In the name of God
Jami Institute of Jami - Mehdi Rasti
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Heat Transfer
Jami Institute of Technology
Heat Exchangers – part 2
Mehdi Rasti
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The Counter-Flow Heat Exchanger
In contrast to the parallel-flow exchanger, this configuration
provides for heat transfer between the hotter portions of the two
fluids at one end, as well as between the colder portions at the
other. For this reason, the change in the temperature difference,
ΔT = Th – Tc , with respect to x is nowhere as large as it is for the
inlet region of the parallel-flow exchanger.
Note that the outlet temperature of the cold fluid may now exceed
the outlet temperature of the hot fluid.
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The Counter-Flow Heat Exchanger
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FIGURE: Temperature distributions for a counter-flow heat exchanger
The Counter-Flow Heat Exchanger
for the counter-flow exchanger the endpoint temperature
differences must now be defined as:
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The Counter-Flow Heat Exchanger
Note that, for the same inlet and outlet temperatures, the log
mean temperature difference for counter-flow exceeds that for
parallel flow,
. Hence the surface area required to
effect a prescribed heat transfer rate q is smaller for the counter-
flow than for the parallel-flow arrangement, assuming the same
value of U. Also note that Tc,o can exceed Th,o for counter-flow
but not for parallel flow.
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Special Operating Conditions
It is useful to note certain special conditions under which heat
exchangers may be operated.
Figure below shows temperature distributions for a heat
exchanger in which the hot fluid has a heat capacity rate,
which is much larger than that of the cold fluid,
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Special Operating Conditions
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Figure: Special heat exchanger conditions. Ch >> Cc or a condensing vapor
Special Operating Conditions
For this case the temperature of the hot fluid remains
approximately constant throughout the heat exchanger, while the
temperature of the cold fluid increases. The same condition is
achieved if the hot fluid is a condensing vapor. Condensation
occurs at constant temperature, and, for all practical purposes.
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Special Operating Conditions
Conversely, in an evaporator or a boiler (Figure below), it is the
cold fluid that experiences a change in phase and remains at a
nearly uniform temperature (
). The same effect is
achieved without phase change if Ch<<Cc. Note that, with
condensation or evaporation, the heat rate is given by below
Equations:
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Special Operating Conditions
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Figure: Special heat exchanger conditions. Ch << Cc or a evaporating liquid
Special Operating Conditions
The third special case (below Figure) involves a counter-flow
heat exchanger for which the heat capacity rates are equal
(Ch=Cc). The temperature difference ΔT must then be constant
throughout the exchanger, in which case
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Special Operating Conditions
Figure: Special heat exchanger conditions. A counter-flow heat exchanger
with equivalent fluid heat capacities (Ch = Cc).
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Example 1
A counter-flow, concentric tube heat exchanger is used to cool
the lubricating oil for a large industrial gas turbine engine. The
flow rate of cooling water through the inner tube (Di = 25 mm) is
0.2 kg/s, while the flow rate of oil through the outer annulus
(Do=45 mm) is 0.1 kg/s. The oil and water enter at temperatures
of 100 and 30C, respectively.
How long must the tube be made if the outlet temperature of the
oil is to be 60C?
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Example 1
Known: Fluid flow rates and inlet temperatures for a counterflow, concentric tube heat exchanger of prescribed inner and
outer diameter.
Find: Tube length to achieve a desired hot fluid outlet
temperature.
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Example 1
Assumptions:
1. Negligible heat loss to the surroundings.
2. Negligible kinetic and potential energy changes.
3. Constant properties.
4. Negligible tube wall thermal resistance and fouling factors.
5. Fully developed conditions for the water and oil (U
independent of x)
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Example 1
Schematic:
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Example 1
Properties: unused engine oil
(Th = 80C = 353 K):
cp = 2131 J/kg K,
μ = 3.25×10^(-2) N.s/m2,
k = 0.138 W/m K.
Water:
(Tc = 35C):
cp = 4178 J/kg K,
μ = 725×10^(-6) N.s/m2,
k = 0.625 W/m K,
Pr 4.85
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Example 1
Analysis: The required heat transfer rate may be obtained from
the overall energy balance for the hot fluid:
The water outlet temperature is:
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Example 1
Accordingly, use of
to evaluate the water properties
was a good choice. The required heat exchanger length may now
be obtained from below Equation:
where
and:
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Example 1
The overall heat transfer coefficient is:
For water flow through the tube:
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Example 1
Accordingly, the flow is turbulent and the convection coefficient
may be computed from below Equation:
Hence
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Example 1
For the flow of oil through the annulus, the hydraulic diameter is:
and the Reynolds number is:
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Example 1
The annular flow is therefore laminar. Assuming uniform
temperature along the inner surface of the annulus and a perfectly
insulated outer surface, the convection coefficient at the inner
surface may be obtained from Table below:
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Example 1
With
, linear interpolation provides:
And :
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Example 1
The overall convection coefficient is then:
and from the rate equation it follows that
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Example 1
Comments:
1. The hot side convection coefficient controls the rate of heat
transfer between the two fluids, and the low value of ho is
responsible for the large value of L. Incorporation of heat transfer
enhancement methods, could be used to decrease the size of the
heat exchanger.
2. Because hi>>ho, the tube wall temperature will follow closely
that of the coolant water. Accordingly, the assumption of uniform
wall temperature, which is inherent in the use of Table 8.2 to27
obtain ho, is reasonable.
Example 2
The counter-flow, concentric tube heat exchanger of Example 1 is
replaced with a compact, plate-type heat exchanger that consists
of a stack of thin metal sheets, separated by N gaps of width a.
The oil and water flows are subdivided into N/2 individual flow
streams, with the oil and water moving in opposite directions
within alternating gaps. It is desirable for the stack to be of a
cubical geometry, with a characteristic exterior dimension L.
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Example 2
Determine the exterior dimensions of the heat exchanger as a
function of the number of gaps if the flow rates, inlet
temperatures, and desired oil outlet temperature are the same as
in Example 1. Compare the pressure drops of the water and oil
streams within the plate-type heat exchanger to the pressure
drops of the flow streams in Example 1, if 60 gaps are specified.
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Example 2
Known:
Configuration of a plate-type heat exchanger. Fluid flow rates,
inlet temperatures, and desired oil outlet temperature.
Find:
1. Exterior dimensions of the heat exchanger.
2. Pressure drops within the plate-type heat exchanger with N 60
gaps, and the concentric tube heat exchanger of Example 1.
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Example 2
Schematic:
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Example 2
Assumptions:
1. Negligible heat loss to the surroundings.
2. Negligible kinetic and potential energy changes.
3. Constant properties.
4. Negligible plate thermal resistance and fouling factors.
5. Fully developed conditions for the water and oil.
6. Identical gap-to-gap heat transfer coefficients.
7. Heat exchanger exterior dimension is large compared to the
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gap width.
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Example 2
Properties:
unused engine oil :
ρ = 852.1 kg/m3
Water:
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Example 2
Analysis:
1. The gap width may be related to the overall dimension of the
heat exchanger by the expression a=L/N, and the total heat
transfer area is
Assuming a<<L and the existence of laminar flow, the Nusselt
number for each interior gap is provided in Table 8.1 and is:
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Example 2
The hydraulic diameter is Dh
2a. Combining the preceding
expressions yields for the water:
Likewise, for the oil:
and the overall convection coefficient is
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Example 2
From Example 1, the required log mean temperature difference
and heat transfer rate are
and q = 8524 W, respectively.
which may be rearranged to yield:
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Example 2
The size of the compact heat exchanger decreases as the number
of gaps is increased, as shown in the figure below.
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Example 2
2. For N = 60 gaps, the stack dimension is L = 0.131 m from the
results of part 1, and the gap width is:
a=L/N=0.131m/60=0.00218 m.
The hydraulic diameter is Dh=0.00436 m, and the mean velocity
in each waterfilled gap is:
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Example 2
providing a Reynolds number of:
For the oil-filled gaps:
yielding a Reynolds number of:
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Example 2
Therefore, the flow is laminar for both fluids, as assumed in part
1. below equations may be used to calculate the pressure drop for
the water:
Similarly, for the oil:
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Example 2
For Example 1, the friction factor associated with the water flow
for a smooth surface condition may be calculated by using below
equation:
The mean velocity is:
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Example 2
and the pressure drop is:
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Example 2
For the oil flowing in the annular region, the mean velocity is:
And the pressure drop is
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Example 2
Comments:
1. Increasing the number of gaps increases the UA product by
simultaneously providing more surface area and increasing the
heat transfer coefficients associated with the flow of the fluids
through smaller passages.
2. The area-to-volume ratio of the N = 60 heat exchanger is
L2(N - 1)/L3 = (N - 1)/L = (60 - 1)/0.131 m = 451 m2/m3.
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Example 2
3. The pressure drops associated with use of the compact heat
exchanger are significantly less than for a conventional
concentric tube configuration. Pressure drops are reduced by
99.9% and 99.5% for the water and oil flows, respectively.
4. Fouling of the heat transfer surfaces may result in a decrease in
the gap width, as well as an associated reduction in heat transfer
rate and increase in pressure drop.
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Example 2
5. Because hc > hh, the temperatures of the thin metal sheets will
follow closely that of the water, and, as in Example 1, the
assumption of uniform temperature conditions to obtain hc and
hh is reasonable.
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Thanks for your nice attention