ISAT 413 - Module V: Industrial Systems

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ISAT 413 - Module V Industrial Systems
Topic 2 Heat Exchanger Fundamentals,
Recuperative Heat Exchangers

• Heat Exchangers
• UA-LMTD Design Method
• e -NTU Design Method
• An Example

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• Heat Exchangers

A heat exchanger is a device for transferring
heat from one fluid to another. There are three
main categories Recuperative, in which the two
fluids are at all times separated by a solid
wall Regenerative, in which each fluid transfers
heat to or from a matrix of material Evaporative
(direct contact), in which the enthalpy of
vaporization of one of the fluids is used to
provide a cooling effect.
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Heat Exchanger (HX) Design Methods
HX designers usually use two well-known methods
for calculating the heat transfer rate between
fluid streamsthe UA-LMTD and the
effectiveness-NTU (number of heat transfer units)
methods. Both methods can be equally employed for
designing HXs. However, the ?-NTU method is
preferred for rating problems where at least one
exit temperature is unknown. If all inlet and
outlet temperatures are known, the UA-LMTD method
does not require an iterative procedure and is
the preferred method.
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LMTD (Log Mean Temperature Difference)
The most commonly used type of heat exchanger is
the recuperative heat exchanger. In this type the
two fluids can flow in counter-flow, in
parallel-flow, or in a combination of these, and
cross-flow.
The true mean temperature difference is the
Logarithmic mean Temperature difference (LMTD),
is defined as
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Heat Transfer Rate of a Heat Exchanger
The heat transferred for any recuperative heat
exchanger can be calculated as(refer to the
diagram shown on the previous slide)
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Heat Exchanger UA-LMTD Design Method
Where U is the overall heat transfer coefficient
(and is assumed to be constant over the whole
surface area of the heat exchanger).
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Heat Exchanger e-NTU Design Method
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6-row, 6-pass plate fin-and-tube cross
counterflow HX
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3-row, 3-pass plate fin-and-tube crossflow HX
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Effectiveness of a 6-row, 6-pass plate
fin-and-tube cross counterflow HX
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A HX Example
A schematic representation of a hybrid central
receiver is shown in the following slide (Slide
12). In this system, molten nitrate salt is
heated in a central receiver to temperature as
high as 1,050oF (565oC). The molten salt is then
passed through a heat exchanger, where it is used
to preheat combustion air for a combined-cycle
power plant. For more information about this
cycle, refer to Bharathan et. Al. (1995) and Bohn
et al. (1995). The heat exchanger used for this
purpose is shown in Slide 13. The plates of the
heat exchanger are made of steel and are 2 mm
thick. The overall flow is counter-flow
arrangement where the air and molten salt both
flow duct-shape passages (unmixed). The shell
side, where the air flow takes place, is baffled
to provide cross flow between the lateral
baffles. The baseline design conditions are
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A hybrid central-receiver concept developed at
the NREL
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Molten-salt-to-air HX used to preheat combustion
air
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A HX Example (continued)
Air flow rate 0.503 kg/s per passage (250
lbm/s) Air inlet temperature 340oC
(650oF) Air outlet temperature 470oC
(880oF) Salt flow rate 0.483 kg/s per
passage (240 lbm/s) Salt inlet temperature
565oC (1050oF) Salt outlet temperature
475oC (890oF) Find the overall heat-transfer
coefficient for this heat exchanger. Ignore the
fouling resistances.
Solution
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Solution (continued)
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Solution (continued)
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Solution (continued)
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Solution (continued)
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Solution (continued)
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• Recuperative Heat Exchangers
• Definition of Recuperative HX
• Types of Recuperative HX
• Design Factors
• Examples

A Recuperative Heat Exchanger (HX) is one in
which the two fluids are separated at all times
by a solid barrier.
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Waste-Heat water-Tube Boiler
Shell Boiler using Waste Gas
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Furnace Gas Air Pre-Heater
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Two-Pass Shell-and-Tube Heat Exchanger
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Gas-to-Gas Heat Recovery with a Plate-Fin Heat
Exchanger
A
A
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Liquid-to-Liquid Plate-Fin Heat Exchanger
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Basic Equations
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Heat Exchanger Configurations
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Extended Surfaces Fins, fpi (fins per inch)
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Example 5.4 (Eastop Croft) Fin Surface
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Example 5.4
A flat surface as shown in the previous slide has
a base temperature of 90oC when the air mean bulk
temperature is 20oC. Air is blown across the
surface and the mean heat transfer coefficient is
30 W/m2-K. The fins are made of an aluminum
alloy the fin thickness is 1.6 mm, the fin
height is 19 mm, and the fin pitch is 13.5 mm.
Calculate the heat loss per m2 of primary surface
with and without the fins assuming that the same
mean heat transfer coefficient applies in each
case. Neglect the heat loss from the fin tips and
take a fin efficiency of 71.
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Example 5.4 (continued)
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?-NTU Method (Effectiveness Number of Thermal
Units Method)
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?-NTU (Effectiveness against NTU) for
shell-and-tube heat exchangers
(with 2 shell passes and 4, 8, 12 tube passes)
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Characteristics of ?-NTU Chart

• For given mass flow rates and specific heats of
  two fluids the value of ? depends on the NTU and
  hence on the product (UAo). Thus for a given
  value of U the NTU is proportional to Ao. It can
  then be seen from the ?-NTU chart that increasing
  Ao increases ? and hence the saving in fuel.
• The capital cost of the heat exchanger increases
  as the area increases and ?-NTU chart shows that
  at high values of ? large increase in area
  produce only a small increase in ?.
• The NTU and hence the effectiveness, ? can be
  increased for a fixed value of area by increasing
  the value of the overall heat transfer
  coefficient, U.

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Increasing HX ? with Fixed Ao (1)
The NTU, and hence ? can be increased for a fixed
value of the area by increasing the value of the
overall heat transfer coefficient, U, which can
be increased by increasing the heat transfer
coefficient for one or both of the individual
fluids.
The heat transfer coefficient can be increased by
reducing the tube diameter, and/or increasing the
mass flow rate per tube.
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Increasing HX ? with Fixed Ao (2)

• Since , for a constant total mass
  flow rate the number of tubes per pass must be
  reduced correspondingly if the mass flow rate per
  tube is increased.
• Also, the heat transfer area is given by
  , where n is the number of tubes per
  pass, and p is the number of tube passes.
  Therefore, to maintain the same total heat
  transfer area for a reduced tube diameter in a
  given type of heat exchanger, it is necessary to
  increase the length of the tubes per pass, L,
  and/or the number of tubes per pass (which will
  reduce the heat transfer rate.)
• The design process is therefore an iterative
  process in order to arrive at the optimum
  arrangement of tube diameter, tube length, and
  number of tubes.

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Overall HX Design Considerations

• Altering the inside diameter of a tube to
  increase the heat transfer coefficient for flow
  through the tube will alter the heat transfer on
  the shell side.
• A full economic analysis also requires
  consideration of the pumping power for both
  fluids. Pressure losses in fluid flow due to
  friction, turbulence, and fittings such as
  valves, bends etc. are proportional to the square
  of the flow velocity. The higher the fluid
  velocity and the more turbulent the flow the
  higher is the heat transfer coefficient but the
  greater the pumping power.

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Example 5.5
(a) A shell-and-tube heat exchanger is used to
recover energy from engine oil and consists of
two shell passes for water and four tube passes
for the engine oil as shown diagrammatically in
the following figure. The effectiveness can be
calculated based on Eastop Equation (3.33). For a
flow of oil of 2.3 kg/s entering at a temperature
of 150oC, and a flow of water of 2.4 kg/s
entering at 40oC, use the data given to
calculate (i) the total number of tubes
required (ii) the length of the tubes (iii) the
exit temperatures of the water and oil (iv) the
fuel cost saving per year if water heating is
currently provided by a gas boiler of efficiency
0.8. (b) What would be the effectiveness and fuel
saving per year with eight tube passes?
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Example 5.5 (continued)
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Use EES for Eastop Example 5.5" hot
oil m_dot_H2.3 oil mass flow rate,
kg/s t_H1150 hot oil inlet temperature,
C cp_H2.19 mean spefici heat of oil,
kJ/kg-K rho_H840 mean oil density,
kg/m3 C_Hm_dot_Hcp_H hot fluid capacity,
kW/K cold water m_dot_C2.4 water mass flow
rate, kg/s t_C240 cold water inlet
temperature, C cp_C4.19 mean specific heat
of water, kJ/kg-K C_Cm_dot_Ccp_C cold fluid
capacity, kW/K data eta_boiler0.8 gas
boiler efficiency v_H0.8 oil velocity in the
tube, m/s eta_Hx0.7 require HX
effectiveness n_pass4 four pass heat
exchanger d_i0.005 tube inside diameter,
m d_o0.007 tube outside diameter,
m U0.400 overall heat transfer coefficient,
kW/m2-K t_hours4000 annual usage,
h cost1.2 cost of water heating, p/kWh
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• a(i) calculate the total number of tubes
  required
• V_dot_Hm_dot_H/rho_H
• A_crossV_dot_h/v_H
• A_1PId_i2/4
• n_tubeA_cross/A_1n_pass
• n_tube697 tubes
• a(ii) calculate the length of the tubes
• Rmin(C_H,C_C)/max(C_H,C_C)
• eta_HX(1-exp(-NTU(1-R)))/(1-Rexp(-NTU(1-R)))
• NTUUA_o/min(C_H,C_C)
• A_oPId_on_tubeL_tube
• L_tube1.27 m
• a(iii) calculate the exit temperature of oil
  and water
• Eta_HXC_H(t_H1-t_H2)/(min(C_h,C_C)(t_H1-t_C2))
• C_H(t_H1-t_H2)C_C(t_C1-t_C2)
• t_C178.6 C t_H273.0 C

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• a(iv) calculate the total heat transfer and
  fuel cost saving per year
• Q_dotC_C(t_C1-t_C2)
• Fuel_savingQ_dott_hourscost/(eta_boiler100)
• Fuel_saving23271 British Pounds
• b(v) calculate eta2_hx if double Ao
• NTU22NTU
• eta2_HX(1-exp(-NTU2(1-R)))/(1-Rexp(-NTU2(1-R))
  )
• eta2_HX0.881
• b(vi) calulate t_H2, t_C2 and fuel_saving
• eta2_HXC_H(t_H1-t2_H2)/(min(C_h,C_C)(t_H1-t_C2)
  )
• C_H(t_H1-t2_H2)C_C(t2_C1-t_C2)
• Q2_dotC_C(t2_C1-t_C2)
• Fuel2_savingQ2_dott_hourscost/(eta_boiler100)
• Fuel2_saving29279 British Pounds