6 HEAT EXCHANGERS
Incropera Chapter 11, Cengel Chapter 11, Welty et al. Chapter 11 THIS WEEK:
• What are heat exchangers and why are they used?
• What types of exchangers are available?
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• How do we determine the size and “duty” of a heat exchanger?
• Combine several aspects of CHE2163 thus far (conduction,
convection, resistances in series) to analyse heat exchangers Learning objectives. By the end of week 6 you should be able to:
• Perform heat exchanger design calculations using both the LMTD and NTU methods, for a range of different heat exchanger configurations (e.g. double pipe, shell and tube, cross-flow heat exchangers);
• Clearly identify the underlying assumptions made when performing heat exchange calculations; identify when these calculations are violated, and explain the implications for your design calculations and heat exchanger operation;
• Determine the overall heat transfer coefficient for a heat exchanger from first principles (based on convection correlations), or from heat exchanger operation (using design calculations), including the effects of fouling.
1.1 TYPES OF HEAT EXCHANGERS
DERIVATION OF KEY HEAT EXCHANGER EQUATIONS
ENERGY BALANCE OVER HEAT EXCHANGER
Rate of heat transferred from hot stream to cold stream at any point in the heat exchanger:
q = UADT Equation H-1 where ΔT =Th-Tc is the driving force for heat transfer.
However, the driving force changes along the heat exchanger:
HOW TO DETERMINE THE RATE OF HEAT TRANSFERRED FROM HOT STREAM TO COLD STREAM?
Consider a small element of a counter-current heat exchanger:
The rate of heat lost from the hot stream if there is no phase change:
Δx Rcond1 o TH OUT
dq=m!hCph(Thin -Thout)=m!hCphTh
Equation H-2
The rate of heat gained by the cold stream (no phase change): dq=m!cCpc(Tcout -Tcin)=m!cCpc____ EquationH-3
dq = UdADT Equation H-4 =
Change in driving force over the element of the heat exchanger:
dDT=DT -DT 21
Equation H-5
Assuming that all heat loss from the hot stream is gained by the cold stream, i.e. no exchange with environment, sub Equations H-34 and H-35 into Equation H-36, then introduce Equation H-41 and integrate over the length of the heat exchanger:
LOG MEAN TEMPERATURE DIFFERENCE METHOD
Rate of heat lost from the hot stream (duty):
q = m! h C p h ( T h i n – T h o u t )
Rate of heat gained by the cold stream (duty):
q = FUADT Equation H-8 lm
• F is calculated from charts, and depends on configuration and temperature gradients within the heat exchanger
• ΔTlm accounts for variation in driving force along the length of the heat exchanger
Equation H-9
Equation H-6
Equation H-7 Rate of heat transferred from hot stream to cold stream:
q=m!C (T -T ) c pc cout cin
æDT ö lnç 2 ÷
Under the following assumptions, Equations H-37 to H-39 are equivalent:
• _____________________________________________ • _____________________________________________ • _____________________________________________ • _____________________________________________
For a counter-current exchanger,
For a co-current exchanger,
ΔT2=Thout-Tcin; ΔT1=
ΔT2=Thout-Tcout; ΔT1=
RESISTANCES IN SERIES
Overall heat transfer coefficient UA=1/Rtotal
Equation H-10
Rconv inside Rcond pipe Rconv outside
qqq A=A=A=
Rtotal = Rconv inside + Rcond pipe + Rconv outside =
Equation H-11
HOW TO INCLUDE ADDITIONAL RESISTANCE DUE TO FOULING
Fouling increase the resistance of the heat exchanger, ie
Rdirty = Rclean + Rfouling Equation H-12
The characteristics of fouling can be calculated from the difference in operation of dirty and clean exchangers:
Rfouling = Rdirty – Rclean
(U A ) d i r t y
– 1 Equation H-13 (U A ) c l e a n
(NB – can use U instead of UA if wall thickness is neglected: This is often a valid assumption).
Note that it is often easier to express problems, first in terms of resistance, before introducing U, because
1=1-1 (UA) fouling (UA)dirty (UA)clean
A common error is to write
(UA) fouling = (UA)dirty – (UA)clean
Which is not the same (eg 1⁄4 = 1⁄2 – 1⁄4 but 4 ≠ 2-4 !!)
Determine the heat transfer area required for a heat exchanger constructed from 25.4-mm outer diameter tubes to cool 25,000 kg/hr of 95% ethanol (Cp = 3.8kJ/(kgK)) from 66°C to 38°C. 25,000 kg/hr of cooling water at 10°C can be supplied to the exchanger. An estimated value of the overall heat transfer coefficient based on the outer tube area is 570 W/(m2K). Compare the following geometries:
a) Parallel-flow shell and tube exchanger with 1 tube pass
b) Counter-current shell and tube exchanger with 1 tube pass c) Cross flow heat exchanger with 1 tube pass and 1 shell pass,
shell side fluid unmixed
NTU METHOD
Maximum heat transfer in a heat exchanger:
qmax = Cmin ΔTmax, ΔTmax= Equation H-14
where Cmin is the thermal mass of the stream with the smaller thermal mass, i.e. Cmin=min(mc Cp c , mh Cp h)
q = eqmax Equation H-15 where heat exchanger effectiveness ε can be determined from charts
or equations:
Heat exchanger effectiveness
ε = f(configuration, NTU, Cmin/Cmax) Equation H-16
where NTU Number of Transfer Units is a measure of the size of the heat transfer unit, i.e. UA relative to the minimum thermal mass:
NTU = UA / Cmin Equation H-17
A well-baffled single shell 2 tube pass heat exchanger has been used under the following conditions. 2500 kg/hr of oil (Cp = 2.09 kJ/(kgK)) entered the tube at 70°C and left at 38°C . Flow of the oil through the tubes was turbulent. Water on the shell side entered at 15°C and left at 27°C. A change in service conditions requires the exchanger to cool a similar oil from a temperature of 93°C. The new oil flow rate is reduced to 1900 kg/hr. What is the temperature of the cooled oil if the same inlet temperature and flow rate of cooling water is used?
HEAT EXCHANGER SUMMARY AND PROBLEM SOLVING GUIDELINES
ALWAYS USE THE SIMPLEST CONCEPTS FIRST
Can you solve the problem with energy balance over either stream (Eqns H-37and H-38) ? If yes use LMTD Method, otherwise use NTU method.
LOG MEAN TEMPERATURE DIFFERENCE (LMTD) METHOD
• If inlet and outlet temperatures are known (or can calculate them from Eqns H-37 and H-38), can determine average driving force ΔTlm and correction factor F.
• From ΔTlm and F, determine UA for known q, or vice versa. NTU METHOD
If outlet temperatures are unknown, can’t determine F and hence
can’t use LMTD method, so use NTU method.
1. Determine Cmin=min(m!cCpc,m!hCph )
2. NTU=UA/Cmin
3. Determine Cmax=max(m!cCpc,m!hCph )
4. ε = f(configuration, NTU, Cmin / Cmax); determine from charts or
from equations.
5. qmax = Cmin (Thin-Tcin )
6. q = εqmax
1.2 SUMMARY OF COMMON ASSUMPTIONS FOR HE PROBLEMS
• Steady state;
• No phase change: heat lost or gained from each stream is due
solely due to temperature change, i.e. sensible heat, and there
is no phase change in either stream;
• Constant and uniform thermal properties;
• All heat lost by the hot stream is gained by the cold stream.
1.3 CHARTS & TABLES
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