Heat transfer Section 01
description
Transcript of Heat transfer Section 01
EDS 2004/EXC 1-1
Training Services
Shell & TubeHeat Exchangers
Presentation on shell and tube exchangers.
EDS 2004/EXC 1-2
Heat Exchangers - Course Content
Introduction to Heat Transfer MechanismsConduction, Convection, and RadiationOverall Heat Transfer CoefficientLog Mean Temperature DifferenceExample Problem - FCC Flue Gas CoolerHydraulicsExample Problems - Tube side Pressure Drop
Section 1 - Introduction to Heat Transfer
Table of Contents of First Section - Introduction and Basics of Heat Transfer
This may be a review of basic concepts but it is useful for establishing nomenclature and symbols.
EDS 2004/EXC 1-3
ConductionConvectionRadiation
Heat Transfer Basics
Three Forms of Heat Transfer
The three forms of heat transfer encountered in industrial situations are generally these.
Radiation heat transfer is of little consequence in most refinery heat exchangers and is generally ignored.
EDS 2004/EXC 1-4
Conduction Heat Transfer
Heat transfer within a material.Fourier’s equation governs.
Q = A k dTdx
Conduction heat transfer is the transfer of heat within a material. The material can be solid, liquid or vapor.
Q is the amount of heat transferred.
A is the area across with the heat is transferred.
k is the thermal conductivity of the material. It is a unique value for each material. It is normally a function of temperature. In this equation, the thermal conductivity at the average temperature should be used.
T is the temperature. dT is the differential temperature.
x is the distance that the heat is transferred. dX is the differential distance.
EDS 2004/EXC 1-5
Convection Heat Transfer
Heat Transfer between solids and liquids/vapors
Q = A h ∆T
where h is a function of physical properties and geometry
Convection heat transfer occurs between solids and either liquids and vapors in contact with the solid.
Q is the amount of heat transferred.
A is the area across with the heat is transferred.
h is the heat transfer coefficient. It is generally empirically found for fluids depending on their physical properties and the geometry of the flowing setup. Therefore, the h value is found using different equations for flow inside a tube and outside a tube, against a flat plate, using non-round shapes etc.
T is the temperature. ∆T is the difference in temperature.
EDS 2004/EXC 1-6
Conduction Heat Transfer
Thermal Conductivity of Various Materials
EXC-Roo-01
This chart shows the thermal conductivity of various materials.
Note that the vapor materials occurs at the bottom of the chart, liquids occur higher up, and solids near the top have the highest values of thermal conductivity.
The higher the value the easier the material transfers heat.
Most materials transfer heat better at higher temperatures, but not all.
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Example - Conduction
T = 100°C
BoilingWater
T = 30°C
Hand
100°C
30°C
dT/dx = - 35000 °C/m
T
x
2 mm
k (steel) = 46 kcal/(hr m°C)
dTdx
(100-30) °C0.002 m= = 35 000 °C/m
EXC-Roo-02
A simple example everyone can relate to - touching a pot of boiling water with your finger.
The inside of the the metal pot has water at its boiling point. The heat is conducted through the metal of the pot to your finger on the outside.
Knowing the thermal conductivity of the pot, the thickness of the metal, and the temperature of the water and your finger -
Continued
EDS 2004/EXC 1-8
Example - Conduction(continued)
Touch boiling kettle with a finger, A ≈ 1 cm x 1 cm = 0.0001 m2
⇒ Q = 0.0001 m2 x 1.61x106 kcal/hr/m2 = 161 kcal/hr = 189 W= 368 BTU/hr
=mC3500
Cm hrkcal46
AQ o
o
The amount of heat transferred to your finger can be calculated.
Using the various sets of units of measure.
It will be noted that the values shown here are actually very low compared to the heat transfer rates in refinery type equipment.
EDS 2004/EXC 1-9
Resistances in SeriesL 1 L 2 L 3
T 1 T 2
k 1 k 2 k 3
T hot T cold
T
x
Tube wall FoulingFouling
Tube metal has higher conductivity than foulingEXC-Roo-03
A slightly more complicated problem involves conduction through multiple layers in series.
In this case, three layers are shown. This could represent the case of a tube with fouling on both sides.
The amount the temperature drops in each layer is a function of the relative values of thermal conductivity. The higher the value the less the temperature drop as there is less resistance to heat transfer.
EDS 2004/EXC 1-10
Resistances in Series (continued)
Heat flux is calculated as follows:
( )cThTeffk
3k3L
2k2L
1k1L
cThTAQ
−=
++
−=
1
3k3L
2k2L
1k1L
effk−
++=so
The last example can be substituted into the conduction equation. By making up a factor called k eff (effective thermal conductivity), the series conduction problem looks just like the single layer problem.
In other words, the effective resistance is the sum of all the resistances. L1/k1 is the “resistance” of the layer 1.
Once you have Q, any intermediate temperatures can be calculated:
Qk
LA
Th T1
1
1
= − get T1, etc.⇒
EDS 2004/EXC 1-11
Convection Heat TransferTubeside heat transfer coefficientUse Dittus-Boelter Equation:
h kD
. C
k D v
ii
p/
i.
=
0 0231 3 0 8µ ρ
µ
collect variables:
( )h . kD
C v i
/
i. p
/ . .= −0 0232 3
0 21 3 0 467 0 8µ ρ
h DD
hioi
oi=
increase this hi ....k increasesDi DECREASESCp increasesµ DECREASESρ increasesv increases
k = Thermal conductivityCp = Specific heatµ = Viscosityρ = Densityv = Velocity
Di = Inside diameter
Correct for Outside Surface Area:
Convection heat transfer is the second type to be discussed.
In convection the heat transfer is found empirically for each type of geometry. This equation covers single phase heat transfer in a circular tube. The heat transfer coefficient is a function of the physical properties of the fluid.
The table on the left shows the effect of changing each of the variables in the equation. Note that the two factors that hurt heat transfer are increasing diameter and increasing viscosity. Heat transfer is worse in larger tubes and with more viscous fluids.
The last equation corrects this coefficient, which is calculated based on inside diameter to the outside diameter of the tubes. All of the heat transfer equations refer to the area across which the heat is transferred. To make that a consistent area is used, all calculations are based on the area of the outside diameter of the tubes. Therefore, this inside coefficient must be corrected.
EDS 2004/EXC 1-12
Convection Heat Transfer(continued)
Shellside CoefficientEquation from Kern:
h . kD
D v C ko
e
e.
p/
w
.
=
0 36
0 55 1 3 0 14ρ
µ
µ µµ
Collect Variables:
( )h . k
D v C o
/
e. .
.p
/
w
.
=
0 36
2 3
0 45 0 22
0 55 1 30 14
µρ
µµ
The last term (viscosity correction) is usually small, and we will ignore it.
increase this ho ....k increasesDe DECREASESCp increasesµ DECREASESρ increasesv increases
This equation is the convection equation for fluid flowing on the outside of a circular tube. The form of the equation is similar. An additional term that corrects for the viscosity of the fluid at the wall temperature is present.
Again, increasing the diameter and viscosity hurts the heat transfer rate.
The diameter here is an equivalent diameter that will be covered later.
EDS 2004/EXC 1-13
Combined Conductionand Convection Coefficients
L
k tube
h 1 h2
T
X
Tube wall
hot cold
EXC-R00-04
Combining the conduction and convection problems.
Consider a flat surface with fluids on each side. The heat is transferred from the hot fluid to the surface by convection. Conduction occurs in the surface. A second convection rate is on the other side of the surface.
This looks similar to the series conduction problem.
EDS 2004/EXC 1-14
Combined Conduction and Convection(continued)
( )QA
T TLk
Lk
Lk
k T Th ceff h c=
−
+ +
= −1
1
2
2
3
3
substitute as shownh1 h2 U for keff (convention)
( )QA
T T
hL
k h
U T Th c
tube
h c=−
+ +
= −1 1
1 2
There may be a few solid layers present (tube, 2 fouling layers).
If you look at the equation for the series conduction problem and substitute the convection heat transfer terms, our effective conductivity term becomes an effective heat transfer term for the overall problem. This is call U or the Overall Heat Transfer Coefficient.
EDS 2004/EXC 1-15
Combined Conduction and Convection(continued)
The overall resistance is the sum of the individual resistances.
( )ch
ofo
fo
tubeif
if
i
ch TTU
h1
kx
kL
k
xh1
TTAQ
−=
++++
−=
or
Q = U A ∆T
The U value term can be expanded to include the fouling layers on both sides of the surface.
The U value becomes an easy way to relate the amount of heat being transferred to the area and the temperature difference.
EDS 2004/EXC 1-16
Tube wall is curved; outside area is different from inside area: tubewall resistance is calculated as:
1
otube
i
oeo
i
o
i h1
krrlogr
rr
h1U
−
+
•
+
=
or if you include fouling resistances1
foi
ofi
otube
i
oeo
i
o
ir
rrr
h1
krrlogr
rr
h1U
−
+
++
•
+
=
Combined Conduction and Convection(continued)
Round Geometry
Up to this point, the surface we showed were flat. Most tubular heat exchangers use round geometries. A slight correction needs to be made due to the curvature of the tubes versus a flat surface.
Therefore, the bottom equation describes the U value for most tubular type heat exchangers.
This equation shows that BEFORE combining resistances they must be based on the same area. The equation shown is based on the OUTER tube area, which is the convention in the heat transfer industry.
EDS 2004/EXC 1-17
When calculating duty, what ∆T should be used here?
t out = 130
t in = 80
T in = 150
T out = 90
Temperature Difference
mean log
inout
outine
inoutoutineff T
tTtTlog
)t(T)t(TT ∆=
−−
−−−=∆
Q=UA∆T is the basic equation for heat transfer in exchangers.
We have just talked about the U value.
The A or area is that of the outside of the tubes.
We still need to discuss the ∆T term.
In a typical exchanger, there are two fluids entering and leaving. Each of these four points has a temperature. Which temperature difference do we use?
For the exchanger shown in the last picture, one where the two fluids flow in precisely opposite direction (counter current flow), the temperature difference is found using this formula. This is referred to as the log mean temperature difference of LMTD.
EDS 2004/EXC 1-18
Temperature Difference(continued)
In practice, pure countercurrent flow situations often do not exist.For example: Most exchangers have more than one tube-pass.
coldincoldout
hotin
hotout
a
b c
d
EXC-Roo-06
However, few if any exchangers are really using counter current flow. This diagram shows a more typical exchanger. The tube side fluid travels the length of the exchanger twice. One trip in the bottom half and the return in the top half. At the same time, the shell side fluid travels once along the length of the exchanger, but many times back and forth across it.
Again, which temperature difference is used here?
EDS 2004/EXC 1-19
Temperature Difference(continued)
∆T in the top half of the bundle > ∆T bottom half.What is the EFFECTIVE TEMPERATURE
difference in the exchanger?What is ∆T at a, b and c and d?
This is a difficult question. The answer varies depending on all 4 inlet/outlet temperatures.
We know that ∆T effective is LESS than ∆Tlm.
⇒
The temperature differences are different at each end of the bundle and the difference between the temperature of the two fluids is different depending where along its journey it is.
The bottom line is that the temperature difference is less than that in a true counter current flow case.
We will return to this problem later.
EDS 2004/EXC 1-20
Example Calculation
FCC Flue Gas Cooler
Flue gas on the tubeside.Boiler feed water is on the shellside.
Flue Gas: 51500 kg/hr, in at 740°C, density = 0.81 kg/m3,Cp = 0.259 kcal/kg/°C, µ = 2.2 x 10-5 kg/s-m,k = 0.029 kcal/hr/m/°C.
Example calculation problem.
The flue gases from an FCC regenerator are often cooled to protect pollution control equipment or merely to conserve energy. The cooling is typically done in a waste heat steam generator. In this case, the flue gas will flow on the inside of the tubes while boiler feed water and steam is on the shell side.
The flue gas conditions and properties are listed here.
EDS 2004/EXC 1-21
Example Calculation(continued)
a) Calculate hio if there are 350 tubes, 1.75" OD (44.45 mm), 0.2" (5.1 mm) wall thickness.
b) Given the BFW coefficient as
ho =8 600kcal/hr/m2/°C, bulk water temperature = 240°C and using tube metal thermal conductivityof = 37 kcal/hr/m/°C, calculate tubewall temperatures.
c) What happens if BFW supply is lost ?
Part A - Calculated the inside heat transfer coefficient for the flue gas stream using the tube data.
Part B - Given the shell side heat transfer coefficient and conditions, calculate the tube wall temperatures.
Part C - Predict what will happen if the supply of boiler feed water is lost.
EDS 2004/EXC 1-22
Solution to Problem - FCC Flue Gas CoolerPart a)1) Calculate velocity:
Volumetric flow ms
ms
=•
=51500
0 81 360017 66
3 3
..
( )tube flow area m=π − •
=3500 04445 2 0 0051
40 323
22. .
.
2) Calculate Reynolds number
Re ( . . ) . ..
=µ
=
− • • •
=−
D vx
i ρ 0 04445 2 0 0051 54 6 0 812 2 10
68 8515
3) Calculate Pr
Pr. .
..=
µ
=
• •
=
−Ck
xp 0 259 2 2 10 36000 029
0 7065
velocity m s= =17 660 323
54 6..
.
Part A -
First calculate the flue gas velocity.
Then calculate the Reynolds Number.
Then find the Prandtl Number.
EDS 2004/EXC 1-23
Solution to Problem - FCC Flue Gas Cooler (continued)
4) Calculate inside heat transfer coefficient
( ) ( )
h kD
Ck
D v
h
h kcal hr m C
ii
p i
i
i
=µ
µ
=−
= °
0 023
0 0290 04445 2 0 0051
0 023 0 706 68851
128 6
1 3 0 8
1 3 0 8
2
.
.. ( . )
. .
. / / /
/ .
/ .
ρ
5) Correct to tube outside diameter
hio = 0.7705 hi = 99.1 kcal/hr/m2/°C
Now calculate the inside heat transfer coefficient and correct it for the outside area of the tubes.
EDS 2004/EXC 1-24
Solution to Problem - FCC Flue Gas Cooler(continued)
Part b)1) Calculate tubewall resistance
resis cer r
rk
Chr mkcal
o eo
i
tubetan
log
. log .. . .
=•
=•
− •
=
°0 04445
20 04445
0 04445 2 0 005137
0 000152
Part B -
Calculate the resistance in the tube wall itself. Remember the tube is round.
EDS 2004/EXC 1-25
Solution to Problem - FCC Flue Gas Cooler(continued)
2) Calculate U
Uh
r rr
k hr r
rr
U kcalhr m C
io
o eo
i
tube ofi
o
ifo= +
•
+ +
+
= + + + +
=°
−
−
1 1
199 1
0 00015 18600
0 0 96 55
1
1
2
log
.. .
QA
kcalhr moutside
= − =96 55 740 240 48 276 2. ( )⇒
Using the inside heat transfer coefficient, the given outside coefficient, the calculated tube wall resistance, and ignoring any fouling resistances, the Overall Heat Transfer Coefficient (U) is now calculated.
EDS 2004/EXC 1-26
Solution to Problem - FCC Flue Gas Cooler(continued)
3) Calculate hot side tubewall temperature:
( )
( ) ( )
( )
QA
U T
QA
kcal h T T
T C
T C
io w hot w hot
w hot
w hot
=
= = − = − =
= − = ° =
= = °
∆
48 276 740 99 1 740
740 487 1
252 9
hr m2 , ,
,
,
.
.
.
Knowing the value of Q/A and U, the temperature differences in each layer can be calculated. Using the inside heat transfer coefficient, and the inside bulk fluid temperature, the temperature at the outside of the tubewall can be found.
EDS 2004/EXC 1-27
Solution to Problem - FCC Flue Gas Cooler(continued)
4) Calculate cold side tubewall temperature:
( )
( )
QA
kcalhr m resis ce
. T
. T kcalhr m
. C hr mkcal
. C
T . . C . C
w,cold
w,cold
w,cold
= = • − =
− = •°
= °
= − ° = °
48 276 1 252 9
252 9 48 276 0 00015 7 3
252 9 7 3 245 6
2
2
2
tan
Summary:hio = 99.1 kcal/hr/m2/°Cho = 8 600 kcal/hr/m2/°C Tubewall temperature: 253°C to 246°C
Using the fact that the Q/A term is a constant, the temperature drop in the tube itself can be found.
In this case, note that the heat transfer coefficient is very high on the outside of the tubes compared to that on the inside. Therefore, most of the temperature drop is in the flue gas itself. The temperature of the tube is much closer to that of the water than the temperature of the flue gas.
EDS 2004/EXC 1-28
Solution to Problem - FCC Flue Gas Cooler (continued)
Part c)
If BFW is lost, tube will soon reach 740°C since ho approaches 0CS tubes cannot tolerate 740°C, and will crack Time to failure is in a few minutesTube failure allows steam into flue gas line
Part C -
It is obvious that if the water on the outside of the tubes is lost, the temperature of the tubes will approach the temperature of the flue gas. This will result in a tube failure.
EDS 2004/EXC 1-29
Hydraulics
Tubeside ∆P is easy to calculate.Calculations similar to pipe flow calculations.
19.0i
t
i
2t
t
vD046.0f where
DnLvf2P
−
µρ
≈
ρ=∆
L = Length of tubesn = Number of passesft = Friction factor
Changing topics:
Pressure drop on the tube side of an exchanger is similar to pipe flow pressure drop. The flow is generally split among many tubes in parallel and may have several directional changes.
The equation for fluid flow in pipes is adequate.
The friction factor is a function of NRe and tube diameter and is found in the literature.
At high NRe, f is constant for a fix tube size.
Tube ID, in. f 0.2 0.0360.25 0.0340.3 0.0320.4 0.0290.5 0.0270.75 0.0241.0 0.0231.5 0.021
EDS 2004/EXC 1-30
Friction Factor Chart
EXC-Roo-07
The friction factor needs to be found using a chart like this or the simplifying equation on the last page.
EDS 2004/EXC 1-31
Hydraulics(continued)
Shellside ∆P Need to know what area to use. Use flow area between tubes at center of shell:
shell
tube-tubeclearance=p-D
bundle-shellclearance
# tubes = D s/p
t
( )A Dp
p D Bfss
t s= −EXC-Roo-08
Ds = Shell diameterp = pitch
Bs = Baffle spacing
Pressure drop on the shell side is a much harder problem. As the flow progresses, it is flowing radially and longitudinally in a cylinder. The number of tubes it is flowing past and the diameter of the cylinder is constantly changing.
This procedure provides a simplified solution to the problem based on using the geometry at the center of the bundle. It assumes a uniform baffle spacing along the axis of the exchanger.
To get more accurate calculations, use a computer simulation such as HTRI or HTFS etc.
EDS 2004/EXC 1-32
Hydraulics(continued)
For small bundles or accurate calculation, allow for bundle-shell clearance.
∆ = +
=
=µ
−
P f D N vD
v WA
f D v
ss s b
e
s
fs
se
( )
..
12
1 79
2
0 19
ρ
ρ
Dp D
De
t
t
=−π
π
44
22
Nb = Number of bafflesWs = Volumetric flow rateDe = Equivalent diameter
Additional parameters for the calculation are given here. Note this is where the equivalent diameter mentioned for the shell side heat transfer coefficient is defined.
EDS 2004/EXC 1-33
Hydraulics in General
∆varies with flow rate, length, and diameter scales:
∆ ∝ ⋅ ⋅ ⋅
⇒∆∆
=
P f geometry fluid flow v LengthScaleDiamScale
PP
vv
LL
DD
( , , ) 2
2
1
2
1
22
1
1
2
ρ
Length Scale (L) is length measured in direction of flow (tube length, or shell diameter).Diameter Scale is a length measured perpendicular to flow (tube ID or equivalent diameter).This general form is valid for both shell and tubesideFor highly turbulent flow, f does not change much with increase of flow rateExceptions: crude and atmospheric residue, asphalt etc. Then variation of f with flow rate must be included.
In general, it is possible to compare the effect on pressure drop by modification to the exchanger quite simply.
The pressure drop ratio is the ratio of the velocity squared, the difference in flowing length, and the difference in flowing diameter.
This is true for either side of an exchanger. It is slightly less precise on the shell side due to the more complicated flow paths.
EDS 2004/EXC 1-34
Quick Estimations
Example 1If existing exchanger has tube ∆P = 70 kPa, and revamp needs extra 10% flow:
∆Prevamp ≈ 70 kPa (1.1)2 = 70 (1.21) = 85 kPa
Example 2A heat exchanger design has 2 tube passes, ∆Ptube = 15 kPa (too low).
Someone proposes changing to 4 tube passesvelocity doubles, number of passes doubles:
∴
∆ = • • = • =P kPa15 2 2 15 8 1202 (too high).
Example 1 - Increase the flow by 10%
This increases the velocity by 10% and has no effect on length or diameter so the result is simply the square of the velocity or 21%.
Example 2 - Going from 2 to 4 tube passes
This doubles the velocity as half the tubes are available for each pass. It also doubles the length as the flow goes along the axis of the exchanger 4 times instead of twice. Therefore, the pressure drop goes up by 2 squared due to velocity times another 2 for length for a total of 8 times.
EDS 2004/EXC 1-35
Quick Estimations(continued)
Example 3Retube existing exchanger from 1" OD 16 BWG (0.065" wall) tubes to 1" OD 14 BWG (0.083" wall) tubes (same flow rate)
Old Tube ID is = 1" - 2 (0.065") = 0.870"New Tube ID is = 1" - 2 (0.083") = 0.834”
Therefore, this retubing will increase the pressure drop by 24%.
∆∆
=
=
=
=
PP
vv
14
16
14
16
2
2 2 5
0 8700 834
0 8700 834
0 8700 834
0 8700 834
1 24
.
.
.
...
.
..
Example 3 - Retubing the exchanger with slightly thicker tubes
The velocity increase by the square of the change in inside diameter. Therefore, the pressure drop increases by the ratio of the diameters to the fourth due to the velocity and once more due to the change in diameter. Therefore, the ratio of pressure drops is the 5th power of the diameter ratio. So the 4% change in diameter results in a 24% change in pressure drop.