Heat Exchanger Lab Experiment

SOLUTIONS

May. 19th, 2017

ME 331 – Introduction to Heat Transfer

Abstract
An experiment was conducted to measure the performance of a simple, concentric tube
heat exchanger operating in both parallel and counter-flow configurations. Separate cold
and hot water supplies were fed into the heat exchanger, and measurements were made of
the inlet temperature, outlet temperature, and flow rate for both streams and for both flow
configurations. For the parallel flow configuration, heat exchanger performance metrics
were calculated to be the following: NTU = 1.20,  = 51%, Ui,NTU = 7.01x103 W/(m2 K). For
the counter-flow configuration, the same metrics were calculated to be: NTU = 1.38,  =
63%, Ui,NTU = 8.17x103 W/(m2 K). The overall heat transfer coefficient U based on the sum
of predicted conductive wall resistance and two convective resistances from empirical
correlations was within 30% of the actual value UNTU for the parallel flow heat exchanger.
For the counter flow heat exchanger, the overall heat transfer coefficient U based on the
sum of predicted conductive and convective resistances was within 15% of the actual
value. As expected, the counter-flow configuration was more effective than the parallel-
flow configuration, for the same heat exchanger geometry.
Introduction

In this heat exchanger laboratory experiment, temperature and flow rate measurements
were taken for a simple concentric tube heat exchanger in order to analyze the heat
exchanger’s performance in both parallel and counter flow configurations. All of the
following equations are taken from the 6th edition of Incropera et. al.’s “Introduction to
Heat Transfer.” The derivation of these equations, along with a more complete description
of heat exchanger design and analysis can be found in the ME 331 textbook [1].

Heat exchanger analysis starts from the application of an energy balance on each of the
fluid streams. The heat gained by the cold fluid is given by
, (Eq. 11.7b)
and the heat lost by the hot fluid is given by
. (Eq. 11.6b)
The subscripts i and o are use to distinguish between inlet and outlet temperatures. In an
ideal heat exchanger, qh and qc are equal to each other for the case of zero heat exchange
with the environment.

The “heat capacity rate” (C) is defined as the product of the mass flow rate and the specific
heat. For the hot fluid,

and for the cold fluid,

.

One measure of the performance of a heat exchanger is its “effectiveness” (), but in order
to define the effectiveness one must first determine the heat exchanger’s “maximum
possible heat transfer rate” (qmax):
. (Eq. 11.18)
Cmin is the minimum of the hot and cold heat capacity rates, since this is the fluid stream
that limits the heat exchange process thereby determining the maximum possible heat
transfer rate.

The heat exchanger effectiveness is then the ratio between the actual heat transfer rate and
the maximum possible heat transfer rate:

. (Eq. 11.19)

The number of transfer units (NTU) is a metric used for quantifying heat exchanger
performance. It is defined as:

(Eq. 11.24)
where U is a overall heat transfer coefficient and A is the corresponding surface area. For
parallel flow heat exchangers, NTU can be found from the following equation:
 , (Eq. 11.28b)
and for counter flow heat exchangers

, (Eq. 11.29b)
where in both equations

An alternative approach for determining the overall heat transfer coefficient for a heat
exchanger is to use a resistance network to model the heat transfer from the hot to the cold
fluid. Solving for U using the following equation for a concentric tube geometry:

. (Eq. 11.5)
In this equation, Rtotal is the total resistance to the heat transfer between the hot and cold
fluids and is equal to the sum of the following (from left to right): convection resistance on
the inner surface of the inner tube, resistance due to fouling on the same inner surface,
conduction resistance through the tube wall separating the hot and cold fluids, resistance
due to fouling on the outer surface of the inner tube, and finally, convection resistance on
the same outer surface of the inside tube.

In order to determine the convection coefficients on the inner and outer surfaces of the
inside tube, convection correlations for internal turbulent flow were used from the
textbook [1]. The hot water stream flows through a standard circular tube geometry. Since
the flow was turbulent (ReD = 1.2 x 104 > 2300) and each of the four straight sections of the
tubing were long enough to model the entire heat exchanger as being fully developed (L/D
= 150 > 10), determining the appropriate correlation was straightforward. The convection
coefficient for the hot water flow on the inner surface of the inner pipe was found using the
Dittus-Boelter equation for cooling since Ts < Tm (See footnote below Eq. 8.60).

The cold water flows through an annular geometry. At the end of Section 8.6 of the
textbook, it states that for fully developed turbulent flows, the convection coefficient for
annular geometries may be found using the hydraulic diameter and the Dittus-Boelter
equation. The hydraulic diameter for annular tubes is:
, (Eq. 8.71)
and for the cold flow, since Ts > Tm, the Dittus-Boelter equation for heating was used.
Procedure
Part A: Parallel-Flow Configuration
1. Verify that the heat exchanger is running in the parallel-flow configuration with the
cold water flowing through the outer tube.
2. Verify that the heat exchanger has reached a steady-state by inspecting the
temperature time history plot on the computer’s data acquisition program.
3. If the heat exchanger is at steady-state, record data: five thermocouple
temperatures (Th,in, Th,out, Tc,in, Tc,out, Tambient) and two volumetric flow rates for both
the hot and cold water streams.

Part B: Switching Flow Configuration

1. Turn off the cold water pump.
2. Unscrew the cold water supply and outlet hoses and let drain.
3. Reattach the hoses to the opposite connections.

Part C: Counter-Flow Configuration

1. Verify that the heat exchanger is running in the counter-flow configuration with the
cold water flowing through the outer tube.
2. Turn on the cold water pump and wait for the heat exchanger to reach steady-state.
3. Verify that the heat exchanger has reached a steady-state by inspecting the
temperature time history plot on the computer’s data acquisition program.
4. If the heat exchanger is at steady-state, record data: five thermocouple
temperatures (Th,in, Th,out, Tc,in, Tc,out, Tambient) and two volumetric flow rates for both
the hot and cold water streams.
Results

Experimental Data:

Flow Tcold,in (℃) Thot,in (℃) Tcold,out (℃) Tℎ𝑜𝑡,out(℃) Tambient (℃) 𝑄cold (gal 𝑄hot (gal
Configuration /min) /min)
Parallel 20.0 43.5 32.1 34.6 24.6 3.81 5.50
Counter 23.7 53.1 42.2 39.7 24.4 3.87 5.54

Dimensions and material properties

ID of inner OD of inner ID of outer OD of outer Length(‘’) k w (𝑊/𝑚 ∙ 𝐾)
tube(‘’) tube(‘’) tube(‘’) tube(‘’)
3/8 ½ 5/8 3/4 225 52

Calculations:

1. Flow properties:
Parallel flow:
Tmean(K) ρ(kg/m3) cp (J/kg ∙ K) μ ∙ 106 (N ∙ s/m2) Pr k(W/m ∙ K)
Cold 299.2 997.2 4179 872 5.96 0.6119
Hot 312.2 992.2 4178 667 4.42 0.6307
Counter flow:
Tmean(K) ρ(kg/m3) cp (J/kg ∙ K) μ ∙ 106 (N ∙ s/m2) Pr k(W/m ∙ K)
Cold 306.1 994.6 4178 752 5.07 0.6218
Hot 319.5 983.4 4180 582 3.81 0.6394

2. Reynolds number for both flows:

Parallel flow configuration:
4 ρh ∀̇h 4 ρc ∀̇c
Re𝐷,ℎ𝑜𝑡 = = 6.90×104 , ReD,cold = = 1.23×104
πDμ π(Do + Di)μ
Counter flow configuration:
4 ρh ∀̇h 4 ρc ∀̇c
Re𝐷,ℎ𝑜𝑡 = = 7.94×104 , ReD,cold = = 1.44×104
πDμ π(Do + Di)μ

3. Calculate q (actual) from the cold flow stream. Calculate q (actual) from the hot flow
stream.
Compare by calculating percent difference.
Parallel flow configuration:

qcold = ρc cp,c ∀̇c (Tc,out − Tc,in ) = 1.20×104 W;

qhot = ρh cp,h ∀̇h (Th,in − Th,out ) = 1.28×104 W; difference = 6%

Counter flow configuration:

qcold = ρc cp,c ∀̇c (Tc,out − Tc,in ) = 1.88×104 W;

 qhot = ρh cp,h ∀̇h (Th,in − Th,out ) = 1.95×104 W; difference = 3%

4. Calculate q (max), heat exchanger effectiveness ε, and number of transfer units NTU.
Parallel flow configuration:

Cmin = Cc = ρc cp,c ∀̇c = 1002W/K

Cmax = Ch = ρh cp,h ∀̇h = 1438W/K
q𝑚𝑎𝑥 = C𝑚𝑖𝑛 (Th,in − Tc,in ) = 2.35×104 W;
q
ε= = 51%, NTU = 1.20
𝑞𝑚𝑎𝑥

Counter flow configuration:

Cmin = Cc = ρc cp,c ∀̇c = 1014W/K
Cmax = Ch = ρh cp,h ∀̇h = 1344W/K
q𝑚𝑎𝑥 = C𝑚𝑖𝑛 (Th,in − Tc,in ) = 2.98×104 W;
q
ε= = 63%, NTU = 1.38
𝑞𝑚𝑎𝑥

5. Calculate the overall heat transfer coefficient U based on NTU.

Parallel flow configuration:
Ui,NTU = 7.01×103 𝑊/𝑚2 ∙ 𝐾, Uo,NTU = 5.26×103 𝑊/𝑚2 ∙ 𝐾
Counter flow configuration:
Ui,NTU = 8.17×103 𝑊/𝑚2 ∙ 𝐾, Uo,NTU = 6.13×103 𝑊/𝑚2 ∙ 𝐾

6. Calculate the overall heat transfer coefficient U based on conduction resistance and
convection resistances from empirical correlations. Compare this overall heat transfer
coefficient with U based on the NTU method.
Equation 11.5 is used, but without fouling resistances.
𝑙𝑛(𝐷2 /𝐷1 ) K
Rw = = 1.54×10−4 - ; k w = 52W/(m2 ∙ K)
2𝜋𝑘𝑤 𝐿 W
Parallel flow configuration:
Nu h(W/m2 ∙ 𝐾) A(m2) U(W/m2 ∙ 𝐾)
Cold(outer) 90.0 17342 0.228 6690
Hot(inner) 355.2 23520 0.171 8920

Ui,∑ resistances = 8.92×103 𝑊/𝑚2 ∙ 𝐾, Uo,∑ resistances = 6.91×103 𝑊/𝑚2 ∙ 𝐾

Counter flow configuration:

Nu h(W/m2 ∙ 𝐾) A(m2) U(W/m2 ∙ 𝐾)
Cold(outer) 97.5 19091 0.228 7107
Hot(inner) 373 25068 0.171 9476

Ui,∑ resistances = 9.47×103 𝑊/𝑚2 ∙ 𝐾, Uo,∑ resistances = 7.11×103 𝑊/𝑚2 ∙ 𝐾

Discussion
1. The accuracy of temperature and flow rate measurements.
Type K thermocouples have standard error of 2.2°C according to the official OMEGA site
[2]. The flow rate uncertainty is estimated to be 0.05 gal/min based on a bucket test
calibration.
2. The accuracy of the q values calculated based on temperature and flow rate
measurement errors.
Assuming no error in the property terms,
2 2
∆q ∆∀̇ √2∆T
= √( ) + ( ) = 0.12, ∆q = 1.4×103 𝑊
𝑞 ∀̇ 𝑇𝑜𝑢𝑡 − 𝑇𝑖𝑛
3. The effect of neglecting heat exchange with the environment. Discuss differences
between q values calculated from the cold and hot streams.
In both cases, qcold < qhot . In parallel flow configuration, qhot − q𝑐𝑜𝑙𝑑 = 761 𝑊; in
counter flow configuration, qhot − q𝑐𝑜𝑙𝑑 = 648 𝑊. Considering the error estimated in
part 2 (above) far exceeds these differences, it cannot be determined whether this
difference is physical or due to measurement error. Therefore the effect of neglecting
heat exchange with the environment cannot be investigated with the data acquired.
However, since the average mean temperature of the cold side is well above ambient
temperature for both flow configurations, the “cold” stream will lose heat to the
ambient environment. Therefore, the magnitude of qcold is expected to be less than the
magnitude of qhot.
4. The accuracy of your estimates of h.
According to the textbook correlations, the estimates of h are expected to be within
30% of the actual values. However, the estimated h values and associated convective
resistances are for clean tubes, and therefore they do not include the effects of fouling
resistance.
5. Discuss differences in the calculated overall heat transfer coefficient U using the NTU
method and the resistance method.
The values of U calculated using the NTU method are based on the actual measured
data; therefore, the values of 1/(UA) from the NTU method represent an actual overall
resistance to heat exchange between the hot and cold streams.
On the other hand, values of 1/(UA) that are based on a sum of resistances account
only for the conductive resistance and two convective resistances on clean tube
surfaces. Using the sum of resistances method, the only way to independently
determine an accurate value of the overall resistance is to specifically account for and
include the resistances due to fouling. Unfortunately, independent determination of the
fouling resistances is not possible because no detailed information about the interior
tube surfaces is available.
The overall resistance 1/(UA), where the value of U is based on the NTU method, is
expected to be greater than the resistance based on the sum of conduction and
convection resistances, since this latter sum does not include fouling resistance.
However, using the sum of resistances, it is possible to estimate the total resistance
that occurs due to fouling. Based on the actual overall resistance from the NTU
 method, the total of the two fouling resistances can be obtained by difference, using
equation 11.5.

. (Eq. 11.5)

In the following equation, fouling resistance is estimated by difference.

′′ ′′
R𝑓,𝑖 R𝑓,o 1 1 1 𝑙𝑛(𝐷2 /𝐷1 )
+ = − − −
𝐴𝑖 𝐴𝑜 (UA)NTU ℎ𝑖𝐴𝑖 ℎ𝑜𝐴𝑜 2𝜋𝑘𝑤 𝐿

For parallel flow, R𝑓,𝑖

′′ ′′
= R𝑓,o = 9.4×10−6 (𝑚2 ∙ 𝐾/𝑊); for counter flow, R𝑓,𝑖′′ ′′
= R𝑓,o =
7.6×10 (𝑚 ∙ 𝐾/𝑊). This could be realistic since the representative fouling factor in
−6 2

Table 11.1 from the text for treated boiler feedwater below 50℃ is 0.0001(𝑚2 ∙ 𝐾/𝑊).
This value from Table 11.1 is about 10~13 times the value of a fouling factor needed to
explain the difference in the two values of (UA).
6. Discuss differences in heat exchanger effectiveness for parallel flow and counter flow
configurations.
The heat exchanger effectiveness for counter flow is approximately 20% greater than
that for parallel flow. This is not surprising since the counter flow configuration yields
greater log mean temperature difference. With U values that are approximately the
same, the counter flow configuration has greater actual q, thus greater effectiveness.

Conclusions

The heat exchanger lab investigated the performance of an unfinned, concentric tube heat
exchanger under parallel and counter flow configurations with NTU method and resistance
method. The temperatures of the inlet and outlet of both fluid steams are measured; the
flow rates are measured. The heat exchanger effectiveness ε and the number of transfer
units NTU are calculated directly from the experimental data. Overall heat transfer
coefficients U are calculated using both the NTU method and the resistance method. In the
latter method, estimates of h from empirical correlations are used. Measured differences in
q values between the cold and hot flows were small and well below the uncertainty in q.
Therefore, heat exchange with the environment was neglected. The counter flow heat
exchanger showed better performance in terms of its effectiveness compared to the
parallel flow.

Differences in the calculated U’s from both methods are discussed. The U values calculated
with the resistance method assume perfectly clean pipes and therefore neglect any fouling
that might be present on the heat exchanger tubing. A fouling resistance is calculated that
accounts for the differences between the U values calculated with the two different
methods and it was found to be 10~13 times smaller than the fouling factors provided in
the textbook. Predicted values of the overall heat transfer coefficient U are based on
empirical correlations and conductive wall resistance. For counter flow configurations, the
predicted values of U were within 15% of the actual values of UNTU; for parallel flow
configurations, the discrepancy is less than 30%. This is consistent with the statement that
convection coefficients calculated from empirical correlations are typically within 30% of
the actual value.
Reference

[1] F. P. Incropera et. al., Introduction to Heat Transfer, 6 th ed. Hoboken, NJ: Wiley, 2011.
[2] Thermocouple. (2017, April 14). Retrieved from
http://www.omega.com/prodinfo/thermocouples.html.