Guides · Thermal

Heat exchanger sizing with the LMTD method

Duty, log mean temperature difference, overall coefficient and the area that follows.

What the calculation gives you

A heat exchanger moves heat from a hot stream to a cold one through a wall. Sizing it means finding the surface area that transfers the required duty with the temperatures you have, or, for an exchanger that exists, finding the duty it can deliver. The log mean temperature difference (LMTD) method answers the first question directly: duty, temperature difference and overall coefficient in, area out.

The duty

Q = ṁ cp (Tin − Tout) ṁ mass flow of one stream · cp its specific heat · the same Q written for the other stream gives its outlet temperature

An energy balance on either stream fixes the duty; the balance on the other stream then fixes the one temperature you do not know. For a phase change, ṁ times the latent heat replaces the sensible term.

The temperature difference

The driving force varies along the exchanger as the streams approach each other. For pure counterflow or parallel flow its correct average is the log mean:

LMTD = (ΔT1 − ΔT2) / ln(ΔT1 / ΔT2) ΔT1, ΔT2 the hot-to-cold differences at the two ends of the exchanger

Counterflow always gives the larger LMTD for the same four temperatures, and it is the only arrangement in which the cold outlet can rise above the hot outlet. For shell-and-tube and crossflow arrangements the counterflow LMTD is multiplied by a correction factor F (from the standard charts, usually 0.8 to 1); an arrangement with F under 0.75 is a poor one and should be changed, not corrected.

The overall coefficient

1 / (U A) = 1 / (hi Ai) + Rwall + 1 / (ho Ao) + Rf hi, ho film coefficients on the two sides · Rwall the conduction resistance of the tube, ln(ro/ri) / (2π k L) · Rf fouling allowance

The resistances add in series, so the smallest h controls U. Each film coefficient comes from a Nusselt number correlation for its geometry and flow, h = Nu k / D: Dittus–Boelter (Nu = 0.023 Re0.8 Pr0.4) inside tubes in turbulent flow, Churchill–Bernstein or Zukauskas for flow across tubes, and coil-specific correlations for helical tubes, where the secondary flow raises h above the straight-tube value. A copper or thin steel wall adds little; a glycol or oil film on the outside, or a gas on either side, adds a lot. Typical overall values: 800 to 1500 W/m2K water to water, 50 to 250 W/m2K with a gas on one side.

The area

A = Q / (U · F · LMTD)

With the area, the number and length of tubes follow from the chosen tube size, and the pressure drop on each side must then be checked: a compact exchanger with a high velocity has a high U and a pressure drop that may be more than the pump can give. When instead the exchanger is fixed and the outlet temperatures are unknown, the effectiveness–NTU method avoids iterating: NTU = U A / Cmin, and the effectiveness follows from NTU and the ratio of the heat capacity rates for each flow arrangement.

Worked example

The worksheet below sizes a double-helix coil exchanger that recovers heat from a sewage stream at 12 °C into a 60% glycol solution at 4 °C, inside a 6 in pipe shell 2 m tall. From the chosen tube size, coil gap and height it builds the length and the inside and outside surface areas of the two helices, then computes the velocity, Reynolds number and friction factor on the sewage side with the coil correction to the friction factor, the head loss through the bends and the straight runs, the Prandtl and Nusselt numbers on each side, the film coefficients, the overall resistance, and finally the heat the exchanger can recover. The sliders set the flow, the coil gap and the glycol velocity; the pipe size and series or parallel flow are selected from lists.

Common mistakes

  • Arithmetic mean instead of log mean. They agree within 5% only when the two end differences are within a factor of about 1.7 of each other. With a close approach at one end the arithmetic mean overstates the driving force badly.
  • Counterflow LMTD for a multi-pass exchanger. Apply the F factor; a 1-2 shell-and-tube with a temperature cross may have F below 0.8.
  • No fouling allowance. Clean U is the day-one value. Sewage, cooling tower water and process streams foul; standard fouling resistances add 20 to 50% to the area.
  • Properties at the wrong temperature. Viscosity drives Re and Pr; evaluate it at the mean film temperature of each stream. Glycol at 4 °C is ten times more viscous than water.
  • Area on the wrong side. U is defined with respect to one surface, usually the outside. Use the same one throughout.
  • Pressure drop checked last, or not at all. A coil with many turns can have a head loss that halves the flow the pump delivers, which then changes h.

Questions

LMTD or effectiveness–NTU?
LMTD when you know all four temperatures and want the area. NTU when you know the exchanger and want the outlet temperatures. They are the same physics; one just avoids iteration in each case.
How much margin should I add?
Beyond the fouling allowance, 10 to 20% on the area is usual for uncertainty in U. Oversizing has a cost in pressure drop and, for a controlled exchanger, in controllability.
Why is the glycol side the bottleneck?
Its film coefficient is low: high viscosity means a low Reynolds number, often laminar, and a low thermal conductivity. Raising its velocity, or putting it inside the tube rather than outside, is where the design effort pays.

The worked example, live

Series Flow Double Helix Heat ExchangerLive example
Series Flow Double Helix Heat Exchanger
 The required calculations for the heat exhcanger include:
  • Head loss due to pipe friction
  • Effective heat transfer surface area
  • potential heat transfer
1
pipeOD ≡
1"
>
 
 
2
FlowType≡
Parallel
>
 
 
3
qsew ≡
90lmn 
>
 
 
 
Flow rate is divided by two if there is parallel flow

4
qsewage ≡
qsew 2 
 if 
FlowType=1 
qsew 
 otherwise 
Add another condition
>
 
0.00075m3s 
 
5
pipethick ≡
0.068in
 if 
πpeOD =10.3mm 
0.088in
 if 
πpeOD =13.7mm 
0.109in
 if 
πpeOD =21.3mm 
0.113in
 if 
πpeOD =26.7mm 
0.133in
 otherwise 
Add another condition
>
 
 
6
pipeID ≡
6.86mm
 if 
πpeOD =10.3mm 
9.14mm
 if 
πpeOD =13.7mm 
15.7mm
 if 
πpeOD =21.3mm 
20.8mm
 if 
πpeOD =26.7mm 
26.7mm
 otherwise 
Add another condition
>
 
 
 The height of the heat exchanger
7
height≡2000mm
>
 
 
 
 The inner radius of a sch. 40 6" pipe:
OD = 168.3 mm
Wall thickness = 7.11 mm

8
XchangeIDradius ≡203mm2 
>
 
0.1015m
 
9

>
 
 
 A chosen gap between the wall of the outer heat exchanger tube and between the two helical coils
10
gap≡
3mm
>
 
 
 The circumference of the outer helix as governed by the gap distance and pipe diameter
11
outerHelixcircumfrence ≡2·(XchangeIDradius −gap−πpeOD 2 )·π
>
 
0.513964558m
 
  The circumference of the inner helix as governed by the gap distance, pipe diameter, and outer Helix circumference.
12
innerHelixcircumfrence ≡2·(XchangeIDradius −(2·gap+πpeOD +πpeOD 2 ))·π
>
 
 
The number of turns is equal to the height of the exchanger divided by the number of coils that can fit in the tube with a vertical gap between each coil ranging from 0 to 2.5 mm
13
numturns ≡heightπpeOD +gap2  
>
 
57.3065903
 
14
lengthinner ≡innerHelixcircumfrence ·numturns 
>
 
 
15
lengthouter ≡outerHelixcircumfrence ·numturns 
>
 
 
The total length of the heat exchanger is equal to the outer run plus the inner run of the helical coils.
16
lengthtotal ≡
lengthinner +lengthouter 
 if 
FlowType=0 
lengthinner +lengthouter 2 
 otherwise 
Add another condition
>
 
22.9003201m
 
 The total surface area of the double helix is equal to the total length multiplied by the circumference of the pipe
17
AreasurfaceOD ≡lengthtotal ·πpeOD ·π
>
 
2.40291214m2
 
18
AreasurfaceID ≡lengthtotal ·pipeID ·π
>
 
1.92089085m2
 
19
Areatotal ≡AreasurfaceOD +AreasurfaceID 
>
 
4.32380299m2
 
 The next section will be used to calculate the head loss due to friction in the pipes
 Minor losses due to fittings are as follows
 
Le/D values
20
elbowled ≡20
>
 
 
21
returnBendled ≡50
>
 
 
 Losses due to pipe bends in 1" pipe
22
RoOuter ≡XchangeIDradius 
>
 
0.1015m
 
23
RoInner ≡XchangeIDradius −pipeOD ·1−gap
>
 
0.0651m
 
24
Do ≡pipeOD 
>
 
0.0334m
 
25
router ≡RoOuter −Do 2 
>
 
0.03405m
 
26
rinner ≡RoInner −Do 2 
>
 
 
27
relRadouter ≡router pipeID  
>
 
1.2752809
 
28
relRadinner ≡rinner pipeID  
>
 
0.593632959
 
29
largeBendouter ≡13·4·numturns 
>
 
 
30
largeBendinner ≡24·4·numturns 
>
 
 
 K values

31
exitloss ≡1
>
 
 
 Friction Factor Calculation

32
μsewage ≡1.15·10−3Pas
>
 
 
33
ρsewage ≡1000kgm3 
>
 
 
34
ε≡0.0015mm
>
 
 
35
g≡9.81ms2 
>
 
 
36
vsewage ≡qsewage (pipeID 2 )2·π 
>
 
1.33951894ms 
 
37
ReNsewage ≡ρsewage ·πpeOD ·vsewage μsewage  
>
 
38904.2893
 
38
ƒs ≡0.25(log10(13.7·(pipeID ε ) +5.74ReNsewage 0.9 ))2 
>
 
0.0221861916
 
 


The friction factor in a helical coil can be calculated as a ratio of the friction factor in a straight pipe
39
Deouter ≡ReNsewage ·(pipeID outerHelixcircumfrence  )12 
>
 
 
40
Deinner ≡ReNsewage ·(pipeID innerHelixcircumfrence  )12 
>
 
 
41
ƒcOuter ≡ƒs ·(1−(1−(11.6Deouter  )0.45)2.2)−1
>
 
 
42
ƒcInner ≡ƒs ·(1−(1−(11.6Deinner  )0.45)2.2)−1
>
 
 
 Total Headloss

43
led≡elbowled +returnBendled 
>
 
 
44
k≡exitloss 
>
 
 
45
hL ≡(ƒs ·(lengthtotal pipeID  +led)+k)·vsewage 22·g 
>
 
1.97373347m
 
46
hlBends ≡(ƒs ·(led+largeBendouter +largeBendinner )+k)·vsewage 22·g 
>
 
17.4421709m
 
47
hl≡((ƒs ·led)+k+ƒcOuter ·(lengthouter pipeID  )+ƒcInner ·(lengthinner pipeID  ))·vsewage 22·g 
>
 
34.1790231m
 
 
Potential Heat transfer


48
Tsewage ≡12°C
>
 
 
49
Tref ≡4°C
>
 
 
50
CpGlycol ≡(0.748·4186.8)Jkg·K 
>
 
 
51
CpWater ≡4.186Jg·K 
>
 
 
52
thermcondGlycol ≡0.206Wm·K 
>
 
 
53
thermcondCopper ≡401Wm·K 
>
 
 
54
thermcondWater ≡0.591Wm·K 
>
 
 
55
μglycolSolution ≡0.009Pa·s
>
 
 
56
ρglycol ≡(1.11·1000)kgm3 
>
 
 
 
Finding h for the glycerol solution.  Assuming a 60% glycol solution at 4 °C.
57
vglycol ≡
1ms 
>
 
 
58
ReNglycol ≡vglycol ·pipeOD ·ρglycol μglycolSolution  
>
 
4119.33333
 
59
Prglycol ≡(μglycolSolution ·CpGlycol )thermcondGlycol  
>
 
136.822998
 
60
Cglycol ≡
0.693·Prglycol 13 
 if 
ReNglycol >40andReNglycol <4000 
0.193·Prglycol 13 
 otherwise 
Add another condition
>
 
0.994512722
 
61
Cglycol ≡0.693·Prglycol (13 )
>
 
3.57097055
 
62
nglycol ≡
0.466
 if 
ReNglycol >40andReNglycol <4000 
0.618
 otherwise 
Add another condition
>
 
0.618
 
63
Nuglycol ≡Cglycol ·(ReNglycol )nglycol 
>
 
611.997031
 
64
hglycol ≡Nuglycol ·thermcondGlycol πpeOD  
>
 
3774.59247WK·m2 
 
 
Finding h for sewage.


65
Prsewage ≡μsewage ·CpWater thermcondWater  
>
 
8.14534687
 
66
NuSewage ≡0.0155·Prsewage 0.6·ReNsewage 0.83
>
 
352.026503
 
67
hsewage ≡NuSewage ·thermcondWater pipeID  
>
 
7792.04731WK·m2 
 
 
Overall R can be calculated using the following formula

68
Ro ≡1AreasurfaceOD ·hsewage  +ln((πpeOD 2 )pipeID 2  )2·π·lengthtotal ·thermcondCopper  +1AreasurfaceID ·hglycol  
>
 
0.000195208889KW 
 
 
Overall heat transfer rate in the system

69
Q≡Tsewage −Tref Ro  
>
 
40981.7403W
 
70

>
 
 
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