8/24/99 Heat Exchanger Design 1
HEAT EXCHANGER DESIGN
GOALS:
u Role of heat exchangers in chemical processing
u Basic concepts and terminology
u Types of heat exchangers
u Design methodology
e Sizing
e Design
e Rating
8/24/99 Heat Exchanger Design 2
PURPOSE OF A HEAT EXHANGER
u To Heat or Cool a stream flowing from item of
equipment to another,The steam may be a:
e A liquid
e A gas
e A multiphase mixture
u To Vaporize a liquid stream
u To Condense a vapor stream
8/24/99 Heat Exchanger Design 3
Shell Side
Inlet Nozzle
Shell Side
Outlet
Nozzle
Tube Side
Inlet Nozzle
Tube Side
Outlet
Nozzle
SINGLE TUBE PASS,SINGLE SHELL
PASS COUNTERCURRENT HEAT
EXCHANGER
8/24/99 Heat Exchanger Design 4
TYPES OF HEAT EXCHANGER SERVICE
u Fluid heated by external utility
e Steam
e Hot oil or molten salt
e Combustion gas (furnace)
e Electricity (resistive,inductive,microwave)
u Fluid cooled by external utility
e Cooling water
e Refrigeration
u Fluid heated or cooled by other process stream
8/24/99 Heat Exchanger Design 5
DUTY Q (kJ/s)
T (K)
Sensible Heat - Heating
Sensible Heat - Cooling
Pure Component Condensation or Vaporization
Multicomponent Cooling/Condensation
TYPES OF HEATING AND COOLING
CURVES
8/24/99 Heat Exchanger Design 6
CALCULATION OF COOLING CURVES
u Sensible Heat,Q = W Cp (Tin - Tout)
(For constant Cp,Q = W (Hin - Hout) otherwise)
u Latent Heat,Q = W l
u Multicomponent Cooling Curves,Requires point-by-
point flash calculations
8/24/99 Heat Exchanger Design 7
FEASIBLE COOLING CURVE
PAIRINGS
u Corollary of the Second Law of Thermodynamics:
Heat can only be transferred from
a higher temperature to a lower one.
u For heat exchangers this means that the higher
temperature cooling curve and the lower temperature
heating curve cannot intersect.
u When this condition is satisfied,the pairing of a heating
and cooling curve is said to be feasible.
8/24/99 Heat Exchanger Design 8
T
T1in
Q
T1out
T2in
T2out
Countercurrent Flow
T2in
T1in
T1out
T2out
Q
T
Cocurrent Flow
FEASIBLE COOLING CURVE PAIRINGS
T
Q
T
Q
INFEASIBLE COOLING CURVE PAIRINGS
8/24/99 Heat Exchanger Design 9
MAXIMUM HEAT EXCHANGER DUTY
u Qmax,the maximum amount of heat that can be
transferred in a heat exchanger,no matter how large it
is,occurs when the heating and cooling curves either
e Intersect at one end of the exchanger or the other or
e Become tangent within the exchanger
u For sensible heating with constant fluid heat capacities,
the curves are straight lines,They will intersect at that
end of the exchanger whose entering fluid has the largest
WCp,call it Wcpmax.
8/24/99 Heat Exchanger Design 10
MAXIMUM HEAT EXCHANGER DUTY
Q
T
T1in
T2out
T1out = T2in
Qmax0
For this example,WCp2 > WCp1
8/24/99 Heat Exchanger Design 11
MAXIMUM DUTY Qmax
uFor WCp1 > WCp2,the exchanger will,pinch” at the
Fluid 1 inlet.
Qmax = WCp2 (T1in - T2in)
u For WCp2 > WCp1 the exchanger will,pinch” at the
Fluid 2 inlet (as in the previous diagram)
Qmax = WCp1 (T1in - T2in)
8/24/99 Heat Exchanger Design 12
BASIC PERFORMANCE EQUATIONS FOR A
HEAT EXCHANGER
For the fluid flowing in the positive z direction an energy
balance on the section dz gives
dT1/dz = (Ua./WCp1)(T2 - T1)
where U = the overall heat transfer coefficient and
a = the heat transfer area per unit length
If the second fluid is a pure component either vaporizing or
condensing,then
dT2/dz = 0
8/24/99 Heat Exchanger Design 13
BASIC PERFORMANCE EQUATIONS FOR A
HEAT EXCHANGER (Cont’d.)
CASE 1,If the second fluid is a pure component either
vaporizing or condensing,then
dT2/dz = 0
CASE 2,If the second fluid is flowing in the negative z
direction,I.e.,in countercurrent flow,an energy balance
on the section dz gives
dT2/dz = (Ua./WCp2)(T2 - T1)
8/24/99 Heat Exchanger Design 14
RATING SOLUTION FOR CASE 1
T1(z) = Exp[- h1 z] T1(0) + (1 - Exp[- h1 z] ) T2
where h1 = U a /WCp1
For z = L,A = aL and N1 = UA/WCp1 where N1 is
defined as the number of heat transfer units (NTU’s) with
respect to Fluid 1,Let T1out = T1(L) and T1in = T1(0).
Then
T1out = Exp[-N1} T1in + (1 - Exp[-N1]) T2
The exchanger efficiency E = (T1out - T1in)/(T2 - T1in)
E = 1 - Exp[-N1]
8/24/99 Heat Exchanger Design 15
DESIGN SOLUTION FOR CASE 1
N1 = - UA/WCp1 = ln {1 - E}
= ln {[T2 - T1out]/[T2 - T1in]}
Note that Q = WCp1(T1out - T1in),
Substituting and rearranging gives
Q = UA LMTD
where
LMTD = [(T2-T1out) - (T2-T1in)]
ln {[T2 - T1out]/[T2 - T1in]}
which is the well-known heat exchanger design equation.
8/24/99 Heat Exchanger Design 16
RATING SOLUTION FOR CASE 2
Without going through the details:
T2out = [Exp(N2-N1) - 1] T1in + [1 - N1/N2] T2in
[Exp(N2-N1) - N1/N2]
where N1 = UA/WCp1
N2 = UA/WCp2
T1out = Exp(N2-N1)[ 1 - N1/N2] T1in
[Exp(N2-N1) - N1/N2]
+ N1/N2[Exp(N2-N1)-1] T2in
[Exp(N2-N1) - N1/N2]
8/24/99 Heat Exchanger Design 17
DESIGN SOLUTION FOR CASE 2
The design solution is essentially the same for Case 2 as for
Case 1,namely,
Q = UA LMTD
where now LMTD = (T2out - T1in) - (T1out - T2in)
ln {(T2out - T1in)/(T1out - T2in)
8/24/99 Heat Exchanger Design 18
REPRISE,ASSUMPTIONS
u The Cp’s are constant over the temperature range
involved,(Reasonable for most exchangers of practical
interest.)
u U is constant over the temperature range involved.
(Reasonable for most exchangers of practical interest.)
u Flow is pure countercurrent or pure cocurrent,If not,a
correction factor F is required to adjust the LMTD.
Q = UA F LMTD
F has been derived for most common heat exchanger
configurations (multiple tube passes,cross flow,etx.).
8/24/99 Heat Exchanger Design 19
RATING CALCULATIONS
u If U and A are known along with the W’s and Cp’s,use
the appropriate performance equation solutions.
uIf only the specifications of the exchanger (number of
tubes,length of tubes,tube diameter,baffle spacing,
baffle cut,etc.) are given,compute A from the geometry
and U from
1/U = 1/hf1 + 1/hwall + 1/hf2 + rfouling
using the appropriate correlations for hf1 and hf2.
8/24/99 Heat Exchanger Design 20
SIZING CALCULATIONS
u,Choose a typical value for U based on the type of
service,[Tables of typical values can be found in Perry’s
and most textbooks on heat exchanger design.]
u Determine the outlet temperatures based on the
performance specifications and the appropriate energy
balances.
u Calculate A from Q = U A LMTD
8/24/99 Heat Exchanger Design 21
RIGOROUS DESIGN
u Determine the basic heat exchanger features such as tube
diameter and wall thickness,tube length,baffle spacing,and
baffle cut.
u Estimate the area required based on a sizing calculation.
u Determine the number of tubes required to provide the esti-
mated area,Check the tube-side fluid velocity,If below the
acceptable range,estimate number of tube passes required.
u Calculate U based on the appropriate correlations and a
reasonable estimate of the fouling resistance.
u Iterate until Uassumed = Ucalculated.
u Check the pressure drops and adjust design if unacceptable.
HEAT EXCHANGER DESIGN
GOALS:
u Role of heat exchangers in chemical processing
u Basic concepts and terminology
u Types of heat exchangers
u Design methodology
e Sizing
e Design
e Rating
8/24/99 Heat Exchanger Design 2
PURPOSE OF A HEAT EXHANGER
u To Heat or Cool a stream flowing from item of
equipment to another,The steam may be a:
e A liquid
e A gas
e A multiphase mixture
u To Vaporize a liquid stream
u To Condense a vapor stream
8/24/99 Heat Exchanger Design 3
Shell Side
Inlet Nozzle
Shell Side
Outlet
Nozzle
Tube Side
Inlet Nozzle
Tube Side
Outlet
Nozzle
SINGLE TUBE PASS,SINGLE SHELL
PASS COUNTERCURRENT HEAT
EXCHANGER
8/24/99 Heat Exchanger Design 4
TYPES OF HEAT EXCHANGER SERVICE
u Fluid heated by external utility
e Steam
e Hot oil or molten salt
e Combustion gas (furnace)
e Electricity (resistive,inductive,microwave)
u Fluid cooled by external utility
e Cooling water
e Refrigeration
u Fluid heated or cooled by other process stream
8/24/99 Heat Exchanger Design 5
DUTY Q (kJ/s)
T (K)
Sensible Heat - Heating
Sensible Heat - Cooling
Pure Component Condensation or Vaporization
Multicomponent Cooling/Condensation
TYPES OF HEATING AND COOLING
CURVES
8/24/99 Heat Exchanger Design 6
CALCULATION OF COOLING CURVES
u Sensible Heat,Q = W Cp (Tin - Tout)
(For constant Cp,Q = W (Hin - Hout) otherwise)
u Latent Heat,Q = W l
u Multicomponent Cooling Curves,Requires point-by-
point flash calculations
8/24/99 Heat Exchanger Design 7
FEASIBLE COOLING CURVE
PAIRINGS
u Corollary of the Second Law of Thermodynamics:
Heat can only be transferred from
a higher temperature to a lower one.
u For heat exchangers this means that the higher
temperature cooling curve and the lower temperature
heating curve cannot intersect.
u When this condition is satisfied,the pairing of a heating
and cooling curve is said to be feasible.
8/24/99 Heat Exchanger Design 8
T
T1in
Q
T1out
T2in
T2out
Countercurrent Flow
T2in
T1in
T1out
T2out
Q
T
Cocurrent Flow
FEASIBLE COOLING CURVE PAIRINGS
T
Q
T
Q
INFEASIBLE COOLING CURVE PAIRINGS
8/24/99 Heat Exchanger Design 9
MAXIMUM HEAT EXCHANGER DUTY
u Qmax,the maximum amount of heat that can be
transferred in a heat exchanger,no matter how large it
is,occurs when the heating and cooling curves either
e Intersect at one end of the exchanger or the other or
e Become tangent within the exchanger
u For sensible heating with constant fluid heat capacities,
the curves are straight lines,They will intersect at that
end of the exchanger whose entering fluid has the largest
WCp,call it Wcpmax.
8/24/99 Heat Exchanger Design 10
MAXIMUM HEAT EXCHANGER DUTY
Q
T
T1in
T2out
T1out = T2in
Qmax0
For this example,WCp2 > WCp1
8/24/99 Heat Exchanger Design 11
MAXIMUM DUTY Qmax
uFor WCp1 > WCp2,the exchanger will,pinch” at the
Fluid 1 inlet.
Qmax = WCp2 (T1in - T2in)
u For WCp2 > WCp1 the exchanger will,pinch” at the
Fluid 2 inlet (as in the previous diagram)
Qmax = WCp1 (T1in - T2in)
8/24/99 Heat Exchanger Design 12
BASIC PERFORMANCE EQUATIONS FOR A
HEAT EXCHANGER
For the fluid flowing in the positive z direction an energy
balance on the section dz gives
dT1/dz = (Ua./WCp1)(T2 - T1)
where U = the overall heat transfer coefficient and
a = the heat transfer area per unit length
If the second fluid is a pure component either vaporizing or
condensing,then
dT2/dz = 0
8/24/99 Heat Exchanger Design 13
BASIC PERFORMANCE EQUATIONS FOR A
HEAT EXCHANGER (Cont’d.)
CASE 1,If the second fluid is a pure component either
vaporizing or condensing,then
dT2/dz = 0
CASE 2,If the second fluid is flowing in the negative z
direction,I.e.,in countercurrent flow,an energy balance
on the section dz gives
dT2/dz = (Ua./WCp2)(T2 - T1)
8/24/99 Heat Exchanger Design 14
RATING SOLUTION FOR CASE 1
T1(z) = Exp[- h1 z] T1(0) + (1 - Exp[- h1 z] ) T2
where h1 = U a /WCp1
For z = L,A = aL and N1 = UA/WCp1 where N1 is
defined as the number of heat transfer units (NTU’s) with
respect to Fluid 1,Let T1out = T1(L) and T1in = T1(0).
Then
T1out = Exp[-N1} T1in + (1 - Exp[-N1]) T2
The exchanger efficiency E = (T1out - T1in)/(T2 - T1in)
E = 1 - Exp[-N1]
8/24/99 Heat Exchanger Design 15
DESIGN SOLUTION FOR CASE 1
N1 = - UA/WCp1 = ln {1 - E}
= ln {[T2 - T1out]/[T2 - T1in]}
Note that Q = WCp1(T1out - T1in),
Substituting and rearranging gives
Q = UA LMTD
where
LMTD = [(T2-T1out) - (T2-T1in)]
ln {[T2 - T1out]/[T2 - T1in]}
which is the well-known heat exchanger design equation.
8/24/99 Heat Exchanger Design 16
RATING SOLUTION FOR CASE 2
Without going through the details:
T2out = [Exp(N2-N1) - 1] T1in + [1 - N1/N2] T2in
[Exp(N2-N1) - N1/N2]
where N1 = UA/WCp1
N2 = UA/WCp2
T1out = Exp(N2-N1)[ 1 - N1/N2] T1in
[Exp(N2-N1) - N1/N2]
+ N1/N2[Exp(N2-N1)-1] T2in
[Exp(N2-N1) - N1/N2]
8/24/99 Heat Exchanger Design 17
DESIGN SOLUTION FOR CASE 2
The design solution is essentially the same for Case 2 as for
Case 1,namely,
Q = UA LMTD
where now LMTD = (T2out - T1in) - (T1out - T2in)
ln {(T2out - T1in)/(T1out - T2in)
8/24/99 Heat Exchanger Design 18
REPRISE,ASSUMPTIONS
u The Cp’s are constant over the temperature range
involved,(Reasonable for most exchangers of practical
interest.)
u U is constant over the temperature range involved.
(Reasonable for most exchangers of practical interest.)
u Flow is pure countercurrent or pure cocurrent,If not,a
correction factor F is required to adjust the LMTD.
Q = UA F LMTD
F has been derived for most common heat exchanger
configurations (multiple tube passes,cross flow,etx.).
8/24/99 Heat Exchanger Design 19
RATING CALCULATIONS
u If U and A are known along with the W’s and Cp’s,use
the appropriate performance equation solutions.
uIf only the specifications of the exchanger (number of
tubes,length of tubes,tube diameter,baffle spacing,
baffle cut,etc.) are given,compute A from the geometry
and U from
1/U = 1/hf1 + 1/hwall + 1/hf2 + rfouling
using the appropriate correlations for hf1 and hf2.
8/24/99 Heat Exchanger Design 20
SIZING CALCULATIONS
u,Choose a typical value for U based on the type of
service,[Tables of typical values can be found in Perry’s
and most textbooks on heat exchanger design.]
u Determine the outlet temperatures based on the
performance specifications and the appropriate energy
balances.
u Calculate A from Q = U A LMTD
8/24/99 Heat Exchanger Design 21
RIGOROUS DESIGN
u Determine the basic heat exchanger features such as tube
diameter and wall thickness,tube length,baffle spacing,and
baffle cut.
u Estimate the area required based on a sizing calculation.
u Determine the number of tubes required to provide the esti-
mated area,Check the tube-side fluid velocity,If below the
acceptable range,estimate number of tube passes required.
u Calculate U based on the appropriate correlations and a
reasonable estimate of the fouling resistance.
u Iterate until Uassumed = Ucalculated.
u Check the pressure drops and adjust design if unacceptable.