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Summer 2005
Double Pipe Heat Exchanger Experiment
OBJECTIVE
- To understand the basic operation of heat exchangers.
- To demonstrate the basic equations of heat exchanger operation.
BACKGROUND
A heat exchanger is a heat transfer device whose purpose is the transfer of energy from one moving fluid stream to another moving fluid stream. It is the most common of heat transfer devices and examples include your car radiator and the condenser units on air conditioning systems. The overall energy transfer is dictated by thermodynamics and the First Law. To perform the thermodynamic analysis on a heat exchanger, we consider the control volume shown in Figure 1.
Figure 1. Control Volume Model
c.v.
Hot Stream
Cold Stream
W = 0, Q = 0
Q (^) int
Note that although there is heat transfer from the hot fluid stream to the cold fluid stream, there is no work or heat transfer from the control volume (c.v.) to the surrounds. The first law for this control volume is then written as
H&^ in =H&out (1)
Considering that we have two flows into the control volume and two flows out of the control volume, we may write a more specific form of the first law as
m& (^) H hˆH,in+m&ChˆC,in=m&HhˆH,out+m&ChˆC,out (2)
or rearranging by grouping the streams
m& H (h ˆH,in-hˆH,out) =m&C(h ˆC,out-hˆC,in) (3)
This, then, is the most general form of the First Law for a heat exchanger. However, for many heat exchangers there is not a phase change occurring for either fluid stream and the fluids are either incompressible liquids or ideal gases. Under these two conditions, we may represent the enthalpies in terms of temperature (a much more
measurable quantity) by using the appropriate equation of state ( dh^ ˆ=cpdT ), which
will introduce the specific heat. Then our First Law becomes in final form
(m & c p )H (T H,in-TH,out) =(m &cp)C (T C,out-TC,in) (4)
Recall that in this transformation from enthalpies to temperatures, we have assumed constant specific heats. To be consistent, we evaluate the specific heat of each fluid at
the linear average between its inlet and outlet temperature, ⎟ ⎠
Tin +Tout .
Unfortunately, thermodynamics does not tell the whole story of a heat exchanger's performance. To achieve the energy transfer predicted by the First Law the principles of convection and conduction heat transfer must be applied. To apply these principles we consider a very small length of the heat exchanger, ∆x, as shown Fig. 2.
∫^ (^ )
L
0
q&^ = UPTH (x)-TC(x)dx (7)
which from our thermodynamics is also equal to
p C(^ C,out C,in)
p H H,in H,out
L
0
H C
=(mc ) T -T
q= UPT (x)-T (x)dx (mc ) T -T
&
We now have a relationship between the heat transfer and thermodynamics. The difficulty with utilizing Eq. (7) lies in evaluating the integral. In order to evaluate the integral, we must know the functional forms of the temperatures, TH and TC. The only way to do this is to write the appropriate differential energy equation for both fluid streams and solve these coupled equations for the temperatures. It proves convenient at this juncture to introduce the concept of an average temperature difference between the two fluid streams. We modify Eq. (7) by noting that by definition
∫ (^ )
∫
L
0
H C avg avg
L
0
H C avg
UPT (x)-T (x)dx=UPL T =UA T
T (x)-T (x)dx= T L
where ∆Tavg is the average temperature difference between the hot and cold fluids as they pass through the heat exchanger. Then our heat transfer is given by
q&^ =UA∆Tavg (10)
The functional form of ∆Tavg can be extracted from the temperature solutions for the differential energy equations noted above. For the simple concentric tube heat exchanger of this experiment, we find that
T
T
ln
T - T
T =
1
2
2 1 avg
⎭
where ∆T 2 is the temperature difference between the two fluid streams at one physical end of the heat exchanger and ∆T 1 is the temperature difference between the two fluid streams at the other physical end of the heat exchanger. For a counterflow heat
exchanger, the hot fluid enters at one physical end and the cold fluid enters at the other physical end so that ∆T 2 and ∆T 1 can be related to hot and cold fluid inlet and outlet temperatures by
T =T -T
T =T -T
2 H,out C,in
1 H,in C,out ∆
Similar expressions may be obtained for a parallel flow heat exchanger.
Unfortunately, the flow in most heat exchangers is so complicated that a simple solution to the differential equation is not possible and we are forced to take another approach. This second approach is based upon the dynamic scaling and dimensionless parameter work you saw in your fluid mechanics course. We begin with some definitions:
Flow Heat Capacity C^ =m&^ cp ,e.g.,CH=(m^ &cp)H (13)
Minimum Heat Capacity Cmin, the smaller of CH and CC Maximum Heat Capacity Cmax, the larger of C (^) H and CC
Heat Capacity Ratio ,(^0 C^1 )
C
C
C = R
max
min R ≤^ ≤ (14)
Effectiveness
C (^ T -T )
C T -T
C T -T
C T -T
q
q
min H,in C,in
C C,out C,in
min H,in C,in
H H,in H,out
maximumpossible
actual &
ε
Number of Transfer Units C
UA
NTU=
min
Our next step would be to employ dynamic similarity to obtain a relationship among our three dimensionless parameters, C (^) R, ε, and NTU. We can partially show this by beginning with Eq. (10), where our heat transfer is given by
q&^ =UA∆Tavg (17)
Figure 3. Effectiveness - NTU Relationship for Counterflow Heat Exchanger
0.0 1.0 2.0 3.0 4.0 5. NTU
Effectiveness Cr = 1. Cr = 0. Cr = 0. Cr = 0. Cr = 0.
A concentric tube or double pipe heat exchanger is one that is composed of two circular tubes. One fluid flows in the inner tube, while the other fluid flows in the annular space between the two tubes. In counterflow, the two fluids flow in parallel, but opposite directions. In parallel flow the two fluids flow in parallel and in the same direction. The above graph may also be represented by an equation as
1 - C exp{- NTU( 1 - C )}
1 - exp-NTU 1 - C
R R
R ⋅
ε (20)
A final note about this equation and its corresponding graph concerns effectiveness behavior when the NTU is small. When the NTU is less that 0.5, all of the C (^) R curves
collapse. Since the graph has a CR = 0 curve, one could take the effectiveness values at CR = 0 to be valid for all CR's when NTU is small. This yields a much simpler equation for cases when CR = 0 or NTU < 0.5 of the form
ε = 1 - exp {− NTU} (21)
PROCEDURE
- Make the appropriate length measurements on the heat exchanger so you can calculate the heat transfer area.
- With the help of your lab instructor turn on the system and set it up for parallel flow.
- Allow the system to come to steady state and record inlet, outlet, and intermediate temperatures of the cold and hot water.
- Repeat the experiment for at least five different flow rates of hot and cold water while maintaining the same C (^) min/Cmax ratio.
- Repeat steps 3-5 with the exchanger in counter-flow configuration.
DATA ANALYSIS
- Each case should be recorded on the Excel spread sheets provided.
- From the experimental data calculate the overall heat transfer coefficient. Plot it versus water velocity.
- Plot the effectiveness versus number of transfer units for this exchanger and compare it to the theoretical relationship.
- Calculate the uncertainty error in your parameters.
- Provide one sample hand calculation of your data processing.
