the heating of liquids in tanks

7/27/2019 The Heating of Liquids in Tanks

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The heating of liquids in tanks is an important requirement in process industries such as the

dairy, metal treatment and textile industries. Water may need to be heated to provide a hot water

utility; alternatively, a liquid may need to be heated as part of the production process itself,whether or not a chemical reaction is involved. Such processes may include boiler feedtanks,

wash tanks, evaporators, boiling pans, coppers, calandrias and reboilers.

Tanks are often used for heating processes, of which there are two major categories:

Totally enclosed tanks, such as those used for storing fuel oil, and where heat load

calculations are generally straightforward.

Open topped tanks, where heat load calculations may be complicated by the introduction

of articles and materials, or by evaporative losses.

Open and closed tanks are used for a large number of process applications:

Boiler feedtanks - The boiler feedtank is at the heart of any steam generation system. It

provides a reservoir of returned condensate and treated make-up water, for feeding the

boiler. One reason for heating the water is to reduce oxygen entering the boiler, with

(theoretically) 0 ppm oxygen at 100C.

Boiler feedtanks are normally operated at between 80C and 90C.

Hot water tanks - Hot water is required for a number of processes in industry. It is often

heated in simple, open or closed tanks which use steam as the heating medium.

The operating temperature can be anywhere between 40C and 85C depending on the

application.

Degreasing tanks - Degreasing is the process where deposits of grease and cooling oil

are removed from metal surfaces, after machining and prior to the final assembly of theproduct.

In a degreasing tank, the material is dipped into a solution, which is heated by coils to a

temperature of between 90C and 95C.

Metal treatment tanks - Metal treatment tanks, which are sometimes called vats, are

used in a number of different processes:

o To remove scale or rust.

o To apply a metallic coating to surfaces.


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The treatment temperatures typically range from 70C to 85C.

Oil storage tanks - Storage tanks are required to hold oils which cannot be pumped at

ambient temperatures, such as heavy fuel oil for boilers. At ambient temperatures, heavy

oil is very thick and must be heated to 30C - 40C in order to reduce its viscosity andallow it to be pumped. This means that all heavy oil storage tanks need to be providedwith heating to facilitate pumping.

Heating tanks used in process industries - Heating tanks are used by a number of

process industries, see Table 2.9.1.

Tabl

e 2.9.1

Process industries which use heating tanks

In some applications the process fluid may have achieved its working temperature, and the onlyheat requirement may be due to losses from the solid surface of the walls and/or the losses from

the liquid surface.

This Tutorial will deal with the calculations which determine the energy requirements of tanks:

the following two Tutorials (2.10 and 2.11) will deal with how this energy may be provided.

When determining the heat requirement of a tank or vat of process fluid, the total heat

requirement may consist of some or all of a number of key components:

The heat required to raise the process fluid temperature from cold to its operating

temperature.

The heat required to raise the vessel material from cold to its operating temperature.

The heat lost from the solid surface of the vessel to the atmosphere.

The heat lost from the liquid surface exposed to the atmosphere.

The heat absorbed by any cold articles dipped into the process fluid.

However, in many applications only some of the above components will be significant. Forexample, in the case of a totally enclosed well-insulated bulk oil storage tank, the total heat

requirement may be made up almost entirely of the heat required to raise the temperature of the

fluid.


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Items 1 and 2, the energy required to raise the temperature of the liquid and the vessel material,

and item 5, the heat absorbed by any cold articles dipped into the process fluid, can be found by

using the Equation 2.6.1. Generally, data can be accurately defined, and hence the calculation ofthe heat requirement is straightforward and precise.

Equation 2.6.1

Items 3 and 4, the heat losses from the vessel and liquid surfaces can be determined by usingEquation 2.5.3.

However, heat loss calculations are much more complex, and usually empirical data, or tables

based on several assumptions have to be relied upon. It follows that heat loss calculations areless accurate.

Equation 2.5.3

Heat loss from the solid surface of the vessel to the atmosphere

Heat will only be transferred provided there is a difference in temperature between the surface

and the ambient air.

Figure 2.9.1 provides some typical overall heat transfer coefficients for heat transfer from bare

steel flat surfaces to ambient air. If the bottom of the tank is not exposed to ambient air, but is

positioned flat on the ground, it is usual to consider this component of the heat loss to be

negligible, and it may safely be ignored.

For 25 mm of insulation, the U value should be multiplied by a factor of 0.2.

For 50 mm of insulation, the U value should be multiplied by a factor of 0.1.

The overall heat transfer coefficients provided in Figure 2.9.1 are for 'still air' conditions only.


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Fig. 2.9.1

Typical overall heat transfer coefficients

from flat steel surfaces

Table 2.9.2 shows multiplication factors which need to be applied to these values if an air

velocity is being taken into account. However, if the surface is well insulated, the air velocity is

not likely to increase the heat loss by more than 10% even in exposed conditions.

Table2.9.2

Effect on heat transfer with air movement

Velocities of less than 1 m/s can be considered as sheltered conditions, whilst 5 m/s may be

thought of as a gentle breeze (about 3 on the Beaufort scale), 10 m/s a fresh breeze (Beaufort 5),and 16 m/s a moderate gale (Beaufort 7).

For bulk oil storage tanks, the overall heat transfer coefficients quoted in Table 2.9.3 may be

used.

Table

2.9.3

Overall heat transfer coefficients for oil tanks

Water tanks: heat loss from the water surface to the atmosphere


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Figure 2.9.2 relates heat loss from a water surface to air velocity and surface temperature. In this

chart 'still' air is considered to have a velocity of 1 m/s, tanks in sheltered positions outdoors

consider velocities at about 4 m/s, whilst tanks in exposed positions outdoors are considered withvelocities at about 8 m/s.

This chart provides the heat loss in W/m rather than the units of the overall heat transfercoefficient of W/mC. This means that this value must be multiplied by the surface area to

provide a rate of heat transfer, as the water to air temperature difference has already been taken

into account.

Heat losses from the water surface, as shown in Figure 2.9.2 are not significantly affected by the

humidity of the air. The full range of humidities likely to be encountered in practice is covered

by the thickness of the curve. However, the graph considers heat losses with an air temperatureof 15.6C and 55% air humidity. Different conditions to these can be calculated from the

Calculators on this website.

To determine the heat loss from the chart, the water surface temperature must be selected fromthe top scale. A line should then be projected vertically downwards to the (bold) heat loss curve.

For indoor tanks a line should be projected horizontally from the intersection to the left-handscale.

For outdoor tanks a horizontal line should be projected either left or right until it intersects therequired location, either sheltered or exposed. A projection vertically downwards will then reveal

the heat loss on the bottom scale.

In most cases, the heat loss from the liquid surface is likely to be the most significant heat losselement. Where practical, heat loss can be limited by covering the liquid surface with a layer of

polystyrene spheres which provide an insulating 'blanket'. Any solution to reduce heat losses

becomes even more important when tanks are located outside in exposed positions as portrayedby the graph in Figure 2.9.2.
http://www.spiraxsarco.com/resources/calculators.asphttp://www.spiraxsarco.com/resources/calculators.asp


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Fig. 2.9.2

Heat loss from water surfaces

Example 2.9.1

For the tank shown in Figure 2.9.3, determine:

Part 1. The mean heat transfer rate required during start-up.

Part 2. The maximum heat transfer rate required during operation.


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Fig. 2.9.3

The tank is unlagged and open topped and is situated on a concrete floor inside a factory.

It is 3 m long by 3 m wide by 2 m high.

Tank total surface area = 24 m (excluding base).

Heat transfer coefficient from tank/air, U1 = 11 W/mC.

The tank is /3 full of a weak acid solution (cp = 3.9 kJ/kgC) which hasthe same density as water (1 000 kg/m)

The tank is fabricated from 15 mm mild steel plate.

(Density = 7 850 kg/m, cp = 0.5 kJ/kgC)

The tank is used on alternate days, when the solution needs to be raised from the lowest

considered ambient temperature of 8C to 60C in 2 hours, and remain at that

temperature during the day.

When the tank is up to temperature, a 500 kg steel article is to be dipped every 20

minutes without the tank overflowing. (cp = 0.5 kJ/kgC)

Part 1 Determine the mean heat transfer rate required during start-up M (start-up)

This is the sum of:

A1. Heating the liquid M (liquid)

A2. Heating the tank material M (tank)

A3. Heat losses from the sides of the tank M (sides)

A4. Heat losses from the liquid surface M (surface)

Part 1.1 Heating the liquid M (liquid)


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Equation 2.6.1

Equation 2.5.3

Where:

T is the mean temperature difference T M


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Part 2 Determine the running load, that is the maximum heat transfer rate required during

operation (operation)


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In operating conditions, the liquid and tank (A1 and A2, from Part 1) are already up to

operating temperature, so the heating components = 0.

In operating conditions, the heat losses from the liquid and tank (A3 and A4, from Part

1) will be greater. This is because of the greater difference between the liquid and tank

temperatures and the surroundings.

Immersing the article in the liquid is clearly the objective of the process, so this heat load

must be calculated and added to the running load heat losses.

Part 2.1 Heat losses from tank sides

Equation 2.5.3

Where:


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Equation 2.6.1

Note that the operational energy requirement (59 kW) is significantly less than the start-up

energy requirement (367 kW). This is typical, and, where possible, the start-up period may

be extended. This will have the effect of reducing the maximum energy flowrate and has

the benefits of levelling demand on the boiler, and making less demand on the temperature

control system.

For tanks that are to operate continuously, it is often only necessary to calculate the

operating requirements i.e. the Part 2 calculations.


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Vessels can be heated in a number of different ways. This tutorial will deal with indirect heating.In these systems, the heat is transferred across a heat transfer surface. Options include:

Submerged steam coils - A widely used form of heat transfer involves the installationinside a tank of a steam coil immersed in a process fluid.

Steam jackets - Steam circulates in the annular space between a jacket and the vessel

walls, and heat is transferred through the wall of the vessel.

Submerged steam coils

The use of tank coils is particularly common in marine applications where cargoes of crude oil,

edible oils, tallow and molasses are heated in deep tanks. Many of these liquids are difficult to

handle at ambient temperatures due to their viscosity. Steam heated coils are used to raise the

temperature of these liquids, lowering their viscosity so that they become easier to pump.

Tank coils are also extensively used in electroplating and metal treatment. Electroplating

involves passing articles through several process tanks so that metallic coatings can be depositedon to their surfaces. One of the first stages in this process is known as pickling, where materials

such as steel and copper are treated by dipping them in tanks of acid or caustic solution to

remove any scale or oxide (e.g. rust) which may have formed.

Steam coil sizing

Having determined the energy required (in Tutorial 2.9), and with knowledge of the steam

pressure/temperature in the coil, the heat transfer surface may be determined using Equation

2.5.3:

Equation 2.5.3

The heat transfer area calculated is equivalent to the surface area of the coil, and will enable anappropriate size and layout to be specified.

Determining the 'U' value

To calculate the heat transfer area, a value for the overall heat transfer coefficient, U, must be

chosen. This will vary considerably with the thermal and transport properties of both fluids and a

range of other conditions.

On the product side of the coil a thermal boundary layer will exist in which there is a temperature

gradient between the surface and the bulk fluid. If this temperature difference is relatively large,then the natural convective currents will be significant and the heat transfer coefficient will be

high.

Assisted circulation (such as stirring) that will induce forced convection, will also result in


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higher coefficients. As convection is partially dependent on the bulk motion of the fluid, the

viscosity (which varies with temperature) also has an important bearing on the thermal boundary

layer.

Additional variations can also occur on the steam side of the coil, especially with long lengths of

pipe. The coil inlet may have a high steam velocity and may be relatively free from water.However, further along the length of the coil the steam velocity may be lower, and the coil may

be running partially full of water. In very long coils, such as those sometimes found in seagoing

tankers or in large bulk storage tanks, a significant pressure drop occurs along the length of thecoil. To acheive the mean coil temperature, an average steam pressure of approximately 75% of

the inlet pressure may be used. In extreme cases the average pressure used may be as low as 40%

of the inlet pressure.

Another variable is the coil material itself. The thermal conductivity of the coil material may

vary considerably. However, overall heat transfer is governed to a large extent by the heat

resistant films, and the thermal conductivity of the coil material is not as significant as their

combined effect. Table 2.10.1 provides typical overall heat transfer coefficients for variousconditions of submerged steam coil application. 'U' values for steam pressures between 2 bar g

and 6 bar g should be found by interpolation of the data in the table.

The range of figures shown in Table 2.10.1 demonstrates the difficulty in providing definitive 'U'

values. Customary figures at the higher end of the scale will apply to installations that aresupplied with clean dry steam, small coils and good condensate drainage. The lower end is more

applicable to poor quality steam, long coils and poor condensate drainage.

The recommended overall heat transfer coefficients will apply to typical conditions and

installations. These recommended rates are empirically derived, and will generally ensure that a

generous safety margin applies to the coil sizing.

In the case of fluids other than water, the heat transfer coefficient will vary even more widely

due to the way in which viscosity varies with temperature. However, the values shown in Table2.10.2 will serve as a guide for some commonly encountered substances, while Table 2.10.3

gives typical surface areas of pipes per metre length.


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Example 2.10.1

Continuing from Example 2.9.1 determine:

Part 1. The average steam mass flowrate during start-up. (Mean heat load = 367 kW)

Part 2. The heat transfer area required.

Part 3. A recommended coil surface area.

Part 4. The maximum steam mass flowrate with the recommended heat transfer area.

Part 5. A recommendation for installation, including coil diameter and layout.

The following additional information has been provided:

Steam pressure onto the control valve = 2.6 bar g (3.6 bar a).

A stainless steel steam coil provides heat.

Heat transfer coefficient from steam/coil/liquid, U = 650 W/mC

Part 1 Calculate the average steam mass flowrate during start-up

Steam pressure onto the control valve = 2.6 bar g (3.6 bar a)

Critical pressure drop (CPD) will occur across the control valve during start-up, therefore the


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minimum steam pressure in the heating coil should be taken as 58% of upstream absolute

pressure. An explanation of this is given in Block 5.

Part 2 Calculate the heat transfer area required.

Part 3 A recommendation for coil surface area

Because of the difficulties in providing accurate 'U' values, and to allow for future fouling of the

heat exchange surface, it is usual to add 10% to the calculated heat transfer area.


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Part 4 The maximum steam mass flowrate with the recommended heat transfer area

Maximum heat transfer (and hence steam demand) will occur when the temperature differencebetween the steam and the process fluid is at its maximum, and should take into consideration

the extra pipe area allowed for fouling.

(a) Consider the maximum heating capacity of the coil (coil)

Using Equation 2.5.3: = UAT

(b) Steam flowrate to deliver 519 kW

Part 5 A recommendation for installation, including coil diameter and layout

(a) Determine coil diameter and length


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From Table 2.10.3, a 100 mm pipe has a surface area of 0.358 m/m run. This application will

require:

It may be difficult to accommodate this length of large bore heating pipe to install in a 3 m 3 m

tank.

One solution would be to run a bank of parallel pipes between steam and condensate manifolds,

set at different heights to encourage condensate to run to the lower (condensate) manifold. The

drain line must fall from the bottom of the condensate manifold down to the steam trap (orpump-trap). See Figure 2.10.1 for a suggested layout.


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Fig. 2.10.1

Possible layout of coils in a rectangular tank

Note the steam supply is situated at one end of its manifold, whilst the trap set is at the other end.

This will help steam to flow and push condensate through the coils.

In the application, the steam and condensate headers would each be 2.8 m long. As the

condensate manifold is holding condensate, the heat from it will be small compared to the steammanifold and this can be ignored in the calculation.

The steam manifold should be 100 mm diameter as determined by the previous velocitycalculation. This will provide a heating area of:

2.8 m x 0.358 m/m = 1.0 m

Consequently 7 m - 1 m = 6 m of heat transfer area is still required, and must be provided by

the connecting pipes.

Arbitrarily selecting 32 mm pipe as a good compromise between robustness and workability:

The lengths of the connecting pipes are 2.5 m.


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CHECK

It is necessary to confirm the steam velocity through the connecting tubes:

On the basis of proportionality of heat transfer area, the steam header will condense:

This leaves 86% of the 850 kg/h = 731 kg/h of steam which must pass through the 18 connecting

pipes and also into the lower (condensate) manifold.

Other steam coil layouts

The design and layout of the steam coil will depend on the process fluid being heated. When the

process fluid to be heated is a corrosive solution, it is normally recommended that the coil inletand outlet connections are taken over the lip of the tank, as it is not normally advisable to drill

through the corrosion resistant linings of the tank side. This will ensure that there are no weak

points in the tank lining, where there is a risk of leakage of corrosive liquids. In these cases thecoil itself may also be made of corrosion resistant material such as lead covered steel or copper,

or alloys such as titanium.

However, where there is no danger of corrosion, lifts over the tank structure should be avoided,

and the steam inlet and outlet connections may be taken through the tank side. The presence of

any lift will result in waterlogging of a proportion of the coil length, and possibly waterhammer,

noise and leaking pipework.

Steam heating coils should generally have a gradual fall from the inlet to the outlet to ensure that

condensate runs toward the outlet and does not collect in the bottom of the coil.

Where a lift is unavoidable, it should be designed to include a seal arrangement at the bottom of

the lift and a small bore dip pipe, as shown in Figure 2.10.2.


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Fig. 2.10.2Tank with a rising discharge pipe

The seal arrangement allows a small amount of condensate to collect to act as a water seal, andprevents the occurrence of steam locking. Without this seal, steam can pass over any condensate

collecting in the bottom of the pipe, and close the steam trap at the top of the riser.

The condensate level would then rise and form a temporary water seal, locking the steam

between the bottom of the riser and the steam trap. The steam trap remains closed until the

locked steam condenses, during which time the coil continues to waterlog.

When the locked steam condenses and the steam trap opens, a slug of water is discharged up theriser. As soon as the water seal is broken, steam will enter the rising pipe and close the trap,

while the broken column of water falls back to lie at the bottom of the heating coil.

The small bore dip pipe will only allow a very small volume of steam to become locked in the

riser. It enables the water column to be easily maintained without steam bubbling through it,ensuring there is a steady and continuous condensate flow to the outlet.

When the seal is ultimately broken, a smaller volume of water will return to the heating coil than

with an unrestricted large bore riser, but as the water seal arrangement requires a smaller volumeof condensate to form a water seal, it will immediately re-form.

If the process involves articles being dipped into the liquid, it may not be convenient to installthe coil at the bottom of the tank - it may be damaged by the objects being immersed in the

solution. Also, during certain processes, heavy deposits will settle at the bottom of the tank and

can quickly cover the heating surface, inhibiting heat transfer.

For these reasons side hung coils are often used in the electroplating industry. In such cases

serpentine or plate-type coils are arranged down the side of a tank, as shown in Figure 2.10.3.


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These coils should also have a fall to the bottom with a water seal and a small bore dip-pipe. This

arrangement has the advantage that it is often easier to install, and also easier to remove for

periodic cleaning if required.

Fig. 2.10.3

Side hung coils

If articles are to be dipped into the tank, it may not be possible to use any sort of agitator toinduce forced convection and prevent temperature gradients occurring throughout the tank.

Whether bottom or side coils are used, it is essential that they are arranged with adequate

coverage so that the heat is distributed evenly throughout the bulk of the liquid.

The diameter of the coil should provide sufficient length of coil for good distribution. A shortlength of coil with a large diameter may not provide adequate temperature distribution. Howevera very long continuous length of coil may experience a temperature gradient due to the pressure

drop from end to end, resulting in uneven heating of the liquid.

Whilst the next two headings, 'Sizing the control valve' and 'The condensate removal

device' are included in this Tutorial, the new reader should refer to later Tutorials for full

and comprehensive information, before attempting sizing and selection of equipment.

Control valve arrangement

The control valve set may be either one or two valves in parallel. A single control valve, largeenough to cope with the maximum flowrate encountered at start-up, may be unable to controlflow accurately at the minimum expected flowrate. This could cause erratic temperature control.

An alternative is to fit two temperature control valves in parallel:

One valve (running valve) sized to control at the lower flowrate.

A second valve (starting valve) to pass the difference between the capacity of the first

valve, and the maximum flowrate.


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The starting valve would have a set-point slightly lower than the running valve, so it would close

first, leaving the running valve to control at low loads.

Sizing the control valve

The control valve set (either one valve or two valves in parallel).

The coil has been sized on mean heat transfer values. However, it may be better to size the

control valve to supply the maximum (start-up) load. With large coils in tanks, this will help to

maintain a degree of steam pressure throughout the length of the coil when the steam is turnedon, helping to push condensate through the coil to the steam trapping device. If the control valve

were sized on mean values, steam pressure in the coil at start-up will tend to be lower and the

coil may flood.

Using one valve

Continuing with Example 2.10.1 the maximum steam load is 850 kg/h and the coil is designed todeliver this at a pressure of 1.1 bar g. A steam valve sizing chart would show that a Kv of about

20 is required to pass 850 kg/h of steam with a pressure of 2.6 bar g at the inlet of the control

valve, and Critical Pressure Drop (CPD) across the valve. (Tutorial 6.4 will show how the valvesize can be determined by calculation).

A DN40 control valve with a larger Kvs of 25 would therefore need to be selected for theapplication.

If one valve is to be used, this valve must ensure the maximum heat load is catered for, whilemaintaining the required steam pressure in the coil to assist the drainage of condensate from it at

start-up. However, for reasons previously explained, two valves may be better.

The running load is 52 kW and with the coil running at 1.1 bar g, the running steam load:

The steam valve sizing chart shows a Kv of 2 is required to pass 85 kg/h with 3.6 bar upstream,

operating at critical pressure drop.

A DN15 KE type valve (Kvs = 4) and a DN25 piston actuated valve (Kvs = 18.6) operating

together will cater for the start-up load. When approaching the control temperature, the larger

valve would be set to shut down, allowing the smaller valve to give good control.

The condensate removal device


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The selection and sizing of the condensate removal device will be very much influenced by the

condensate backpressure. For the purpose of this example, it is assumed the backpressure is

atmospheric pressure. The device should be sized so it is able to satisfy both of the followingconditions:

Pass 850 kg/h of condensate with 1.1 bar g in the coil, i.e. the full-load condition. Pass the condensate load when steam pressure in the coil equals the condensate

backpressure, i.e. the stall load condition.

If the steam trap is only sized on the first condition, it is possible that it may not pass the stallload (the condition where the product approaches its required temperature and the control valve

modulates to reduce steam pressure). The stall load may be considerable. With respect to non-

flow type applications such as tanks, this may not be too serious from a thermal viewpointbecause the contents of the tank will almost be at the required temperature, and have a huge

reservoir of heat.

Any reduction in heat transfer at this part of the heating process may therefore have littleimmediate effect on the tank contents.

However, condensate will back up into the coil and waterhammer will occur, along with itsassociated symptoms and mechanical stresses. Tank coils in large circular tanks tend to be of

robust construction, and are often able to withstand such stresses. Problems can however occur in

rectangular tanks (which tend to be smaller), where vibration in the coil will have more of aneffect on the tank structure. Here, the energy dissipated by the waterhammer causes vibration,

which can be detrimental to the life of the coil, the tank, and the steam trap, as well as creating

unpleasant noise.

With respect to flow-type applications such as plate heat exchangers, a failure to consider thestall condition will usually have serious implications. This is mainly due to the small volume inthe heat exchanger.

For heat exchangers, any unwanted reduction in the heating surface area, such as that caused by

condensate backing up into the steam space, can affect the flow of heat through the heatingsurface. This can cause the control system to become erratic and unstable, and processes

requiring stable or accurate control can suffer with poor performance.

If heat exchangers are oversized, sufficient heating surface may remain when condensate backs

up into the steam space, and reduction of thermal performance may not always occur. However,

with heat exchangers not designed to cope with the effects of waterlogging, this can lead tocorrosion of the heating surface, inevitably reducing the service life of the exchanger.

Waterlogging can, in some applications, be costly. Consider a waterlogging air heater frost coil.

Cold air at 4C flowing at 3 m/s can soon freeze condensate locked in the coils, resulting in

premature and unwarranted failure. Proper drainage of condensate is essential to maintain theservice life of any heat exchanger and air heater.

Steam traps are devices which modulate to allow varying amounts of condensate to drain from


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applications under varying conditions. Float traps are steam traps designed to modulate and

release condensate close to steam temperature, offering maximum plant performance, maximum

plant life, and maximum return on plant investment.

When stall conditions occur, and a steam trap cannot be used, an automatic pump-trap or pump

and trap in combination will ensure correct condensate drainage at all times, thus maximising thethermal capability and lifetime costs of the plant.

Steam jackets

The most commonly used type of steam jacket consists simply of an outer cylinder surrounding

the vessel, as shown in Figure 2.10.4. Steam circulates in the outer jacket, and condenses on the

wall of the vessel. Jacketed vessels may also be lagged, or may contain an internal air spacesurrounding the jacket. This is to ensure that as little steam as possible condenses on the outer

jacket wall, and that the heat is transferred inwards to the vessel.

Fig. 2.10.4A conventional jacketed vessel

The heat transfer area (the vessel wall surface area), can be calculated in the same manner as

with a steam coil, using Equation 2.5.3 and the overall heat transfer coefficients provided in

Table 2.10.4.

Although steam jackets may generally be less thermally efficient than submerged coils, due to

radiation losses to the surroundings, they do allow space for the vessels to be agitated so that heat

transfer is promoted. The U values listed in Table 2.10.4. are for moderate non-proximityagitation.

Commonly the vessel walls are made from stainless steel or glass lined carbon steel. The glass

lining will offer an additional corrosion resistant layer. The size of the steam jacket space will

depend on the size of the vessel, but typically the width may be between 50 mm and 300 mm.


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the heating of liquids in tanks

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