Laboratory 2 Pre lab
Heat Exchanger Design
The pre lab report has several major sections:
I. Overall Energy Balance
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Application to the hot (h) and cold (c) fluids
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Assume negligible heat transfer between the exchanger and its surroundings and
negligible potential and kinetic energy changes for each fluid
q h = m& h (ih ,i − ih ,o )
q h = m& c (ic ,i − ic ,o )
where I is fluid enthalpy.
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Assume no phase change and constant specific heat
q h = m& h C p ,h (Th ,i − Th ,o ) = C h (Th ,i − Th ,o )
q h = m& h C p ,c (Th ,i − Th ,o ) = C c (Th ,i − Th ,o )
With the energy balance, we have q h = q c .
II. Heat Exchanger Analysis: Log Mean Temperature Difference
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Use of the Log Mean Temperature Difference
A form of Newton’s Law of cooling may be applied to heat exchangers by using
Log Mean Exchanger Analysis (LMEA) (See the detailed derivation 11.3 in the
textbook)
q = UAΔTlm
ΔTlm =
ΔT2 − ΔT1
ln (ΔT2 ΔT1 )
For counter flow heat exchanger in following figure, we have
ΔT1 ≡ Th ,1 − Tc ,1 = Th ,i − Tc ,o
ΔT2 ≡ Th , 2 − Tc , 2 = Th ,o − Tc ,i
For parallel flow heat exchanger in following figure, we have
ΔT1 ≡ Th ,1 − Tc ,1 = Th ,i − Tc ,i
ΔT2 ≡ Th , 2 − Tc , 2 = Th ,o − Tc ,o
Note that, for the same inlet and outlet temperatures, the log mean temperature
difference for counter-flow exceeds that for parallel flow, ΔTlm ,CF > ΔTlm , PF .
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Computational Features/Limitations of the LMTD Method
(1) LMTD method may be applied to design problems for which the fluid flow rates
and Ti as well as To are known
(2) If the LMTD method is used in performance calculations for which both outlet
temperatures must be determined from knowledge of the inlet temperatures,
solution procedure is iterative.
(3) For both design and performance calculations, the effectiveness-NTU (Number
of Transfer Units) method may be used without iteration.
III. Effectiveness-NTU Method
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Number of Transfer Units
NTU =
UA
C min
Note that a dimentionless parameter which magnitude affects HX performance.
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Heat Exchanger Effectiveness
ε=
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q
q max
Maximum Heat Rate
q min = C min (Th ,i − Tc ,i )
⎧C h if C h < C c
C min = ⎨
⎩C c if C c < C h
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Performance Calculations
ε = f (NTU , C min / C max )
where C r = C min C max . See the table 11.3 and Figure 11.10 and 11.11.
IV. Prelab Analysis:
1.
Analysis two case: Parallel flow and Counter flow configuration with two kinds of
flow configurations. That means that you have four cases in total: parallel flow with
7, counter flow with 19 tubes, parallel flow with 19 and counter flow with 19 tubes.
The flow rate of cooling water through each inner tube (6.35 mm) is 1.0 litre/min,
while the flow rate of hot water through the outer annulus (64 mm) is 0.5 litre/min.
The hot and cooling water enter at temperature 60 ºC and 20ºC, respectively and the
hot water leaves at a temperature of 40 ºC. The tubes are all 65.9 m long. There are
the following assumptions: Negligible heat loss to the surroundings, Negligible
kinetic and potential energy changes, constant properties, negligible tube wall
thermal resistance and fouling factors and fully developed conditions for the cooling
water and hot water. See the Table A 6 for all properties of water. And one litre is
equal to 0.001 cubic metre.
2.
Compute the overall convection coefficient U coming from the energy balance
equation (11.14) for each case (four cases in total).
3.
Conduct an analysis to determine the effects of the number of tubes, flow
configuration (parallel v.s. counter flow) on performance by using effectivenessNTU relation (use 11.28a and 11.29a only in your textbook). Prepare two plots
(effectiveness-NTU relation) for parallel and counter flow arrangements using
Matlab or Excel. You must indicate in your report which case you would like to use
to remove the heat most efficiently.
4.
Prepare a plan for the experiments you will conduct in lab#2 (Note: groups will share
data.)
5.
Each group is to submit a brief description of their analysis, prediction, and plan at
the beginning of lab section (b).