Use a by-pass strip if tubes are removed under a nozzle. Removing tubes leaves an open area where the shell fluid can flow either over or under the bundle.
If the bundle entrance area is equal to or greater than the inlet nozzle flow area, use a pressure loss coefficient of 1.0. If the bundle exit area is equal to or greater than the exit nozzle area, us a pressure loss coefficient of 0.58. There are indications that it should be larger.The following procedure is for the situation where the nozzle flow area is greater than the entrance or exit area and the bundles do not have an impingement plate. If there is an impingement plate, there will have to be added a turning loss to the calculation below.If the two shell side nozzles are not the same size, calculate the inlet pressure drop and take 2/3 of it and make a separate calculated pressure drop for the outlet and take 1/3 of it.
Shell Entrance or Exit Area
1. Calculate the bundle bypass area Sb = π x Dn x h
2. Calculate the slot area Aslot = 0.7854Dn2 (Pt -Dt)/(F2 x Pt)
3. Calculate the shell entrance and exit area.(As)
As = Sb + Aslot
(refer TEMA RGP-RCB-4.621 & 4.622)
4. Calculate ratio of Sb to total area FR = Sb/As
5. Kn = 0.65 +2.14 (FR -0.4)
(minimum Kn = 0.8, maximum = 1.8)
6. ΔPn = Kn x .000108Vs2 x density
(ΔPn = total of both nozzles)
ΔPn = Total nozzle pressure drop (lb/ft2)
Dn = Nozzle ID (in)
Ds = Shell ID (in)
Dt = Tube outside diameter (in)
F2 = 0.707 for 45 degree pitch, all others use 1.0
h = 0.5(Ds-OTL)(in)
Kn = Pressure loss coefficient
OTL= Outer tube limit diameter (in)
Pt = Tube center to center pitch (in)
Vs = velocity in the entrance/exit area (ft/sec)
990,000 lb/hr with a density of 62.4 lb/ft3 is flowing through a 13.25 in. ID nozzle. The shell ID is 23.25 in. and the OTL is 22.375 in. The tube OD is 0.75 in. on a tube pitch of 0.9375 in. with 30 degree layout.
h = 0.50(23.25-22.375)= 0.4375
Sb = π x 13.25 x 0.4375 = 18.23
Aslot = 0.7854(13.252) (0.9375-0.75)/(1.00 x .9375)= 27.58
Calculate total area As
As = Sb + Aslot = 18.23 + 27.58 = 45.81
FR = 18.23/45.81 = 0.4
Kn = 0.65 +2.14(0.4 -0.4) = 0.65 (use minimum 0.8)
Calculate nozzle pressure drop
Vs = (990000 x 0.04)/(45.81 x 62.4)= 13.85
ΔPn = 0.8 x 0.000108 x 13.852 x 62.4 = 1.03 psi
Comment - Using 1.5 total pressure loss coefficient and the nozzle flow area gives only 0.21 PSI
An estimate for maximum tube velocity inside steel tubes
Vmax = 80./sqrt(density)
where Vmax = maximum velocity (ft/sec)
density = lb/cu ft.
This type of service has steam condensing out from a non-condensible gas which is mostly CO2. The condensing curve has a hump which will give a LMTD higher than one calculated from a straight line condensing plot. An equation which makes a quick estimate for the LMTD is: Standard LMTD x Factor
For outlet process temperatures below 153.5 F.
Then LMTD Factor = 1.4 -0.0092(T -110)
Where T = outlet temperature and air inlet temperature is 100 F.
When a fluid has a high surface tension, the fluid doesn't readily flow from the gap between the fins. This lowers the heat transfer. The types of fluids that are to be avoided are those whose surface tension is above 30 to 40 dynes/cm. This includes such fluids as condensing steam, aqueous solutions with a high % of water, amines and glycols.
Has performance declined after the bundle has been pulled and later installed back in the shell? If the longitudinal long baffle is sealed on the sides with leaf seals, they are probably the problem. These thin flexible strips should be positioned so that they form a concave pattern and flex upward. Then when the shell fluid puts pressure on the leafs, they will press harder against the sides of the shell. If there is too much pressure or if the bundle is installed upside down, the leafs will flex downward and the shell fluid will bypass the bundle. Another possibility is that the leaf seals were damaged when the bundle was out of the shell.
One of the fluid by-pass streams that lowers the shell-side heat transfer is the stream that flows around the bundle. To evaluate, calculate a heat transfer varable named FSBP. It is the ratio of the bundle by-pass area to the crossflow area. The by-pass ratio for triangular and square tube pitch is normally:
FSBP =  Bs(Ds -OTL)
Where shell ID = inside diameter of shell (in)
Bs = distance between baffles (in)
Do = tube OD (in)
Ds = shell ID (in)
OTL = outer diameter of the tube bundle (in)
P = tube spacing (in)
If FSBP is more that 0.15, then seal bars are needed.
If the heat exchanger is counter-current flow, the steam with the highest factor as calculated below goes inside the tubes:
Factor = (flow)2 / density
You can also use this if the molecular weight and temperature are about the same on both sides:
Factor = (flow)2 / pressure
Density (#/cu. ft.)
Four thin(.oo8") stainless strips are normally used to seal the sides of the long baffle. Because of their flexibility they are not able to withstand large shell side pressure drops. It is best to limit the pressure drop to 5 psi with 7.5 psi being the maximum.
Use bare tubes if the bundle is quite small or the gas temperature is greater than 1350 to 1400 F.
Usually when shell-and-tube heat exchangers are designed, the tube layout is made so that the shell entrance area is approximately equal to the shell nozzle flow area. The average distance to the 1st tube row is Dn/4 where Dn is the inside diameter of the shell nozzle. In this case the pressure loss coefficient is 1.0 for the pressure drop calculation for the shell nozzle entrance.
If the shell nozzles are greater than 2" and tubes are not omitted from the tube layout, the nozzle entrance pressure drop can be significantly higher than the normal calculation based on the nozzle flow area. In a case of a 6" shell nozzle and no tubes were omitted in a BEM type heat exchanger, the pressure drop was 3 times higher than that calculated with just the nozzle flow area. For more information you can refer to the tip "Calculate Shell Nozzle Pressure Drop" on this page.
An equation from the Bureau of Standards Miscellaneous Publication No. 97 can be used when the Specific Gravity is greater than 0.67 or less than 0.93. It is:
Lat heat = (111 -.09T)/SG60
Lat heat = Latent Heat in Btu/lb
T = temperature in F.
SG60 = Specific gravity @ 60 F
For hdrocarbons below a Specific Gravity of 0.67 and pressures below 50 psia. Use
Lat heat = 172 -0.195T
1. If there is excess surface and the liquid is clean - lower the liquid level.
2. Add a mist eliminator
3. Add more vapor outlet nozzles.
For Reynolds numbers below 10,000 there is a L/D effect on the heat transfer coefficient inside tubing. If you use the full tube length for L, you may be too conservative. There will be turbulation at the tube entrance before laminar flow is fully developed. The turbulent length needs to be subtracted from the full tube length.
Use the following for tube sizes 1.0 inch or less.
Boil h = 22(Δt)1.25
Δt is tube wall temperature - liquid temperature
Where Boil h = heat transfer coefficient, Btu/(hr)(ft2)(OF)
t = temperature, OF
Cond h = 4.15W0.8
Where Cond h = condensing heat transfer coefficient, Btu/(hr)(ft2)(OF)
W = lbs/hr/tube
For good operation of a sulfur condenser the design velocities inside the tubes should be within certain limits. The velocity range is between 1.5 and 6.0 lb/sq ft-sec. Below this range there will be slugging. Above this range there is sulfur fogging.
What to ask the vendor if his quote is undersurfaced.
1. Are there seal strips? If so, how many?
2. What tube hole clearance was used in the baffles
The conventional way of determining the minimum shells in series when it is non-counterflow is to calculate 2 variables and use the TEMA curves. Then chose the chart with an LMTD correction factor that exceeds a value greater than 0.80. Another method is to use the temperature cross in the proposed operating temperatures. The amount of temperature cross determines the minimum number of shells in series. Temperature cross is the amount the cold side outlet temperature exceeds the hot side outlet temperature. The maximum temperature cross for shells in series are as follows:
1 Shell Tcross = 0 - 3 F.
2 Shells in series Tcross = Range 1.8+0.11Range 3 Shells in series Tcross = Range 1.49+0.0032RangeWhere:
Tcross = temperature cross (t2 - T2) = F.
Range = temperature drop of hot fluid =F.
The temperature range on the hot side is from 180 to 100F.
The cooling water enters at 85F. If the cooling water outlet temperature is 115F.
Can this be done with 2 shells in series?
Tcross = (180 -100)
Tcross = 29.9 F.
Since the actual temperature cross is 15 F. and the maximum is 29.9 F., the minimum number of shells in series is 2.
The shell side pressure drop calculation can be improved by better equations for the baffle window and the nozzle pressure drops. Both of these methods can be found elsewhere on this web page.
The baffle window pressure drop in the open literature is a function only of the number of tubes crossed and the velocity in the window. It does not take into account a friction factor, type of tube pattern or fluid eddies.
When there are no tubes removed under the shell nozzles and the nozzles are large, using the nozzle flow area can result in wrong pressure drop calculations.
This is taken from the first experimental case in "A Reappraisal of Shellside Flow in Heat Exchangers HTD-Vol. 36". Average flow of 990,000 lb/hr with a density of 62.4 lb/ft3 is flowing through a 13.25 ID nozzle. The shell ID is 23.25 in. and the OTL is 22.375 in. The effective tube length is 11.729 ft. The tube OD is 0.75 in. on a tube pitch of 0.9375 in. with 30 degree layout. There are 7 baffles and 26% baffle cut
From the following the cross flow pressure drop is calculated:
Bs = 17.6 in
fi = 0.1025 Ideal tube bank correlation ( J. Taborek)
Nc = 13.75
Rb = 0.536
Re = 40,249
Rl = 0.615
ΔPc = 6.41 psi
ΔPshell = ΔPc + ΔPw + ΔPn
From other tips: ΔPw = 12.78
ΔPn = 1.03
ΔPshell = 6.41 +12.78 +1.03 = 20.2 psi
Experimental = 20.3 psi
Excess surface = 100. x Aactual -Acalculated
Aactual = actual heat transfer surface
Acalulated = surface calculated from design overall heat transfer coefficient
To calculate overdesign surface use the clean overall heat transfer coefficient for Acalculated
For single phase liquids and no impingement plate
Minimum area = Flow(#/hr) x .04
For boilng liquids and no impingement plate
Minimum area = Flow(#/hr) x .04
Minimum area = minimum nozzle area at shell entry = sq. inches
ρ = density (lb/cu.ft.)
The minimum velocity for slurries inside tubes for shell-and-tube is 4 ft/sec. This is for a fine material like a catalyst. For slurries there is a special Reynolds number used for calculating the settling velocity. For more information on slurries refer to chapter C11 in the piping handbook.
After a heat exchanger goes into operation it may develope leaks in the tube walls. The following procedure calculates the new heat load and new overall heat transfer coefficient.
1. Using the actual overall heat transfer coefficient (U). calculate the heat transfer resistances that exclude the tubeside resistance
Rother = 1/U -1/hio
2. Calculate new hio and new surface using usable number of tubes
3. Calculate new U
Unew = 1/(1/hio + Rother)
4. Calculate new heat load from new surface and new U
Since the design of heat exchangers is a trial and error solution, a good starting point is desired. Usually the design starts with an estimated overall heat transfer coefficient. If you don't know a good starting value for this coefficient the equations presented here give this starting point with simple equations. In the design of heat exchangers using up the maximum allowable pressure drops gives the highest heat transfer for single phase fluids. The equations below estimates the tube velocity(W)for a gas that will meet the maximum allowable pressure drop. From W you can calculate the tube count or heat transfer coefficient. For a given tube length the following equation gives the optimum tube velocity for turbulent flow. Gases will be in turbulent flow more than 99% of the time. If your calculated tubeside velocity is below what the following equation calculates, you need more tube travel where tube travel is in the form of number of tube passes or total tube length(s) for countercurrent flow. These equations can be used for two phase flow as long as the two phase viscosity is less than 0.015 cp,
For 3/4 inch tubes with 0.06 tube wall
W = 1600(ΔPρ/L)0.555
For 1.0 inch tubes with 0.06 tube wall
W = 3500(ΔPρ/L)0.555
L = total tube lengths in ft.
(Add 8 x tube ID for turning losses for each tube pass)
W = lb/hr/tube
ΔP = allowable pressure drop inside tubes in psi (deduct 15% for nozzle pressure drops)
ρ = density in lb/cu.ft.
Use 3/4 inch tubes and 16 foot tubes. The maximum allowable pressure drop inside the tubes is 7 psi (after nozzle deduction) and the gas density is 2.66 lb/cu.ft. The tube side flow is 195,000 lb/hr. What should be the starting tube count?
W = 1600(7 x 2.66/(16+5))0.555
W = 1500 lb/hr/tube
Tube count = 195,000/1500 = 130
For a shell-and-tube heat exchanger there is a tip on this site that calculates the shell diameter when given the tube count. It has the description "Calculate S & T diameter from number of tubes".
A simple equation is presented for a kettle reboiler. It is conservative for very small bundles. The crital heat flux depends on the geometry of the bundle. The following estimate is based on 3/4 inch tubes on 15/16 inch pitch. It is actually good for any tube diameter with a tube pitch/tube diameter ratio of 1.25 and triangular tube pitch. A boiling temperature of -30 F. is assumed for the propane.
CHF = 32500
CHF = crital heat flux in Btu/(hr)(ft)2
Ds = shell bundle diameter in inches
What is the critical heat flux for a 41 inch diameter bundle?
CHF = 32500
CHF = 12,850
With a longitudinal baffle and a long temperature range there can be a problem with heat conduction through the longitudinal baffle. There will be a loss of thermal efficiency due to the heat conduction. The longitudinal baffle can be fabricated in one of two ways.
1. Leaving an small enclosed air gap between two longitudinal baffles.
2. Spray an insulating material like Ryton on the longitudinal baffle.
To convert from diameter cut to area cut
% area cut = -4.3 +0.816Dcut + 0.00563Dcut2
To convert from area cut to diameter cut
% diameter cut = 5.6 +1.06Acut -0.00367Acut2
Where baffle cuts are expressed as a percent
0.0005 steam,steam condensate,engine jacket water
0.0010 boiler feed water
0.0015 clean water,moutain water,etc.
0.0020 normal cooling tower water
For cooling water when velocity is 3 -8 ft/sec
Fouling = 0.025/V1.67
Where V =ft/sec
0.0010 If sp. gravity At 60F less than 0.80, lube oil and heating oils
0.0020 If sp. gravity At 60F 0.80 -0.87
0.0030 If sp. gravity At 60F 0.87 -1.00
0.0050 Heavy fuel oils
For kettle reboilers use segmental baffles instead of full supports if shell fouling factor is greater Than 0.002(hr-ft2-F/Btu)
Use the higher of the shell-side and tube-side design temperatures up to 650 F. At higher design temperatures use the arithmetic average of the 2 design temperatures.
High viscosity fluids can have a problem achieving the design heat transfer. The fluids are usually petroleum based and have an API of 20 or less.
Low pressure drops can cause maldistribution of the tubeside flow which in turn reduces the heat transfer. That is why you can see allowable pressure drops 2 or 3 times higher than usual. There is a method by A.C. Mueller for calculating this minimum allowable pressure drop. Another thing that can help is to use more tube passes and shorter tubes than normal. Also the fluid could be placed in the shell side if cleanig isn't a problem.
Water in the shell-side
Use the arithmetic average of the shell-side and tube-side design temperatures.
Water in the tube-side
Use the higher of the tube-side design temperature or tube-side outlet temperature + 1/3 of the LMTD.
At low operating pressures there will be a sensible heat liquid zone with relatively low heat transfer. This is caused by the fact that a small pressure change will cause a large increase in the boiling point. There has been a case where 90% of the tube length was in the sub-cooled phase. What can you change that will decrease the size of the liquid preheat zone and increase the overall heat transfer?
One answer is to evaluate the piping system above the top tubesheet. In order to make an evaluation check the pressure drop at the outlet. There is on this section of the website equations to calculate the pressure drop of a nozzle that is at right angle to the top channel. Most vertical thermosyphons have the outlet nozzle at right angles to the top channel. There may be a simple change of enlarging the outlet nozzle that would be the cure. But there needs to be a check to make sure the nozzle and connecting piping are not so large that there is liquid slip. If enlarging the right angle nozzle and piping is not the answer then there are other configerations that will use less outlet pressure drop. Next the pressure drop of using a B type channel with a long radius ell could be tried. If this doesn't do it, try a mitered channel design.
Another solution to the problem is to investigate inserts such as swisted tape, wire matrix , or helically coiled.
A rule of thumb is that the pressure drop at the outlet nozzle should not be greater than 30% of the total static head. There is another tip in this boiling section about choking the flow with a small outlet nozzle. The inside flow area of the outlet nozzle should be the same or greater than the total flow area insde the tubing. For a channel with a side outlet the pressure drop is composed of a turning loss and a contraction loss The following equations calculate the pressure drop at the outlet. The pressure drop for expansion into the channel is not included here but is with the tube pressure drop.
Ktr = ___1______
(If Ktr less than 0.40, use 0.40)
Kc = 0.5(1 - (No/Ds)2)
KT = Ktr + Kc
ΔPn = KT = 0.000108 x Vn2 x ρtp
Ds = Top channel ID (inches)
Ktr = pressure loss coefficient for turning loss
Kc = pressure loss coefficient for contraction into nozzle
KT = total pressure loss coefficient
No = Outlet nozzle ID (inches)
Vn = velocity thru nozzle (ft/sec)
ρtp = two-phase density (lb/ft3)
ΔPn = pressure drop thru channel and outlet nozzle (Psi)
Its difficult to estimate a gas heat transfer coefficient in the shell side because of the many variables. The following will give you a value within 25%.
Ho = 430.Cp(ΔP/L x ρ)1/3
Cp = specific heat (Btu/lb-F)
L = tube length (ft)
ΔP = shell side pressure drop (Psi)
(subtract nozzle losses)
In designing a shell-and-tube heat exchanger, use a 30o triangular tube pitch if possible. This will lower the vortex shedding frequency which is a direct function of something called a Strouhal number. The Strouhal number is a constant composed of the vortex shedding frequency, shell side velocity and tube OD.
The 30o triangular tube pitch has a significantly lower Strouhal number than other tube pitch types. Using Barrington as a source, for 3/4 inch tubes on 30o triangular tube pitch the Strouhal number is 0.21. But for 60o rotated triangular tube pitch the Strouhal number is 0.81.
Too small of a tubeside inlet nozzle can cause maldistribution of the fluid into the tubes and cause lower heat transfer. This will cause the tubeside fluid to jet into a relatively small amount of tubes. This lowers the flow to a majority of the tubes. To improve the flow distribution you could install a larger nozzle, an enlarger or a distribution plate.
To calculate the amount of surface in the u-bends would be difficult if you had to calculate the surface of each tube row and then add up all the rows for a total. The following is an equation that gives the u-bend surface. It is based on the typical bend radius of 1.5 x tube OD
U-bend area = (Count1.44 Do P0.78) /73.5
Count = total number of tubes
If the boiling temperature is too low there is not full boiling. The following give the boiling temperature difference where full boiling decreases to partial boiling. This temperature difference depends upon the nature of the fluid being nucleated.
Water Δ T = 8.0F.
Lt. HC Δ T = 5.0F.
(lowfins Δ T = 1.8F.)
Gas turbine exhaust provides an economic solution to waste heat recovery. One system consists of vertical rows of fin tubes with a steam drum on top. The following is the typical number of tube rows to start the calculation of a gas turbine heat recovery system:
Economizer 3 - 4 rows
Boiler 8 - 10 rows
Superheater 2 - 3 rows