83 m Surface area of one tube: A = πDL = 4.83 x 20 x 10-3π = 0.303 m2 Numbers of tubes
Total outside surface area of tubes (Provisional area) 278 918 Outside surface area of one tube 0.303 20
Step 4 (Cont’d) An estimate of the bundle diameter Db can be obtained from equation below which is an empirical equation based on standard tube layouts. The constants for use in this equation, for triangular and square patterns, are given in Table 3.
N t 1n1 Db do ( ) K1 where Db = bundle diameter in mm, do = tube outside diameter in mm., Nt = number of tubes. As the shell-side fluid is relatively clean use 1.25 triangular pitch. So, for this example:
918 Bundle diameter D b 20 ( ) 0.249
1
2.207 826 mm 21
Step 4 (Cont’d) Table 3: Constants K1 and n1
22
Step 4 (Cont’d) Use a split-ring floating head type for Fig. 6. From Fig. 6, bundle diametrical clearance is 68 mm. Shell diameter (Ds): Ds= Bundle diameter + Clearance = 826 + 68 = 894 mm. Note 1: nearest standard pipe size are 863.6 or 914.4 mm. Note 2: Shell size could be read from standard tube count tables [Kern (1950), Ludwig (2001), Perry et al. (1997), and Saunders (1988)].
23
Step 4 (Cont’d)
Fig. 6: Shell-bundle clearance 24
Step 6 Estimate tube- and shell-side heat transfer coefficient Tube-side heat transfer coefficient:
40 + 25 Mean water temperature (Tavg ) = = 33 C ⇒ ρ = 995 kg m3 2 π π Tube cross - sectional area (a) = D2 = × 16 2 = 201 mm2 4 4 Since we have two tubes pass, we divide the total numbers of tubes by two to find the numbers of tubes per pass, that is:
918 Tubes per pass = = 459 2 Total flow area is equal to numbers of tubes per pass multiply by tube cross sectional area:
Total flow area = 459 × (201 × 106 ) = 0.092 m2 25
Step 6 (Cont’d)
Fig. 7: Equivalent diameter, cross-sectional areas and wetted perimeters. 26
Step 6 (Cont’d) Cooling water flow 68.9 Water mass velocity = = = 749 kg s m2 Total flow area 0.092 Water mass velocity (Gt ) 749 Water linear velocity (ut ) = = = 0.75 m s Water density (ρ) 995
Coefficients for water: a more accurate estimate can be made by using equations developed specifically for water. The physical properties are conveniently incorporated into the correlation. The equation below has been adapted from data given by Eagle and Ferguson (1930):
4200 (1.35 + 0.02t) u 0.8 hi = di0.2
where hi = inside coefficient, for water, W/m2 °C, t = water temperature, °C, ut = water linear velocity, m/s, di = tube inside diameter, mm.
27
Step 6 (Cont’d) 4200 (1.35 + 0.02t) u 0.8 4200 (1.35 + 0.02 × 33) 0.75 hi = = 0.2 di 16 0.2
0.8
= 3852 W/m2 C
The equation can also be calculated using equation below; this is done to illustrate use of this method. hi di 0.33 μ 0.14 = jh Re Pr ( ) kf μw where hi = inside coefficient, for water, W/m2 °C, di = tube inside diameter, mm kf = fluid thermal conductivity, W/m2 °C jh = heat transfer factor, dimensionless Re = Reynolds number, dimensionless Pr = Prandtl number, dimensionless μ = viscosity of water, N s/m2 μw = viscosity of water at wall temperature, N s/m2 28
Step 6 (Cont’d) Viscosity of water (μ) from Table 1 = 0.8 mNs m2 Fluid thermal conductivity from Table 1 = 0.59 W m C ρudi 995 × 0.75 × 16 × 10 3 Re = = = 14925 _ 3 μ 8 × 10 _ Cpμ 4.2 × 103 × 0.8 × 10 3 Pr = = = 5.7 kf 0.59 μ
Neglect (μ ) w _ L 4.83 × 103 = = 302 ⇒ From Fig. 8, jh = 3.9 × 10 3 di 16 hi =
_ kf μ 0.59 3 0.33 0.14 2 jh Re Pr 0.33 ( )0.14 = × 3.9 × 10 × 14925 × 5.7 × 1 = 3812 W m C _ 3 di μw 16 × 10
Check reasonably the previously calculated value 3812 W/m2°C with value calculated, 3852 W/m2°C. 29
Step 6 (Cont’d)
Fig. 8: Tube-side heat-transfer factor 30
Step 6 (Cont’d) Shell-side heat transfer coefficient: Baffle spacing: The baffle spacings used range from 0.2 to 1.0 shell diameters. A close baffle spacing will give higher heat transfer coefficients but at the expense of higher pressure drop. Area for cross-flow: calculate the area for cross-flow As for the hypothetical row at the shell equator, given by: (pt _ do )Dslb As = pt
Where pt = tube pitch (distance between the centers of two tubes, Fig. 7). do = tube outside diameter, m, Ds = shell inside diameter, m, lb = baffle spacing, m. (p d ) p is the ratio of the clearance between tubes and Note: the term the total distance between tube centers. t
_
o
t
31
Step 6 (Cont’d) Baffle spacing: Choose baffle spacing = 0.2 Ds=0.2
894 = 178 mm
Tube pitch: Pt = 1.25 do= 1.25
20 = 25 mm
Cross-flow area: _ (p t _ do ) (25 _ 20) As = Dslb = × 894 × 178 × 10 6 = 0.032 m2 pt 25
32
Step 6 (Cont’d) Shell-side mass velocity Gs and the linear velocity ut: Ws As G us = s ρ
Gs =
Where Ws = fluid flow-rate on the shell-side, kg/s, ρ = shell-side fluid density, kg/m3. Shell equivalent diameter (hydraulic diameter): calculate the shellside equivalent diameter, see Fig. 7. For an equilateral triangular pitch arrangement: 2 4( de =
pt 1 d × 0.87p t _ π o ) 1.10 2 2 4 = (p2t πdo do 2
_
0.917 d2o )
Where de = equivalent diameter, m. 33
Step 6 (Cont’d) Shell-side mass velocity Gs: Mass velocity, Gs =
Ws 100000 1 kg = × = 868 s m2 As 3600 0.032
Shell equivalent diameter (hydraulic diameter):
de =
1.10 2 (p t do
_
0.917 d2o ) =
1.1 (25 2 20
_
0.917 × 202 ) = 14.4 mm
34
Step 6 (Cont’d) 95 + 40 Mean shell side temperatur e = = 68 C 2 Methanol density (ρ) from Table 1 = 750 kg m3 ) Viscosity of methanol (μ from Table 1 = 0.34 mNs m2 ) Heat capacity from Table 1 = 2.84 kJ kgC Thermal conductivi ty Table 1 = 0.19 W mC ρu sde G sde 868 × 14.4 × 10 - 3 Re = = = = 36762 _ μ μ 3 0.34 × 10 3
_
2.84 × 10 × 0.34 × 10 3 Pr = = = 5.1 kf 0.19 Cpμ
Choose 25 per cent baffle cut, from Fig. 9
jh = 3.3 × 10
_
3
35
Step 6 (Cont’d)
Fig. 9: Shell-side heat-transfer factors, segmental baffles 36
Step 6 (Cont’d) For the calculated Reynolds number, the read value of jh from Fig. 9 for 25 per cent baffle cut and the tube arrangement, we can now calculate the shell-side heat transfer coefficient hs from:
h sde μ 0.14 0.19 = jh Re Pr 1 3 ( ) (without viscosity correction term) → hs = × 3.3 × 10 - 3 × 36762 × 5.1 1 3 3 kf μw 1.44 × 10 = 2740
Nu =
The tube wall temperature can be estimated using the following method: Mean temperature difference across all resistance: 68 -33 =35 °C U 600 across methanol film = × ΔT = × 35 = 8 C ho
2740
Mean wall temperature = 68 – 8 = 60 °C μ = 0.37 mNs/m2 μ 0.14 ) = 0.99 μw Which shows that the correction for low-viscosity fluid is not significant. (
37
Step 7 (Cont’d) Pressure drop Tube side: From Fig. 10, for Re = 14925 jf = 4.3 Neglecting the viscosity correction term:
10-3
L μ -m ρu2t ΔPt = Np [8jf ( )( ) + 2.5] di μw 3 995 × 0.75 2 - 3 4.83 × 10 = 2 (8 × 4.3 × 10 ( ) + 2.5) 16 2 2 = 7211N m = 7.2 kPa (1.1 psi)
low, could consider increasing the number of tube passes. Shell side G 868 Linear velocity = s = = 1.16 m/s ρ 750 From Fig. 11, for Re = 36762 jf = 4
10-2
Neglect viscosity correction 38
Step 7 (Cont’d)
Fig. 10: Tube-side friction factors 39
Step 7 (Cont’d)
Fig. 11: Shell-side friction factors, segmental baffles 40
Step 7 (Cont’d) Ds L ρu2t 4.83 × 103 750 × 1.162 - 2 894 ΔPs = 8jf ( )( ) = 8 × 4 × 10 ( )( ) de L s 2 14.4 178 2 = 272019 N m2 = 272 kPa (39 psi) too high,
could be reduced by increasing the baffle pitch. Doubling the pitch halves the shell side velocity, which reduces the pressure drop by a factor of approximately (1/2)2 272 ΔPs = = 68 kPa (10psi), acceptable 4 This will reduce the shell-side heat-transfer coefficient by a factor of (1/2)0.8(ho∝ Re0.8 ∝ us0.8) ho = 2740 (1/2)0.8 = 1573 W/m2°C This gives an overall coefficient of 615 W/m2°C – still above assumed value of 600 W/m2°C
41
Step 8 (Cont’d) Take the thermal conductivity of cupro-nickel alloys from Table 1, 50 W/m°C, the fouling coefficients from Table 3; methanol (light organic) 5000 Wm-2°C-1, brackish water (sea water), take as highest value, 3000 Wm-2°C-1 do do ln 1 1 1 1 do 1 di do = + + + × + × Uo ho hod 2K w di hid di hi 20 -3 20 × 10 ln 1 1 1 20 1 20 1 16 = + + + × + × Uo 2740 5000 2 × 50 16 3000 16 3812
Uo = 738 W m2 C
Well above assumed value of 600 Wm-2°C 42
References 1.
EAGLE, A. and FERGUSON, R. M. (1930) Proc. Roy. Soc. A. 127, 540. On the coefficient of heat transfer fromthe internal surfaces of tube walls.
2. 3.
KERN, D. Q. (1950) Process Heat Transfer (McGraw-Hill).
4.
PERRY, R. H., GREEN, D.W. and MALONEY, J. O. (1997) Perry’s Chemical Engineers Handbook, 7th edn (McGrawHill).
5.
SAUNDERS, (Longmans).
LUDWIG, E. E. (2001) Applied Process Design for Chemical and Petroleum Plants, Vol. 3, 3rd edn (Gulf).
E.
A.
D.
(1988)
Heat
Exchangers 43
2/1/2009 8:11:44 AM
44
Step 5 Table. 3: Fouling factors (coefficients), typical values
45
