design of shell and tube heat exchanger

1. DESIGN OF HEAT EXCHANGER
SHELL & TUBE HEAT EXCHANGER

2. PROBLEM
Benzene is to be cooled from 60 to 35oC at a rate of 50,000 kg/hr. Cooling
water is available at 20oC and the maximum exit temperature must be
limited to 32oC. Design a suitable heat exchanger.

3. DATA GIVEN
Benzene is to be cooled from 60 to 35oC at a rate of 50,000 kg/hr. Cooling
water is available at 20oC and the maximum exit temperature must be
limited to 32oC.
Design a suitable heat exchanger.
 ask for a suitable heat exchanger, always – shell & tube
Hot fluid Cold fluid
benzene Water
T1 : 60oC t1 : 20oC
T2 : 35OC t2 : 32OC
W : 50,000 kg/hr

4. STEP 1 : ROOTING OF FLUIDS
 Tube side fluid
 Cooling water
 The more fouling, erosive or corrosive fluid
 Less viscous fluid
 The fluid under higher pressure
 The hotter fluid
 The smaller volumetric flow rate
 Condensing steam
 Shell side fluid
 Condensing vapours
 If fluid temperature change is > 150°C (300°F)
 Very small volumetric flow rates
 High viscosity streams
Here,
Tube side fluid – Water
Shell side fluid - Benzene

5. STEP 2 : HEAT BALANCE TO DETERMINE UNKNOWN QUANTITY
Mean temperature –
= 47.5 oC = 320.5 K
= 26 oC = 299 K
Shell side fluid
(hot fluid)
Tube side fluid
(cold fluid)
W, Cp,shell , T1, T2,
Benzene
T1 = 60oC T2 = 35oC
50,000 kg/hr
w, Cp,tube , t1 , t2
Water
t1 = 20oC t2 = 32oC

6. To determine Cp – refer page no: 2-165, table 2-153 – 8e
refer page no: 2-170, table 2-196 -7e
Write down the values of C1, C2, C3, C4 and C5 for both benzene and water
For Benzene, T = Tmean and for water T = tmean
After substituting
Cp, benzene = 141658.45 J/kmol.K
= 1813.53 J/kg.K
Cp, water = 75370.74 J/kmol.K
= 4183.78 J/kg.K
J / (Kmol K)
J / (Kg K)
÷ mol. wt
(kg/kmol)

7. W Cp, benzene (T1 – T2) = w Cp, water (t1 – t2)
Substitute all values and calculate unknown w
w = 45,152.79 kg/hr
W = 50,000 kg/hr
W = 13.89 kg/s
w = 12.54 kg/s
Kg/ hr
Kg/ s
÷ 3600

8. STEP 3 : DETERMINATION OF LMTD

9. Here,
Countercurrent heat exchanger is used
LMTD = 20.83OC
T1
t2
T2
t1
ΔT2 = T1-t2
ΔT1 = T2-t1

10. STEP 4 : MINIMUM FLOW AREA PER TUBE PASS
1. Allowable velocity
If tube side fluid is liquid  1 m/s
If tube side fluid is gas  20 m/s
here, tube side fluid is liquid, so v = 1 m/s
2. Mass flow rate, w
tube side fluid is considered
w = 45152.79 kg/hr
= 12.54 kg/s

11. 3. Density – page no. 2-98, table – 2-32 7e & 8e
Here, tube side fluid is water and temp = 299 K,
ρ = 55.219 kmol/m3
= 994.77 Kg/ m3
Minimum flow area per tube pass = 0.0126 m2
Kmol / m3
Kg / m3
x Mol. Wt
(kg/kmol)

12. STEP 5 : MINIMUM NUMBER OF TUBES IN EACH PASS
Page no: 11-42, table 11-12  8e PCH
Assume, Out side dia of tube = ¾”
From table corresponding to ¾” OD, 14 BWG metal, 8ft length, we get
Inside dia = 0.584” = 0.584 x 0.0254 = 0.0148 m
Outside dia = ¾” = 0.75” = 0.01905 m
1 inch = 0.0254 m

13. Inside area of tube = 1.719 x 10-4 m2
Min. flow area = 0.0126 m2
No. of tubes/ pass = 73

14. STEP 6 : TOTAL NO. OF TUBES
Total no. of tubes = (no. of tubes/ pass) x no. of pass
Always opt for a 1-2 exchanger (1 shell pass and 2 tube pass)
So, No. of tube pass = 2
Check for no. of pass : page no. 11-6, fig 11-4 (a)
R = (T1 – T2) /(t2 – t1) S = (t2 – t1) / (T1 – t1)

15. Using chart to find FT
• Find the S (0.3) on X axis
and draw a normal line to
the R curve (R= 2.08).
• Then extend the line from
the point of intersection
with Y axis.
• The Y axis will give the FT
• From that can calculate FT
R = 2.08
S = 0.3
FT = 0.88
Check for no. of pass
If FT > 0.8, assumption is correct
So, No. of pass = 2
Total no. of tubes = 2 x 73 = 146

16. STEP 7: INSIDE DIAMETER OF SHELL
SHELL – contains shell fluid & the tube bundle
• Shell diameter should be selected to give a close fit of the tube bundle.
• The clearance between the tube bundle and inner shell wall depends
on the type of exchanger.
SHELL DIAMETER = BUNDLE DIAMETER + (2 X CLEARANCE)
• Industrially accepted clearance = 7/8 “
• The tubes are arranged in Triangular, Square and Rotated square
pattern
• Generally select triangular pitch

17. Page no- 11-43
For triangular pitch, PT = 1.25 OD = 0.0238m
By trial and error method, we can find C, C = -21.5
Substitute in eqn, C = 0.75 (D/d) – 36  D = 14.5” = 368 mm
Shell ID = 14.5 + 2x 7/8 = 16.25” = 412 mm
Compare it with Std shell thickness in 6th edition PCH
Shell ID = 438 mm
No. of tubes = 228

18. STEP 8: REYNOLDS NUMBER – TUBE SIDE FLUID
di = 0.0148 m
ρ = 994.77 kg/m3
v = ?
μ = ?
To determine ‘v’
w- Flow rate of tube side fluid
a – inside area of tube
Cumulative velocity = 73.35 m/s w = 45152.79kg/hr = 12.54 kg/s
a = 1.719 x 10-4 m2
ρ =994.77 kg/ m3
v = 0.64 m/s
Re = 10469.4

19. To determine ‘μ’ - page no 2-449, fig 2- 32
To calculate μ using chart
• From table 2-318, identify the X & Y
coordinates of liquid (water: 10.2,13)
• Mark that point on fig 2-32
• On LHS Y-axis mark corresponding
temp in degree Celsius (26 C)
• Join both points and extend to RHS Y-
axis. The point on RHS Y-axis give
viscosity in cP.
μ = 0.9cP = 0.9x10-3P=0.9x10-3
μ = 9 x 10-4
Poise
kg/ms

20. STEP 9 : jH FACTOR
Page no : 5 -16, eqn 5-50c 7e PCH
jH = 3.6119 x 10-3

21. STEP 10: HEAT TRANSFER COEFFICIENT
To determine hi , eqn 5-50c is used
Nst = Stanton number
Npr = Prandtl number
Thermal conductivity, k is to be calculated to calculate Npr
Here we are using water as coolant , page no. 2-451, table 2-322 give Npr
of liquid refrigerants
From table we get, Npr of water at 300K = 5.69

22. Heat transfer coefficient
Heat transfer coefficient, hi = 3018.59 W/m2K
Npr

23. STEP 11: CALCULATION OF hio
hio= 2345.16 W/m2K

24. Reynolds number, Re = 26806.37
I. = 0.0138 m
II. Gs = v . ρ = mass flow rate (W) / area (as)
= 0.013 m C = PT – OD of tube = 4.75 x 10-3
Gs = v. ρ = 1068.37 kg/m2s B = 150 mm (assumption)
III. μ from page no. 2-449 (as shown earlier, here check for benzene)
μ = 0.55 cP = 5.5 x 10-4 kg/ms
STEP 12: REYNOLDS NUMBER – SHELL SIDE FLUID

25. STEP 13 : jH FACTOR
Page no : 5 -16, eqn 5-50c 7e PCH
jH = 2.9928 x 10-3

26. STEP 14: HEAT TRANSFER COEFFICIENT
To determine hi , eqn 5-50c is used
Nst = Stanton number
Npr = Prandtl number
To determine hi , eqn 5-50c is used
Nst = Stanton number
Npr = Prandtl number
Heat transfer coefficient,
ho = 7199.88 W/m2K

27. • Thermal conductivity,
k is to be calculated
to calculate Npr
• Page no. 2-450
• Fig 2-33
1. Note temp of shell
side fluid on LHS y-
axis (47.5 C)
2. Identify the
compound number
and mark that in the
fig (Benzene-13)
3. Join these two
points (Temp and
compound number)
and extend to RHS
Y-axis (k)
4. k= 138 mW/mK
= 0.138 W/mK

28. STEP 15 : CALCULATE tw
• Tmean = mean temperature of benzene = 320.5 K
• tmean = mean temperature of water = 299 K
tw = 304.28 K = 31.28oC

29. STEP 16: CALCULATION OF (μw/μb)0.14
• μb = bulk viscosity at mean temp
• μw = viscosity at tw
From page no. 2 -449 calculate μb,benzene, μw, benzene, μb, water, μw, water
(μb)benzene = 5.5 x 10-4 (μb)water = 9x10-4
(μw)benzene = 5.6 x10-4 (μw)water = 8.3 x10-4
hi, corrected = hi/(μw/μb)0.14
water = 3053.09 W/m2K
ho, corrected = ho /(μw/μb)0.14
benzene = 7181.74 W/m2K

30. = 0.002125 m
= 0.0168 m
kw  page no. 2-461, table 2-328- 8e
page no. 2-335, table 2-375- 7e
kw = 9.4 Btu/h.ft2.(oF/ft) = 18.91 W/mK
1/Ui = 5.347 x 10-4 m2K/W
Ui = 1870.17 W/m2K
STEP 17: OVERALL HEAT TRANSFER COEFFICIENT, Ui
Btu/h.ft2.(oF/ft)
Kcal/hr.m.oC
W/mK
X 1.73
X 1.163

31. Q = Ui. Ai. FT. ΔTLMTD
• Q = W Cp (T1 – T2) = w Cp (t2 – t1) = 6.29 x105 J/s
• FT = 0.88
• ΔTLMTD = 293.83 K
• Ui = 1870.17 W/m2K
Ai = 1.3 m2
STEP 18: HEAT TRANSFER AREA, Ai

32. Calculated Dirt factor,
Ui = Uc = 1870.17 W/m2K
n = 228
di = 0.0148 m
L = 8ft = 2.45 m
UDa = 74.55 W/m2K
Rdc = 0.0129 m2K/W
STEP 19: CALCULATION OF DIRT FACTOR, Rdc

33. Page no. 11-25, table 11-3 – 8e
Dirt factor, Rd = 0.03 (oF . Ft2 . H)/ Btu
= 5.283 x 10-4 m2Ks/J
Rdc > Rd : so our assumptions are
correct
(oF . Ft2 . H)/ Btu
M2.K. s/ J
X 0.1761

34. STEP 20: PRESSURE DROP – TUBE SIDE / TUBE PASS
Pressure drop through tube side fluid (ΔP)T = (ΔP)t + (ΔP)r
1. Pressure drop (ΔP)t for std length of tube is given by Fanning eqn:
= 1187.13 N/m2
2. Pressure drop through expansion & contraction, (ΔP)r
= 1629.83 N/m2
(ΔP)T = 2816.96 N/m2
For 1 tube pass- 2 expansion
& 2 contraction losses. At
entrance and exit there is
contraction and expansion.
( 2x0.5)+(2x1)+(2x0.5)=4
So, 4 velocity heads per pass

35. nT = 2
V = 0.64 m/s
ρ = 994.77 kg/m3
L = 2.45m
d = 0.0148 m
f  page no. 6-10, figure 6-9
From table 6-1, surface roughness, ε, for particular material, here assume
commercial steel or wrought iron = 0.0457

37. X axis = NRe = 1.04 x 104
Y axis (RHS) = f = 0.0088
Yaxis (LHS)= ε/di = 0.00309
Use chart to calculate ‘f’
• Find the NRe (1.04x104) on X axis and draw a normal line to the ε/di
(0.00309) curve.
• Then extend the line from the point of intersection with Y axis.
• The Y axis will give the f (0.0088)

38. = 0.0203
• ρ of benzene is measured from page no. 2-98, table – 2-32 7e & 8e
STEP 20: PRESSURE DROP – SHELL SIDE
(ΔP)shell = 56388.20 N/m2

39. Assume allowable pressure drop inside tube = 10psi
= 10 x 6894.757 N/m2
= 68947.57 N/m2
(ΔP) allowable > (ΔP) shell > (ΔP) tube
68947.57 N/m2 > 56388.20 N/m2 > 2816.96 N/m2
So, check for shell side and tube side fluid is correct
Our design is correct.