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Changes shown by ➧
CONTENTS
Section
Page
SCOPE ............................................................................................................................................................ 2
REFERENCES ................................................................................................................................................ 2
GENERAL ....................................................................................................................................................... 2
THEORY AND DESIGN CONSIDERATIONS................................................................................................. 2
ZONES.................................................................................................................................................... 2
THE BASIC STEPS ................................................................................................................................ 3
SPECIAL DESIGN CONSIDERATIONS FOR DRUMLESS CONDENSERS ......................................... 8
DETAILED DESIGN PROCEDURE........................................................................................................ 8
VACUUM PIPESTILL OVERHEAD CONDENSER DESIGN PROCEDURE ........................................ 14
NOMENCLATURE ........................................................................................................................................ 15
TABLES
Table 1
Table 2
Table 3
Table 4
FIGURES
Figure 1
Figure 2A
Figure 3B
Figure 4
Figure 5
Figure 6A
Figure 6B
Figure 7A
Figure 7B
Figure 8
T-Q Curve Calculation Procedure....................................................................................... 19
Calculation Procedure For Condenser Design.................................................................... 21
Wide-Cut Condenser Design Sample Calculation .............................................................. 51
A Sample Condenser Design ............................................................................................. 55
Typical Heat Release (“T-Q") Curve (for Wide-Cut Hydrocarbons
in the Presence of Steam) .................................................................................................. 70
Sample Flash Curve and Tds Plot ...................................................................................... 71
Vapor Pressure of Hydrocarbons (METRIC)....................................................................... 74
Average Vapor Quantity in Condensers (or Reboilers)....................................................... 75
Shell-Side Heat Transfer for Partial Condensation of Wide Cuts
(Not Recommended for Definitive Designs)........................................................................ 76
Flash Curve Corrections ..................................................................................................... 77
Flash Curve Corrections ..................................................................................................... 77
Enthalpy of Water Above 32°F ........................................................................................... 78
Enthalpy of Water Above 0°C ............................................................................................. 78
Heat Transfer Coefficient for Fluids in Tubes ..................................................................... 79
Revision Memo
12/99
Minor revisions and updated information throughout the section.
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
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This section presents design considerations and the recommended manual design procedure for shell and tube condensers.
REFERENCES
4.
5.
Gloyer, W., Industrial and Engr. Chemistry, 42, 7, 1361, July, 1950.
Devore, A., Petroleum Refiner, 38, 6, 205, June, 1959.
Akers, Deans, Crosser, Condensing Heat Transfer Within Horizontal Tubes, Presented at the Second National Heat Transfer
Conference, AIChE-ASME, Chicago, Illinois, August 18 - 21, 1958.
Diehl, J. E. and Uhruh, C. H., Petroleum Refiner, 36, 10, 124, October 1959.
Chenoweth, J. M. and Martin, M. W., Petroleum Refiner, 34, 151, October 1955.
6.
7.
Heat Exchange Institute (HEI) Standards for Steam Surface Condensers.
Perry's Chemical Engineers' Handbook, 6th Ed.
1.
2.
3.
GENERAL
➧
➧
The same basic approach that is used to design liquid-liquid exchangers is also used to design condensers. This approach uses
a trial-and-error calculation to balance an assumed exchanger size against a calculated size, while staying within the prescribed
limits of pressure drop. The basic difference lies in the so-called “zoning" of the condenser.
Zoning is necessary because the concept of LMTD (based on the entire heat exchanger) is invalid whenever the heat transfer
mechanisms vary from one region to another within the exchanger. The LMTD concept assumes that the heat transfer
coefficients and the specific heats remain constant for both fluids. This is not the case for condensers. However, we may break
down the exchanger into a number of zones and the LMTD concept can be applied separately in each of these zones. These
zones are selected in a manner that the heat transfer coefficients and the specific heats (represented by the slope of heat
release (T-Q) curve may be assumed to be constant. Therefore, all phase-change points, such as dew points (both hydrocarbon
and steam) and bubble point define a start of a new zone. Additional zones may be required if the slope of the T-Q curve varies
appreciably.
It should be noted here that this discussion is primarily concerned with the more difficult case of condensing a wide-cut
hydrocarbon in the presence of steam. For narrow cuts [15 to 20°F (8 to 11°C)], zoning is usually unnecessary and the following
method simplifies considerably.
HTRI has developed updated empirical correlations and procedures for computer analysis, which are explained in their design
manual and reports. For design and rating of condensers, the CST-3 Computer Program is recommended. This program may
be used stand-alone, or through the heat exchanger network analysis program HEXTRAN.
THEORY AND DESIGN CONSIDERATIONS
ZONES
In setting up the zones, one assumes that the heat transfer coefficient is constant within the zone and that heat release is directly
proportional to temperature change within the zone.
One can see from this that the larger the number of condenser zones, the more accurate would be the condenser design.
Unfortunately, the calculation of each zone is so time consuming that, for hand calculation, it is impractical to set up more than
two or three zones. As shown in Figure 1, the usual zones are:
1. Vapor cooling.
2. Vapor cooling + hydrocarbon condensation and subcooling.
3. Vapor cooling + water condensation and subcooling + hydrocarbon condensation and subcooling.
Figure 1 is somewhat simplified, in that the coolant is shown for a one-tube-pass unit. In this typical sequence, the hydrocarbon
dew point is above the steam dew point. If the steam dew point should occur first, Zone 2 is essentially eliminated.
When better accuracy is desirable, Zone 3 is sometimes divided into two “sub-zones" of roughly equal heat duty.
In the absence of desuperheating, steam condensation or any other sharp break in the heat release curve, one condenser zone
is adequate.
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THE BASIC STEPS
The following steps are basic to the design of a unit for condensing wide-cut hydrocarbons in the presence of steam.
1. Determine the hydrocarbon dew point.
2. Determine the steam dew point.
3. Determine the heat released in each zone.
4. Determine the simple LMTD for each zone, then ∆tew (weighted effective log mean temperature deference, °F or °C).
5. Estimate an overall coefficient (i.e., total area) for the condenser; determine mechanical features.
6. Calculate area required for each zone; sum the area.
7. Adjust mechanical features of the unit until the assumed total area is equal to the sum of the areas calculated for each zone.
These basic steps will next be discussed generally.
Dew Points; The Flash Curve - For wide-cut condensation, the hydrocarbon dew point is easily found by trial-and-error
reductions in temperature to find the point where hydrocarbon vapor pressure equals its known partial pressure. (For the usual
case of a fractionator overhead, the hydrocarbon is already at its dew point.)
The steam dew point is more difficult, since the total number of moles of vapor changes between the hydrocarbon dew point and
the steam dew point. This change in composition must be determined before the steam dew point can be calculated.
It is usual practice to assume that as condensation proceeds, the vapor and liquid maintain an equilibrium composition. Hence,
the composition and thermal properties of the system are obtained from equilibrium flash calculations. These calculations predict
a “flash curve," which is a plot of temperature vs. “percent off" (percent not condensed). The design manual method for
estimating the flash curve is a simplification of the Exxon Blue Book method.
The entire flash curve is not constructed. Instead, several points along the volumetric flash curve are calculated. These are
plotted and connected with straight lines. The molal flash curve lies slightly below the volumetric curve, and the weight flash
curve lies above the volumetric curve. For the sake of simplification, this separation of the curves is assumed proportional to the
width of certain boiling ranges in the cut. The three curves are plotted on the same graph and used to predict composition and
temperature of the vapor and liquid phases as condensation proceeds. An example is shown in Figure 2.
Since these curves represent conditions at one atmosphere partial pressure, an alignment chart is included to convert to the
pressure of the condensing system (Figure 3).
The steam dew point is then found by a trial-and-error reduction in the system temperature to find the temperature, where water
vapor pressure equals its calculated partial pressure. The method is described in detail in Table 1.
Heat Release in Each Zone - The following individual heat duties are calculated in each zone in which they occur. All duties
occur simultaneously in the third zone only.
1. Cool entering liquid.
2. Cool vapor and gas that do not condense.
3. Cool condensing HC vapor.
4. Cool hydrocarbon condensate.
5. Cool condensing steam.
6. Cool steam condensate.
7. Remove latent heat of hydrocarbon.
8. Remove latent heat of steam.
To calculate Duties 3, 4, 5, and 6, one assumes that 50% (or the average quantity) of the material is cooled through 100% of the
temperature change in the zone.
Although enthalpy charts can be used for these calculations, the use of average specific heats and average latent heats is
adequate for most cases. Note that when enthalpy charts are used, Items 7 and 8 must be calculated at the average
temperature (rather than inlet to outlet conditions) to avoid duplicating the sensible-heat portion of the duty.
Temperature Driving Force - After the zone heat duties have been calculated, the weighted average LMTD (∆tew) can be
calculated. This is done by calculating the LMTD for each zone and then “weighting" the LMTD according to the amount of heat
transferred in the zone. The conventional equation is as follows:
∆t ew =
Q
q dh
q ds
q
+
+ sc
∆t dh
∆t ds
∆t sc
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
Eq. (1)
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For explanation of symbols, see NOMENCLATURE.
Each zone ∆t is corrected for noncountercurrent flow by multiplying by Fn from Figure 2 of Section IX-D. Fn is calculated from
the zone terminal temperatures.1 This use of Fn is not theoretically sound but is justified, as a matter of convention, to make sure
that enough shell passes are specified for the design temperature cross.
The use of this equation for weighting ∆t's is strictly a matter of industrial convention. It gives the true value of ∆tew only when the
coefficient is constant over the entire condenser; and this condition, of course, almost never prevails. But since the equation has
been widely adopted, the use of other methods for weighting ∆t's might be misleading to exchanger suppliers.
In reality, ∆tew does not enter into the condenser design and its value is of no real importance. The design is based on a
summation of zone areas which are calculated from the individual zone ∆t's. The overall ∆tew is estimated only so that an overall
average Uo can be reported.
Heat Transfer Coefficients, Shell-Side Condensation - From an analytical viewpoint, a condenser is an extremely complex
piece of equipment. The approach used in this manual is sometimes called the “Gloyer" (Reference 1) analysis, named after the
first person to describe it in the literature. This method is not a theoretical analysis, but an empirical attempt to consider the
various phenomena which occur simultaneously in a condenser. This approach considers the following mechanisms:
1. Vapor cooled by forced convection.
2. Liquid cooled by “dripping" from tube to tube.
3. Liquid cooled by forced convection along the bottom of the shell.
4. Condensation.
In each zone of the condenser, a heat transfer coefficient “h" is calculated for each of the above mechanisms. The heat transfer
coefficient for the zone is then evaluated by weighting and averaging the individual coefficients:
Q zone
q1
q
+ ... + 4
h1
h4
h zone =
Eq. (2)
(Subscripts refer to above mechanisms.)
The overall zone coefficient is then calculated from the usual resistance equation:
1
U zone
=
1
+ ro + rw + rio + R io
h zone
Eq. (3)
Q zone
Uzone ∆t zone
Eq. (4)
and the area from
A zone =
The total area of the condenser is simply the sum of the zone areas.
To calculate the coefficients for vapor cooling and “bottom flow" liquid cooling, one uses the average vapor and liquid quantities
in the zone. These quantities are determined with the aid of Figure 4. This chart corrects for the variations in ∆T (between tubeside and shell side) with tube length. The usual relations for single-phase heat transfer are then used to calculate coefficients.
The coefficient for “drip cooling" is taken as 1.5 times the condensing coefficient. It is arbitrarily assumed that half of the liquid
cooling duty is by “drip cooling" and half by “bottom flow" (for zones with all vapor entering).
To calculate the condensing coefficient, the following modifications of the Nusselt equation recommended by Devore should be
used (Reference 2).
1 For hydrofiner effluent-hydrofiner feed exchangers and other similar services where there is a “pinch” in the T-Q curves of the two streams,
each shell of the final arrangement should be checked graphically for a temperature cross (i.e., plot shell-side and tube-side temperatures vs.
duty curves on a single graph and compare inlet vs. outlet temperatures for each shell). If a temperature cross occurs, either the shell areas or
the number of shells should be adjusted to remove the cross.
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THEORY AND DESIGN CONSIDERATIONS (Cont)
For Vertical Tubes
1/ 3
 1.04 x 10 4 
 2
 k f  sf 
hλ = 
1
/
3
 Γ

 µf 


 
(Customary)
Eq. (5)
(Metric)
Eq. (5)M
or
1/ 3
 s2 
 254 
hλ =  1 / 3  k f  f 
 µf 
Γ

 
where:
Γ =
12 Wc
π Nt do
(Customary)
Eq. (6)
Γ =
1000 Wc
π Nt do
(Metric)
Eq. (6)M
or
This term is usually called “tube loading." The equation is restricted to:
Γ
≤ 1090
µf
(Customary)
Γ
≤ 450
µf
(Metric)
or
For Horizontal Tubes
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 
 Γ 1/ 3





1/ 3
(Customary)
Eq. (7)
(Metric)
Eq. (7)M
or
 s2 
 204 
 k f  f 
h′λ = 
 µf 
 Γ 1/ 3 


1/ 3
where:
Γ =
Wc
L c ns
Eq. (8)
The term ns is the number of “condensate streams" in the bundle.
ns
=
1.29 Nt0.480 for square tube layout, 90°.
ns
=
2.08 Nt0.495 for rotated triangular layout, 60°.
ns
=
1.02 Nt0.519 for triangular layout, 30°.
ns
=
1.37 Nt0.518 for rotated square layout, 45°.
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The term Lc is the length of tube on which condensation occurs. If a condenser consists of one zone, Lc equals the tube length.
Otherwise Lc is the fraction of the total tube length included in the zone in question. (If there are two shells in series, note that
one zone might include 0.5L, the next zone 1.5L.)
The paper by Devore suggests that the value of h′λ be increased by a multiplier which depends essentially on the number of
tubes in the bundle. For a typical bundle, the value of this so-called “turbulence factor" is 1.5 to 2.5. This has not been included
in this procedure. Instead, a modification is recommended which will relate the condenser design to vapor mass velocity. (An
effect of mass velocity has been recognized for some time.)
A great deal of Exxon data on wide-cut condensation show that the overall outside coefficient varies linearly with Gx, where x is
between 0.7 and 0.9. A similar effect probably holds for narrow cuts as well. Based on this, the condensing coefficient for
horizontal bundles should be adjusted as follows:
G 
h λ = h′λ  v 
 5 
0.70
(Customary)
Eq. (9)
(Metric)
Eq. (9)M
or
 G 
h λ = h ′λ  v 
 24.4 
0.70
and
h λ max = 2.0 h′λ
Eq. (10)
Where Gv is the average vapor mass velocity in the zone. Note that this mass velocity correction is not recommended for purecomponent condensation.
In zones where both steam and hydrocarbon are condensing, the hydrocarbon wets the tubes preferentially. This forces the
steam to condense on the hydrocarbon condensate film and the condensing coefficient for steam equals the condensing
coefficient for the hydrocarbon.
Empirical Equation for Shell-Side Condensation - Not Recommended for Definitive Designs - Exxon has analyzed a
considerable amount of data on wide-cut condensation. These analyses have resulted in an equation which fits the original data
with an average accuracy of 10%. So far, however, the equation has not been tested on enough additional data to substantiate
its use for definitive designs. The equation can be used for estimating coefficients and for checking condenser performance
under various operating conditions. Because of its simplicity, it is hoped that an equation of this type will eventually be backed up
by enough data to warrant use in design work.
The equation for the shell-side condensing coefficient, which includes the shell-side fouling factor, is as follows:
ln (h oc ) = 0.82 ln (g)G + f (α) − 0.002 Pi
(Customary)
Eq. (11)
(Metric)
Eq. (11)M
or
ln (h oc ) = 0.82 ln(G) + f (α) − 0.00029 Pi + 0.436
The function f(α) is shown in Figure 5. This curve shows that the shell-side coefficient first decreases sharply with increasing
nocondensables, then slowly increases. “G" is inlet mass velocity and “Pi" is inlet pressure (absolute). The data analyzed
include condensers in the following services:
1. Fluid coking - scrubber overhead.
2. Cat cracking - fractionator overhead.
3. Fluid hydroforming - scrubber overhead.
4. Pipestill overhead.
5. Prefractionator overhead.
The total range of the variables is as follows:
1.47 < G < 20.9 lb/sec-ft2
0.008 < α < 0.896
19.2 < Pi < 205 psia
(Customary)
or
7.18 < G < 102 kg/s-m2
(Metric)
(Customary)
or
132 < Pi < 1410 kPa abs
(Metric)
The equation includes no data on narrow-cut condensation and will give answers much too low for such cases.
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Heat Transfer Coefficient: Condensation on the Tube Side - Condensation inside the tubes is seldom used in industry.
Hence, little is known about the heat transfer coefficients in such service. In general, the coefficients are lower than for shell-side
condensation.
If the condenser is vertical, Nusselt's analysis is undoubtedly as valid inside the tubes as on the outside. But in the usual case,
such as an air-fin condenser, the unit would be horizontal. This tends to fill part of the tube cross-section with condensate,
blanking some of the condensing area. This mechanism is completely outside of Nusselt's work.
The following empirical method has been proposed by Akers, Deans and Crosser (Reference 3). The method is based on data
for condensing propane and Freon-12 in horizontal tubes. The authors correlated their data within about 20% by using an
“equivalent" liquid mass velocity and the usual single-phase heat transfer relationships. Neither the Nusselt equation nor the
Carpenter-Colburn equation would correlate the data.
The basic equations are as follows:
htλ = 5.03 Re1/3 Pr1/3 (12 kf / di)
Re < 50,000 (Customary)
Eq. (12)
or
htλ = 5.03 Re1/3 Pr1/3 (1000 kf / di)
htλ = 0.0265 Re0.8 Pr1/3 (12 kf / di)
(Metric)
Re > 50,000 (Customary)
Eq. (12)M
Eq. (13)
or
htλ = 0.0265 Re0.8 Pr1/3 (1000 kf / di)
(Metric)
Eq. (13)M
Re =
Ge di
29 µ
(Customary)
Eq. (14)
Re =
Ge di
1000 µ
(Metric)
Eq. (14)M
or
ρ 
G e = G′l + G′v  l 
 ρv 
1/ 2
Eq. (15)
G′l and G′v are superficial average mass velocities of the liquid and vapor, respectively, based on the total flow area. The
Nusselt, Prandtl and Reynolds numbers are in dimensionless forms. Physical properties used are those of the condensate film at
average film temperature.
It was found that these equations correlated the data for vertical, inclined, and horizontal tubes.
Eqs. (12) through (15) are for the condensing coefficient only, and they would be applicable by themselves to narrow-cut
condensation. When condensing a wide cut, the designer should follow the principles used on the shell side for wide cuts and
calculate coefficients for gas and liquid cooling. For gas cooling, use the average vapor velocity in the zone to calculate a singlephase vapor heat transfer coefficient. The liquid cooling coefficient may be assumed to be equal to the condensing coefficient.
There is, of course, no drip cooling and hence no required breakdown of the liquid cooling duty.
Pressure Drop - Pressure drop in condensers is a complex variation of two-phase flow that has not been successfully analyzed
theoretically. The data are amenable to correlation, however, and such correlations are frequently reported in the literature.
The recommended procedure treats the two-phase system as a single phase, using the viscosity of the liquid and a density
corresponding to the average density for both phases. The weighted average density is evaluated for each zone based on the
inlet and outlet volume flow rates, or
Average density =
2 x Mass rate
( Vol. rate in) + ( Vol. rate out )
This calculation method is applicable for both tube-side and shell-side pressure drops using weighted average values of density
and velocity. Both tube-side and shell-side pressure drops are evaluated by the “no change of phase" procedure as outlined in
Section IX-D.
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➧
The two-phase pressure drop calculated for each zone is multiplied by the ratio (Zone Area/Total Area) to estimate the zone
pressure drop. The overall pressure drop is obtained by summing the zone pressure drops.
This procedure is believed to be conservative, i.e., actual pressure drops should be equal to or less than the calculated values.
In cases where pressure drop is expensive or where a much lower than expected pressure drop would cause process difficulties,
an independent check can be made by use of the Lockhart & Martinelli correlation (see Perry's Chemical Engineers' Handbook,
6th Ed., pages 5-40 to 5-42). This method is considered to be the best available generalized correlation for two-phase pressure
drop in a system with heat (mass) transfer taking place. However, the accuracy of this correlation is only ± 50%. Thus, the
actual pressure drop could be anywhere between 50% and 150% of the calculated value. Also, the HTRI CST-3 computer
program is incorporated into HEXTRAN program and is available for problems requiring detailed analysis.
SPECIAL DESIGN CONSIDERATIONS FOR DRUMLESS CONDENSERS
➧
When drumless condensers are designed, the following design criteria should be followed:
1. The condenser surface should be 110% of the surface required for condensing in order to account for surface normally
covered by liquid.
2. The condenser should be equipped with a 2 in. (50 mm) vent located as close to the liquid outlet end as practicable.
3. A pot for liquid-vapor separation should be located on the condenser outlet. Pots up to 14 in. (356 mm) diameter should be
sized for a liquid velocity of 1 ft (0.3 m) per second. The velocity in pots 16 in. (406 mm) and larger should be limited to 1.5 ft
(0.5 m) per second. The pot should be 3 to 5 ft (0.9 to 1.5 m) long.
4. The condenser shell should be equipped with a gauge glass covering the entire shell diameter and pot.
5. The condenser should be elevated sufficiently to satisfy pump NPSH requirements with the condenser pot empty. In any
event, the bottom of the condenser shell should be a minimum of 20 ft (6 m) above grade.
6. An anti-vortex baffle should be installed in the condenser pot.
7. The condenser should be located with its liquid outlet end on the pump side of the structure. All pump suction piping should
be continuously sloped downwards towards the pump. Horizontal runs on the pump suction piping should be sloped at least
2 in. per 100 ft (50 mm per 30 m).
8. Pump suction lines up to and including 3 in. (76 mm) diameter should not have a liquid velocity in excess of 1.5 ft/sec
(0.5 m/s). Four to 8 in. lines should be limited to 2.25 ft/sec (0.7 m/s). For 10 in. (254 mm) diameter and larger sizes, the
liquid velocity should not exceed 3.5 ft/sec (1.1 m/s).
9. Pumps should be heavy-duty refinery type.
10. Pumps should be equipped with a recirculation line returning upstream of the condenser. The recirculation line should be
equipped with a block valve and a restriction orifice sized for 25% of the normal pump capacity.
DETAILED DESIGN PROCEDURE
Table 2 lists the step-by-step procedure in the form of a calculation sheet. Each step is discussed below.
The condensation calculation procedure presented in this section is based on hydrocarbon (with or without small amounts of
steam, less than 15% by weight) condensation. The procedure should not be used for the design of vacuum pipestill overhead
condensers or steam surface condensers. A sizing procedure for vacuum pipestill overhead condensers is presented later in this
section. Steam surface condensers should be sized using the procedures presented in the Heat Exchange Institute (HEI)
Standards for Steam Surface Condensers.
1. Determine the inlet and outlet temperatures, t1, t2, T1 and T2.
2. Determine the minimum number of shells in series; Fn (Figure 2, Section IX-D) based on exchanger inlet and outlet
temperatures equal to or greater than 0.8.
3.
Determine the overall duty Q and the weighted overall LMTD, ∆tew.
a.
If a T-Q curve (temperature vs. heat release) has already been prepared, determine ∆tew as follows:
(1) From the curve, determine Q, Tdh and qdh, Tds and qds and qsc.
(2) Calculate tdh and tds.
t dh = t 2 − ( t 2 − t1)
qdh
Q
Eq. (16)
t ds = t 2 − ( t 2 − t1)
qdh + qds
Q
Eq. (17)
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THEORY AND DESIGN CONSIDERATIONS (Cont)
(3) Calculate ∆tdh, ∆tds and ∆tsc. These are the LMTDs in each of the three zones. Apply individual Fn's (Figure 2,
Section IX-D) to each zone, as calculated from zone terminal conditions.
(4) Calculate ∆tew
∆t ew =
4.
Q
qdh
qds
q
+
+ sc
∆t dh
∆t ds
∆t sc
b. If a T-Q curve is not available, the method outlined in Table 1 may be used.
Estimate the total area required for the condenser. See Table 1, Section IX-C for typical values of Uo.
A = Q / Uo ∆tew
5.
6.
7.
Eq. (19)
For the condensing side, estimate the pressure drop so that the baffle pitch (or number of tube passes) can be established.
a.
Calculate average mixture density ρ based on inlet and outlet flow volumes.
b.
Assume a value of Pb (or Np) and calculate ∆P. Use liquid viscosity at average condensing temperature with equations
for no change of phase.
Adjust Pb or Np to obtain desired approximate pressure drop. Note that this is an approximate pressure drop, used only
in setting up Pb or Np. A definitive ∆P will be calculated later.
c.
9.
Eq. (18)
Calculate area per shell, number of shells and number of tubes according to the usual method.
Use the appropriate section of the manual to calculate the transfer coefficient and pressure drop for the noncondensing side.
Calculate Uox, the overall coefficient neglecting the resistance on the condensing side.
1
= ro + rw + rio + (R o or R io )
U ox
8.
From Eq. (1)
Calculate Adh, the area required for the desuperheating zone of the condenser.
a. Calculate the gas cooling coefficient (hgc) using the usual method of no change of phase.
b. Calculate Udh and Adh.
Udh =
hgc Uox
hgc + Uox
Adh = qdh / Udh ∆tdh
Eq. (20)
Eq. (21)
10. Calculate Ads, the area required for the hydrocarbon condensation and steam desuperheating zone.
a. Estimate the condensate film temperature in the zone, assuming no temperature drop across the vapor film.
For Shell-Side Condensation


U
t f = t s − (1 / 2) ( t s − t1) 1 − zone 
U
ox 

Eq. (22)
For Tube-Side Condensation


U
t f = t t − (1 / 2) ( t t − t s ) 1 − zone 
U
ox 

All temperatures refer to the zone only, note the overall condenser.
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Eq. (23)
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b.
Calculate W v, the average vapor flow rate in the zone. Use Figure 4.
c.
Calculate Wl , average liquid flow rate.
Wl = W – W v
Eq. (24)
d.
Determine µf for the vapor; and µf, and sf for the liquid.
e.
Calculate h′ λ the condensing coefficient.
(1) If on the tube side, calculate Ge and Re, obtain f, and calculate htλ.
G′v =
Np Wv
19.6 N t di2
(Customary)
Eq. (25)
(Metric)
Eq. (25)M
(Customary)
Eq. (26)
(Metric)
Eq. (26)M
or
G′v = 1.273 x 106
G′l =
Np Wv
N t di2
Np Wv
19.6 N t di2
or
G′l =
Np Wv
N t di2
Ge = G′l + G′v ( ρl / ρv) 1/2
From Eq. (15)
Re =
Ge di
29 µ
(Customary)
From Eq. (14)
Re =
Ge di
1000 µ
(Metric)
From Eq. (14)M
or
Calculate htλ from Eq. (12) or (13).
(2) If on the shell side,
A. Vertical bundle
Γ =
12 Wc
π Nt do
(Customary)
From Eq. (6)
Γ =
1000 Wc
π Nt do
(Metric)
From Eq. (6)M
or
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THEORY AND DESIGN CONSIDERATIONS (Cont)
1/ 3
 1.04 x 10 4 
 2
 k f  sf 
hλ = 
1
/
3
 Γ

 µf 


 
(Customary)
From Eq. (5)
(Metric)
From Eq. (5)M
or
1/ 3
hλ
 s2 
 254 
=  1/ 3  k f  f 
 µf 
Γ 
 
If Γ / µ f is larger than 1,100 (or 450), the value of hλ is quite conservative.
B.
Horizontal bundle
Calculate ns, the number of condensate streams in the bundle.
ns = 1.29 Nt0.480 for square tube layout, 90°
ns = 2.08 Nt0.495 for rotated triangular layout, 60°
ns = 1.02 Nt0.519 for triangular layout, 30°
ns = 1.37 Nt0.518 for rotated square layout, 45°
Calculate Γ, the tube loading. For this step Ads must be assumed. The final calculated value of Ads should be checked
back for agreement (10 to 15%) with the assumed value.
Γ =
Wc
L c ns
Lc =
A ds
(L − 0.5) Ns
A
(Customary)
Eq. (27)
Lc =
A ds
(L − 0.152) Ns
A
(Metric)
Eq. (27)M
From Eq. (8)
or
Calculate h′λ , the condensing coefficient uncorrected for vapor velocity.
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 
 Γ 1/ 3





1/ 3
(Customary)
From Eq. (7)
(Metric)
From Eq. (7)M
or
 s2 
 204 
 k f  f 
h′λ = 
 µf 
 Γ 1/ 3 


1/ 3
Calculate xv, the fraction of shell free–flow area occupied by the vapor.
xv = 1 − xl
W  µ 
1
= 1 +  v   v 
xl
 Wl   µl 
Eq. (28)
0.111
 ρl 
 
ρ 
 v
0.555
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Eq. (29)
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Calculate average mass velocity of the vapor in the zone
Gv =
Wv
25 A x x v
(Customary)
Eq. (30)
Gv =
106 Wv
A x xv
(Metric)
Eq. (30)M
or
Calculate hλ
G 
h λ = h′λ  v 
 5 
0.70
(Customary)
From Eq. (9)
(Metric)
From Eq. (9)M
or
 G 
h λ = h′λ  v 
 24.4 
0.70
h λ max = 2.0 h′λ
f.
From Eq. (10)
Calculate hgc, the vapor cooling coefficient.
(1) If on the shell side:
Use Gv, the average vapor mass velocity, to calculate the uncorrected vapor cooling coefficient hgc. Use
correlations for no change of phase. Use the same film temperature as calculated for the condensing coefficient.
Correct for the fact that heat flowing through the gas film must also overcome resistance of the condensate film.
C
1
1
=
+ 1
h gc
hλ
h′gc
Eq. (31)
where:

 T1 + T2


− tf  

2
 Wc 

C1 = 1 − 

Wv1
T1 − T2




Eq. (32)
The factor “C1" helps account for the additional gas cooling surface available from liquid droplets, ripples, etc. The
temperatures T1 and T2 refer to the zone inlet and outlet temperatures.
The value of C1 should fall in the range of 0.5 to 1.0. If the calculated value of C1 is less than 0.5, set C1 equal to
0.5; if the calculated value of C1 is greater than 1.0, set C1 equal to 1.0.
(2) If on the tube side:
Calculate Gv.
A. Calculate xv [see Eq. (28)]
B.
Gv = G′v / xv
Eq. (33)
Use Gv with the procedure for no change of phase and calculate h′ gc .
Calculate hgc.
C1
1
1
=
+
h gc
hλ
h ′gc
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Eq. (34)
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THEORY AND DESIGN CONSIDERATIONS (Cont)
g.
Calculate hlc , the liquid cooling coefficient. For vertical bundles, assume hlc = hλ.
For horizontal bundles:
(1) If on the shell side
Calculate Gl
Gl =
Wl
25 A x x l
(Customary)
Eq. (35)
Gl =
106 Wl
A x xl
(Metric)
Eq. (35)M
or
Using Gl , with the procedure for no change of phase, calculate h'lc , liquid cooling obtained by “bottom flow."
Calculate the liquid cooling obtained by “drip cooling."
hdc = 1.5 hλ
Eq. (36)
Calculate hlc
For all vapor entering;
h lc =
2 h dc h′lc
h dc + h′lc
Eq. (37)
For liquid + vapor to the zone;
hlc =
( Wl + Wc ) hdc h′lc
Wc
Wc 

h′lc +  Wl +
 hdc
2
2 

Eq. (38)
(2) If on the tube side
hlc
h.
= hλ
Calculate hds, the weighted zone coefficient on the condensing side.
hds =
i.
qλ
q
+
+ lc
hλ
hgc
hl c
Eq. (40)
hds Uox
hds + Uox
Eq. (41)
qds
Uds ∆t e( ds )
Eq. (42)
Calculate Ads
A ds =
k.
qds
qgc
Calculate Uds, the zone coefficient.
Uds =
j.
Eq. (39)
Compare Ads calculated here with Ads assumed to calculate Γ. If necessary, repeat calculation to obtain agreement.
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11. Calculate Ads, the area required for the steam condensing zone. Calculations for this zone are performed in the same way
as for the previous zone. Note the following:
a. The condensing coefficient for steam equals the hydrocarbon condensing coefficient.
b. The cooling coefficient for condensed steam equals the hydrocarbon liquid cooling coefficient..
12. Calculate the total condenser area.
A = Adh + Ads + Asc
Eq. (43)
13. If the area calculated in the previous step equals the area estimated in Step (4), proceed to the next step.
If the calculated area is not equal to the estimated area, repeat the calculation with a better estimated area until agreement is
reached.
14. Calculate the overall coefficient, Uo.
Uo =
Q
A ∆t ew
Eq. (44)
15. Calculate the clean coefficient, Uc
1
1
=
− rio − ro + 0.001
Uc
Uo
(Customary)
Eq. (45)
1
1
=
− rio − ro + 0.00018
Uc
Uo
(Metric)
Eq. (45)M
or
16. Calculate pressure drop on the condensing side.
a. Refer to Step 8 and calculate two-phase pressure drops for the overall condenser at the average conditions of each
zone.
b. Calculate zone pressure drops
c.
A

∆Pzone = ∆Pat zone conditions  zone 
 A total 
Eq. (46)
∆P = ∆Pdh + ∆Pds + ∆Psc
Eq. (47)
Calculate ∆P
VACUUM PIPESTILL OVERHEAD CONDENSER DESIGN PROCEDURE
➧
Since a vacuum pipestill overhead condenser heat duty primarily results from condensing steam, the detailed condenser
calculation procedure presented in this section is not applicable for design of these units. However, the HEXTRAN Program with
HTRI CST-3 program may be used. For preliminary hand calculations, when sizing a vacuum pipestill overhead condenser, the
following calculation procedure is recommended.
1.
2.
3.
➧
4.
Use an overall heat transfer coefficient of 130 Btu/hr-ft2-°F or 738 W/m2-K. (Values between 110 to 130 Btu/hr-ft2-°F
(625 and 738 W/m2-K) have been used for past designs.)
Use the steam dew point temperature, rather than the hydrocarbon dew point temperature, as the condensing zone inlet
temperature for calculating the effective temperature difference.
Design for 3 to 12 mm Hg (0.4 to 1.6 kPa) pressure drop. The pressure drop should be estimated based on 1/2 the
calculated pressure drop using inlet vapor conditions. (Past designs frequently employed divided flow, TEMA Type J, shells
with double segmental, modified disk and donut, baffles at or near maximum baffle spacing to obtain low pressure drop
values.)
Estimate nozzle pressure drop based on three velocity heads for inlet and outlet nozzles.
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NOMENCLATURE
A
Adh
Ads
Asc
As
Ax
B1
B2
=
=
=
=
=
=
=
=
Total exchanger area, ft2 (m2)
c
=
Specific heat at caloric temperature, Btu/lb-°F (kJ/kg-K)
cf
D
Dt
di
do
f
Fbt
Fn
FF
FJ
Fs
Ft
Fv
G
Ge
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
Specific heat of the shell-side fluid at average film temperature, Btu/lb-°F (kJ/kg-K)
Shell I.D., in. (mm)
Diameter of tube bundle (“outer tube limit"), in. (mm)
Tube I.D., in. (mm)
Tube O.D., in. (mm)
Fanning friction factor, dimensionless
Correction factor to account for baffle type, dimensionless
Correction factor for log mean temperature difference (due to partially concurrent flow), dimensionless
F shell correction factor, dimensionless
J shell correction factor, dimensionless
Shell-side pressure drop correction factor, dimensionless
Tube-side pressure drop correction factor, dimensionless
Correlation factor used to determine average vapor flow, dimensionless
Mass velocity, lb/sec-ft2 (kg/s-m2)
Equivalent mass velocity of liquid and vapor in 2-phase flow (see Figure 4), lb/sec-ft2 (kg/s-m2)
Gl
=
Average mass velocity of liquid, lb/sec-ft2 (kg/s-m2)
G′l
=
Superficial mass velocity of liquid, lb/sec-ft2 (kg/s-m2)
Gv
=
Average mass velocity of vapor, lb/sec-ft2 (kg/s-m2)
G′v
=
Superficial mass velocity of vapor, lb/sec-ft2 (kg/s-m2)
hdc
=
Liquid cooling coefficient resulting from “dripping" in the condenser, Btu/hr-ft2-°F (W/m2-K)
hdh
=
Film coefficient for condensing side of desuperheating zone, Btu/hr-ft2-°F (W/m2-K)
hds
=
Film coefficient for condensing side of hydrocarbon condensing zone, Btu/hr-ft2-°F (W/m2-K)
hsc
=
Film coefficient for condensing side of steam condensing zone, Btu/hr-ft2-°F (W/m2-K)
h′gc
=
Gas cooling coefficient uncorrected for condensate film resistance, Btu/hr-ft2-°F (W/m2-K)
hgc
=
Gas cooling coefficient, Btu/hr-ft2-°F (W/m2-K)
hio
=
Inside film coefficient corrected to outside area, Btu/hr-ft2-°F (W/m2-K)
h′l c
=
Liquid cooling coefficient resulting from “bottom flow" in the condenser, Btu/hr-ft2-°F (W/m2-K)
hl c
=
Weighted liquid cooling coefficient, Btu/hr-ft2-°F (W/m2-K), assuming half of liquid cooling is “drip" cooling and half is
ho
=
Shell-side film coefficient no change of phase, Btu/hr-ft2-°F (W/m2-K)
hoc
=
Shell-side condensing coefficient including fouling factor, Btu/hr-ft2-°F (W/m2-K)
h′λ
=
Shell-side condensing film coefficient uncorrected for effect of vapor mass velocity, Btu/hr-ft2-°F (W/m2-K)
hλ
=
Shell-side condensing film coefficient, Btu/hr-ft2-°F (W/m2-K)
htλ
=
Tube-side condensing film coefficient, Btu/hr-ft2-°F (W/m2-K)
Area of desuperheating zone, ft2 (m2)
Area of hydrocarbon condensation and steam desuperheating zone, ft2 (m2)
Area of steam condensation zone, ft2 (m2)
Area/shell, ft2 (m2)
Free flow area between shell baffles, in.2 (mm2)
Bundle factor for shell-side heat transfer, dimensionless
Bundle factor for shell-side pressure drop, dimensionless
“bottom flow" cooling
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NOMENCLATURE (Cont)
k
=
Thermal conductivity at caloric temperature, Btu/hr-ft-°F (W/m-K)
kf
=
Thermal conductivity of fluid at average film temperature, Btu/hr-ft-°F (W/m-K)
kw
=
Thermal conductivity of the tube metal at average tube temperature, Btu/hr-ft-°F (W/m-K)
l
=
Tube thickness, ft (m)
L
Lc
=
=
Tube length, ft (m)
Length of tube exposed to condensation, ft (m)
LMTD
MW
Nb
Np
Ns
Nt
Ntc
Nu
ns
Pb
Pi
=
=
=
=
=
=
=
=
=
=
=
Log mean temperature difference, °F (K)
Molecular weight, lb/mol (kg/kmol)
Number of shell baffles
Number of tube passes per shell
Number of shells in series
Number of tubes in the bundle
Number of tubes across the center line of the bundle
Nusselt number, hd/k, dimensionless
Number of condensate streams in the shell
Baffle pitch, in. (mm)
Inlet pressure, psia
Pr
Pt
=
=
Prandtl number, cµ/k, dimensionless
Tube pitch, in. (mm)
∆Pdh
=
Pressure drop in desuperheating zone, psi (kPa)
∆Pds
=
Pressure drop in hydrocarbon condensing zone, psi (kPa)
∆Psc
=
Pressure drop in steam condensing zone, psi (kPa)
∆Pt
=
Total tube-side pressure drop, psi (kPa)
∆Ptf
=
Tube pressure drop due to friction, psi/tube pass (kPa/tube pass)
∆Ptr
=
Tube pressure drop due to turns, psi/tube pass (kPa/tube pass)
∆Psf
=
Shell-side pressure drop due to friction, psi/shell (kPa/shell)
∆Psr
=
Shell-side pressure drop due to turns, psi/shell (kPa/shell)
∆Ps
=
Total shell-side pressure drop, psi (kPa)
Q
qdh
qds
qsc
qgc
=
=
=
=
=
Total heat duty, Btu/hr (W)
Heat duty in desuperheating zone, Btu/hr (W)
Heat duty in hydrocarbon condensing zone, Btu/hr (W)
Heat duty in steam condensing zone, Btu/hr (W)
Heat duty from gas cooling in a zone, Btu/hr (W)
qlc
=
Heat duty from liquid cooling in a zone, Btu/hr (W)
qλ
=
Heat duty from condensation in a zone, Btu/hr (W)
Rc
=
Total resistance (clean) to heat transfer, hr-ft2-°F/Btu (m2-K/W)
Rio
=
Inside film resistance to heat transfer corrected to outside area, hr-ft2-°F/Btu (m2-K/W)
Ro
=
Outside film resistance to heat transfer, hr-ft2-°F/Btu (m2-K/W)
Rt
=
Total resistance (duty) to heat transfer, hr-ft2-°F/Btu (m2-K/W)
ri
=
Tube-side fouling factor referred to tube inside surface area, hr-ft2-°F/Btu (m2-K/W)
rio
=
Tube-side fouling factor referred to tube outside surface area, hr-ft2-°F/Btu (m2-K/W)
ro
=
Outside fouling factor, hr-ft2-°F/Btu (m2-K/W)
rw
Re
S
=
=
=
Resistance of tube wall metal at average wall temperature, hr-ft2-°F/Btu (m2-K/W)
Reynolds number
Free width between baffles, in. (m)
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NOMENCLATURE (Cont)
sf
=
Specific gravity of the condensate film
Tdh
=
Hydrocarbon dewpoint, °F (°C)
Tds
tdh
=
=
tds
=
Steam dewpoint, °F (°C)
Temperature of the fluid being heated, after it gains enough heat to have cooled the condensing stream from Tdh to
T2, °F (°C)
Temperature of the fluid being heated, after it gains enough heat to have cooled the condensing stream from Tds to
T2,°F (°C)
Tm
=
Tube sheet design temperature, ° F (°C)
T1
=
Inlet temperature of fluid being cooled, °F (°C)
T2
=
Outlet temperature of fluid being cooled, °F (°C)
t1
=
Inlet temperature of fluid being heated, °F (°C)
t2
=
Outlet temperature of fluid being heated, °F (°C)
tf
=
Average shell-side film temperature, °F (°C)
ts
=
Caloric temperature of the shell fluid, °F (°C)
tt
=
Caloric temperature of the tube fluid, °F (°C)
tw
=
Average tube wall temperature, °F (°C)
∆tdh
=
Log mean temperature difference in desuperheating zone, °F (°C)
∆tds
=
Log mean temperature difference in hydrocarbon condensing zone, °F (°C)
∆tsc
=
Log mean temperature difference in steam condensing zone, °F (°C)
∆te
=
Log mean temperature difference corrected for nonideal countercurrent flow (Effective temperature difference),
°F (°C)
∆tew
=
Weighted effective log mean difference, °F (°C)
∆tm
=
Log mean temperature difference for true countercurrent flow, °F (°C)
Uc
=
Overall clean coefficient of heat transfer, Btu/hr-ft2-°F (W/m2-K)
Udh
=
Overall heat transfer coefficient in desuperheating zone, Btu/hr-ft2-°F (W/m2-K)
Uds
=
Overall heat transfer coefficient in hydrocarbon condensing zone, Btu/hr-ft2-°F (W/m2-K)
Usc
=
Overall heat transfer coefficient in steam condensing zone, Btu/hr-ft2-°F (W/m2-K)
Uo
=
Overall duty coefficient of heat transfer, Btu/hr-ft2-°F (W/m2-K)
Uox
=
Overall condenser heat transfer coefficient, neglecting the resistance of condensate film, Btu/hr-ft2-°F (W/m2-K)
u
V
=
=
Compressibility factor of the vapor
Velocity, ft/sec (m/s)
Vl1
=
Volume liquid entering a zone, ft3/hr (m3/s)
Vl 2
=
Outlet liquid volume, ft3/hr (m3/s)
Vv
Vv1
Vv2
W
Wc
=
=
=
=
=
Volume vapor entering a zone, ft3/hr (m3/s)
Inlet vapor volume, ft3/hr (m3/s)
Outlet vapor volume, ft3/hr (m3/s)
Mass flow rate, lb/hr (kg/s)
Quantity condensed in a zone (or overall), lb/hr (kg/s)
Wl1
=
Inlet liquid rate, lb/hr (kg/s)
Wl 2
=
Outlet liquid rate, lb/hr (kg/s)
Wl
=
Average liquid flow rate, lb/hr (kg/s)
Wv
Wv1
Wv2
=
=
=
Average vapor flow rate, lb/hr (kg/s)
Inlet vapor flow rate, lb/hr (kg/s)
Outlet vapor flow rate, lb/hr (kg/s)
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NOMENCLATURE (Cont)
xv
=
Fraction of cross-sectional flow area occupied by vapor
xl
=
Fraction of cross-sectional flow area occupied by liquid
Yth
=
Tube-side heat transfer correlation factor
µ
=
Viscosity at caloric temperature, cP (Pa•s)
µf
=
Viscosity of fluid at average film temperature, cP (Pa•s)
µl
=
Average liquid viscosity, cP (Pa•s)
µv
=
Average vapor viscosity, cP (Pa•s)
µw
=
Viscosity of the tube-side fluid at tube wall temperature, cP (Pa•s)
α
=
Ratio vapor rate to inlet vapor rate, shell-side condensation
β
=
In a condenser: Weight fraction condensed
Γ
=
Tube loading, lb/hr-ft (kg/s-m)
∆3
=
Correlation factors used in calculating equilibrium flash curve
λ
=
Average latent heat of vaporization, Btu/lb (kJ/kg)
ρ
=
Fluid density (No change of phase), lb/ft3(kg/m3)
ρl
=
Average liquid density, lb/ft3 (kg/m3)
ρl1
=
Inlet density of liquid, lb/ft3 (kg/m3)
ρl 2
=
Outlet density of liquid, lb/ft3 (kg/m3)
ρv
=
Average vapor density, lb/ft3 (kg/m3)
ρv1
=
Inlet density of vapor, lb/ft3 (kg/m3)
ρv2
=
Outlet density of vapor, lb/ft3 (kg/m3)
∆1,
∆2,
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TABLE 1
T-Q CURVE CALCULATION PROCEDURE
I. CALCULATION PROCEDURE
1.
From the predicted ASTM distillation of the condensing hydrocarbon, determine t10, t30, t50, t70, and t90. Then calculate ∆1
from:
∆1 =
2
( t70 − t10 )
3
2.
From Figure 6, determine ∆2 and ∆3.
3.
Calculate the 10, 30, 50, 70, and 90 Vol. % points on the atmospheric flash curve.
4.
5.
a.
t10f = t10 + (∆1 – ∆3) – ∆2
b.
t30 f = t10 +
c.
t50f = t10 + 0.48 (t50 – t10 – ∆1) + (∆1 – ∆3)
d.
t70 f = t10 f +
e.
t90f = t10 + 0.23 (t90 – t10 – 2∆1) + (∆1 – ∆3) + ∆2
∆2 
2
 ∆1
+
( t 30 − t10 ) + ( ∆1 − ∆3 ) − 

3
3
2 

3
∆2
2
Estimate the temperatures at 10, 30, 50, 70, and 90 wt% flashed and at the same mol% flashed.
t10w = t10f + 0.05 (t50f – t10f)
t30w = t30f + 0.10 (t50f – t10f)
t50w = t50f + 0.10 (t70f – t30f)
t90w = t90f + 0.05 (t90f – t50f)
t70w = t70f + 0.10 (t90f – t50f)
t30m = t30f – 0.04 (t50f – t10f)
t10m = t10f – 0.02 (t50f – t10f)
t50m = t50f – 0.04 (t70f – t30f)
t90m = t90f – 0.02 (t90f – t50f)
t70m = t70f – 0.04 (t90f – t50f)
Prepare a graph (see Figure 2) with percent off on the X-axis and temperature on the Y-axis.
a.
6.
Plot the five points calculated for the volume flash curve. Connect the points as follows:
(1) A straight line from t30f through t10f, intersecting the 0% axis.
(2) A straight line from t70f through t90f, intersecting the 100% axis.
(3) A straight line between t50f and t30f.
(4) A straight line between t50f and t70f.
b. Plot the five weight percent flashed points and connect each adjacent point with a straight line. Connect t10w with the 0
vol. % intercept, and t90w with the 100 vol. % intercept.
Repeat this process with the five mol percent flashed points.
Estimate Tdh, the dew point of the hydrocarbon. (For the overhead of the fractionation tower, T1 = Tdh.)
a. On the flash curve prepared above, the atmospheric-pressure dew point is the point at 100% NOT condensed.
b. Calculate the partial pressure of the condensable hydrocarbon and adjust the atmospheric–pressure dew point to this
partial pressure, using the alignment chart, Figure 3.
HC Partial Pr ess. =
7.
mol of Condensable HC
x P1
Total mol of vapor
Estimate Tds, the dew point of steam in the system. This is the temperature at which the vapor pressure of water is equal to
the partial pressure of the water in the system.
a. This calculation is done in tabular form, as shown below.
b. The calculations for the table should be carried out at increasing values of weight percent condensed until the first point
is found where the vapor pressure of water @ system temperature is less than the calculated partial pressure of water in
the system.
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T-Q CURVE CALCULATION PROCEDURE
c.
Prepare a graph with temperature on Y-axis and pressure of the X-axis. From the table just completed, plot the last 3 or
4 steam partial pressure vs. system temperature points. On the same graph, plot the last 3 or 4 water-vapor-pressure
vs. system temperature points. The intersection of the two curves gives the approximate steam dew point of the system
(see Figure 2).
Wt. % Hydrocarbon Condensed(1)
Vapor Temp. @ Atm. Press.(2)
Mol % HC not Condensed(2)
Mol of HC not Condensed
Mol of Non-Condensables (“NC")
Mol of Steam
Total Mol of Vapor
Total Pressure, psia or kPa(3)
Partial Pressure of Steam, psia or kPa
Partial Press. of Condensables, psia or kPa
System Temperature, °F or °C(4)
Vapor Press. of H2O at Sys. Temp.(5), psia or kPa
Notes:
8.
(1)
Arbitrary values.
(2)
Read from flash curves.
(3)
Assume percent pressure drop = mol percent condensed.
(4)
Vapor temperature at 1 atm. adjusted to the partial pressure of condensables by means of the alignment chart, Figure 3.
(5)
From steam tables or from alignment chart Figure 3. (Use the point marked H2O as the “normal boiling point".)
Determine the amount and type of heat duty in each zone. Evaluate specific heat (c) and latent heat of vaporization (λ) at
the arithmetic average zone temperature.
a. Sensible heat transferred from material which passes through the zone without changing phase.
q = W c (Temp Change)
b.
c.
d.
Sensible heat transferred from vapor or steam which condenses in the zone.
Sensible heat transferred from hydrocarbon or water which has condensed in the zone.
q =
Wc
c (Temp Change )
2
(c of vapor)
q =
Wc
c (Temp Change)
2
(c of liquid)
Latent heat transferred from condensing hydrocarbon or steam.
Q = λ Wc
Note that Item (8a) may include as many as 4 calculations: steam; non-condensables; vapor which does not condense;
and any entering liquid.
Item (8b) may include two calculations: vapor and steam.
Items (8c) and (8d) may each include two calculations: hydrocarbon and water.
The above calculations, performed for each zone, will give qdh, qds, and qsc. These values permit the calculation of the
LMTD's and overall temperature difference (∆tm) – also referred to as the Mean Temperature Difference (MTD).
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TABLE 2
CALCULATION PROCEDURE FOR CONDENSER DESIGN
I.
Exchanger Number E -
Refinery
Service
Calc By/Date
ZONE DUTIES AND TEMPERATURES*
A.
Vapor Desuperheating Zone (If Present); 1st Zone
CUSTOMARY UNITS
METRIC UNITS
qdh
Fluid Being Condensed (Shell/Tube)
=
Btu/hr
W
T1
=
°F
°C
Tdh
Fluid Being Heated (Shell/Tube)
=
°F
°C
t2
=
°F
°C
tdh
=
°F
°C
T −T
R = 1 dh =
t 2 − t dh
=
t 2 − t dh
=
T1 − t dh
=
=
°F
°C
=
°F
°C
ts = 0.4 (T1 – Tdh) + Tdh
=
°F
°C
tt = 0.4 (t2 – tdh) + tdh
=
°F
°C
tt = 0.4 (T1 – Tdh) + Tdh
=
°F
°C
ts = 0.4 (t2 – tdh) + tdh
=
°F
°C
=
Btu/hr-°F
W/K
j=
∆t m =
(T1 − t 2 ) − (Tdh − t dh )
=
 T − t2 

ln  1

 T dh − t dh 
From charts given in Section IX-D
Fn =
for this zone**
∆te(dh) = ∆tmFn
For shell-side condensation:
For tube-side condensation:
qdh
∆t e
*
**
Use LMTD correction factor for each zone, based on zone temperatures.
Use LMTD corrections factor, based on terminal temperatures to evaluate minimum number of shells in series. Fn must
be equal to or greater than 0.8.
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CALCULATION PROCEDURE FOR CONDENSER DESIGN
B.
Hydrocarbon Condensing Steam Desuperheating Zone (2nd Zone)
CUSTOMARY UNITS
METRIC UNITS
qds
Fluid Being Condensed (Shell/Tube)
=
Btu/hr
W
Tdh
=
°F
°C
Tds
=
°F
°C
tdh
=
°F
°C
tds
=
°F
°C
=
°F
°C
∆te(ds) = ∆tm Fn
For shell-side condensation:
=
°F
°C
ts = 0.4 (Tdh – Tds) + Tds
=
°F
°C
tt = 0.4 (tdh – tds) + tds
For tube-side condensation:
=
°F
°C
tt = 0.4 (Tdh – Tds) + Tds
=
°F
°C
ts = 0.4 (tdh – tds) + tds
=
°F
°C
=
Btu/hr-°F
W/K
qsc
Fluid Being Condensed (Shell/Tube)
=
Btu/hr
W
Tds
=
°F
°C
T2
=
°F
°C
tds
=
°F
°C
t1
=
°F
°C
T − T2
=
R = ds
t ds − t1
=
t ds − t1
=
Tds − t1
=
R=
Tdh − Tds
=
t dh − t ds
=
j=
t dh − t ds
=
Tdh − t ds
=
∆tm =
(Tdh − t dh ) − ( Tds − t ds )
T −t 
ln  dh dh 
 Tds − t ds 
From charts given in Section IX-D,
Fn =
for
shells
qds
∆Te
C.
Steam Condensing Zone (3rd Zone)
j=
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
(Tds − t ds ) − (T2 − t1 )
METRIC UNITS
=
°F
°C
∆te(sc) = ∆tmFn
For shell-side condensation:
=
°F
°C
ts = 0.4 (Tds – T2) + T2
=
°F
°C
tt = 0.4 (tds – t1) + t1
For tube-side condensation:
=
°F
°C
tt = 0.4 (Tds – T2) + T2
=
°F
°C
ts = 0.4 (tds – t1) + t1
=
°F
°C
=
Btu/hr-°F
WK
=
°F
°C
=
Btu/hr-ft2-°F
W/m2-K
m2
∆t m =
T −t 
ln  ds ds 
 T 2 − t1 
From charts given in Section IX-D,
Fn =
for
shells
qsc
∆t e
D.
Weighted Temperature Difference
∆t ew =
Q
qdh
qds
qsc
+
+
∆t e( dh )
∆t e( ds )
∆t e( sc )
where: Q = qdh + qds + qsc
II.
*
=
SHELL-SIDE CONDENSATION
A. Overall Tube-Side Calculation (No Change of Phase)
1.
Assumed value of Uo
2.
3.
A = Q / Uo ∆tew
Tube metal: ___________________
=
ft2
Thermal conductivity, kw
Tube outer diameter, do
=
=
Btu/hr-ft-°F
in.
W/m-K
mm
Wall thickness, l
=
ft
m
Tube insider diameter, di
Tube length, L
=
=
in.
ft
mm
m
4.
Tube pitch (Pt) and layout (Note 1)*
=
in.
mm
5.
As = A / Ns (Note 2)*
=
ft2
m2
6.
Nt = 3.82 As / (L – 0.5) do
Nt = 318 As / (L – 0.152) do (Metric)
=
General notes contained on last page of calculation form.
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CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
7.
Assumed Np
=
8.
ts overall = 0.4 (T1 – T2) + T2
=
°F
°C
9. tt overall = 0.5 (t1 + t2)
10. Tube-side properties at tt
=
°F
°C
a.
Density, ρ
=
lb/ft3
kg/m3
b.
Viscosity, µ
=
cP
Pa•s
c.
Specific Heat, c
=
Btu/lb-°F
kJ/kg-K
d.
Thermal Conductivity, k
=
Btu/hr-ft-°F
W/m-K
11. Tube-side flow rate, W = Q / c (t2 – t1)
=
lb/hr
kg/s
=
ft/sec
12. V =
Np W
(19.6) ρ N t di2
(1.273 x 10 6 ) Np W
V =
(Metric)
ρ Nt di2
m/s
13. If tube-side fluid is water
hio =
hio =
 t 
368
( Vdi ) 0.7  t 
do
 100 
0.26
1.27 x 104
 1.8 t t + 32 
( Vdi )0.7 

do
 100

=
Btu/hr-ft-°F
0.26

( V )1.73 
∆P = 0.020 Ft N s Np  V 2 + 0.158 L

(di )1.27 


( V )1.73 
∆P = 1.48 Ft Ns Np  V 2 + 22.9 L

(di )1.27 

W/m2-K
(Metric)
=
psi
(Metric)
kPa
14. For fluids other than water, evaluate overall tube-side heat transfer coefficient and pressure drop by the procedure
outlined in Section IX-D, tube-side iteration.
Btu/hr-ft-°F
W/m2-K
=
psi
kPa
=
hr-ft2-°F/Btu
m2-K/W
16. Tube-side fouling factor, ri
=
hr-ft2-°F/Btu
m2-K/W
17. rio = (do / di) ri
=
hr-ft2-°F/Btu
m2-K/W
=
hr-ft2-°F/Btu
m2-K/W
hio
=
∆Pt
15. Rio = 1 / hio
18. Shell-side fouling factor, ro
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
=
hr-ft2-°F/Btu
m2-K/W
20. 1 / Uox = Rio + rio + rw + ro
=
hr-ft2-°F/Btu
m2-K/W
21. Uox
=
Btu/hr-ft2-°F
W/m2-K
=
°F
°C
19. rw = l / kw
B.
METRIC UNITS
Overall Shell-Side Calculation

1
U 
( t s − t1) 1 − o 
2
U
ox 

1.
t f = ts −
2.
For fluids other than hydrocarbons, determine vapor film properties. For hydrocarbons, determine only liquid
viscosity µf
Viscosity, µf
=
cP
Pa•s
Specific Heat, cf
=
Btu/lb-°F
kJ/kg-K
Thermal Conductivity, kf
=
Btu/hr-ft-°F
W/m-K
=
=
in.
in.
mm
mm
=
in.
mm
Assumed Pb
=
NB = FF (10 L / Pb) (Note 4)
=
NB = FF (833 L / Pb) (Note 4) (Metric)
9. Ax = S (Pb – 0.375) (Segmental Baffles)
=
Ax = S (Pb – 9.53) (Metric)
Ax = 0.85S (Pb – 0.375) (Double Segmental Baffles) =
Ax = 0.85S (Pb – 9.53) (Metric)
10. Total shell-side flow rate, W
=
in.
mm
3.
NTC = 1.19
= 1.10
Nt square layout
Nt triangular layout
4.
5.
Dt = (NTC – 1) Pt + do
D = Dt / 0.9 (Note 3)
6.
S=
7.
8.
D − (do NTC )
(Note 4)
FF
=
in.2
mm2
in.2
lb/hr
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CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
11. Inlet vapor density and volume
ρ v1 =
Inlet MW  Pi (psia)   1 

  
10.7
 T1 + 460   u 
=
lb/ft3
or
ρ v1 =
Inlet MW  Pi (kPa abs)   1 

  
8.29
 T1 + 273   u 
(Metric)
Inlet vapor rate, W v1
Vv1 = W v1 / ρv1
12. Outlet vapor density and volume
ρv 2 =
Outlet MW  Pi (psia)   1 

  
10.7
 T2 + 460   u 
kg/m3
=
=
lb/hr
kg/s
=
ft3/hr
m3/s
=
lb/ft3
or
ρv 2 =
Outlet MW  Pi (kPa abs )   1 

  
8.29
 T2 + 273   u 
Outlet vapor rate, W v2
(Metric)
kg/m3
=
=
lb/hr
kg/s
=
ft3/hr
m3/s
ρl1
=
lb/ft3
kg/m3
Inlet liquid rate, Wl1
=
lb/hr
kg/s
Vl1 = Wl1 / ρl1
=
ft3/hr
m3/s
ρl 2
=
lb/ft3
kg/m3
Inlet liquid rate, Wl 2
=
lb/hr
kg/s
Vl 2 = Wl 2 / ρl 2
=
ft3/hr
m3/s
=
lb/ft3
kg/m3
psi
kPa
Vv2 = W v2 / ρv2
13. Inlet liquid density and volume
14. Inlet liquid density and volume
15. Average mixture density
ρ =
2W
Vv1 + Vl1 + Vv 2 + Vl 2
16. Calculate shell-side pressure drop (∆Ps) using
=
procedures outlined in Section IX-D, Shell-Side Iteration
The above-assumed geometry will now be checked using the following detailed procedures.
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
C.
METRIC UNITS
Vapor Desuperheating Zone (Shell-Side Condensation)
1.
2.
t +t  U
t f =  s t  + o (Rio + rio + rw + ro ) ( t s − t t )
2
 2 
=
°F
=
lb/ft3
°C
(ts and tt for this zone were calculated in Step I.A.)
Inlet vapor density and volume
ρ v1 =
Inlet MW  Pi (psia)   1 

  
10.7
 T1 + 460   u 
or
ρ v1 =
3.
Inlet MW  Pi (kPa abs)   1 

 
8.31  T1 + 273   u 
(Metric)
kg/m3
=
For fluids other than hydrocarbons, determine film properties. For hydrocarbons, determine only µf
Viscosity, µf
=
cP
Pa•s
Specific Heat, cf
=
Btu/lb-°F
kJ/kg-K
Thermal Conductivity, kf
=
Btu/hr-ft-°F
W/m-K
4.
Calculate shell-side coefficient (hgc) using
Section IX-D, Shell-Side iteration
=
Btu/hr-ft2-°F
W/m2-K
5.
Udh =
=
Btu/hr-ft2-°F
W/m2-K
=
ft2
m2
hgc Uox
hgc + Uox
(Uox was evaluated in Step II.A.21.)
6.
A dh =
qdh
∆t e Udh
(qdh and ∆te were evaluated in Step I.A.)
D.
Hydrocarbon Condensing - Steam Desuperheating Zone (Shell-Side Condensation)
1.
Assume a zone area, Ads
=
ft2
m2
2.
Uds assumed =
qds
A ds ∆t e
=
Btu/hr-ft2-°F
W/m2-K
(qds and ∆te for this zone were evaluated in Step I.B.)
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3.
tf = ts −

U 
1
( t s − t t ) 1 − ds 
2
Uox 

METRIC UNITS
°F
=
°C
(ts and tt for this zone were evaluated in Step I.B.)
4.
∆Tzone inlet
=
∆Tzone outlet
Tdh − t dh
Tds − t ds
=
5.
Fv (Figure 4)
=
6.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone, W c)
+ (Vapor not condensed in zone, W v2)
= Wv
=
=
=
Average Mol. weight of vapor
=
lb / hr
mol / hr
=
lb/ft3
7.
lb/hr
mol/hr
kg/s
kmol/s
kg / s
kmol / hr
Average vapor properties at film temperature
a.
Density, ρ v =
Avg MW
10.7
 P (psia)   1 

  
 t f + 460   u 
or
ρv =
b.
Avg MW  P (kPa abs)   1 

 
8.31  t f + 273   u 
(Metric)
Viscosity, µf
(Used for hgc calc.)
kg/m3
=
=
cP
Pa•s
lb/hr
kg/s
8.
Wl = W − Wv
=
9.
Average condensate properties at film temp.
a. Specific gravity, sf
=
b.
=
Btu/hr-ft-°F
W/m-K
=
cP
Pa•s
Thermal Conductivity, kf
c. Viscosity, µf
10. Condensing Coefficient (Horizontal Bundle)
a.
ns = 1.29 Nt 0.48 (90° Square Tube Layout)
=
ns = 1.02 Nt 0.519 (30° Triangular Layout)
=
(For other layouts, see text)
b.
Lc =
A ds
(L − 0.5 ) Ns
A
=
ft
or
A ds
=
(L − 0.152) Ns (Metric)
A
(Note that if a zone occupies more than one shell,
Lc will be a multiple of the tube length.)
Lc =
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Wc
L c ns
c.
Γ=
d.
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 
 Γ 1/ 3




 condensate
METRIC UNITS
=
lb/hr-ft
=
Btu/hr-ft2-°F
kg/s-m
1/ 3
or
1/ 3
 s2 
 204 
 k f  f 
h ′λ = 
(Metric)
µ 
 Γ 1/ 3 
 f  condensate
0.111
0.555
e.
Wv  µ f (v ) 
1
=1+


xl
W l  µ f (l) 
f.
xv = 1− x l
g.
Gv =
 FJ 
Wv

 , (Note 4), or
25 A x x v  FBT 
=
Gv =
10 6 Wv  FJ 


A x x v  FBT 
=
h.
 ρl 


ρ 
 v
W/m2-K
=
=
=
G 
h λ = h′λ  v 
 5 
(Metric)
lb/sec-ft2
kg/sm2
0.70
=
Btu/hr-ft2-°F
or
 G 
h λ = h′λ  v 
 24.4 
0.70
(Metric)
W/m2-K
=
If hλ > 2 h′λ , use 2.0 h′λ
(For pure components, h λ = h′λ )
11. Condensing Coefficient (Vertical Bundles)
a.
Γ=
12 Wc
π N t do
=
lb/hr-ft
or
Γ=
1000 Wc
π N t do
(Metric)
=
kg/s-m
1/ 3
b.
 1.04 x 10 4 
 2
 k f  sf 
hλ = 
1
/
3
 Γ

 µf 


 condensate
=
Btu/hr-ft2-°F
or
1/ 3
 s2 
 254 
hλ =  1 / 3  k f  f 
(Metric)
 µf 
Γ

 condensate
=
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12. Vapor cooling coefficient
Calculate h′gc using procedures outlined in Section
IX-D, shell-side iteration (use W v as the flowrate)
h′gc
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
=
Btu/hr-ft2-°F
lb/hr
W/m2-K
kg/s
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
=
Btu/hr
Btu/hr
W
W
=
Btu/hr
W
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K

 Tdh + Tds

− tf  
 Wc 
2


C1 = 1 − 
 Wv1 (Tdh − Tds ) 




1
h gc
=
C
1
+ 1 =
h λ h ′gc
hgc
13. Liquid cooling coefficient (Horizontal Bundles)
a. Bottom Cooling Coefficient
Calculate h′lc using procedures outlined in Section
IX-D, shell-side iteration (use Wl as condensate flow rate)
h′lc
b.
Drip Cooling Coefficient
c.
hdc = 1.5hλ
Total liquid cooling coefficient (Horizontal)
Bundles), Liquid entering zone, Wl
h lc =
2 h dc h′lc
(total vapor entering zone)
h dc + h′lc
or
h lc =
( Wl + Wc ) h dc h′lc
Wc
W 

h′lc +  Wl + c  h dc
2
2 

(For liquid + vapor entering zone)
14. Liquid cooling coefficient (Vertical Bundles)
hlc = hλ
15. Weighted hydrocarbon condensing zone coefficient
a. Zone condensing duty, qλ
b. Zone vapor cooling duty, qgc
c.
Zone liquid cooling duty, qlc
hds =
Uds =
qds
qλ qgc qlc
+
+
hλ hgc hlc
hds Uox
hds + Uox
(Uox was evaluated in Step II.A.21.)
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16. Zone Area
A ds =
qds
Uds ∆t e( ds )
=
ft2
m2
=
Btu/hr-ft2-°F
W/m2-K
=
°F
°C
(qds and ∆te for this zone were evaluated in Step I.B.)
E.
Steam Condensing Zone (Shell-Side Condensation)
qsc
( A − A ds ) ∆t e( sc )
1.
Usc =
2.
t f = ts −
 U 
1
( t s − t t ) 1 − sc 
2
 Uox 
(ts and tt for this zone were evaluated in Step I.C.)
3.
∆Tzone inlet
∆Tzone outlet
=
Tds − t ds
= _____________
T2 − t 1
=
4.
Fv (Figure 4)
=
5.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone, W c)
+ Non Condensable
= Wv
=
=
=
Average Mol. weight of vapor
=
lb / hr
mol / hr
=
lb/ft3
6.
lb/hr
mol/hr
kg/s
kmol/s
kg / s
kmol / hr
Average vapor properties at film temperature
a.
Density, ρ v =
Avg MW
10.7
 P (psia)   1 


 t + 460   u 
 f
 
or
ρv =
b.
7.
8.
Avg MW
8.31
 P (kPa abs)   1 


 t + 273   u 
 f
 
kg/m3
(Metric) =
Viscosity, µf
(Used for hgc calc.)
=
cP
Pa•s
Wl = W – W v
=
lb/hr
kg/s
Average condensate properties at film temperature
a. Specific gravity, sf
=
b.
=
Btu/hr-ft-°F
W/m-K
cP
Pa•s
Average zone liquid rate, Wl
Thermal conductivity, kf
c. Viscosity, µf
=
(Note that hydrocarbon film properties should be used)
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9.
METRIC UNITS
Condensing Coefficient (Horizontal Bundle)
ns = 1.29 Nt 0.48 (90° Square Tube Layout)
=
ns = 1.02 Nt 0.519 (30° Triangular Layout)
=
b.
(For other layouts, see text)
Asc = A – Adh – Ads
=
ft2
c.
Lc =
=
ft
a.
A sc
(L − 0.5 ) N s
A
m2
or
Lc =
A sc
(L − 0.152) N s
A
(Metric)
=
m
(Note that if a zone occupies more than one shell,
Lc will be a multiple of the tube length.)
Wc
L c ns
d.
Γ=
e.
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 
 Γ 1/ 3




 condensate
=
lb/hr-ft
=
Btu/hr-ft2-°F
kg/s-m
1/ 3
or
1/ 3
 s2 
 204 
 k f  f 
(Metric)
h′λ = 
1/ 3
 µf 
Γ


 condensate
f.
Wv  µ f (v ) 
1
=1+


xl
W l  µ f (l) 
g.
xv = 1− x l
h.
Gv =
0.111
 ρl 


ρ 
 v
W/m2-K
=
0.555
=
=
 FJ 
Wv

 , (Note 4),
25 A x x v  FBT 
=
lb/sec-ft2
or
Gv =
i.
10 6 Wv  FJ 

 , (Note 4)
A x x v  FBT 
G 
h λ = h′λ  v 
 5 
(Metric)
kg/sm2
=
0.7
=
Btu/hr-ft2-°F
or
 G 
h λ = h′λ  v 
 24.4 
0.7
(Metric)
=
If hλ > 2 h′λ , use 2 h′λ
(For pure components, h λ = h′λ )
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10. Condensing Coefficient (Vertical Bundles)
a.
Γ=
12 Wc
π Nt do
=
lb/hr-ft
or
Γ=
1000 Wc
π Nt do
(Metric)
=
kg/s-m
1/ 3
b.
 1.04 x 10 4 
 2
 k f  sf 
hλ = 
 Γ1/ 3

 µf 


 condensate
=
Btu/hr-ft2-°F
or
1/ 3
 s2 
 254 
hλ =  1 / 3  k f  f 
(Metric)
 µf 
Γ

 condensate
W/m2-K
=
11. Vapor cooling coefficient
Calculate h′gc using procedures outlined in Section
IX-D, shell-side iteration (use W v as the flowrate)
h′gc
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
=
Btu/hr-ft2-°F
lb/hr
W/m2-K
kg/s
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K

 Tdh + Tds

− tf  
 Wc 
2

 =
12. C1 = 1 − 
 Wv1 (Tdh − Tds ) 




C
1
1
=
+ 1 =
h gc
hλ
h′gc
hgc
13. Liquid cooling coefficient (Horizontal Bundles)
a. Bottom Cooling Coefficient
Calculate h′lc using procedures outlined in Section
IX-D, shell-side iteration (use Wl as condensate flow rate)
h′lc
b.
Drip Cooling Coefficient
c.
hdc = 1.5hλ
Total liquid cooling coefficient (Horizontal)
Bundles), Liquid entering zone, Wl
h lc =
( Wl + Wc ) h dc h ′lc
Wc
W 

h′lc +  Wl + c  h dc
2
2 

14. Liquid cooling coefficient (Vertical Bundles)
hlc = hλ
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15. Weighted steam condensing coefficient
a.
Zone condensing duty, qλ
=
Btu/hr-ft2-°F
W/m2-K
W/m2-K
b.
Zone vapor cooling duty, qgc
=
Btu/hr-ft2-°F
c.
Zone liquid cooling duty, qlc
=
Btu/hr-ft2-°F
W/m2-K
q sc
q gc
=
Btu/hr-ft2-°F
W/m2-K
hsc Uox
hsc + Uox
=
Btu/hr-ft2-°F
W/m2-K
=
ft2
m2
=
A = Adh + Ads + Asc
If total area calculated is not reasonably close to total
area assumed, reiterate the various zone calculations.
G. Condensing Pressure Drop (Shell-Side Condensation)
ft2
m2
h sc =
Usc =
qλ
q
+
+ lc
hλ
h gc
h lc
16. Zone Area
A sc =
F.
qsc
Usc ∆t e( sc )
(qsc and ∆te for this zone were evaluated in Step I.C.)
Total Area
1.
1.
Hydrocarbon condensing zone inlet volumes
a. Vapor
(1) ρ v1( dh ) =
Vapor MW
10.7
 P (psia)   1 


 T + 460   u 
 dh
 
lb/ft3
=
or
ρ v1( dh ) =
Vapor MW
8.31
 P (kPa abs)   1 


 T + 273   u  (Metric)
 dh
 
(2) Vapor rate, W v1(dh)
=
lb/hr
kg/s
=
ft3/hr
m3/s
(1) ρl1( dh )
=
lb/ft3
kg/m3
(2) Liquid rate, Wl1( dh )
=
lb/hr
kg/s
(3)
=
ft3/hr
m3/s
(3) Vv1(dh) = W v1(dh) / ρv1(dh)
b.
kg/m3
=
Liquid (if present)
Vl1( dh )
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2.
METRIC UNITS
Steam condensing zone inlet volumes
a. Vapor
(1) ρ v1( dh ) =
Vapor MW
10.7
 P (psia)   1 


 T + 460   u 
 ds
 
lb/ft3
=
or
ρ v1( dh ) =
Vapor MW  P (kPa abs )   1 


 T + 273   u  (Metric)
8.31
 ds
 
(2) Vapor rate, W v1(ds)
=
lb/hr
kg/s
=
ft3/hr
m3/s
(1) ρl1( ds )
=
lb/ft3
kg/m3
(2) Liquid rate, Wl1( ds )
=
lb/hr
kg/s
Vl1( ds ) = Wl1( ds ) / ρl1( ds )
=
ft3/hr
m3/s
Steam condensing zone outlet volumes
a. Vapor (if present)
Vv2 (evaluated in Step II.B.12.)
=
ft3/hr
m3/s
Vl 2 (evaluated in Step II.B.14.)
=
ft3/hr
m3/s
=
psi
kPa
=
lb/ft3
kg/m3
=
psi
kPa
=
lb/ft3
kg/m3
=
psi
kPa
=
psi
kPa
(3) Vv1(ds) = W v1(ds) / ρv1(ds)
b.
Liquid
(3)
3.
4.
Vapor desuperheating zone ∆Pdh
Calculate ∆Pdh using procedures outlined in
Section IX-D, shell-side iteration, ∆Pdh
5.
6.
7.
kg/m3
=
Hydrocarbon condensing zone, ∆Pds
2W
Vv1( dh ) + Vl1( dh ) + Vv1( ds ) + Vl1( ds )
a.
ρds =
b.
Calculate ∆Pds using procedures outlined in
Section IX-D, shell-side iteration
Steam condensing zone ∆Psc
2W
Vv1( ds ) + Vl1( ds ) + Vv 2 + Vl 2
a.
ρsc =
b.
Calculate ∆Psc using procedures outlined in
Section IX-D, shell-side iteration
Total condensing-side pressure drop
∆P = ∆Pdh + ∆Pds + ∆Psc
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III. TUBE-SIDE CONDENSATION
A. Overall Shell-Side Calculation (No Change of Phase)
1.
Assumed value of Uo
=
Btu/hr-ft2-°F
W/m2-K
2.
Assumed value of Rio
=
hr-ft2-°F/Btu
m2-K/W
3.
4.
A = Q / Uo ∆tew
Tube metal:
=
ft2
m2
Thermal conductivity, kw
Tube outer diameter, do
=
=
Btu/hr-ft-°F
in.
W/m-K
mm
Wall thickness, l
=
ft
m
Tube inside diameter, di
Tube length, L
Tube pitch (Pt) and layout (Note 1)
As = A / Ns (Note 2)
Nt = 3.82 As / (L – 0.5) do
or
Nt = 318 As / (L – 0.152) do (Metric)
=
=
=
=
=
in.
ft
in.
ft2
mm
mm
mm
m2
tt overall = 0.4 (T1 – T2) + T2
=
°F
°C
=
°F
°C
=
hr-ft2-°F/Btu
m2-K/W
=
hr-ft2-°F/Btu
m2-K/W
=
hr-ft2-°F/Btu
m2-K/W
U
t +t 
12. t f =  s t  + o (Rio + rio + rw + ro ) ( t s − t t )
2
 2 
=
°F
°C
13. Shell-side density, ρ
=
lb/ft3
kg/m3
Viscosity, µf
=
cP
Pa •s
Specific heat, cf
=
Btu/lb-°F
kJ/kg-K
Thermal conductivity, kf
=
Btu/hr-ft-°F
W/m-K
5.
6.
7.
8.
9. ts overall = 0.4 (t2 – t1) +t1
10. Fouling factors
a.
rio =
do
(ri ) =
di
b. ro
11. Tube wall resistance
rw = l / k w
=
14. For fluids other than hydrocarbons, determine film
properties. For hydrocarbons, determine only µ
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15. Shell-side flow rate, W = Q / c (t2 – t1)
16. NTC = 1.19
lb/hr
kg/s
=
=
in.
in.
mm
mm
=
in.
mm
=
=
in.
mm
Nt square layout
= 1.10 N t triangular layout
17. Dt = (NTC – 1) Pt + do
18. D = Dt / 0.9 (Note 3)
19. S =
=
METRIC UNITS
D − (do NTC )
(Note 4)
FF
20. Assumed Pb
21. NB = FF (10L / Pb) (Note 4),
or
NB = FF (833 L / Pb) (Note 4) (Metric)
22. Ax = S (Pb – 0.375) (Segmental Baffles)
or
Ax = S (Pb – 9.53) (Metric)
Ax = 0.85S (Pb – 0.375) (Double Segmental Baffles)
or
Ax = 0.85S (Pb – 9.53) (Metric)
=
=
=
=
=
in.2
mm2
in.2
mm2
=
23. Calculate ∆Ps using procedures outlines in
=
Section IX-D, shell-side iteration (if pressure drop is
unreasonable, adjust Pb and reiterate.)
24. Calculate shell-side coefficient using procedures
outlines in Section IX-D, shell-side iteration
psi
kPa
=
Btu/hr-ft2-°F
W/m2-K
=
hr-ft2-°F/Btu
m2-K/W
26. 1 / Uox = Rio + rio + rw + ro
=
hr-ft2-°F/Btu
m2-K/W
27. Uox
=
Btu/hr-ft2-°F
W/m2-K
ho
25. Ro = 1 / ho
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B.
METRIC UNITS
Overall Tube-Side Calculation
1.
2.
Total shell-side flow rate, W
Inlet vapor density and volume
ρ v1 =
Inlet MW
10.7
 Pi (psia)   1 


 T + 460   u 
 1
 
Inlet MW
8.31
 Pi (kPa abs)   1 


 T + 273   u 
1

 
=
lb/hr
=
lb/ft3
kg/s
or
ρ v1 =
3.
(Metric)
kg/m3
=
Inlet vapor rate, W v1
=
lb/hr
kg/s
Vv1 = W v1 / ρv1
Outlet vapor density and volume
=
ft3/hr
m3/s
=
lb/ft3
ρ v2 =
Outlet MW
10.7
 Pi (psia)   1 


 T + 460   u 
 2
 
Outet MW
8.31
 Pi (kPa abs)   1 


 T + 273   u 
2

 
or
ρ v2 =
4.
5.
6.
kg/m3
=
Outlet vapor rate, W v2
=
lb/hr
kg/s
Vv2 = W v2 / ρv2
Inlet liquid density and volume
=
ft3/hr
m3/s
ρl1
=
lb/ft3
kg/m3
Inlet liquid rate, Wl1
=
lb/hr
kg/s
Vl1 = Wl1 / ρl1
=
ft3/hr
m3/s
ρl 2
=
lb/ft3
kg/m3
Outlet liquid rate, Wl 2
=
lb/hr
kg/s
Vl 2 = Wl 2 / ρl 2
=
ft3/hr
m3/s
=
lb/ft3
kg/m3
Outlet liquid density and volume
Average density
ρ=
7.
(Metric)
2W
Vv1 + Vl1 + Vv 2 + Vl 2
Assumed Np
=
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CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
8.
V=
Np W
=
19.6 ρ N t di2
METRIC UNITS
ft/sec
or
1.273 x 10 6 Np W
V =
9.
(Metric)
ρ N t di2
=
G = ρV
10. Re =
lb/sec-ft2
kg/s-m2
=
psi
kPa
=
psi
=
di G
di G
or
29 µ
1000 µ
(Metric)
m/s
=
11. Calculate Fanning friction factor f
f
= 16 / Re
= 10–4 Re 0.575
(Re ≤ 2000)
(2000 < Re < 3800)
=
=
(Re ≥ 3800)
=
= 0.0035 + 0.264 / Re0.42
12.
ρV2 / 9270
13. ∆Ptf = 48 f
ρV2 / 2000
or
2
L ρV  µ w 


di 9270  µ 
(Metric)
0.14 or 0.25
or
= 4000 f
L ρV 2  µ w 


di 2000  µ 
0.14 or 0.25
(Metric)
=
kPa
(use 0.25 for laminar flow)
14. ∆Ptr = 3 (ρV2 / 9270)
∆Ptr
= 3 (ρV2 / 2000)
=
(Metric)
psi
=
15. ∆Pt = Ft Ns Np (∆Ptf + ∆Ptr)
=
kPa
psi
kPa
(See Table 3, Section IX-D)
(If ∆Pt is unreasonable change Np and reiterate tube-side calculation.)
The above assumed geometry will now be checked using the following detailed procedure.
C.
Vapor Desuperheating Zone (Shell-Side Condensation)
1.
ρ v1 =
Inlet MW
10.7
 Pi (psia)   1 


 t + 460   u 
 t
 
Inlet MW
8.31
 Pi (kPa abs)   1 


 t + 273   u 
t

 
=
lb/ft3
or
ρ v1 =
(Metric)
=
(tt was calculated in Step I.A.)
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2.
3.
4.
Total tube-side inlet vapor, W
At tt, determine for fluids other than hydrocarbons:
=
METRIC UNITS
lb/hr
kg/s
Viscosity, µf
=
cP
Pa•s
Specific Heat, c
=
Btu/lb-°F
kJ/kg-K
Thermal Conductivity, k
=
Btu/hr-ft-°F
W/m-K
=
ft/sec
V=
Np W
19.6 ρ v N t di2
1.273 x 10 6 Np W
V =
(Metric)
ρ N t di2
5.
G = Vρv
6.
Re =
7.
From Figure 8, determine Yth
a.
di G
29 µ
or
di G
(Metric)
1000 µ
k (cµ / k)1/3
* = Yth k  cµ 
hio
do
 k 
*
= 1 / Rio
m/s
lb/sec-ft2
kg/s-m2
=
Btu/hr-ft-°F
W/m-K
=
Btu/hr-ft2-°F
W/m2-K
°F
°C
=
1/ 3
b.
=
=
=
* Uncorrected for viscosity.
c.
* + r ) (t – t )
tw = tt + Uo ( Rio
io
s
t
=
d.
(µ / µw)0.14
=
8.
* (µ / µ )0.14
hgc = hio
w
=
Btu/hr-ft2-°F
W/m2-K
9.
Udh =
=
Btu/hr-ft2-°F
W/m2-K
=
ft2
m2
hgc Uox
hgc + Uox
(Uox was evaluated in Step II.A.27.)
10. A dh =
qdh
∆t e( dh ) Udh
(qdh and ∆te were evaluated in Step I.A.)
D.
Hydrocarbon Condensing - Steam Desuperheating Zone (Shell-Side Condensation)
1.
Assume a zone area, Ads
=
ft2
m2
2.
Uds =
qds
A ds ∆t e( ds )
=
Btu/hr-ft2-°F
W/m2-K
(qds and ∆te for this zone were evaluated in Step I.B.)
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CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
3.
tf = tt −
 U 
1
( t t − t s ) 1 − ds 
2
 Uox 
METRIC UNITS
°F
=
°C
(ts and tt for this zone were evaluated in Step I.B.)
4.
∆Tzone inlet
∆Tzone outlet
=
Tdh − t dh
Tds − t ds
=
5.
Fv (Figure 4)
=
6.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone)
+ (Vapor not condensed in zone)
= Wv
=
=
=
Average Mol. weight of vapor
=
lb / hr
mol / hr
=
lb/ft3
7.
lb/hr
mol/hr
kg/s
kmol/s
kg / s
kmol / hr
Average vapor properties at film temperature
a.
Density, ρ v =
Avg MW
10.7
 P (psia)   1 


 t + 460   u 
 f
 
or
ρv =
b.
Avg MW  P (kPa abs)   1 

 
8.31  t f + 273   u 
(Metric)
Viscosity, µf
(Used for hgc calc.)
kg/m3
=
=
cP
Pa•s
lb/hr
kg/s
8.
Wl = W − Wv
=
9.
Average condensate properties at film temperature
a. Specific gravity, sf
=
b.
=
Btu/hr-ft-°F
W/m-K
=
cP
Pa•s
=
lb/sec-ft2
Thermal Conductivity, kf
c. Viscosity, µf
10. Condensing Coefficient
a.
G′v =
Np W v
19.6 N t di2
or
G′v =
1.273 x 10 6 Np W v
N t di2
(Metric)
=
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b.
G′l =
G′l =
Np W l
or
19.6 N t di2
1.273 x 10 6 Np Wl
N t di2
(Metric)
c.
G e = G′l + G′v (ρ l / ρ v )1/ 2
d.
Re =
e.
f.
kf Prf1/3
If Re < 50,000
or
htλ = 5.03 Re1/3 kf Prf1/3 [12 / di]
htλ = 5.03 Re1/3 kf Prf1/3 [1000 / di] (Metric)
If Re > 50,000
di Ge
di Ge
or
29 µ f ( l )
103 µ f ( l )
(Metric)
or htλ = 0.0265 Re0.8 kf Prf1/3 [1000 / di ] (Metric)
11. Vapor Cooling Coefficient
 ρl 


ρ 
 v
kg/s-m2
lb/sec-ft2
kg/s-m2
=
Btu/hr-ft-°F
W/m-K
=
=
Btu/hr-ft-°F
=
Btu/hr-ft-°F
=
W/m-K
=
W/m-K
0.555
a.
W v  µ f (v ) 
1
=1+


xl
W l  µ f (l ) 
b.
xv = 1− x l
=
c.
Gv =
G ′v
xv
=
d.
Re =
Gv di
lb/sec-ft2
=
=
htλ = 0.0265 Re0.8 kf Prf1/3 [12 / di]
0.111
=
METRIC UNITS
=
lb/sec-ft2
kg/sm2
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
29 µ f ( l )
or
Re = 10 −3
Gv di
(Metric)
µ f (l )
e.
From Figure 8, determine Yth
f.
k (cµ / k)1/3
g.
* = Yth k  cµ 
hio
do
 k 
1/ 3
*
= 1 / Rio
=
* Uncorrected for viscosity.
h.
* + r ) (t – t )
tw = tt + Uo ( Rio
io
s
t
i.
(µ / µw)0.14
=
j.
* (µ / µ )0.14
h′gc = hio
w
=
Btu/hr-ft2-°F
W/m2-K
1
1
1
=
+
h gc
h tλ
h′gc
=
Btu/hr-ft2-°F
W/m2-K
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12. Liquid cooling coefficient
hlc = htλ
13. Weighted zone coefficient
a. Zone condensing duty, qλ
b. Zone vapor cooling duty, qgc
c.
Zone liquid cooling duty, qlc
qds
qgc
qλ
q
+
+ lc
hλ
hgc
h lc
hds =
Usc =
hds Uox
h ds + Uox
=
Btu/hr-ft2-°F
W/m2-K
=
=
Btu/hr
Btu/hr
W
W
=
Btu/hr
W
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
ft2
m2
=
ft2
m2
=
Btu/hr-ft2-°F
W/m2-K
°F
°C
(Uox was evaluated in Step III.A.27.)
14. Zone Area
A ds =
qds
Uds ∆t e( ds )
(qds and ∆te for this zone were evaluated in Step I.B.)
E.
Steam Condensing Zone (Tube-Side Condensation)
1.
Asc = A – Adh – Ads
2.
Usc =
qsc
A sc ∆t e( sc )
(qsc and ∆te(sc) for this zone were evaluated in Step I.C.)
3.
tf = tt +

U 
1
( t t − t s ) 1 − sc 
2
Uox 

=
(ts and tt for this zone were evaluated in Step I.C.)
4.
∆Tzone inlet
∆Tzone outlet
=
Tds − t ds
T2 − t1
=
5.
Fv (Figure 4)
=
6.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone, W)
+ Vapor not condensed in zone
= Wv
=
=
=
Average Mol. weight of vapor
=
lb/hr
mol/hr
kg/s
lb / hr
mol / hr
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CUSTOMARY UNITS
7.
Average zone liquid rate, Wl
Wl = W – W v
8.
=
lb/hr
kg/s
=
Btu/hr-ft2-°F
W/m-K
=
cP
Pa•s
=
lb/sec-ft2
Average H.C. condensate properties at film temperature
a. Specific gravity, sf
=
b.
9.
METRIC UNITS
Thermal conductivity, kf
c. Viscosity, µf
Condensing Coefficient
a.
G′v =
Np Wv
19.6 N t di2
or
G′v =
b.
Gl =
1.273 x 10 6 Np W v
N t di2
(Metric)
Np Wl
=
19.6 Nt di2
kg/s-m2
=
lb/sec-ft2
or
Gl =
1.273 x 10 6 Np Wl
Nt di2
c.
G e = G′l + G ′v (ρ l / ρ v )1/ 2
d.
Re =
(Metric)
di Ge
,
29 µ f ( l )
kg/s-m2
=
=
lb/sec-ft2
kg/s-m2
W/m-K
=
or
Re =
e.
f.
or
or
di Ge
1000 µ f ( l )
(Metric)
=
kf Prf1/3
If Re < 50,000
htλ = 5.03 Re1/3 kf Prf1/3 [12 / di]
htλ = 5.03 Re1/3 kf Prf1/3 [1000 / di ] (Metric)
If Re > 50,000
=
Btu/hr-ft-°F
htλ = 0.0265 Re0.8 kf Prf1/3 [12 / di]
htλ = 0.0265 Re0.8 kf Prf1/3 [ 1000 / di ] (Metric)
=
Btu/hr-ft-°F
1/ 3
W/m-K
g.
* = Yth k  cµ 
hio
do
 k 
h.
* + r ) (t – t )
tw = tt + Uo ( Rio
io
s
t
i.
(µ / µw)0.14
=
j.
* (µ / µ )0.14
h′gc = hio
w
1
1
1
=
+
h gc
h tλ
h′gc
*
= 1 / Rio
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
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METRIC UNITS
10. Vapor Cooling Coefficient
0.111
1
W  µ (v) 
= 1+ v  f

xl
Wl  µ f ( l ) 
b.
xv = 1− x l
=
c.
Gv =
G ′v
xv
=
d.
Re =
Gv di
29 µ f ( l )
Re = 10 − 3
 ρl 
 
ρ 
 v
0.555
a.
, or
G v di
lb/sec-ft2
kg/sm2
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
(Metric)
µ f (l )
=
e.
From Figure 8, read Yth
f.
k (cµ / k)1/3
g.
* = Yth k  cµ 
*
= 1 / Rio
hio
do
k
 vap.
=
1/ 3
* Uncorrected for viscosity.
h.
* + r ) (t – t )
tw = tt + Uo ( Rio
io
s
t
i.
(µ / µw)0.14
=
j.
* (µ / µ )0.14
h′gc = hio
w
=
Btu/hr-ft2-°F
W/m2-K
1
1
1
=
+
h gc
h tλ
h′gc
=
hr-ft2-°F/Btu
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
=
=
Btu/hr
Btu/hr
W
W
=
Btu/hr
W
=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
11. Liquid cooling coefficient
hlc = htλ
12. Weighted zone coefficient
a. Zone condensing duty, qλ
b. Zone vapor cooling duty, qgc
c.
Zone liquid cooling duty, qlc
hsc =
Usc =
qsc
qλ qgc qlc
+
+
hλ hgc hlc
hsc Uox
hsc + Uox
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CUSTOMARY UNITS
METRIC UNITS
13. Zone Area
A sc =
qsc
Usc ∆t e( sc )
=
ft2
m2
=
A = Adh + Ads + Asc
If total area calculated is not reasonably close to total
area assumed, reiterate the various zone calculations.
G. Condensing Pressure Drop (Tube-Side Condensation)
ft2
m2
F.
(qsc and ∆te for this zone were evaluated in Step I.C.)
Total Area
1.
1.
All vapor flow mass velocity
Gv =
Gv =
2.
Np Wv
19.6 Nt di2
, or
1.273 x 10 6 Np Wv
Nt di2
lb/sec-ft2
=
(Metric)
kg/s-m2
=
(W v = Total vapor rate to condenser.)
This mass velocity will be used to calculate the ∆P
for each zone of the condenser.
Hydrocarbon condensing zone inlet volumes
a. Vapor
(1) ρ v1 =
Inlet MW  Pi (psia)   1 

  
10.7
 T1 + 460   u 
lb/ft3
=
or
ρ v1 =
b.
Inlet MW  Pi (kPa abs)   1 

 
8.31  T1 + 273   u 
(Metric)
kg/m3
=
(2) Vapor rate, W v1(dh)
=
lb/hr
kg/s
(3) Vv1(dh) = W v1(dh) / ρv1(dh)
=
ft3/hr
m3/s
(1) ρl1( dh )
=
lb/ft3
kg/m3
(2) Liquid rate, Wl1( dh )
=
lb/hr
kg/s
(3)
=
ft3/hr
m3/s
Liquid (if present)
Vl1( dh )
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
3.
METRIC UNITS
Steam condensing zone inlet volumes
a. Vapor
Vapor MW  Pi (psia)   1 


 T + 460   u 
10.7
 ds
 
=
Vapor MW  Pi (kPa abs)   1 


 T + 273   u 
8.32
 ds
 
(Metric)
(1) ρ v1( ds ) =
lb/ft3
or
ρ v1( ds ) =
b.
(2) Vapor rate, W v1(ds)
=
lb/hr
kg/s
(3) Vv1(dh) = W v1(ds) / ρv1(ds)
Liquid (if present)
=
ft3/hr
m3/s
(1) ρl1( ds )
=
lb/ft3
kg/m3
(2) Liquid rate, Wl1( ds )
=
lb/hr
kg/s
=
ft3/hr
m3/s
=
ft3/hr
m3/s
=
ft3/hr
m3/s
=
=
ft/sec
m/s
=
psi
Vl1( ds ) = Wl1( ds ) / ρl1( ds )
(3)
4.
Steam condensing zone outlet volumes
a. Vapor (if present)
Vv2 (evaluated in Step III.B.3.)
b. Liquid
Vl 2 (evaluated in Step III.B.5.)
5.
kg/m3
=
Vapor desuperheating zone ∆P
di G
di G
or
29 µ
1000 µ
a.
Re =
b.
Calculate Fanning friction factor f
f
=
=
(Metric)
(Re ≤ 2000) =
(2000 < Re < 3800) =
16 / Re
10-4 Re 0.575
= 0.0035 + 0.264 / Re0.42
From Step III.C.4., V
c.
∆Ptf = 48 f
=
L ρV 2  µ w 


di 9270  µ 
(Re ≥ 3800)
0.14 or 0.25
or
= 4000 f
L ρV 2  µ w 


di 2000  µ 
0.14 or 0.25
(Metric)
=
(Use 0.25 for laminar flow)
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d.
∆Ptr = 3 (ρV2 / 9270)
=
METRIC UNITS
psi
or
∆Ptr = 3 (ρV2 / 2000)
e.
(Metric)
=
A
∆Pdh = Ft Ns Np (∆Ptf + ∆Ptr) dh
A
=
kPa
psi
kPa
=
lb/ft3
kg/m3
=
ft/sec
m/s
=
psi
kPa
(For Ft, see Table 3, Section IX-D)
6.
Hydrocarbon condensing zone ∆P
a.
Re =
di Ge
di Ge
or
29 µ f ( l )
1000 µ f ( l )
(Metric)
=
( uf ( l ) was evaluated in Step III.E.8.c.)
b.
Calculate the average density, ρds
ρ ds =
2W
Vv1( dh ) + Vl1( dh ) + Vv1( ds ) + Vl1( ds )
c.
V = G / ρds
d.
ρds
V2 / 9270
e.
f
= 16 / Re
= 10–4 Re 0.575
or ρds
V2 / 2000
(Metric)
(Re ≤ 2000) =
(2000 < Re < 3800) =
= 0.0035 + 0.264 / Re0.42
f.
∆Ptf = 48 f
L ρds V 2  µ w 


di 9270  µ 
(Re ≥ 3800)
=
0.14 or 0.25
=
psi
or
= 4000 f
L ρds V 2  µ w 


di 2000  µ 
0.14 or 0.25
(Metric)
=
kPa
(Use 0.25 for laminar flow)
g.
∆Ptr = 3 (ρds V2 / 9270)
or
∆Ptr = 3 (ρds V2 / 2000)
h.
=
(Metric)
=
A
∆Pds = Ft Ns Np (∆Ptf + ∆Ptr) ds
A
=
psi
kPa
psi
(For Ft, see Table 3, Section IX-D)
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
CUSTOMARY UNITS
7.
METRIC UNITS
Steam condensing zone ∆P
a.
Re =
di G
29 µ f ( l )
=
or
di G
(Metric)
1000 µ f ( l )
( uf ( l ) was evaluated in Step III.D.9.c.)
b.
Calculate the average density, ρds
ρds =
2W
Vv1( dh ) + Vl1( dh ) + Vv1( ds ) + Vl1( ds )
=
lb/ft3
kg/m3
m/s
c.
V = G / ρds
=
ft/sec
d.
ρds V2 / 9270
=
psi
or
ρds V2 / 2000
e.
f
(Metric)
kPa
(Re ≤ 2000)
=
(2000 < Re < 3800)
=
(Re ≥ 3800)
=
= 16 / Re
= 10-4 Re 0.575
= 0.0035 + 0.264 / Re0.42
f.
∆Ptf = 48 f
2
L ρsc V
di 9270
 µw 


 µ 
0.14 or 0.25
=
psi
or
= 4000 f
L ρsc V 2  µ w 


di 2000  µ 
0.14 or 0.25
(Metric)
=
kPa
(Use 0.25 for laminar flow)
g.
∆Ptr = 3 (ρsc V2 / 9270)
=
psi
or
∆Ptr = 3 (ρds V2 / 2000)
h.
(Metric)
=
A ds
A
=
psi
kPa
=
psi
kPa
∆Psc = Ft Ns Np (∆Ptf + ∆Ptr)
kPa
(For Ft, see Table 3, Section IX-D)
8.
Total condensing-side pressure drop
∆P = ∆Pdh + ∆Pds + ∆Psc
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TABLE 2 (Cont)
CALCULATION PROCEDURE FOR CONDENSER DESIGN
Notes:
Note 1
Pt
Note 2
Max. Area Per Shell (Square Pitch)*
16 ft Tubes
20 ft Tubes
3/4 in. tubes
—
1.000 in.
4780 ft2
5950 ft2
1 in. tubes
—
1.250 in.
4140 ft2
5150 ft2
1 1/2 in. tubes
—
1.875 in.
2660 ft2
3300 ft2
Note 1
Pt
Note 2
Max. Area Per Shell (Square Pitch)*
4.88 m Tubes
6.1 m Tubes
19.05 mm tubes
—
25.4 mm
444 m2
553 m2
25.4 mm tubes
—
31.75 mm
385 m2
478 m2
38.1 mm tubes
—
47.63 mm
247 m2
307 m2
* For triangular pitch, use the area of same size of tubes divided by 0.866 and (Pt of triangle /
Pt of square)2 if a different tube pitch is used.
Note 3 Do not make “D" smaller than (Dt + 1 in.) or (Dt + 25.4 mm) nor larger than (Dt + 3 in.)
or (Dt + 76.2 mm).
1 for E & J shells
FBT =
1 for Segmental Baffles
Note 4 FF =
FJ
=
2 for F shells
=
0.5 for J shells
=
1 for E & F shells
=
2 for Double Segmental Baffles
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TABLE 3
WIDE-CUT CONDENSER DESIGN SAMPLE CALCULATION
CUSTOMARY
A.
METRIC
Design Data
Inlet vapor is pipestill overhead
1. Condensable HC
2.
3.
4.
5.
= 20,000 lb/hr
2.52 kg/s
Mean ABP
= 231°F
110.6°C
M.W.
= 100
100
ρ
= 49.9 lb/ft3
799 kg/m3
Noncondensables
(Hydrocarbon M.W. = 58)
Steam
ASTM Distillation of Condensable Hydrocarbon:
= 580 lb/hr
0.073 kg/s
= 720 lb/hr
0.0907 kg/s
IBP - 165°F
IBP - 73.9°C
60 - 127.8°C
10 - 183
70 - 269
20 - 211
80 - 274
30 - 231
90 - 283
40 - 245
FBP - 300
50 - 255
Tower overhead pressure - 5.3 psig
10 - 83.9
20 - 99.4
30 - 110.6
40 - 118.3
50 - 123.9
= 20.0 psia
70 - 131.7
80 - 134.4
90 - 139.4
FBP - 148.9
Tower overhead temperature
= 263°F 128.3°C
60 - 262°F
36.5 kPa gage = 137.8 kPa abs
6.
Assume condensate cooled to 95°F (35°C); water rise is 80 to 95°F (27 to 35°C)
B.
Calculations
1.
Flash Curves
From ASTM Distillation
t10 = 183°F
t70 = 269°F
t10 = 83.9°C
t70 = 131.7°C
t30 = 231°F
t90 = 283°F
t30 = 110.6°C
t90 = 139.4°C
t50 = 255°F
∆1 =
t50 = 123.9°C
2
(269 − 183) = 57
3
∆1 =
2
(131.7 − 83.9 ) = 31.9
3
From Figure 6, ∆2 = 31, ∆3 = 16, ∆1 – ∆3 = 41
∆2 = 18, ∆3 = 9.5, ∆1 – ∆3 = 22.4
t10f = t10 + (∆1 – ∆3) – ∆2 = 183 + 41 – 31 = 193°F
t10f = t10 + (∆1 – ∆3) – ∆2 = 83.9 + 22.4 – 18 = 88.3°C
t30 f = t10 +
2
 ∆1 ∆2 
+
( t30 − t10 ) + ( ∆1 − ∆3 ) − 
 = 220o F
3
2 
 3
t30 f = t10 +
t50f = 231°F
= 110.2°C
t70f = 240°F
= 115.3°C
2
 ∆1 ∆2 
+
( t30 − t10 ) + ( ∆1 − ∆3 ) − 
 = 104.5o C
3
2 
 3
= 122.6°C
t90f = 252°F
The above are points on the volumetric flash curve. To estimate the molal and weight flash curves, use the equations
given in Section IX-F and obtain:
t10w = 195°F
t10m = 192°F
t10w = 89.4°C
t10m = 87.9°C
t30w = 226°F
t30m = 220°F
t30w = 106.7°C
t30m = 103.6°C
t50w = 233°F
t50m = 230°F
t50w = 111.3°C
t50m = 109.8°C
t70w = 242°F
t70m = 239°F
t70w = 116.5°C
t70m = 114.8°C
t90m = 252°F
t90w = 123.2°C
t90w = 253°F
The three sets of points are plotted and connected as shown in Figure 2.
t90m = 122.4°C
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WIDE-CUT CONDENSER DESIGN SAMPLE CALCULATION
2.
Calculate Tds (Since this is overhead from a fractionator, Tdh = t1)
Molar composition of inlet vapor:
Condensables:
20,000 lb / hr
= 200 mol/hr
100 lb / hr
NC: 580/58 = 10 (NC: Noncondensables)
Steam 720/18 =
40
250 mol / hr total
2.520 kg / s
= 0.0252 kmol/s
100 kg / kmol
0.0731/58 = 0.0013
0.0907/18 =
0.0050
kmol/s
0.0315
Partial pressure, condensables + NC =
210
x 20.0 = 16.8 psia
250
=
0.0265
x 137.8 = 115.9 kPa abs
0.0315
Partial pressure, condensables
200
x 20.0 = 16.0 psia
250
=
0.0252
x 137.8 = 110.2 kPa abs
0.0315
=
Partial pressure of steam
= 20.0 – 16.8 = 3.2 psia
= 137.8 – 115.9 = 21.9 kPa abs
Determine conditions at 20 wt% HC condensed (see Table on following page).
From the flash curves, the atmospheric-pressure temperature would be 248°F (119.9°C) at 20 wt% condensed. Move
horizontally at 248°F (119.9°C) to the molal flash curve, and note that 16.8 mol% has been condensed.
Calculate the remaining condensate moles:
200 – (0.168 x 200)
= 166 mol
Total remaining
mols of vapor
= 166 +10 + 40 = 216
Pressure drop is
assumed negligible,
so that P
= 20.0 psia
0.0252 – (0.166 – 0.0252) = 0.021 kmol
= 0.0210 + 0.0013 + 0.005 = 0.0273
= 137.8 kPa abs
Partial pressure steam =
40
x 20.0 = 3.7 psia
216
=
0.005
x 137.8 = 25.2 kPa abs
0.0273
Partial pressure,
condensables
166
x 20.0 = 15.4 psia
216
=
0.021
x 137.8 = 106 kPa abs
0.0273
=
WIDE-CUT CONDENSER DESIGN SAMPLE CALCULATION
From the alignment chart, Figure 3, the temperature of the hydrocarbon [which normally boils at 248°F (119.9°C)] is
250°F (121.1°C) at 15.4 psia (105.9 kPa abs). From the same chart, the vapor pressure of water at 250°F (121.1°C) is
29.0 psia (177 kPa abs).
Since the partial pressure of the water is less than its vapor pressure (3.7 < 29.0 psia) (25.2 < 177 kPa), steam is not yet
condensing. Therefore, it is necessary to make another trial at a higher percentage HC condensed. Each successive
trial is tabulated.
At 85% HC condensed, the partial pressure of water was found to be 913 psia (64.2 kPa abs), while its vapor pressure
should have been 6.3 psia (45.5 kPa abs). Hence, steam is condensing, and Tds has been passed.
The exact value of Tds is found by plotting the vapor pressure and partial pressure of water on the same graph, as
shown in Figure 2. The composition at Tds is then determined and entered in the last column of the table. For this
case, Tds = 184°F (82.8°C).
Vapor Pressure of Water @ 184°F (82.8°C) = Partial Pressure = 8.3 psia (57.7 kPa abs)
Partial Pressure of
condensables + NC
= 20.0 – 8.3 = 11.7 psia
= 137.8 – 57.7 = 80.1 kPa abs
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TABLE 3 (Cont)
PARAMETER
CUSTOMARY UNITS
Wt.% Hydrocarbon Condensed(1)
Vapor Temp. @ Atm. Pressure(2)
Mol % HC not Condensed(2)
Mol of HC not Condensed
Mol of noncondensables (“NC")
Mol of Steam
Total Mols of Vapor
Total Pressure, psia(3)
Partial Pressure of Steam
Partial Pressure of Condensables
System Temperature(4)
Vapor Pressure of H20 at System Temp.(5)
0
258
100
200
10
40
250
20.0
3.2
16.0
263
35.5
20
248
83.2
166
10
40
216
20.0
3.7
15.4
250
29.0
70
226
42.5
85
10
40
135
20.0
5.9
12.6
215
15.2
PARAMETER
80
210
23.5
47.0
10
40
97
20.0
8.2
9.7
186
8.6
85
203
18.1
36.2
10
40
86.2
20.0
9.3
8.4
173
6.3
81
—
23
46
10
40
96
20.0
8.3
—
184
8.3
85
93.8
17.6
0.00444
0.0013
0.0050
0.01074
137.8
64.2
57.0
75.6
45.5
80.5
—
22.4
0.00564
0.0013
0.0050
0.01194
137.8
57.7
—
82.8
57.7
METRIC UNITS
Wt.% Hydrocarbon Condensed(1)
Vapor Temp. @ Atm. Pressure(2)
Mol % HC not Condensed(2)
Mol of HC not Condensed
Kmol of noncondensables (“NC")
Kmol of Steam
Total Kmol of Vapor
Total Pressure, kPw abs(3)
Partial Pressure of Steam
Partial Pressure of Condensables
System Temperature(4)
Vapor Pressure of H20 at System Temp.(5)
0
126.2
100
0.0252
0.0013
0.0050
0.0315
137.8
21.9
110.2
128.3
210
20
119.9
83.4
0.0210
0.0013
0.0050
0.0273
137.8
25.2
106.0
121.1
177
70
106.7
39.9
0.0101
0.0013
0.0050
0.0164
137.8
42.0
84.9
101.1
103
80
98.0
22.9
0.00577
0.0013
0.0050
0.01207
137.8
57.1
65.9
83.9
59.3
Notes:
(1)
Arbitrary values.
(2)
Read from flash curves, Figure 2.
(3)
(4)
Assume percent pressure drop = condensed (neglected in this example).
Vapor temp. @ 1 atm. adjusted to the partial pressure of condensables by means of the alignment chart, Figure 3.
(5)
From steam tables or from alignment chart, Figure 3 (use the point marked H20 as the “normal boiling point").
Remaining Condensable moles = x; moles of condensables + NC = x + 10 (= x + 0.0013).
Total moles = x + 10 + 40 (= x + 0.0013 + 0.0050).
x + 10
x + 50
x + 0.0013
x 20.0 = 11.7
x + 0.0063
Solving for x, remaining condensable moles = 46
Mol% NOT condensed =
( 46) (100 )
200
= 23%
x 137.8 = 80.1
= 0.0056
=
(0.0056 ) (100 )
0.0252
= 22%
From the flash curves, 23 mol% (22 mol%) not condensed corresponds to 19 wt% not condensed.
First zone condensate = 20,000 (1 – 0.19) = 16,400 lb/hr
= 2.52 (1 – 0.19) = 2.04 kg/s
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TABLE 3 (Cont)
WIDE-CUT CONDENSER DESIGN SAMPLE CALCULATION
CUSTOMARY
3.
METRIC
Heat Release
a.
From T1 to Tds (Zone 1)
Temp change
= 263 – 184 = 79°F
Average %
From Blue Book,
= 224°F
c (vapor + N.C.)
c (condensate)
c (steam)
λ (condensate)
= 128.3 – 82.8 = 45.5°C
= 105.6°C
= 2.05
= 1.88
= 1.90
= 333
= 0.49
= 0.45
= 0.453
= 143
Vapor Cooling
Steam
Vapor + N.C.
(720)
(4,180)
(79)
(79)
(0.453) = 0.026 x 106
(0.49) = 0.162
Condens. Vapor
(16,400 )
2
(79)
(0.49)
=
(16,400 )
2
(79)
(0.45)
= 0.292 x 106 Btu/hr
0.318
0.506 x 10
6
(103) (0.0907)
(103) (0.5531)
(45.5)
(45.5)
(1.90)
(2.05)
= 7.8 x 103
= 51.6
(103) (2.040/2) (45.5)
(2.05)
=
(2.040)
2
(1.88)
= 87.3 x 103 W
95.1
154.5 x 103
Liquid Cooling
Condens. Liquid
(103)
(45.5)
LATENT HEAT
Hydrocarbon (16,400) (143)
qds = Zone Total
b. From Tds to T2 (Zone 2)
Temp Change
Average T
From Blue Book:
c (vapor + N.C.)
c (Steam)
c (Condensate)
λ (Condensate)
λ (Water)
Vapor Cooling
N.C.
(580)
Condens. Steam (720/2)
(89)
(89)
Condens. Vapor (3,600/2) (89)
= 2.34 x 106 Btu/hr
= 3.14 x 106 Btu/hr
= (103) (2.04000) (333) = 679.3 x 103 W
= 921.1 x 103 W
= 184 – 95 = 89°F
= 82.8 – 35 = 47.8°C
= 140°F
= 58.9°C
= 0.44 Btu/lb°F
= 0.448 Btu/lb°F
= 0.405 Btu/lb°F
= 143 Btu/lb
= 1,020 Btu/lb
= 1.84 kJ/kg-K
= 1.88 kJ/kg-K
= 1.70 kJ/kg-K
= 333 kJ/kg
= 2,370 kJ/kg
(0.44) = 0.0023 x 106
(0.448) = 0.014
(103) (0.0731) (47.8)
(103) (0.0907/2) (47.8)
(1.84)
(1.88)
= 6.4 x 103
= 4.1
(103) (0.4536/2) (47.8)
(1.84)
=
(103) (2.0400) (47.8)
(103) (0.4536/2) (47.8)
(1.70)
(1.70)
= 165.8 x 103
= 18.4
(103) (0.0907/2) (47.8)
(4.19)
=
(0.44)
=
0.071
0.108 x 106
19.9
30.4 x 103
Liquid Cooling
Entering Liquid (16,400) (89)
Condens. Liquid (3,600/2) (89)
(0.405) = 0.591 x 106
(0.405) = 0.064
Condens. Steam (720/2)
(1.0)
(89)
=
0.032
0.687 x 10 6
9 .1
193.3 x 103
Latent Heat
Hydrocarbon
(3,600) (143)
Steam
(720)
qsc = Zone Total
Total Condenser Duty
= 0.515 x 106
(1,020) =
0.735
1.250 x 106 Btu / hr
(103) (0.4536)
(333)
= 151 x 103
(103) (0.0907)
(2370)
=
215 x 103
366 x 103 W
= 2.05 Btu/hr
= 589.7 x 103 W
= 3.14 + 2.05 = 5.19 MMBtu/hr = 921.1 x 103 + 589.7 x 103 = 1.511 x 106 W
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TABLE 4
A SAMPLE CONDENSER DESIGN
I.
Exchanger Number E - 100
Refinery
Service
Calc By/Date
Pipestill Overhead Condenser
ZONE DUTIES AND TEMPERATURES*
A.
Vapor Desuperheating Zone (If Present); 1st Zone
CUSTOMARY UNITS
METRIC UNITS
qdh
Fluid Being Condensed (Shell/Tube)
=
Btu/hr
W
T1
=
°F
°C
Tdh
Fluid Being Heated (Shell/Tube)
=
°F
°C
t2
=
°F
°C
tdh
=
°F
°C
T −T
R = 1 dh =
t 2 − t dh
=
t 2 − t dh
=
T1 − t dh
=
=
°F
°C
=
°F
°C
ts = 0.4 (T1 – Tdh) + Tdh
=
°F
°C
tt = 0.4 (t2 – tdh) + tdh
=
°F
°C
tt = 0.4 (T1 – Tdh) + Tdh
=
°F
°C
ts = 0.4 (t2 – tdh) + tdh
=
°F
°C
=
Btu/hr-°F
W/K
j=
∆tm =
(T1 − t 2 ) − (Tdh − t dh )
 T −t 
ln  1 2 
 Tdh − t dh 
From charts given in Section IX-D
Fn =
for this zone**
∆te(dh) = ∆tmFn
For shell-side condensation:
For tube-side condensation:
qdh
∆t e
*
**
Use LMTD correction factor for each zone, based on zone temperatures
Use LMTD corrections factor, based on terminal temperatures to evaluate minimum number of shells in series. Fn must
be equal to or greater than 0.8.
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
B.
Hydrocarbon Condensing Steam Desuperheating Zone (2nd Zone)
CUSTOMARY UNITS
qds
Fluid Being Condensed (Shell/Tube)
=
3.14 x 106
Tdh
=
263
Tds
=
184
tdh
=
tds
=
METRIC UNITS
921.1 x 103
W
°F
128.3
°C
°F
84.4
°C
95
°F
35.0
°C
86
°F
30.0
°C
Btu/hr
R=
Tdh − Tds
=
t dh − t ds
=
8.78
8.78
j=
t dh − t ds
=
Tdh − t ds
=
0.051
0.051
=
130
°F
72.2
°C
∆te(ds) = ∆tm Fn
For shell-side condensation:
=
130
°F
72.2
°C
ts = 0.4 (Tdh – Tds) + Tds
=
216
°F
102
°C
tt = 0.4 (tdh – tds) + tds
For tube-side condensation:
=
90
°F
32
°C
tt = 0.4 (Tdh – Tds) + Tds
=
°F
°C
ts = 0.4 (tdh – tds) + tds
=
°F
°C
∆tm =
(Tdh − t dh ) − ( Tds − t ds )
T −t 
ln  dh dh 
 Tds − t ds 
From charts given in Section IX-D,
Fn =
1.0
for
1
shells
qds
∆Te
C.
=
24,200
Btu/hr-°F
12,800
W/K
qsc
Fluid Being Condensed (Shell/Tube)
=
2.05 x 106
Btu/hr
589.7 x 103
W
Tds
=
184
°F
84.4
°C
T2
=
95
°F
35.0
°C
tds
=
86
°F
30.0
°C
t1
=
80
°F
26.7
°C
T − T2
=
R = ds
t ds − t1
=
14.8
14.8
t ds − t1
=
Tds − t1
=
0.058
0.058
Steam Condensing Zone (3rd Zone)
j=
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
∆tm =
(Tds − t ds ) − (T2 − t1)
T −t 
ln  ds ds 
 T 2 − t1 
CUSTOMARY UNITS
METRIC UNITS
=
______________
=
44.2
°F
24.6
°C
∆te(sc) = ∆tmFn
For shell-side condensation:
=
42.4
°F
23.6
°C
ts = 0.4 (Tds – T2) + T2
=
131
°F
55.0
°C
tt = 0.4 (tds – t1) + t1
For tube-side condensation:
=
82
°F
28.0
°C
tt = 0.4 (Tds – T2) + T2
=
°F
°C
ts= 0.4 (tds – t1) + t1
=
°F
°C
From charts given in Section IX-D,
Fn =
0.96
for
1
shells
qsc
∆t e
D.
=
Q
qdh
qds
qsc
+
+
∆t e( dh )
∆t e( ds )
∆t e( sc )
where: Q = qdh + qds + qsc
*
Btu/hr-°F
25,000
WK
Weighted Temperature Difference
∆t ew =
II.
48,500
=
5.19
3.14 / 130 + 2.05 / 42.4
15.11
9.21 / 72.2 + 5.9 / 23.6
=
71.3
°F
40
°C
=
48.5
Btu/hr-ft2-°F
272
W/m2-K
139
m2
SHELL-SIDE CONDENSATION
A. Overall Tube-Side Calculation (No Change of Phase)
1.
Assumed value of Uo
2.
3.
A = Q / Uo ∆tew
Tube metal: ___________________
=
1500
ft2
Thermal conductivity, kw
Tube outer diameter, do
=
=
64
0.75
Btu/hr-ft-°F
in.
110.8
19.05
W/m-K
mm
Wall thickness, l
=
0.0054
ft
0.00165
m
Tube insider diameter, di
Tube length, L
=
=
0.62
20
in.
ft
15.75
6.1
mm
m
4.
Tube pitch (Pt) and layout (Note 1)*
=
1.0
in.
25.4
mm
5.
As = A / Ns (Note 2)*
=
1500
ft2
139
m2
6.
Nt = 3.82 As / (L – 0.5) do, or
Nt = 318 As / (L – 0.152) do (Metric)
=
=
392
General notes contained on last page of calculation form.
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
7.
Assumed Np
=
2
8.
ts overall = 0.4 (T1 – T2) + T2
=
162
°F
723
°C
9. tt overall = 0.5 (t1 + t2)
10. Tube-side properties at tt
=
87.5
°F
30.8
°C
a.
Density, ρ
=
62.4
lb/ft3
999.5
kg/m3
b.
Viscosity, µ
=
0.8
cP
0.0008
Pa•s
c.
Specific Heat, c
=
1.006
Btu/lb-°F
d.
Thermal Conductivity, k
=
11. Tube-side flow rate, W = Q / c (t2 – t1)
=
346,000
lb/hr
=
3.8
ft/sec
12. V =
V =
Np W
(19.6) ρ N t di2
or
(1.273 x 10 6 ) Np W
ρ N t di2
2
4.2
Btu/hr-ft-°F
(Metric)
kJ/kg-K
W/m-K
43.3
kg/s
1.14
m/s
4860
W/m2-K
22.8
kPa
13. If tube-side fluid is water
hio =
368
 t 
( Vdi )0.7  t 
do
 100 
0.26
=
863
Btu/hr-ft2-°F
or
h io =
 1.8 t t + 32 
1.27 x 10 4

( Vd i ) 0.7 
do
100


0.26
(Metric) =

( V )1.73 
∆P = 0.020 Ft N s Np  V 2 + 0.158 L

(d i )1.27 

=
3.4
psi
or

( V )1.73 
∆P = 1.48 Ft N s Np  V 2 + 22.9 L

(d i )1.27 

(Metric)
=
14. For fluids other than water, evaluate overall tube-side heat transfer coefficient and pressure drop by the procedure
outlined in Section IX-D, tube-side iteration.
hio
=
Btu/hr-ft2-°F
W/m2-K
∆Pt
=
psi
kPa
15. Rio = 1 / hio
=
0.00116
hr-ft2-°F/Btu
0.000206
m2-K/W
16. Tube-side fouling factor, ri
=
0.00150
hr-ft2-°F/Btu
0.000264
m2-K/W
0.000320
m2-K/W
0.000176
m2-K/W
17. rio = (do / di) ri
=
0.00181
hr-ft2-°F/Btu
18. Shell-side fouling factor, ro
=
0.001
hr-ft2-°F/Btu
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
19. rw = l / kw
B.
=
METRIC UNITS
0.00008
hr-ft2-°F/Btu
0.000015
m2-K/W
0.000717
m2-K/W
20. 1 / Uox = Rio + rio + rw + ro
=
0.00405
hr-ft2-°F/Btu
21. Uox
=
247
Btu/hr-ft2-°F
1395
W/m2-K
=
132
°F
55.6
°C
Overall Shell-Side Calculation

1
U 
( t s − t1) 1 − o 
2
U
ox 

1.
t f = ts −
2.
For fluids other than hydrocarbons, determine vapor film properties. For hydrocarbons, determine only liquid
viscosity µf
3.
Viscosity, µf
=
Specific Heat, cf
=
Btu/lb-°F
kJ/kg-K
Thermal Conductivity, kf
=
Btu/hr-ft-°F
W/m-K
NTC = 1.19
= 1.10
cP
0.00035
Pa•s
Nt square layout
Nt triangular layout
4.
5.
Dt = (NTC – 1) Pt + do
D = Dt / 0.9 (Note 3)
6.
S=
7.
8.
0.35
D − (do NTC )
(Note 4)
FF
Assumed Pb
NB = FF (10L / Pb) (Note 4),
or
NB = FF (833 L / Pb) (Note 4) (Metric)
9. Ax = S (Pb – 0.375) (Segmental Baffles)
or
Ax = S (Pb – 9.53) (Metric)
Ax = 0.85S (Pb – 0.375) (Double Segmental Baffles)
or
Ax = 0.85S (Pb – 9.53) (Metric)
10. Total shell-side flow rate, W
=
23.6
=
=
23.3
26.0
in.
in.
590.6
656.2
mm
mm
=
8.3
in.
208.5
mm
=
=
13.375
in.
340
mm
107.9
in.2
68903
mm2
2.684
mm2
kg/s
=
=
=
=
=
=
23.5
in.2
21,300
lb/hr
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
11. Inlet vapor density and volume
ρ v1 =
Inlet MW
10.7
 Pi (psia)   1 

  
 T1 + 460   u 
=
0.256
lb/ft3
or
ρ v1 =
Inlet MW  Pi (kPa abs)   1 


 T + 273   u 
8.29
 1
 
(Metric)
Inlet vapor rate, W v1
=
Vv1 = W v1 / ρv1
12. Outlet vapor density and volume
ρv2 =
=
Outlet MW  Pi (psia)   1 


 T + 460   u 
10.7
 
 2
4.1
kg/m3
21,300
lb/hr
2.684
kg/s
=
83,200
ft3/hr
0.655
m3/s
=
0.204
lb/ft3
3.26
kg/m3
or
ρv2 =
Outet MW  Pi (kPa abs)   1 


 T + 273   u 
8.29
 2
 
Outlet vapor rate, W v2
Vv2 = W v2 / ρv2
13. Inlet liquid density and volume
(Metric)
=
=
=
580
lb/hr
0.0731
kg/s
2840
ft3/hr
0.0224
m3/s
ρl1
=
lb/ft3
kg/m3
Inlet liquid rate, Wl1
=
lb/hr
kg/s
Vl1 = Wl1 / ρl1
=
ft3/hr
m3/s
14. Inlet liquid density and volume
ρl 2
=
38.7
lb/ft3
619
kg/m3
Inlet liquid rate, Wl 2
=
20,720
lb/hr
2.61
kg/s
Vl 2 = Wl 2 / ρl 2
=
535
ft3/hr
0.0042
m3/s
=
0.491
lb/ft3
7.88
kg/m3
1.25
psi
15. Average mixture density
ρ=
2W
Vv1 + Vl1 + Vv 2 + Vl 2
16. Calculate shell-side pressure drop (∆Ps) using
=
procedures outlined in Section IX-D, Shell-Side Iteration
The above-assumed geometry will now be checked using the following detailed procedures.
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
C.
METRIC UNITS
Vapor Desuperheating Zone (Shell-Side Condensation)
1.
2.
t +t  U
t f =  s t  + o (Rio + rio + rw + ro ) ( t s − t t )
 2  2
=
°F
=
lb/ft3
°C
(ts and tt for this zone were calculated in Step I.A.)
Inlet vapor density and volume
ρ v1 =
Inlet MW
10.7
 Pi (psia)   1 

  
 T1 + 460   u 
or
ρv1 =
3.
Inlet MW  Pi (kPa abs)   1 

  
8.31  T1 + 273   u 
(Metric)
kg/m3
=
For fluids other than hydrocarbons, determine film properties. For hydrocarbons, determine only µf
Viscosity, µf
=
cP
Pa•s
Specific Heat, cf
=
Btu/lb-°F
kJ/kg-K
Thermal Conductivity, kf
=
Btu/hr-ft-°F
W/m-K
4.
Calculate shell-side coefficient (hgc) using
Section IX-D, Shell-Side iteration
=
Btu/hr-ft2-°F
W/m2-K
5.
Udh =
=
Btu/hr-ft2-°F
W/m2-K
=
ft2
m2
hgc Uox
hgc + Uox
(Uox was evaluated in Step II.A.21.)
6.
A dh =
qdh
∆t e Udh
(qdh and ∆te were evaluated in Step I.A.)
D.
Hydrocarbon Condensing - Steam Desuperheating Zone (Shell-Side Condensation)
1.
Assume a zone area, Ads
=
400
ft2
37.2
m2
2.
Uds assumed =
qds
A ds ∆t e
=
60.4
Btu/hr-ft2-°F
342.9
W/m2-K
(qds and ∆te for this zone were evaluated in Step I.B.)
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
3.
t f = ts −
 U 
1
( t s − t t ) 1 − ds 
2
 Uox 
=
168.4
=
1.71
0.46
METRIC UNITS
°F
°C
75.8
(ts and tt for this zone were evaluated in Step I.B.)
4.
∆Tzone inlet
=
∆Tzone outlet
Tdh − t dh
Tds − t ds
1.71
5.
Fv (Figure 4)
=
6.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone, W c)
+ (Vapor not condensed in zone, W v2)
= Wv
lb/hr
= 7550
= 4900
=12450
Average Mol. weight of vapor
=
74.6
lb / hr
mol / hr
=
0.25
lb/ft3
7.
0.46
mol/hr
71
96
167
kg/s
0.938
0.643
1.587
kmol/s
0.009
0.0119
0.0209
75.6
kg / s
kmol / hr
Average vapor properties at film temperature
a.
Density, ρ v =
Avg MW  P (psia)   1 


 t + 460   u 
10.7
 
 f
or
Density, ρ v =
b.
Avg MW
8.31
 P (kPa abs)   1 

   (Metric)
 t f + 273   u 
Viscosity, µf
(Used for hgc calc.)
=
4.0
kg/m3
=
0.0088
cP
0.88 x 10-5
Pa•s
8850
lb/hr
1.1
kg/s
8.
Wl = W − Wv
=
9.
Average condensate properties at film temp.
a. Specific gravity, sf
=
0.65
b.
=
0.077
Btu/hr-ft-°F
0.133
W/m-K
=
0.3
cP
0.0003
Pa•s
=
22.7
Thermal Conductivity, kf
c. Viscosity, µf
10. Condensing Coefficient (Horizontal Bundle)
a.
ns = 1.29 Nt 0.48 (90° Square Tube Layout)
ns = 1.02 Nt 0.519 (30° Triangular Layout)
(For other layouts, see text)
b.
Lc =
A ds
(L − 0.5 ) Ns
A
0.65
22.6
=
=
5.2
ft
or
Lc =
A ds
(L − 0.152) Ns
A
(Metric)
=
(Note that if zone occupies more than one shell,
Lc will be a multiple of the tube length.)
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
Wc
L c ns
c.
Γ=
d.
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 


Γ 1/ 3



 condensate
METRIC UNITS
=
139
lb/hr-ft
=
139
Btu/hr-ft2-°F
0.0568
kg/s-m
766.4
W/m2-K
1/ 3
or
1/ 3
 s2 
 204 
h′λ = 
 k f  f 
 µf 
 Γ1 / 3 

 condensate
e.
1
W  µ (v) 
= 1+ v  f

xl
Wl  µ f ( l ) 
f.
xv = 1− x l
g.
Gv =
0.111
 ρl 
 
ρ 
 v
(Metric)
=
0.555
=
17
17
=
0.941
0.941
 FJ 
Wv

 , (Note 4),
25 A x x v  FBT 
=
4.9
10 6 Wv  FJ 


A x x v  FBT 
=
lb/sec-ft2
or
Gv =
h.
G 
h λ = h′λ  v 
 5 
(Metric)
24.5
kg/sm2
768.1
W/m2-K
0.70
=
137
Btu/hr-ft2-°F
or
 G 
h λ = h′λ  v 
 24.4 
0.70
(Metric)
=
If hλ > 2 h′λ , use 2.0 h′λ
(For pure components, h λ = h′λ )
11. Condensing Coefficient (Vertical Bundles)
a.
Γ=
12 Wc
π Nt do
=
lb/hr-ft
or
Γ=
1000 Wc
π Nt do
(Metric)
=
kg/s-m
1/ 3
b.
 2
 1.04 x 10 4 
 k f  sf 
hλ = 
1
/
3


 µf 
Γ


 condensate
=
Btu/hr-ft2-°F
or
1/ 3
 s2 
 254 
hλ =  1 / 3  k f  f 
 µf 
Γ

 condensate
(Metric)
=
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
12. Vapor cooling coefficient
Calculate h′gc using procedures outlined in Section
IX-D, shell-side iteration (use W v as the flowrate)
h′gc
=
16.0
Btu/hr-ft2-°F
89.3
W/m2-K
=
22.6
Btu/hr-ft2-°F
126.4
W/m2-K

 Tdh + Tds

− tf  
 Wc 
2

  = 0.59
C1 = 1 − 
 Wv1 (Tdh − Tds ) 




C
1
1
=
+ 1
h gc h λ h ′gc
hgc
13. Liquid cooling coefficient (Horizontal Bundles)
a. Bottom Cooling Coefficient
Calculate h′lc using procedures outlined in Section
IX-D, shell-side iteration (use Wl as condensate flow rate)
h′lc
b.
Drip Cooling Coefficient
c.
hdc = 1.5hλ
Total liquid cooling coefficient (Horizontal)
Bundles), Liquid entering zone, Wl
h lc =
2 h dc h ′lc
h dc + h′lc
=
17.6
Btu/hr-ft2-°F
99.5
W/m2-K
=
=
205.5
Btu/hr-ft2-°F
lb/hr
1152.2
W/m2-K
kg/s
=
32.4
Btu/hr-ft2-°F
183.2
W/m2-K
or
h lc =
( Wl + Wc ) h dc h ′lc
Wc
W 

h′lc +  Wl + c  h dc
2
2 

=
Btu/hr-ft2-°F
W/m2-K
=
Btu/hr-ft2-°F
W/m2-K
= 2.34 x 106
= 0.506 x 106
Btu/hr
Btu/hr
679.3 x 103
154.5 x 103
W
W
= 0.292 x 106
Btu/hr
87.3 x 103
W
=
64.8
Btu/hr-ft2-°F
356.6
W/m2-K
=
51.3
Btu/hr-ft2-°F
284
W/m2-K
(For liquid + vapor entering zone)
14. Liquid cooling coefficient (Vertical Bundles)
hlc = hλ
15. Weighted hydrocarbon condensing zone coefficient
a. Zone condensing duty, qλ
b. Zone vapor cooling duty, qgc
c.
Zone liquid cooling duty, qlc
hds =
Uds =
qds
qλ qgc qlc
+
+
hλ hgc hlc
hds Uox
hds + Uox
(Uox was evaluated in Step II.A.21.)
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
16. Zone Area
A ds =
qds
Uds ∆t e( ds )
=
471
ft2
44.9
m2
=
47.1
Btu/hr-ft2-°F
265.5
W/m2-K
=
111
°F
44.1
°C
=
6.53
0.36
(qds and ∆te for this zone were evaluated in Step I.B.)
E.
Steam Condensing Zone (Shell-Side Condensation)
qsc
( A − A ds ) ∆t e( sc )
1.
Usc =
2.
t f = ts −
 U 
1
( t s − t t ) 1 − sc 
2
 Uox 
(ts and tt for this zone were evaluated in Step I.C.)
3.
∆Tzone inlet
∆Tzone outlet
=
Tds − t ds
= _____________
T2 − t1
6.53
4.
Fv (Figure 4)
=
5.
Average zone vapor rate, W v
(Fv) (Vapor condensed in zone, W c)
+ Non Condensable
= Wv
lb/hr
= 1550
= 580
= 2135
Average Mol. weight of vapor
=
52.0
lb / hr
mol / hr
=
0.183
lb/ft3
6.
0.36
mol/hr
31
10
41
kg/s
0.196
0.073
0.269
kmol/s
0.0036
0.0013
0.0049
54.2
kg / s
kmol / hr
2.93
kg/m3
Average vapor properties at film temperature
a.
Density, ρ v =
Avg MW  P (psia)   1 

 
10.7  t f + 460   u 
or
ρv =
b.
7.
8.
Avg MW  P (kPa abs)   1 

 
8.31  t f + 273   u 
(Metric) =
Viscosity, µf
(Used for hgc calc.)
=
0.0068
cP
0.0000068
Pa•s
Wl = W – W v
=
19165
lb/hr
2.415
kg/s
Average condensate properties at film temperature
a. Specific gravity, sf
=
0.62
b.
=
0.078
Btu/hr-ft-°F
0.135
W/m-K
=
0.38
cP
0.00038
Pa•s
Average zone liquid rate, Wl
Thermal conductivity, kf
c. Viscosity, µf
(Note that hydrocarbon film properties should be used)
0.62
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
9.
METRIC UNITS
Condensing Coefficient (Horizontal Bundle)
ns = 1.29 Nt 0.48 (90° Square Tube Layout)
=
ns = 1.02 Nt 0.519 (30° Triangular Layout)
=
b.
(For other layouts, see text)
Asc = A – Adh – Ads
=
1083
ft2
c.
Lc =
=
14.08
ft
a.
A sc
(L − 0.5) Ns
A
22.7
22.6
94.1
m2
4.03
m
0.006
kg/s-m
1522.5
W/m2-K
or
Lc =
A sc
(L − 0.152) Ns
A
(Metric)
=
(Note that if zone occupies more than one shell,
Lc will be a multiple of the tube length.)
Wc
L c ns
d.
Γ=
e.
 s2 
 8.33 x 10 3 
 kf  f 
h′λ = 
 µf 


Γ 1/ 3



 condensate
=
13.6
lb/hr-ft
=
273.4
Btu/hr-ft2-°F
1/ 3
or
 s2 
 204 
h′λ = 
 kf  f 

 µf 
 Γ1 / 3 


f.
1
W  µ (v) 
= 1+ v  f

xl
Wl  µ f ( l ) 
g.
xv = 1− x l
h.
Gv =
1/ 3
(Metric)
=
condensate
0.111
 ρl 
 
ρ 
 v
0.555
 FJ 
Wv

 , (Note 4),
25 A x x v  FBT 
=
2.39
2.39
=
0.581
0.581
=
1.36
lb/sec-ft2
or
Gv =
i.
10 6 Wv  FJ 

 , (Note 4)
A x x v  FBT 
G 
h λ = h′λ  v 
 5 
(Metric)
=
0.72
kg/sm2
617.4
W/m2-K
0.7
=
110.3
Btu/hr-ft2-°F
or
 G 
h λ = h ′λ  v 
 24.4 
0.7
(Metric)
=
If hλ > 2 h′λ , use 2 h′λ
(For pure components, h λ = h′λ )
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
10. Condensing Coefficient (Vertical Bundles)
a.
Γ=
12 Wc
π Nt do
=
1000 Wc
π Nt do
=
lb/hr-ft
or
Γ=
kg/s-m
1/ 3
b.
 2
 1.04 x 10 4 
 k f  sf 
hλ = 


 µf 
Γ1/ 3


 condensate
Btu/hr-ft2-°F
=
or
1/ 3
 s2 
 254 
hλ =  1 / 3  k f  f 
 µf 
Γ

 condensate
(Metric)
W/m2-K
=
11. Vapor cooling coefficient
Calculate h′gc using procedures outlined in Section
IX-D, shell-side iteration (use W v as the flowrate)
h′gc
=
5.4
Btu/hr-ft2-°F
30.9
W/m2-K
=
6.7
Btu/hr-ft2-°F
38.1
W/m2-K

 Tdh + Tds

− tf  
 Wc 
2

  = 0.76
12. C1 = 1 − 
 Wv1 (Tdh − Tds ) 




C
1
1
=
+ 1 =
h gc
hλ
h′gc
hgc
13. Liquid cooling coefficient (Horizontal Bundles)
a. Bottom Cooling Coefficient
Calculate h′lc using procedures outlined in Section
IX-D, shell-side iteration (use Wl as condensate flow rate)
h′lc
b.
Drip Cooling Coefficient
c.
hdc = 1.5hλ
Total liquid cooling coefficient (Horizontal)
Bundles), Liquid entering zone, Wl
h lc =
( Wl + Wc ) h dc h′lc
Wc
W 

h′lc +  Wl + c  h dc
2
2 

=
28.0
Btu/hr-ft2-°F
160.2
W/m2-K
=
=
165.5
Btu/hr-ft2-°F
lb/hr
955
W/m2-K
kg/s
=
30.8
Btu/hr-ft2-°F
175.6
W/m2-K
14. Liquid cooling coefficient (Vertical Bundles)
hlc = hλ
=
Btu/hr-ft2-°F
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
METRIC UNITS
15. Weighted steam condensing coefficient
a.
Zone condensing duty, qλ
=
1.25 x 106
Btu/hr-ft2-°F
366 x 103
W/m2-K
0.108 x 106
Btu/hr-ft2-°F
30.4 x 103
W/m2-K
b.
Zone vapor cooling duty, qgc
=
c.
Zone liquid cooling duty, qlc
= 0.687 x 106
Btu/hr-ft2-°F
193.3 x 103
W/m2-K
=
41.2
Btu/hr-ft2-°F
236.9
W/m2-K
=
35.3
Btu/hr-ft2-°F
202.5
W/m2-K
=
1373
ft2
123.4
m2
hsc =
Usc =
qsc
qλ qgc qlc
+
+
hλ hgc hlc
hsc Uox
hsc + Uox
16. Zone Area
A sc =
F.
qsc
Usc ∆t e( sc )
(qsc and ∆te for this zone were evaluated in Step I.C.)
Total Area
ft2
=
A = Adh + Ads + Asc
If total area calculated is not reasonably close to total
area assumed, reiterate the various zone calculations.
G. Condensing Pressure Drop (Shell-Side Condensation)
1.
1.
m2
Hydrocarbon condensing zone inlet volumes
a. Vapor
Vapor MW  P (psia)   1 


 T + 460   u 
10.7
 dh
 
=
0.256
Vapor MW  P (kPa abs)   1 


 T + 273   u 
8.31
 dh
 
(Metric)
=
(1) ρ v1( dh ) =
lb/ft3
or
ρ v1( dh ) =
b.
4.1
kg/m3
(2) Vapor rate, W v1(dh)
=
21,300
lb/hr
2.684
kg/s
(3) Vv1(dh) = W v1(dh) / ρv1(dh)
=
83,200
ft3/hr
0.655
m3/s
Liquid (if present)
(1) ρl1( dh )
=
(2) Liquid rate, Wl1( dh )
=
(3)
=
Vl1( dh )
lb/ft3
0
lb/hr
ft3/hr
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TABLE 4 (Cont)
A SAMPLE CONDENSER DESIGN
CUSTOMARY UNITS
2.
METRIC UNITS
Steam condensing zone inlet volumes
a. Vapor
Vapor MW  P (psia)   1 


 T + 460   u 
10.7
 ds
 
=
0.171
Vapor MW  P (kPa abs)   1 


 T + 273   u 
8.31
 ds
 
(Metric)
=
(1) ρ v1( dh ) =
lb/ft3
or
ρ v1( dh ) =
=
4900
lb/hr
0.644
kg/s
(3) Vv1(ds) = W v1(ds) / ρv1(ds)
Liquid
=
28,700
ft3/hr
0.235
m3/s
(1) ρl1( ds )
=
40.5
lb/ft3
648
kg/m3
(2) Liquid rate, Wl1( ds )
=
16,400
lb/hr
2.04
kg/s
Vl1( ds ) = Wl1( ds ) / ρl1( ds )
=
405
ft3/hr
0.00315
m3/s
Steam condensing zone outlet volumes
a. Vapor (if present)
Vv2 (evaluated in Step II.B.12.)
=
2840
ft3/hr
0.0224
m3/s
Vl 2 (evaluated in Step II.B.14.)
=
535
ft3/hr
0.0042
m3/s
(3)
4.
Vapor desuperheating zone ∆Pdh
Calculate ∆Pdh using procedures outlined in
Section IX-D, shell-side iteration, ∆Pdh
5.
6.
7.
kg/m3
(2) Vapor rate, W v1(ds)
b.
3.
2.74
=
psi
kPa
Hydrocarbon condensing zone, ∆Pds
2W
Vv1( dh ) + Vl1( dh ) + Vv1( ds ) + Vl1( ds )
a.
ρds =
b.
Calculate ∆Pds using procedures outlined in
Section IX-D, shell-side iteration
=
0.38
lb/ft3
6.01
kg/m3
=
0.41
psi
2.9
kPa
=
1.31
lb/ft3
20.3
kg/m3
=
0.35
psi
2.5
kPa
=
0.76
psi
5.4
kPa
Steam condensing zone ∆Psc
2W
Vv1( ds ) + Vl1( ds ) + Vv 2 + Vl 2
a.
ρsc =
b.
Calculate ∆Psc using procedures outlined in
Section IX-D, shell-side iteration
Total condensing-side pressure drop
∆P = ∆Pdh + ∆Pds + ∆Psc
The area calculated in this sample (Item F) is not close enough to the assumed. Making the following modifications and
repeating the calculation gives an acceptable area: Nt = 496, NTC = 26.5, Dt = 26.3 in. (668 mm), D = 29.0 in. (737 mm), S = 9.1
in. (231 mm), Ax = 102.1 in.2 (65871 mm2), Pb = 11.6 in. (294.6 mm), and NB = 17.
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
DESIGN PRACTICES
Section
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
Page
IX-F
70 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
EXXON
ENGINEERING
FIGURE 1
TYPICAL HEAT RELEASE (“T-Q") CURVE
(FOR WIDE-CUT HYDROCARBONS IN THE PRESENCE OF STEAM)
Q, MW
0.2
420
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
200
380
T1
180
340
160
140
Vapor Cooling + HC
Tdh = HC Dewpoint
Condensation + Liquid Cooling
260
120
+ Steam Condensation
+ Water Cooling
220
100
Vapor
Tds = Steam
Cooling
180
Dewpoint
Vapor Cooling + HC
80
Condensation + Liquid
Cooling
140
100
60
t2
60
t1
tdh
T2
40
tds
0
1
qdh
DP9FF01
2
3
4
20
5
qds
6
qsc
Q, MM Btu/hr
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
6.3
Temperature, °C
Temperature, °F
300
DESIGN PRACTICES
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
EXXON
ENGINEERING
Section
Page
IX-F
71 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
FIGURE 2A
SAMPLE FLASH CURVE AND Tds PLOT
260
255
250
245
240
235
230
%
ht
eig
W
%
e
um
l
Vo
M
ol
e
220
215
220
210
Partial Pressure
of Steam
210
205
200
200
Vapor
Pressure
of H2O
195
190
180
Tds = 184 F
190
170
186
4
6
8
10
12
14
80
90
Pressure, psia
180
0
DP9FFO2A
10
20
30
40
50
60
70
Percent Not Condensed
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
100
Temperature, F
Temperature, F
%
225
DESIGN PRACTICES
Section
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
Page
IX-F
72 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
EXXON
ENGINEERING
FIGURE 2B
SAMPLE FLASH CURVE AND Tds PLOT
125
120
115
%
ht
eig
W
110
%
M
ol
e
105
120
Partial Pressure
of Steam
110
100
100
95
90
Vapor
Pressure
of H2O
90
80
Tds = 82.8 °C
85
70
20
40
60
80
100
120
Pressure, kPa abs
0
DP9FF02B
10
20
30
40
50
60
70
80
90
Percent Not Condensed
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
100
Temperature, C
Temperature, C
%
e
m
lu
Vo
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
EXXON
ENGINEERING
DESIGN PRACTICES
Section
Page
IX-F
PROPRIETARY INFORMATION - For Authorized Company Use Only
73 of 79
Date
December, 1999
FIGURE 3A
VAPOR PRESSURE OF HYDROCARBONS
1.0
1200
1100
2.0
1000
0
120
0
110
900
3.0
0
100
800
4.0
900
700
5.0
600
500
500
400
400
20
40
300
50
H2O
60
200
70
80
90
100
150
250
100
50
200
200
300
150
400
500
600
100
700
800
900
1000
1500
50
DP9FF03A
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
Vapor Pressure, psia
30
Temperature, °F
300
t, °F
oin
P
iling
l Bo
ma
r
o
N
700
600
350
6.0
7.0
8.0
9.0
10
800
DESIGN PRACTICES
Section
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
Page
IX-F
74 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
EXXON
ENGINEERING
FIGURE 3B
VAPOR PRESSURE OF HYDROCARBONS (METRIC)
650
10
600
650
600
500
20
500
30
400
400
40
300
200
150
Temperature, °C
200
60
70
80
90
100
150
200
150
300
100
400
500
600
50
100
10
1000
1500
2000
50
3000
4000
5000
6000
10,000
10
DP9FF03B
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
Vapor Pressure, kPa
300
°C
nt,
Poi
g
ilin
l Bo
ma
r
o
N
50
DESIGN PRACTICES
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
EXXON
ENGINEERING
Section
Page
IX-F
PROPRIETARY INFORMATION - For Authorized Company Use Only
75 of 79
Date
December, 1999
FIGURE 4
AVERAGE VAPOR QUANTITY IN CONDENSERS (OR REBOILERS)
0.7
0.6
Wv = (Vapor not condensed) + Fv (Vapor condensed)
Wv = Fv (Liquid vaporized) + (Initial vapor if any)
(Condensing)
(Vaporizing)
0.5
Fv
0.4
0.3
0.2
0.1
DP9FF04
0.2
0.3
0.5
0.8 1
2
Ratio of
3
5
Inlet ∆ T
T1 − t 2 *
=
Outlet ∆ T
T2 − t1
8 10
20
30
50
80
* Or Corresponding Zone Temperature
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
DESIGN PRACTICES
Section
CALCULATION PROCEDURE, CONDENSATION
Page
IX-F
HEAT EXCHANGE EQUIPMENT
76 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
EXXON
ENGINEERING
FIGURE 5
SHELL-SIDE HEAT TRANSFER FOR PARTIAL CONDENSATION OF WIDE CUTS
(NOT RECOMMENDED FOR DEFINITIVE DESIGNS)
In(hoc) = 0.82 In(G) + f(α ) - 0.002 Pi, or
In(hoc) = 0.82 In(G) + f(α ) - 0.0029 Pi + 0.436 (Metric)
4.0
3.5
f(α )
3.0
2.5
2.0
0.0
DP9FF05
0.1
0.2
0.3
0.4
α =
0.5
0.6
0.7
lb/hr (or kg/s) vapor out
lb/hr (or kg/s) vapor in
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
0.8
0.9
1.0
DESIGN PRACTICES
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
EXXON
ENGINEERING
Section
Page
IX-F
77 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
FIGURE 6A
FLASH CURVE CORRECTIONS
60
0
20
40
40
0
200
30
20
D3, °F
D2, °F
300
0
60
150
100
°F
50
50
0
250
10
0
-10
t 10
00
=7
1
+D
-20
50
-30
-40
0
50
100
150
200
0
250 300
50
100
150 200
250
300
D1, °F
D1, °F
DP9FF06A
FIGURE 6B
FLASH CURVE CORRECTIONS
140
30
120
1
+D
20
=
150
°C
95
0
20
t 10
0
26
100
10
0
32
D3, °C
D2, °C
80
0
0
37
60
-10
40
-20
20
0
DP9FF06B
20 40 60 80 100 120 140 160 180
D1, °C
0
20
40
60
80
D1, °C
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
100 120 140
DESIGN PRACTICES
Section
CALCULATION PROCEDURE, CONDENSATION
Page
IX-F
HEAT EXCHANGE EQUIPMENT
78 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
EXXON
ENGINEERING
FIGURE 7A
ENTHALPY OF WATER ABOVE 32°F
1600
0-1 Atm.
1400
Enthalpy (Btu/lb)
Sat. Vapor
1200
1000
Critical
Point
800
600
Sat. Liquid
400
200
0
100
200
300
400
500
600
700
800
900
1000
Temperature, F
DP9FF07A
FIGURE 7B
ENTHALPY OF WATER ABOVE 0°C
0-1 Atm.
Sat. Vapor
Enthalpy (kJ/kg)
3000
Critical
Point
2000
Sat. Liquid
1000
0
DP9FF07B
100
200
300
400
500
Temperature, C
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
600
DESIGN PRACTICES
HEAT EXCHANGE EQUIPMENT
CALCULATION PROCEDURE, CONDENSATION
EXXON
ENGINEERING
Section
Page
IX-F
79 of 79
Date
PROPRIETARY INFORMATION - For Authorized Company Use Only
December, 1999
FIGURE 8
HEAT TRANSFER COEFFICIENT FOR FLUIDS IN TUBES
107
5
106
5
yth
hio = 0 .012
5
10
12 L/di or 1000 L/di =
5
25
35
60
120
180
350
600
 µ 
y th

k Pr 1 / 3 

di
 µw 
or
 µ 
y

hio = th k Pr 1 / 3 

di
 µw 
0 .14
0.14
(Metric)
104
102
DP9FF08
5
103
5
104
Reynolds Number, Re =
5
105
d iG
29 µ
or
5
d iG
10 3 µ
106
(Metric)
EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.
5
107