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Heat Exchangers:
Design Considerations
Chapter 11
Sections 11.1 through 11.3
Types
Heat Exchanger Types
Heat exchangers are ubiquitous to energy conversion and utilization. They involve
heat exchange between two fluids separated by a solid and encompass a wide
range of flow configurations.
• Concentric-Tube Heat Exchangers
Parallel Flow Counterflow
 Simplest configuration.
 Superior performance associated with counter flow.
Types (cont.)
• Cross-flow Heat Exchangers
Finned-Both Fluids
Unmixed
Unfinned-One Fluid Mixed
the Other Unmixed
 For cross-flow over the tubes, fluid motion, and hence mixing, in the transverse
direction (y) is prevented for the finned tubes, but occurs for the unfinned condition.
 Heat exchanger performance is influenced by mixing.
Types (cont.)
• Shell-and-Tube Heat Exchangers
One Shell Pass and One Tube Pass
 Baffles are used to establish a cross-flow and to induce turbulent mixing of the
shell-side fluid, both of which enhance convection.
 The number of tube and shell passes may be varied, e.g.:
One Shell Pass,
Two Tube Passes
Two Shell Passes,
Four Tube Passes
Types (cont.)
• Compact Heat Exchangers
 Widely used to achieve large heat rates per unit volume, particularly when
one or both fluids is a gas.
 Characterized by large heat transfer surface areas per unit volume, small
flow passages, and laminar flow.
(a) Fin-tube (flat tubes, continuous plate fins)
(b) Fin-tube (circular tubes, continuous plate fins)
(c) Fin-tube (circular tubes, circular fins)
(d) Plate-fin (single pass)
(e) Plate-fin (multipass)
Overall Coefficient
Overall Heat Transfer Coefficient
• An essential requirement for heat exchanger design or performance calculations.
• Contributing factors include convection and conduction associated with the
two fluids and the intermediate solid, as well as the potential use of fins on both
sides and the effects of time-dependent surface fouling.
• With subscripts c and h used to designate the hot and cold fluids, respectively,
the most general expression for the overall coefficient is:
( ) ( )
( ) ( ) ( ) ( )
, ,
1 1 1
1 1
c h
f c f h
w
o o o oc c h h
UA UA UA
R R
R
hA A A hAη η η η
= =
′′ ′′
= + + + +
Overall Coefficient

( )o,
Overall surface efficiency of fin array (Section 3.6.5)
1 1
o
f
c or h f
c or h
A
A
η
η η
→
 
= − − ÷
 
total surface area (fins and exposed base)
surface area of fins only
t
f
A A
A
= →
→
Assuming an adiabatic tip, the fin efficiency is
( )
,
tanh
f cor h
c or h
mL
mL
η
 
=  ÷
 
( )2 /c or h p w c or h
m U k t=
, partial overall coe
1
fficientp c or h
f c or h
hU
hR
 
= → ÷ ÷′′+ 
 2
for a unit surfFouling fact ace area (m W)or K/fR′′ → ×
Table 11.1→
conduction resistanWall (K/Wce )wR →
LMTD Method
A Methodology for Heat Exchanger
Design Calculations
- The Log Mean Temperature Difference (LMTD) Method -
• A form of Newton’s Law of Cooling may be applied to heat exchangers by
using a log-mean value of the temperature difference between the two fluids:
1mq U A T∆=
( )
1 2
1
1 21n /
m
T T
T
T T
∆ ∆
∆
∆ ∆
−
=
Evaluation of depends on the heat exchanger type.1 2andT T∆ ∆
• Counter-Flow Heat Exchanger:
1 ,1 ,1
, ,
h c
h i c o
T T T
T T
∆ ≡ −
= −
2 ,2 ,2
, ,
h c
h o c i
T T T
T T
∆ ≡ −
= −
LMTD Method (cont.)
• Parallel-Flow Heat Exchanger:
1 ,1 ,1
, ,
h c
h i c i
T T T
T T
∆ ≡ −
= −
2 ,2 ,2
, ,
h c
h o c o
T T T
T T
∆ ≡ −
= −
 Note that Tc,o can not exceed Th,o for a PF HX, but can do so for a CF HX.
 For equivalent values of UA and inlet temperatures,
1 , 1 ,m CF m PFT T∆ ∆>
• Shell-and-Tube and Cross-Flow Heat Exchangers:
1 1 ,m m CFT F T∆ ∆=
Figures 11.10 - 11.13F →
Energy Balance
Overall Energy Balance
• Assume negligible heat transfer between the exchanger and its surroundings
and negligible potential and kinetic energy changes for each fluid.
( ), ,h i h ohq m i i
×
= −
( ), ,c c o c iq m i i
×
= −
fluid enthalpyi →
• Assuming no l/v phase change and constant specific heats,
( ), , ,p h h i h ohq m c T T
×
= − ( ), ,h h i h oC T T= −
( ), , ,c p c c o c iq m c T T
×
= − ( ), ,c c o c iC T T= −
, Heat capacity r sateh cC C →
• Application to the hot (h) and cold (c) fluids:
Special Conditions
Special Operating Conditions
 Case (a): Ch>>Cc or h is a condensing vapor ( ).hC → ∞
– Negligible or no change in ( ), , .h h o h iT T T=
 Case (b): Cc>>Ch or c is an evaporating liquid ( ).cC → ∞
– Negligible or no change in ( ), , .c c o c iT T T=
 Case (c): Ch=Cc.
1 2 1mT T T∆ ∆ ∆= =–
Problem: Overall Heat Transfer Coefficient
em 11.5: Determination of heat transfer per unit length for heat recovery
device involving hot flue gases and water.
KNOWN: Geometry of finned, annular heat exchanger. Gas-side temperature and
convection coefficient. Water-side flowrate and temperature.
FIND: Heat rate per unit length.
SCHEMATIC:
Do = 60 mm
Di,1 = 24 mm
Di,2 = 30 mm
t = 3 mm = 0.003m
L = (60-30)/2 mm = 0.015m
Problem: Overall Heat Transfer Coefficient
(cont.)
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional
conduction in strut, (4) Adiabatic outer surface conditions, (5) Negligible gas-side radiation,
(6) Fully-developed internal flow, (7) Negligible fouling.
PROPERTIES: Table A-6, Water (300 K): k = 0.613 W/m⋅K, Pr = 5.83, µ = 855 × 10-6
N⋅s/m2
.
ANALYSIS: The heat rate is
where
( ) ( ) ( )w oc c h
1/ UA 1/ hA R 1/ hAη= + +
( ) ( )
( )
i,2 i,1 4
w
ln D / D ln 30/ 24
R 7.10 10 K / W.
2 kL 2 50 W / m K lmπ π
−= = = ×
⋅
Problem: Overall Heat Transfer Coefficient
(cont.)
the internal flow is turbulent and the Dittus-Boelter correlation gives
The overall fin efficiency is
From Eq. 11.4,
Problem: Overall Heat Transfer Coefficient
(cont.)
Hence
It follows that
and
<
for a 1m long section.
Problem: Overall Heat Transfer Coefficient
(cont.)
COMMENTS: (1) The gas-side resistance is substantially decreased by using the fins
and q is increased.
(2) Heat transfer enhancement by the fins could be increased further by using a material of
larger k, but material selection would be limited by the large value of Tm,h.
Problem: Ocean Thermal Energy Conversion
m 11.47: Design of a two-pass, shell-and-tube heat exchanger to supply
vapor for the turbine of an ocean thermal energy conversion
system based on a standard (Rankine) power cycle. The power
cycle is to generate 2 MWe at an efficiency of 3%. Ocean
water enters the tubes of the exchanger at 300K, and its desired
outlet temperature is 292K. The working fluid of the power
cycle is evaporated in the tubes of the exchanger at its
phase change temperature of 290K, and the overall heat transfer
coefficient is known.
FIND: (a) Evaporator area, (b) Water flow rate.
SCHEMATIC:
Problem: Ocean Thermal
Energy Conversion (cont)
ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and
potential energy changes, (3) Constant properties.
PROPERTIES: Table A-6, Water ( = 296 K): cp = 4181 J/kg⋅K.
ANALYSIS: (a) The efficiency is
Hence the required heat transfer rate is
Also
and, with P = 0 and R = ∞, from Fig. 11.10 it follows that F = 1. Hence
Problem: Ocean Thermal
Energy Conversion (cont)
b) The water flow rate through the evaporator is
COMMENTS: (1) The required heat exchanger size is enormous due to the small
temperature differences involved,
(2) The concept was considered during the energy crisis of the mid 1970s but has not since
been implemented.

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Heat Exchanger Design and LMTD Method

  • 1. Heat Exchangers: Design Considerations Chapter 11 Sections 11.1 through 11.3
  • 2. Types Heat Exchanger Types Heat exchangers are ubiquitous to energy conversion and utilization. They involve heat exchange between two fluids separated by a solid and encompass a wide range of flow configurations. • Concentric-Tube Heat Exchangers Parallel Flow Counterflow  Simplest configuration.  Superior performance associated with counter flow.
  • 3. Types (cont.) • Cross-flow Heat Exchangers Finned-Both Fluids Unmixed Unfinned-One Fluid Mixed the Other Unmixed  For cross-flow over the tubes, fluid motion, and hence mixing, in the transverse direction (y) is prevented for the finned tubes, but occurs for the unfinned condition.  Heat exchanger performance is influenced by mixing.
  • 4. Types (cont.) • Shell-and-Tube Heat Exchangers One Shell Pass and One Tube Pass  Baffles are used to establish a cross-flow and to induce turbulent mixing of the shell-side fluid, both of which enhance convection.  The number of tube and shell passes may be varied, e.g.: One Shell Pass, Two Tube Passes Two Shell Passes, Four Tube Passes
  • 5. Types (cont.) • Compact Heat Exchangers  Widely used to achieve large heat rates per unit volume, particularly when one or both fluids is a gas.  Characterized by large heat transfer surface areas per unit volume, small flow passages, and laminar flow. (a) Fin-tube (flat tubes, continuous plate fins) (b) Fin-tube (circular tubes, continuous plate fins) (c) Fin-tube (circular tubes, circular fins) (d) Plate-fin (single pass) (e) Plate-fin (multipass)
  • 6. Overall Coefficient Overall Heat Transfer Coefficient • An essential requirement for heat exchanger design or performance calculations. • Contributing factors include convection and conduction associated with the two fluids and the intermediate solid, as well as the potential use of fins on both sides and the effects of time-dependent surface fouling. • With subscripts c and h used to designate the hot and cold fluids, respectively, the most general expression for the overall coefficient is: ( ) ( ) ( ) ( ) ( ) ( ) , , 1 1 1 1 1 c h f c f h w o o o oc c h h UA UA UA R R R hA A A hAη η η η = = ′′ ′′ = + + + +
  • 7. Overall Coefficient  ( )o, Overall surface efficiency of fin array (Section 3.6.5) 1 1 o f c or h f c or h A A η η η →   = − − ÷   total surface area (fins and exposed base) surface area of fins only t f A A A = → → Assuming an adiabatic tip, the fin efficiency is ( ) , tanh f cor h c or h mL mL η   =  ÷   ( )2 /c or h p w c or h m U k t= , partial overall coe 1 fficientp c or h f c or h hU hR   = → ÷ ÷′′+   2 for a unit surfFouling fact ace area (m W)or K/fR′′ → × Table 11.1→ conduction resistanWall (K/Wce )wR →
  • 8. LMTD Method A Methodology for Heat Exchanger Design Calculations - The Log Mean Temperature Difference (LMTD) Method - • A form of Newton’s Law of Cooling may be applied to heat exchangers by using a log-mean value of the temperature difference between the two fluids: 1mq U A T∆= ( ) 1 2 1 1 21n / m T T T T T ∆ ∆ ∆ ∆ ∆ − = Evaluation of depends on the heat exchanger type.1 2andT T∆ ∆ • Counter-Flow Heat Exchanger: 1 ,1 ,1 , , h c h i c o T T T T T ∆ ≡ − = − 2 ,2 ,2 , , h c h o c i T T T T T ∆ ≡ − = −
  • 9. LMTD Method (cont.) • Parallel-Flow Heat Exchanger: 1 ,1 ,1 , , h c h i c i T T T T T ∆ ≡ − = − 2 ,2 ,2 , , h c h o c o T T T T T ∆ ≡ − = −  Note that Tc,o can not exceed Th,o for a PF HX, but can do so for a CF HX.  For equivalent values of UA and inlet temperatures, 1 , 1 ,m CF m PFT T∆ ∆> • Shell-and-Tube and Cross-Flow Heat Exchangers: 1 1 ,m m CFT F T∆ ∆= Figures 11.10 - 11.13F →
  • 10. Energy Balance Overall Energy Balance • Assume negligible heat transfer between the exchanger and its surroundings and negligible potential and kinetic energy changes for each fluid. ( ), ,h i h ohq m i i × = − ( ), ,c c o c iq m i i × = − fluid enthalpyi → • Assuming no l/v phase change and constant specific heats, ( ), , ,p h h i h ohq m c T T × = − ( ), ,h h i h oC T T= − ( ), , ,c p c c o c iq m c T T × = − ( ), ,c c o c iC T T= − , Heat capacity r sateh cC C → • Application to the hot (h) and cold (c) fluids:
  • 11. Special Conditions Special Operating Conditions  Case (a): Ch>>Cc or h is a condensing vapor ( ).hC → ∞ – Negligible or no change in ( ), , .h h o h iT T T=  Case (b): Cc>>Ch or c is an evaporating liquid ( ).cC → ∞ – Negligible or no change in ( ), , .c c o c iT T T=  Case (c): Ch=Cc. 1 2 1mT T T∆ ∆ ∆= =–
  • 12. Problem: Overall Heat Transfer Coefficient em 11.5: Determination of heat transfer per unit length for heat recovery device involving hot flue gases and water. KNOWN: Geometry of finned, annular heat exchanger. Gas-side temperature and convection coefficient. Water-side flowrate and temperature. FIND: Heat rate per unit length. SCHEMATIC: Do = 60 mm Di,1 = 24 mm Di,2 = 30 mm t = 3 mm = 0.003m L = (60-30)/2 mm = 0.015m
  • 13. Problem: Overall Heat Transfer Coefficient (cont.) ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional conduction in strut, (4) Adiabatic outer surface conditions, (5) Negligible gas-side radiation, (6) Fully-developed internal flow, (7) Negligible fouling. PROPERTIES: Table A-6, Water (300 K): k = 0.613 W/m⋅K, Pr = 5.83, µ = 855 × 10-6 N⋅s/m2 . ANALYSIS: The heat rate is where ( ) ( ) ( )w oc c h 1/ UA 1/ hA R 1/ hAη= + + ( ) ( ) ( ) i,2 i,1 4 w ln D / D ln 30/ 24 R 7.10 10 K / W. 2 kL 2 50 W / m K lmπ π −= = = × ⋅
  • 14. Problem: Overall Heat Transfer Coefficient (cont.) the internal flow is turbulent and the Dittus-Boelter correlation gives The overall fin efficiency is From Eq. 11.4,
  • 15. Problem: Overall Heat Transfer Coefficient (cont.) Hence It follows that and < for a 1m long section.
  • 16. Problem: Overall Heat Transfer Coefficient (cont.) COMMENTS: (1) The gas-side resistance is substantially decreased by using the fins and q is increased. (2) Heat transfer enhancement by the fins could be increased further by using a material of larger k, but material selection would be limited by the large value of Tm,h.
  • 17. Problem: Ocean Thermal Energy Conversion m 11.47: Design of a two-pass, shell-and-tube heat exchanger to supply vapor for the turbine of an ocean thermal energy conversion system based on a standard (Rankine) power cycle. The power cycle is to generate 2 MWe at an efficiency of 3%. Ocean water enters the tubes of the exchanger at 300K, and its desired outlet temperature is 292K. The working fluid of the power cycle is evaporated in the tubes of the exchanger at its phase change temperature of 290K, and the overall heat transfer coefficient is known. FIND: (a) Evaporator area, (b) Water flow rate. SCHEMATIC:
  • 18. Problem: Ocean Thermal Energy Conversion (cont) ASSUMPTIONS: (1) Negligible heat loss to surroundings, (2) Negligible kinetic and potential energy changes, (3) Constant properties. PROPERTIES: Table A-6, Water ( = 296 K): cp = 4181 J/kg⋅K. ANALYSIS: (a) The efficiency is Hence the required heat transfer rate is Also and, with P = 0 and R = ∞, from Fig. 11.10 it follows that F = 1. Hence
  • 19. Problem: Ocean Thermal Energy Conversion (cont) b) The water flow rate through the evaporator is COMMENTS: (1) The required heat exchanger size is enormous due to the small temperature differences involved, (2) The concept was considered during the energy crisis of the mid 1970s but has not since been implemented.