HRSG Design

1

Scope and Definition of HRSGs

Scope of HRSGs to be discussed.

The Heat Recovery Steam Generator, or HRSG, comes in numerous shapes, designs, configurations, arrangements, etc. To simplify our discussion herein, we will first state that the type of HRSG we are reviewing is what may be referred to as a water tube(as opposed to a fire tube) type heat recovery unit. This refers to the process fluid, i.e., the steam or water being on the inside of the tube with the products of combustion being on the outside of the tube. The products of combustion are normally at or close to atmospheric pressure, therefore, the shell side is generally not considered to be a pressure vessel.

Definition of concepts and terminology used in discussions.

In the design of an HRSG, the first step normally is to perform a theoretical heat balance which will give us the relationship between the tube side and shell side process. Before we can compute this heat balance, we must decide the tube side components which will make up our HRSG unit. Even though these components may include other heat exchange services, at this time we will only consider the three primary coil types that may be present, i.e., Evaporator, Superheater, Economizer. When we refer to an Evaporator Section, this includes all the evaporator coils making up the total evaporator for a Pressure System. A pressure system includes all the components included in the various streams associated with that pressure level.

Evaporator Section

The most important component would, of course, be the Evaporator Section, since without this coil(or coils), the unit would not be an HRSG. Throughout our discussion, we will refer to a main heat transfer component as a "section".

2

When the section is broken into more than one segment, i.e., such as for a change in tube size, material, extended surface, location, etc., we will refer to the segments as coils. So an evaporator section may consist of one or more coils. In these coils, the effluent(water), passing through the tubes is heated to the saturation point for the pressure it is flowing.

Superheater Section

The Superheater Section of the HRSG is used to dry the saturated vapor being separated in the steam drum. In some units it may only be heated to little above the saturation point where in other units it may be superheated to a significant temperature for additional energy storage. The Superheater Section is normally located in the hotter gas stream, in front of the evaporator.

Economizer Section

The Economizer Section, sometimes called a preheater or preheat coil, is used to preheat the feedwater being introduced to the system to replace the steam(vapor) being removed from the system via the superheater or steam outlet and the water loss through blowdown. It is normally located in the colder gas downstream of the evaporator. Since the evaporator inlet and outlet temperatures are both close to the saturation temperature for the system pressure, the amount of heat that may be removed from the flue gas is limited due to the approach to the evaporator, known as the pinch which is discussed later, whereas the economizer inlet temperature is low, allowing the flue gas temperature to be taken lower.

3

Types and Configurations of HRSGs The evaporator section type is very important since it generally defines the overall configuration of the HRSG unit. For this discussion, we will use the word "type" to refer to the general configuration of the evaporator. Even though there are many types, or configurations of HRSGs, we will define five general types for our discussion.

D-Frame evaporator layout

O-Frame evaporator layout

A-Frame evaporator layout

I-Frame evaporator layout

Horizontal tube evaporator

layout

Superheater configurations

Economizer configurations

D-Frame evaporator layout. This configuration is very popular for HRSG units recovering heat from small gas turbines and diesel engines. It is a very compact design and can be shipped totally assembled. It is limited, however, since the bent tube arrangement quickly causes the module to exceed shipping limitations for units having a large gas flow.

4

O-Frame evaporator layout. This configuration has probably been used for more years than any of the others. It has the advantage of the upper header being configured as the steam separation drum. Or, the upper header can be connected to the steam drum by risers, allowing more than one O-Frame evaporator to be connected to the same steam drum, resulting in shipable modules being able to handle very large gas flows.

A-Frame evaporator layout. This configuration is simply a variation of the O-Frame Evaporator. It was popular for services with a large amount of ash, since the center area between the lower drums could be configured as a hopper to collect and remove solid particles.

5

I-Frame evaporator layout. In the past twenty years, this configuration has become the most popular of all the Evaporator designs. This type module can be built in multiple axial modules or in multiple lateral modules, allowing it to be designed to accept any gas flow. There are numerous variations of this design where tube bundles may contain one, two, or three rows of tubes per header. It is also, normally, more economical to manufacture, ship and field construct. The tube bundles may be shipped to field installed in the modules, or as loose bundles which are installed into a field erected shell.

Horizontal tube evaporator layout. The horizontal tube evaporator is used, not only for heat recovery from Gas Turbine exhaust, but for recovery from flue gases in Refinery and Petrochemical furnaces also. It has similar size limitations due to shipping restrictions similar to the O-frame modules. It is generally a less expensive unit to manufacture than the other configurations, but if it is a natural circulation design with large tubes, such as in some CO Boilers, or very long tubes, special consideration needs to be given to assure all tubes are provided with sufficient effluent. These considerations will be discussed later on in this document.

6

Superheater configurations.

Superheater designs would normally follow along with the evaporator type that is being used. Three basic superheater designs are shown below, Horizontal Tube, Vertical Tube, and I-Frame. The Horizontal Tube design would normally be used for the D-Frame Evaporator if gas flow is vertical up at the outlet. This horizontal design would be expected to be used also on a horizontal evaporator design. The Vertical Tube design would generally be used with the A-Frame or O-Frame Evaporator and with the D-Frame if the gas exits horizontally. The I-Frame Superheater would be used with the I-Frame Evaporator, but may also be used with the other evaporator designs.

Economizer configurations.

Economizer designs would normally follow along with the evaporator type that is being used and be similar in design to the superheater. The configurations would be similar to the ones shown above for the superheaters.

7

Preparing a flow schematic for the HRSG

After deciding on the evaporator type to be used for the unit, the next important step in the design of an HRSG unit is to decide the arrangement of the various coils in the unit. Of course, if only an evaporator is present, this may consist of a very simple schematic, but if, as in most cases, there are more than one coil, then consideration needs to be given as to their position in the gas stream.

Arrangement of coils.

Obviously, the best place to put the highest temperature coil, the superheater, would be in the hottest part of the gas stream. Since, this is where it would take the least amount of surface to exchange the heat, and would allow a stepped heat recovery for maximum heat exchange. The curve below shows this relationship between the heat given up and the three primary coils found in an HRSG.

In viewing this generalized sketch showing the relationship between the heat absorbed and the heat given up, it is easy to see the area referred to as the "pinch" at the evaporator outlet. By laying a straight edge on the heat given up line and rotating it while holding it at the pinch, it is also, easy to see that, at a very high inlet temperature, there may be a critical approach temperature occur at the economizer inlet, and going the other way, at a lower inlet temperature, this may occur at the superheater outlet.

Of course, modern HRSG units are not always this simple. The components can and are placed in many configurations to achieve desired results. The range of arrangements that the coils may be placed, is only limited by the users imagination and the constraints of the temperature approaches. Shown below are just a few examples of various arrangements.

8

Typical single pressure arrangements.

Typical dual pressure arrangements.

9

Typical triple pressure arrangements.

Preparing the schematic.

Now that we have a general idea of how to arrange the coils, we prepare the flow schematic. This flow schematic gives us a preliminary picture of how the HRSG will look. Also, we can use the sketch to perform the preliminary heat balance which we will review in Section 4.

For our example flow schematic, we will assume a single pressure HRSG with a superheater and economizer section.

10

It is not important that you necessarily use this style schematic, but it is important to be consistent in the style you use. If you always present an evaporator in the same way, and a superheater always looks like "your" superheater, the flow schematics become very recognizable to anyone needing to refer to them. Remember, the flow schematic does not need to represent the actual mechanical design of the HRSG, neither in looks, or direction of flow, hot to cold, etc.

Now, using a similar approach to above, let's construct a flow schematic to represent a triple pressure unit with an integral deaerator.

11

Heat Balance Evaporator Pinch Design:

The evaporator pinch, or approach temperature, is what limits the amount of heat that can be recovered in most HRSG designs. As was discussed in the previous section, Schematics, the limiting effect of this approach is important. For many general purpose HRSG's such as those found in refineries and chemical plants, a pinch of 50 °F provides an economical design with a realistic payout. But in the more competetive markets of combined cycle or co-generation plants, it is not uncommon to see pinch points below 30 °F. And as a practice, a 30 °F pinch design for these HRSG's should be considered.

It should, however, be remembered that the closer the pinch, or approach, the less reliable the results will be. In other words, it would be easy to calculate the steam generated in a unit at a 5 °F pinch, but the probability of achieving this result with the actual equipment would be almost nil. If you look at the added amount of surface required to go from a 10 °F to a 5 °F pinch versus the change in surface to go from 50 °F to 45 °F, you will quickly see the why this is true.

Other process approach temperatures:

Other process appoach temperatures are similar to the special case of the "pinch" discussed above. But, they do not, except in some situations, control the overall design of the HRSG. A 50 °F approach is a good minimum to consider for coils such as hot oil, superheaters, economizers, etc. Of course, the same is true with these coils, the higher the approach temperature, the less surface it will take to exchange the heat. This is why most of the flow in these coils are counter current to the gas flow, which provides a higher approach temperature.

Economizer water approach:

The economizer water approach temperature to the evaporator satuaration temperature is very important and should be selected with care. If too close an approach is used in the design, vaporization may occur in the coil during off design cases which may cause severe upsets in the unit. It should be noted, however, that just because the economizer does vaporize at some operating condition, it does not necessarily mean a problem, since the design can be such that it can handle this condition. But, for most designs, it is better to avoid this condition. A normal design approach temperature is 20 °F. This approach gives significant safety factor for load swings. But, again, you should rate HRSG at all expected operating conditions.

Superheated Steam Desuperheating:

Superheat desuperheating is the best way to control the outlet temperature of the HRSG superheater. It is not, though, the only way. Steam bypassing around all or part of the superheating coil and then remixing it to control the temperature is done with great success. If a spray desuperheater is used, it can be placed at the outlet, or at an itermediate point in the superheater coil. Placing

12

it at an intermediate point gives the added protection of preventing accidental water slugs which may damage downstream equipment.

Blowdown requirements:

The boiler blowdown requirement is set by the condition of the feedwater. Primarily it is used to control solids build up in the steam separation drum. If nothing is known of the feedwater at time HRSG is being designed, an allowance should be used in design. For normal modern facilities, a 2% allowance should be sufficient. For others, a 5% allowance should be provided for in the design. But, you should keep in mind that somewhere along the route from design to production, this must be revisited to assure proper operating conditions in the HRSG.

Developing the Heat Balance for an HRSG:

We begin with the first sample schematic that we prepared in Section 3, a single pressure HRSG with a superheater, evaporator, and economizer.

For our process conditions, we will assume the following: Gas Side : 800,000 lbs/hr of Gas Turbine Exhaust at 980 °F Setting Loss To Atmosphere, 2% of Heat Absorbed Maximum Back Pressure at Gas Turbine Exhaust Flange, 8" H2O

Gas Properties : Volume %

Nitrogen, N2 72.55

Oxygen, O2 12.34

Carbon Dioxide, CO2 3.72

Water, H2O 10.52

Argon, Ar 0.87

13

Sulphur Dioxide, SO2 0.0

Carbon Monoxide, CO 0.0

Tube Side : Steam at outlet, Maximum Flow at 600 psig and 750 °F Feedwater at 227 °F and pressure required at inlet.

For our example, we will make the following assumtions :

Pinch At Evaporator, °F 50.0

Economizer Water Approach, °F 20.0

Blowdown, % of Steam Out 2.0

Pressure Drop In Superheater, psi 15.0

Pressure Drop In Economizer, psi 10.0

Now, we have set all of our conditions, so we can proceed with a heat balance. For these calculations, we will need a calculator to provide us with the properties of the flue gas, water, and steam. We can start these in separate windows so we can keep them available as we work out our solution.

We can now populate our schematic with all known values.

Now we can calculate the missing data. Heat Available To Evaporator And Superheater:

Havail = Wg (hin - hpinch = 800000 (244.735 - 124.836) = 95,919,200 Btu/hr Resulting in a net heat available of

Hnet = Havail / (1 + SL/100) = 95919200 / (1 + 2/100) = 94,038,431 Btu/hr

14

Heat Required By Steam Flow (To Pinch Point): Hreqd = Ws (hs - hl) + (Ws + Ws * Bldwn/100) ( hl - hecon)

But, since Hnet is equal to Hreqd, we can restate the equation as,

Ws = Hnet / [ (hs - hl) + (1 + Bldwn/100) ( hl - hecon)] = 94038431 / [(1379.598-477.876) + (1 + 2/100) (477.876 - 454.662)]

= 101,619 lb/hr Now that we have the steam flow at the Superheater, 101,619 lb/hr, we can calculate the Superheater heat required, QSH

QSH = Ws (hs - hv) = 101619 (1379.598 - 1203.188) = 17,926,608 Btu/hr And the gas enthalpy at the outlet of the superheater coil, hg2

hg2 = hg1 - QSH * (1 + SL/100) / Wg = 244.735 - (17926608*1.02/800000) = 221.878 Btu/lb

Which results in a gas temperature leaving the superheater of 898.134 °F. The evaporator duty, QEvap, is equal to,

QEvap = Ws * (hv - hl) + Ws (1 + Bldwn/100) (hl - hecon) = 101619 (1203.188 - 477.876) + 101619 (1.02) (477.876 - 454.662) =

76,111,643 Btu/hr and the steam generated in the evaporator coil, Wevap, is equal to,

Wevap = QEvap / (hv - hl) = 76111643 / (1203.188 - 477.876) = 104,936 lbs/hr Now, we can calculate the Economizer duty, QEcon, as equal to,

QEcon = Ws (1 + Bldwn/100) (hecon - hbfw) = 101619 (1.02) (454.662 - 196.644) = 26,743,922 Btu/hr

And the gas enthalpy at the outlet of the economizer coil, hg4

hg4 = hg3 - QEcon * (1 + SL/100) / Wg = 124.836 - (26743922*1.02/800000) = 90.737 Btu/lb

Which results in a stack gas temperature leaving the economizer of 412.522 °F.

We can now complete our schematic with all known values.

15

Heat Transfer

Tube material and selection

Selecting the tube material and size to use in a HRSG design is really a matter of experience. As you work with different HRSG's for different services, you develop a knowledge of what fit before in a similar design, so you know where to start with a new design. But a few general rules can be used to start the selection.

For the typical, general purpose HRSG, using standard tubing sizes, the 2" tube size will normally work out to be the most economical tube size. The cost will generally go up with a smaller or larger tube size. Most HRSG units recover heat from a relatively low temperature gas, i.e., less than 1,000 °F. Of course, many of the modern HRSG's are supplementary fired to achieve even greater efficiencies. But, with the exception of the superheater, you can normally assume that carbon steel tubes will work for the evaporator and the economizer. If the superheater outlet temperatures are low, such as 600 °F and below, you should be able to assume carbon steel tubes to start. If higher than 600 °F, you may want to start with T11 tubes.

In a similar manner, you can make some preliminary estimates to determine what the design metal temperature for the HRSG tubes need to be. With this temperature, you would select the least material that is good for the temperature. Eventual analysis may show that a higher alloy and a thinner wall may be more economical, so running calculations with several materials is always wise.

Typical generic, pipe and tube specifications used for HRSG tubes:

Generic Specification Pipe Specification Tube Specification

Carbon Steel SA 106 Gr B SA 178 A

1¼ Cr ½ Mo SA 335 Gr P11 SA 213 T11

2¼ Cr 1 Mo SA 335 Gr P22 SA 213 T22

5 Cr ½ Mo SA 335 Gr P5 SA 213 T5

9 Cr 1 Mo SA 335 Gr P9 SA 213 T9

18 Cr 8 Ni SA 312 TP 304 SA 213 TP 304

16 Cr 12 Ni 2 Mo SA 312 TP 316 SA 213 TP 316

18 Cr 10 Ni Ti SA 312 TP 321 SA 213 TP 321

18 Cr 10 Ni Ti SA 312 TP 321H SA 213 TP 321H

And other, more exotic materials for special services are used as may be needed. The wall thickness required, for the heat absorbing tubes, is calculated by using the ASME Section 1. For heat absorbing tubes, there are two formulas that may apply, so it is normal practice to check the required thickness and

16

maximum allowed working pressure, MAWP, using both formulas, then using the more appropriate.

Tube Wall Thickness: Using ASME, Section 1, PG 27.2.1

t = (P * D) / (2 * S1 + P) + 0.005 * D + e And using ASME, Section 1, PG 27.2.1

t = (P * D) / (2 * S2 *E + 2*y*P) + C Where,

t = Minimum required thickness, in

P = Maximum allowable working pressure, psia

D = Outside diameter of cylinder, in

S1 = Maximum allowable stress value (PG-23), psi

S2 = Maximum allowable stress value (PG-23), psi

e = Thickness factor for expanded tube ends

y = Temperature coefficient

E = efficiency

C = Minumum allowance for threading and structural stability, in

When using tubes with the OD of a standard pipe size or using piping specifications and the HRSG uses returns, you would normally select a standard return bend to return the flow to the next tube. These returns bends are normally manufactured in two turning radii, called "short radius" and "long radius". The short radius return refers to a 180° return bend using a radius of one nominal diameter, ie, a 4" pipe size return has a radius of 4", and a 6" has a radius of 6", etc. The "long" radius bend has a radius equal to 1.5 nominal diameters so a 4" return has a radius of 6" and a 6" has a radius of 9". These standard returns are manufactured in most of the pipe schedules and are also available in "minimum wall" specifications.

Typical generic and pipe specifications used for return bends:

Generic Specification Pipe Specification

Carbon Steel SA 234 WPB

1¼ Cr ½ Mo SA 234 WP11

2¼ Cr 1 Mo SA 234 WP22

5 Cr ½ Mo SA 234 WP5

9 Cr 1 Mo SA 234 WP9

18 Cr 8 Ni SA 403 WP304

16 Cr 12 Ni 2 Mo SA 403 WP316

18 Cr 10 Ni Ti SA 403 WP321

18 Cr 10 Ni Ti SA 403 WP321H

17

If using standard pipe fittings manufactured to standard pipe schedules, you would assume 80% of the standard wall thickness. If you are bending tubes or pipe for the application, you would need to calculate the thinning in the bend. The following calculator estimates the ratio of that thinning.

Tube Length Selection

Now that we have selected a tube diameter, material, wall thickness, and tube spacing, we need to decide what length the tubes should be. Pipe and tubes are manufactured in random lengths, ie, since the billet size varies, the actual length of the tube that is extruded, from a billet, varies from one tube to the next. For lower cost materials, it is usually cheaper to scrap pieces of tube, then it is to make center welds to try and use all the material. But another high cost factor involved with the length is the supports and guides for the tube in the HRSG.

For vertical tubes, usually the overall HRSG shape and size dictate the best tube length. It is necessary to consider the maximaum shipping width and length in setting the tube length. The support and guide requirement varies depending on whether the tubes are supported from the top(hung) or bottom of the tubes.

In the horizontal tube HRSG's, the overall shape and size also figure into the equation. But, within these constraints, the span between supports must be considered. If the user has not specified a maximum span, then generally you would not want to exceed 35 tube OD's. This has been a general industry "not to exceed" rule of thumb used for many HRSG designs. But care should be taken to consider the service and wall temperature of the tubes. Once you have determined the span between supports, the tube length would be selected to use the minimum number of supports, while avoiding unnecessary centerwelds, if centerwelds are allowed by user. All of this must be balanced with the fact that the pressure loss in the tubes is increased dramatically in the returns, so generally you want the longest straight tube possible. The pressure loss in the returns is reviewed in the "Process" section , under "Intube Pressure Drop".

Now, using the single pressure HRSG that we demonstrated in developing the heat balance in Section 4, we can select the heat transfer tubes necessary to proceed with the thermal design. We are going to select an HRSG unit using 2.000" od tubes, so the following selections will be based on this tube od.

18

For the superheater, we can assume that the tube wall temperature will be above 750 °F and less than 850 °F, so we will use the 850°F for the design temperature. We will select for our HRSG design, the O-Frame evaporator. So for the superheater we could use a Vertical Tube or an I-Frame type. We will choose the I-Frame.

This design can use either bent tubes, i.e., three rows per bundle, or two straight tubes in a bundle. For this sample, we decide we want three rows per bundle using an 8" pipe header, then we we will decide that the bend radius of the bend in the first and third rows is 6". Using a design pressure of 700 psia, we can check to see if a standard 0.120" minimum wall thickness will be sufficient for this design. Using our bend thinning calculator, we see that the wall thickness after bending will be 0.1013". Then using the tube wall thickness calculator, we see that using the PG 27.2.1 method, that with our 1/32" corrosion allowance, the tube thickness is okay.

Now, we look at our evaporator, the O-Frame design that we selected above.

19

Okay, here again we have bent tubes. We will assume that the bend radius of these tubes is 12", so the wall thickness using 0.120" to start with, will be 0.1091" after bending. Using SA 178 Gr A tubes, we check the required wall thickness and find that this tube will be fine. Note that setting the design temperature below 700° F really doesn't have any affect on the PG 27.3.1 calculation, since this formula requires a minimum of 700 °F for the stress value selection. When we check the required wall, we get 0.1102 if we use a 1/32" corrosion allowance. but since no corrosion allowance was indicated, we will use this tube.

For our economizer, we will use the I-Frame design, but use two tubes per bundle with 6" pipe headers, so we will once again need bent tubes, using a 6" bend radius. This results in a 0.1013" wall thickness at the bends. For this economizer, the 0.120' wall thickness will be okay without corrosion allowance.

Extended surface material and selection

The heat transfer sections of the HRSG frequently use extended surface to improve the overall heat exchange between the hot gases and the steam or water in the tubes. These extended surfaces are usually either a thin plate fin wraped helically around the tube or round or eliptical shaped studs. Following is a description of the more popular extended surfaces.

Segmented Fins: These are usually one of the two types shown below.

High Frequency Continuously Welded

Standard Frequency Spot Welded

The standard frequency, spot welded, design is not used as often since this design is normally selected when using very thin, high density finning such as in a large heat recovery boiler. Most HRSG designs use fins less than 0.049 inch thick. The standard frequency, spot welded fin also has a foot which presents a place where corrosion can occur if flue gases are corrosive or moisture is present.

20

Solid Fins: These are the most popular fins for modern HRSG's.

High Frequency Continuously Welded

Stud Fins: These are used generally when the fuel is No. 6 or higher.

Resistance Welded

Thermal rating procedures for all these extended surface types are presented in the following pages. Both segmented fin types are rated using the same formulas.

For our sample boiler, that we have been developing throughout this material, we will choose to use 0.049" thick fins with a density of 6 fins per inch. We will use an 11% chrome alloy material for the superheater, A-176 TP409. For the evaporator and the economizer, we will use carbon steel, A 366, which should be good for all the temperatures in these sections. We will recheck our selections after doing the thermal calculations to confirm they are okay. It is important to note, that since the source of our gas is a gas turbine, it is a very clean service and we could have used a thinner fin at a higher density.

Indirect, non-luminous, radiation

In-direct, non-luminous radiation in an HRSG design is only significant to consider when the flue gases are above 1,000 °F. This normally would occur only following supplementary firing of the flue gas. It should be noted that there may also be some direct radiation occur in an HRSG if the duct burners are located directly in front of the HRSG heat transfer sections. For most designs, however, simply considering the radiation using the following method will suffice. If the HRSG is designed with an evaporator shield section, i. e., there is an evaporator coil placed between the burner and a superheater coil, there usually is no problem with tube wall temperature due to indirect, non-luminous, radiation. But if there is no shield coil, the superheater coil needs to be checked carefully to make sure that maximum tube wall temperature is not exceeded.

21

In an HRSG similar to this one, the superheater tubes are directly exposed to the hot gases and flame in the combustion chamber. They are also exposed to the heat radiated off the refractory lined combustion chamber in front of the tubes. To calculate the heat transfered to these tubes by radiation, we will use the methods described below.

qn = σαAcpF(Tg4 - Tw

4) Where,

qn = Non-luminous radiant heat transfer to the tubes, Btu/hr

σ= Stefan-Boltzman constant, 0.173E-8 Btu/ft2-hr-R4

α = Relative effectiveness factor of the tube bank

Acp = Cold plane area of the tube bank, ft2

F = Exchange factor

Tg = Effective gas temperature in firebox, °R

Tw = Average tube wall temperature, °R

Relative Effectiveness Factor, α :

For a single row in front of a refractory wall, use Total One Row. For two rows in front of a refractory wall, use Total Two Rows. For double sided firing, use Direct One Row.

22

Since all the radiant heat directed toward this bank of tubes is absorbed by the tubes in the convection, the relative absorption effectiveness factor, , for the tubes can be taken to be equal to one.

Cold Plane Area, Acp :

The cold plane area for the tube section is equal to the cold plane area of the first row of tubes.

Acp = Ntube*Stube*Ltube Where, Ntube = Number of tubes wide Stube = Tube spacing, ft Ltube = Tube length, ft Exchange Factor, F :

Because the flue gas in the firebox is a poor radiator, the equation must be corrected using an exchange factor which is dependent on the emissivity of the gas and the ratio of refractory area to cold plane area. Since the tubes themselves are not perfect absorbers, the curves are based on a tube-surface absorptivity of 0.9. This is a value considered typical for oxidized metal surfaces. The overall radiant exchange factor, F, can be taken from the curve below as presented by Mekler & Fairall in Petroleum Refiner, June 1952.

Where, Aw/αAcp :

The equivalent cold plane area,αAcp, is the product of the effectiveness factor and the cold plane area as described above. The Aw can be described as follows,

23

Aw = Ar - Acp

and,

Aw = Effective refractory area, ft2

Ar = Total refractory area, ft2

αAcp = Equivalent cold plane area, ft2

The Ar factor is the inside area of the plenum in front of (or below, in some designs) the tubes. The openings where the flue gas enters are normally ignored, since the ducting connecting them perform the same reflective purpose.

Flue Gas Emissivity :

The gas emissivity can be described by the curve presented by Lobo and Evans, at AICHE, 32nd Annual Meeting, November 1939. The tube wall temperature has only a minor effect. Therefore, the emissivity can be correlated as a function of PL product and the gas temperature, Tg. Variations in tube wall temperatures between 600 and 1200°F cause less than 1% deviation from these curves.

And,

PL = Product of the Partial Pressure of the carbon dioxide and water times the Beam Length, in atm-ft.

Where, Partial Pressure Of CO2 & H2O : The only constituents normally in the flue gas that contribute significantly to the radiant emission are the carbon dioxide and the water, the sum of these are all that are considered. The Partial pressure of a gas component in atm's is the mole volume fraction percent of that component.

24

Mean Beam Length : In computing the mean beam length, placement of the tubes must be taken into account. If the firebox is a rectangular shape with the tubes down the center, the beam length would be based on half the box. Beam lengths for other configurations, such as a cylindrical heater with an octagonal tube or cross tube layout, must be calculated with consideration for those cavities. The mean beam length for heaters can be accounted for according to Wimpress in Hydrocarbon Processing, October 1963, as follows:

For Box Type Heaters

Dimension Ratio Mean Beam Length

1-1-1 to 1-1-3 1-2-1 to 1-2-4

2/3(Furnace Volume)1/3

1-1-4 to 1-1-inf 1 x Smallest Dimension

1-2-5 to 1-2-inf 1.3 x Smallest Dimension

1-3-3 to 1-inf-inf 1.8 x Smallest Dimension

With the box dimensions, length, width, and height being in any order

For Vertical Cylindrical Heaters

Length/Diameter < 2 (((L/D)-1)*0.33 + 0.67)*D

Length/Diameter >= 2 Diameter

For most HRSG designs, the beam length can be taken as the width of the duct or superheter section.

Effective gas temperature in firebox, Tg

This temperature is the calculated gas temperature entering the duct or plenum in front of the tube coil.

Average tube wall temperature, Tw

The tube wall temperature calculation is discussed later in this section.

The radiant heat from this calculation will reduce the gas temperature used in the convection transfer calculation.

25

Indirect, non-luminous, radiation

To demonstrate this routine, we will assume that the gas temperature after firing is 1400 °F, and the tube wall temperature is 800 °F. We will assume that the gas prior to firing is the gas turbine exhaust we used in our example for calculating the heat balance. Gas Side : 800,000 lbs/hr of Gas Turbine Exhaust at 980 °F


Nitrogen, N2 72.55

Oxygen, O2 12.34


Water, H2O 10.52

Argon, Ar 0.87

Sulphur Dioxide, SO2 0.0

Carbon Monoxide, CO 0.0

To obtain the flue gas properties, we must do a combustion calculation. For this, we will assume fuel gas that is defaulted in the calculator below. The combustion process is covered in detail in Section 6, so for this exercise, we will just use a combustion calculator, such as the one below. Notice, that instead of firing with air as the source of oxygen, we use the exhaust gas for the oxygen source. For this case, we are not adding any additional air.

Gas Properties After Firing: Volume %

Nitrogen, N2 71.8121

Oxygen, O2 10.1589


Water, H2O 12.4283

Argon, Ar 0.861

The flue gas flow from our combustion calculation above is 805,012 lbs/hr.

Now we can perform the calculations as follows, From above, we determine α is equal to 1. And, Ar = 8 * 14 * 2 + 4 * 14 * 2 + 8 * 14.6 * 2 = 569.6 ft2 Acp = 23 * 16 = 368 ft2 Aw = 569.6 - 368 = 201.6 ft2 So, Aw/αAcp = 201.6 / 368 = 0.548

26

Partial Pressure of H2O and CO2 = 0.047397 + 0.124283 = 0.17168 atm Beam length = width = 8 ft PL = BL * PP = 8 * 0.17168 = 1.37344

Emissivity from curve = 0.342 So, Exchange Factor, F = 0.421 Now, using our radiant heat equation,

qr = σαAcpF(Tg4 - Tw

4) = 0.173E-8 * 1.0 * 368 * 0.421 (18604 - 12604)

= 2,523,400 Btu/hr

We now need to calculate the theoretical exit temp, then using the average of the 1400 and the exit temperature, recalculate the heat given up again, so it requires a few iterations to get a valid answer.

Enthalpy of flue gas after burner = 369.880 Btu/lb So, Enthalpy after radiant heat release = 369.880 - 2523400/805012 = 366.734 Btu/lb Which results in an exit temperature of 1389.448 °F And the corrected temperature for the calculation is (1400 + 1389.448)/2 = 1394.724 °F

To do this iteration in an efficient manner, we can set up a JavaScript to let our browser do this tedious work.

But before we do finalize our heat balance, we need to look at the distribution of this heat to the tubes. Since most of this radiant heat is absorbed by the first two rows of tubes, we will assume that only the first two rows are absorbing this heat. In actuality, some of the radiant heat would go to the third row, how much would depend on the tube diameter and spacing. But the amount to the third row is insignificant and can simply be added to the second row which will be more conservative. For the distribution to the first row of tubes, we can use the alpha, from the curves above, for direct to one row.

27

Convection Heat Transfer

The HRSG primarily absorbs heat from the hot flue gases by convection heat transfer. In the convection section, heat is transferred by both radiation and convection. The convection transfer coefficients for fin and stud tubes are explored here as well as bare tube transfer. The short beam radiation is treated separately from the convection transfer below. This section of the HRSG Design is divided into five main areas:

Convection Transfer, Bare Tubes Convection Transfer, Fin Tubes

Convection Transfer, Stud Tubes

Short Beam, Reflective Radiation Convection Section Design

Convection Transfer, Bare Tubes

Overall Heat Transfer Coefficient, Uo:

Uo = 1/Rto Where,

Uo = Overall heat transfer coefficient, Btu/hr-ft2-F

Rto = Total outside thermal resistance, hr-ft2-F/Btu

And,

Rto = Ro + Rwo + Rio

Ro = Outside thermal resistance, hr-ft2-F/Btu

Rwo = Tube wall thermal resistance, hr-ft2-F/Btu

Rio = Inside thermal resistance, hr-ft2-F/Btu

And the resistances are computed as,

Ro = 1/he

Rwo = (tw/12*kw)(Ao/Aw)

Rio = ((1/hi)+Rfi)(Ao/Ai)

Where,

he = Effective outside heat transfer coefficient, Btu/hr-ft2-F

hi = Inside film heat transfer coefficient, Btu/hr-ft2-F

tw = Tubewall thickness, in

kw = Tube wall thermal conductivity, Btu/hr-ft-F

Ao = Outside tube surface area, ft2/ft

Aw = Mean area of tube wall, ft2/ft

Ai = Inside tube surface area, ft2/ft

Rfi = Inside fouling resistance, hr-ft2-F/Btu

Inside film heat transfer coefficient, hi: The inside heat transfer coefficient calculation procedure is covered in detail, elsewhere in this course. Effective outside heat transfer coefficient, he

he = 1/(1/(hc+hr)+Rfo)

28

Where,

hc = Outside heat transfer coefficient, Btu/hr-ft2-F

hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F

Rfo = Outside fouling resistance, hr-ft2-F/Btu

Outside film heat transfer coefficient, hc: The bare tube heat transfer film coefficient, hc, can be described by the following equations. For a staggered tube arrangement,

hc = 0.33*kb(12/do)((cp*µb)/kb)

1/3((do/12)(Gn/µb)))

0.6

And for an inline tube arrangement,

hc = 0.26*kb(12/do)((cp*µb)/kb)

1/3((do/12)(Gn/µb)))

0.6

Where,

hc = Convection heat transfer coefficient, Btu/hr-ft2-F

do = Tube outside diameter, in

kb = Gas thermal conductivity, Btu/hr-ft-F

cp = Gas heat capacity, Btu/lb-F

µb = Gas dynamic viscosity, lb/hr-ft

Gn = Mass velocity of gas, lb/hr-ft2

We can use this typical superheater coil as a sample bare tube bank. We will assume the first two rows are bare:

Process Conditions: Gas flow, lb/hr = 800,000 Gas temperature in, °F = 980 Gas temperature out, °F = 968.9 Compostion, moles N2, % = 72.55 O2, % = 12.34 CO2, % = 3.72 H2O, % = 10.52 Ar, % = 0.87 Mechanical Conditions: Tube Diameter, in = 2.000 Tube Spacing, in = 4 Number Tubes Wide = 28 Tube Effective Length, ft = 28.000 Number Of Tubes = 56 Tube Arrangement = Inline Pitch

29

kb, Btu/hr-ft-F = 0.0324 cp, Btu/lb-F = 0.2805 µb, cp = 0.0359 = 0.0359*2.42 = 0.0869 lb/hr-ft To calculate the mass velocity, Gn, we need to first calculate the net free area of the tube bank. For these calculations, we are going to assume the tube rows are corbelled, so the net free area, NFA: NFA = Nwide*tspc/12*tlgth-Nwide*tOd/12*tlgth = 28*4/12*28-28*2/12*28 = 130.667 ft2 Therefore, Gn = Wgas / NFA = 800000/130.667 = 6122.448 And using our formula for hc,

hc = 0.26*0.0324 (12/2)((0.2805*0.0869 )/0.0324)1/3((2/12)(6122.448/0.0869

)))0.6 = 12.7152

Convection Transfer, Fin Tubes

You will notice that the heat transfer equations for the fin tubes are basically the same as for the bare tubes untill you reach the he factor, where a new concept is introduced to account for the fin or extended surface. The procedure presented herein are taken from the Escoa manual which can be downloaded in full from the internet.

Overall Heat Transfer Coefficient, Uo:

Uo = 1/Rto Where,



And,






Ro = 1/he

Rwo = (tw/12*kw)(Ao/Aw)


Where,




30


Ao = Total outside surface area, ft2/ft




Inside film heat transfer coefficient, hi: The inside heat transfer coefficient calculation procedure is covered in detail, elsewhere in this course. Effective outside heat transfer coefficient, he:

he = ho(E*Afo+Apo)/Ao Where,

ho = Average outside heat transfer coefficient, Btu/hr-ft2-F

E = Fin efficiency


Afo = Fin outside surface area, ft2/ft

Apo = Outside tube surface area, ft2/ft

And, Average outside heat transfer coefficient, ho:

ho = 1/(1/(hc+hr)+Rfo) Where,

hc = Outside heat transfer coefficient, Btu/hr-ft2-F

hr = Outside radiation heat transfer coefficient, Btu/hr-ft2-F

Rfo = Outside fouling resistance, hr-ft2-F/Btu

Outside film heat transfer coefficient, hc:

hc = j*Gn*cp(kb/(cp*b))0.67

Where,

j = Colburn heat transfer factor

Gn = Mass velocity based on net free area, lb/hr-ft2

cp = Heat capacity, Btu/lb-F

kb = Gas thermal conductivity, Btu/hr-ft-F

b = Gas dynamic viscosity, lb/hr-ft

Colburn heat transfer factor, j:

j = C1*C3*C5(df/do)0.5

((Tb+460)/(Ts+460))0.25

Where,

C1 = Reynolds number correction

C3 = Geometry correction

C5 = Non-equilateral & row correction

df = Outside diameter of fin, in

31

do = Outside diameter of tube, in

Tb = Average gas temperature, F

Ts = Average fin temperature, F

Reynolds number correction, C1:

C1 = 0.25*Re-0.35

Where,

Re = Reynolds number

Geometry correction, C3: For segmented fin tubes arranged in,

a staggered pattern,

C3 = 0.55+0.45*e(-0.35*lf/Sf)

an inline pattern,

C3 = 0.35+0.50*e(-0.35*lf/Sf)

For solid fin tubes arranged in,


C3 = 0.35+0.65*e(-0.25*lf/Sf)

an inline pattern,

C3 = 0.20+0.65*e(-0.25*lf/Sf)

Where,

lf = Fin height, in

sf = Fin spacing, in

Non-equilateral & row correction, C5: For fin tubes arranged in,


C5 = 0.7+(0.70-0.8*e(-0.15*Nr^2)

)*e(-1.0*Pl/Pt)

an inline pattern,

C5 = 1.1-(0.75-1.5*e(-0.70*Nr))*e(-2.0*Pl/Pt)

Where,

Nr = Number of tube rows

Pl = Longitudinal tube pitch, in

Pt = Transverse tube pitch, in

Mass Velocity, Gn:

32

Gn = Wg/An Where,

Wg = Mass gas flow, lb/hr

An = Net free area, ft2

Net Free Area, An:

An = Ad - Ac * Le * Nt Where,

Ad = Cross sectional area of box, ft2

Ac = Fin tube cross sectional area/ft, ft2/ft

Le = Effective tube length, ft

Nt = Number tubes wide

And,

Ad = Nt * Le * Pt / 12

Ac = (do + 2 * lf * tf * nf) / 12

tf = fin thickness, in

nf = number of fins, fins/in

Surface Area Calculations: For the prime tube,

Apo = Pi * do (1- nf * tf) / 12 And for solid fins,

Ao = Pi*do(1-nf* tf)/12+Pi*nf(2*lf(do+lf)+tf(do+2*lf))/12

And for segmented fins,

Ao = Pi*do(1-nf* tf)/12+0.4*Pi*nf(do+0.2)/12+Pi*nf (do+0.2)((2*lf-

0.4)(wn+tf)+ws*tf)/(12*ws) And then,

Afo = Ao - Apo Where,

ws = Width of fin segment, in

In the O-Frame Evaporator example we are using here, we should point out, that some of the tube rows can be used for downcomers instead of risers. Or , alternatively, the downcomers may be outside the gas pass. If they are part of the tube bank, they normally would not be finned, even if the riser tubes are finned. The sketch shown here would indicate the center two rows are downcomers since they are outside the collection baffle which directs the water/vapor mixter, coming from the riser tubes, through the primary separators. In this example, we are not going to consider any of the tubes as downcomers, but if we did, they would absorb heat, but would need to be rated separately

33

since they would have a different inside rate and overall heat transfer coefficient. We can describe a sample fin tube bank as follows:

Process Conditions: Gas flow, lb/hr = 800,000 Gas temperature in, °F = 898 Gas temperature out, °F = 533 Average fin temperature, °F = 701 Compostion, moles N2, % = 72.55 O2, % = 12.34 CO2, % = 3.72 H2O, % = 10.52 Ar, % = 0.87

Mechanical Conditions: Tube Diameter, in = 2.00 Tube Spacing, in = 4 Number Tubes Wide = 28 Tube Effective Length, ft = 28.00 Number Of Tubes = 392

Tube Arrangement = Inline Pitch Fin Height, in = 0.75 Fin Thickness, in = 0.049 Fin Density, fins/in = 6 Fin Type = Segmented Fin Segment Width, in = 0.3125

kb, Btu/hr-ft-F = 0.0278 cp, Btu/lb-F = 0.2719 µb, cp = 0.0314 = 0.0314*2.42 = 0.0760 lb/hr-ft To calculate the mass velocity, Gn, we need to first calculate the net free area of the tube bank. For these calculations, we are going to assume the tube rows are corbelled, so the net free area, An: Ad = 28*28*4/12 = 261.333 Ac = (2+2*0.75*0.049*6)/12 = 0.2034 So,

An = 261.333 - 0.2034 * 28 * 28 = 101.8674

And, Gn = 800000 / 101.8674 = 7853.3466

Now we can calculate the reynolds number, Re,

Re = 7853.3466*2/(12*0.0760) = 17222.251

And,

C1 = 0.25*17222.251-0.35 = 0.00823

34

For, sf = 1/6-.049=0.1177

C3 = 0.35+0.50*e(-0.35*0.75/0.1177) = 0.4037 And, Pl = 4.0

C5 = 1.1-(0.75-1.50*e(-0.70*14))*e(-2.0*4/4) = .9985 Now we can calculate the Colburn factor,

j = 0.00823*0.4037*.9985(3.5/2)0.5((715.5+460)/(701+460))0.25 = 0.0044 And finally,

hc = 0.0044*7853.3466*0.2719(0.0278/(0.2719*0.0760))0.67 = 11.475

The radiation transfer coefficient, hr is described later in this section. Fouling resistances, Rfi and Rfo are allowances that depend upon the process or service of the HRSG and the fuels that are being burned.

Fin Efficiency, E: For segmented fins,

E = x * (0.9 + 0.1 * x) And for solid fins,

E = y * (0.45 * ln(df / do) * (y - 1) + 1) Where,

y = x * (0.7 + 0.3 * x) And,

x = tanh(m * B) / (m * B) Where,

B = lf + (tf /2) For segmented fins,

m = (ho (tf + ws) / (6 * kf * tf * ws))0.5

And for solid fins, m = (ho / (6 * kf * tf))

0.5 Fin Tip Temperature, Ts: The average fin tip temperature is calculated as follows,

Ts = Tg + (Tw - Tg) * 1/((e1.4142mB+e-1.4142mB)/2) Maximum Fin Tip Temperature, Tfm: The maximum fin tip temperature is calculated as follows,

Tsm = Twm + θ(Tgm - Twm) Where,

Tsm = Maximum Fin Tip Temperature, F

Tgm = Maximum Gas Temperature, F

35

Twm = Maximum Tube Wall Temperature, F

And, The value for theta, θ, can be described by the following curve.

Convection Transfer, Stud Tubes For studded tubes, the correlations used are as provided by Birwelco, Ltd. Overall Heat Transfer Coefficient, Uo:

Uo = 1/Rto Where,



And,






Ro = 1/he

Rwo = (tw/(12*kw))(Ao/Aw)


Where,





36

Ao = Outside surface area, ft2/ft




Effective outside heat transfer coefficient, he: For staggered and inline pitch,

he = (hso*E*Afo+ht*Apo)/Ao Where,

ht = Base tube outside heat transfer coefficient, Btu/hr-ft2-F

hso = Stud outside heat transfer coefficient, Btu/hr-ft2-F


Afo = Stud outside surface area, ft2/ft

Apo = Tube outside surface area, ft2/ft

Inline pitch correction,

he = he*(do/Pl)0.333

Where,

do = Outside tube diameter, in

Pl = Longitudinal pitch of tubes, in

Base tube outside heat transfer coefficient, ht:

ht = (0.717/do

0.333)(Gn/1000)

0.67(Tb+460)

0.3

And the stud coefficient,

hs = 0.936*(Gn/1000)0.67

(Tb+460)0.3

With fouling,

hso = 1/(1/hs+Rfo) Where,

hs = Stud outside heat transfer coefficient, Btu/hr-ft2-F

Gn = Mass velocity of flue gas, lb/hr-ft2

Tb = Average gas temperature, F

Stud efficiency, E:

E = 1/((ex+e

-x)/1.950)

Where,

X = Ls/12((2*hso)/(ks*Ds/12))0.5

And,

Ls = Length of stud, in

Ds = Diameter of stud, in

ks = Conductivity of stud, Btu/hr-ft-F

37

Short Beam, Reflective Radiation

The gas radiation factor, hr, can be calculated from the following correlations. This factor is used in calculating the overall heat transfer coefficient for bare tubes and fin tubes. The formulas for the stud tubes has this factor built into the equations.

For bare tubes,

hr = 2.2*γr*(pp*mbl)0.50

And for fin tubes,

hr = 2.2*γr*(pp*mbl)0.50

(Apo/Ao)0.75

Where,

hr = Average outside radiation heat transfer coefficient, Btu/hr-ft2-F

γr = Outside radiation factor, Btu/hr-ft2-F

pp = Partial pressure of CO2 & H2O, , atm

mbl = Mean beam length, ft

Apo = Bare tube exposed surface area, ft2/ft

Ao = Total outside surface area, ft2 Outside radiation factor, γ r:

The outside radiation factor can be described by the following curves:

38

Thermal Conductivity Of Metals

The thermal conductivity of the tube material and the extended surface is needed for calculating the heat transfer coefficients. The conductivity of some of the popular materials can be described by the following curves.

39

Tube Wall Temperature Calculation The temperature of the tube wall may be calculated using the following equations. This method does not take coking into account.

Tw = Flux*do/di*Rfi+Flux*do/di*1/hi+Flux*do/(do-tw)*tw/(kw*12)+Tf

40

Where,

Tw = Tube wall temperature, F

Flux = Flux rate, Btu/hr-ft2 of bare tube


di = Inside tube diameter, in

tw = Tube wall thickness, in

Rfi = Inside fouling factor, hr-ft2-F/Btu

hi = Fluid film coefficient, Btu/hr-ft2-F

kw = Thermal conductivity of tube wall, Btu/hr-ft-F

Tf = Bulk process fluid temperature, F

Process

Thermal properties of water and steam

The properties of water and steam have been described by the “International Association for the Properties of Water and Steam”. The latest release is the IAPWS97 document, which is available at this site under publications. This document includes formulas for the various properties which make it fairly simple to develop software for use in HRSG design and rating. The formulation suggested for industrial use is the IAPWS-IF97 which replaces the 1967 document, IFC67. It should be noted that there is no significant differences in these two sets of formulations, in the pressure and temperature ranges normally used in HRSGs.

Thermal properties of flue gas

The thermal properties for flue gas used in these procedures are curve fits of the following families of curves. The source for these curves is not currently known, as they were originally constructed some thirty years ago as we were developing some of the software proceedures presented herein. However, the reliability of these values seem to be very suitable for HRSG design.

41

42

43

Combustion, supplementary firing

The supplemental combustion in HRSG systems takes place in a burner which usually is located in the flue gas stream going to the HRSG. However, some systems because of high firing requirements and low O2 in the flue gas, may require to be fired externally to the flue gas stream and may reuire additional combustion air. The types of burners and how they function are not covered in detail in this section. The amount of heat released can be easily calculated for a gas when we know the composition of the fuel and the heating values of the various components. For liquid fuels, the heating values are obtained by a calorimeter test. From these values and using the standard combustion equation, we can determine the composition of the flue gas leaving the burner. As an example, the combustion of methane could be stated :

CH4 + 2O2 --- > CO2 + 2H2O Of course for fuel gases containing many more components and burning in flue gas rather than pure oxygen, the equation gets more complicated. Therefore, a task that in itself is quite simple, becomes a burden to do by hand, but can be easily accomplished by a simple computer program. The heating values normally used in HRSG design are the LHV, lower heating values.

44

The following Lower Heating Values(LHV) are typical:

Component Btu/lb Component Btu/lb Component Btu/lb

CH4 21,520 C2H6 20,432 C3H8 19,944

N-C4H10 19,680 I-C4H10 19,629 N-C5H12 19,517

I-C5H12 19,478 C6H14 19,403 CO 4,347

H2 51,623 N2 0 CO2 0

C 14,093 S 3,983 C2H4 20,295

C3H6 19,691 C4H8 19,496 C6H6 17,480

H2O 0 O2 0 H2S 6,545

Heat Transfer Coefficients

The inside film coefficient needed for the thermal calculations may be estimated by several different methods. The API RP530, Appendix C provides the following methods,

For liquid flow with Re =>10,000,

hl = 0.023(k/di)Re0.8*Pr

0.33(µb/µw)0.14 And for vapor flow with Re =>15,000,

hv = 0.021(k/di)Re

0.8*Pr0.4(Tb/Tw)0.5

Where the Reynolds number is,

Re = di*G/µb And the Prandtl number is,

Pr = Cp*µb/k Where,

hl = Heat transfer coefficient, liquid phase, Btu/hr-ft2-°F

k = Thermal conductivity, Btu/hr-ft-°F

di = Inside diameter of tube, ft

µb = Absolute viscosity at bulk temperature, lb/ft-hr

µw = Absolute viscosity at wall temperature, lb/ft-hr

hv = Heat transfer coefficient, vapor phase, Btu/hr-ft2-°F

Tb = Bulk temperature of vapor, °R

Tw = Wall Temperature of vapor, °R

G = Mass flow of fluid, lb/hr-ft2

Cp = Heat capacity of fluid at bulk temperature, Btu/lb-°F

45

For two-phase flow, htp = hlWl + hvWv

Where,

htp = Heat transfer coefficient, two-phase, Btu/hr-ft2-°F

Wl = Weight fraction of liquid

Wv = Weight fraction of vapor

It should be stressed at this time, that there are many ways to calculate the inside heat transfer coefficient, and a lot of care should be taken in the procedure selected for use in HRSG design. Other methods, such as HTRI, Maxwell, Dittus-Boelzer, or others may be more appropriate for a particular HRSG design. You will notice that we have not offered a method for film boiling coefficient. The reason for this ommision is that in the evaporator, when the hi is high, which it is for the boiling case, it does not have any significant effect on the calcuations. We would recommend that a value of 1,500 to 2,000 Btu/hr-ft2-°F be used for these cases.

Intube Pressure Drop

The intube pressure drop may be calculated by any number of methods available today, but the following procedures should give sufficient results for HRSG design. The pressure loss in HRSG tubes and fittings is normally calculated by first converting the fittings to an equivalent length of pipe. Then the average properties for a segment of piping and fittings can be used to calculate a pressure drop per foot to apply to the overall equivalent length. This pressure drop per foot value can be improved by correcting it for inlet and outlet specific volumes.

Friction Loss: ∆p = 0.00517/di*G

2*Vlm*F*Lequiv Where,

∆p = Pressure drop, psi

di = Inside diameter of tube, in

G = Mass velocity of fluid, lb/sec-ft2

Vlm = Log mean specific volume correction

F = Fanning friction factor

Lequiv = Equivalent length of pipe run, ft

And, Vlm = (V2-V1)/ln(V2/V1)

For single phase flow,

V1 = Specific volume at start of run, ft3/lb

V2 = Specific volume at end of run, ft3/lb

46

For mixed phase flow,

Vi = 10.73*(Tf/(Pv*MWv)*Vfrac+(1-Vfrac)/ ρl Where,

Vi = Specific volume at point, ft3/lb

Tf = Fluid temperature, °R

Pv = Press. of fluid at point, psia

MWv = Molecular weight of vapor

Vfrac = Weight fraction of vapor %/100

ρl = Density of liquid, lb/ft3

Fanning Friction Factor: The Moody friction factor, for a non-laminar flow, may be calculated by using the Colebrook equation relating the friction factor to the Reynolds number and relative roughness. And the Fanning friction factor is 1/4 the Moody factor. For a clean pipe or tube, the relative roughness value for an inside diameter given in inches is normally 0.0018 inch.

Equivalent Length Of Return Bends:

The equivalent length of a return bend may be obtained from the following curves based on Maxwell table and can be corrected using the Reynolds number correction factor.

Lequiv = FactNre*Lrb Where,

FactNre = Reynolds number correction

Lrb = Equivalent length of return bend, ft

Return Bend Equivalent Length:

47

Reynolds Correction:

Where,

G = Mass velocity, lb/sec-ft2

Di = Inside tube diameter, in

Visc = Viscosity, cp

Gas Side Pressure Drop Across Tubes

The gas side pressure drop may be calculated by any number of methods available today, but the following procedures should give sufficient results for heater design.

Bare Tube Pressure Loss: For bare tubes we can use the method presented by Winpress(Hydrocarbon Processing, 1963),

∆p = Pv /2 * Nr Where,

∆p = Pressure drop, inH2O

Pv = Velocity head of gas, inH2O


48

And the velocity head can be described as,

Pv = 0.0002307 * (Gn /1000)2 / ρg Where,

Gn = Mass velocity of gas, lb/hr-ft2

ρg = Density of gas, lb/ft3

The Mass velocity is described as,

Gn = Wg / An Where,

Wg = Mas gas flow, lb/hr

An = Net free area, ft2

And, An = Ad - do/12 * Lt * Nt

For staggered tubes without corbels,

Ad = ((Nt +0.5) * Pt/12) * Le For staggered tubes with corbels or inlune tubes,

Ad = (Nt * Pt/12) * Le Where,

Ad = Convection box area, ft2


Le = Tube length, ft

Pt = Transverse pitch of tubes, in

Nt = Number of tubes per row

Fin Tube Pressure Loss: For the fin tube pressure drop, we will use the Escoa method.

∆p = ((f+a)*Gn

2*Nr)/(b*1.083E+109) And, For staggered layouts,

f = C2 * C4 * C6 * (df/do)0.5 For inline layouts,

f = C2 * C4 * C6 * (df/do)1.0 And,

a = ((1+B2)/(4*Nr))*ρb*((1/ρout)-(1/ρin)) Where,


49

ρb = Density of bulk gas, lb/ft3

ρout = Density of outlet gas, lb/ft3

ρin = Density of inlet gas, lb/ft3

Gn = Mass gas flow, lb/hr-ft2



df = Outside fin diameter, in

And,

B = An / Ad For staggered tubes without corbels,

Ad = ((Nt +0.5) * Pt/12) * Le

For staggered tubes with corbels or inlune tubes,

Ad = (Nt * Pt/12) * Le

Net Free Area, An:

An = Ad - Ac * Le * Nt

50

Where,

Ad = Cross sectional area of box, ft2

Ac = Fin tube cross sectional area/ft, ft2/ft

Le = Effective tube length, ft

Nt = Number tubes wide

And,

Ac = (do + 2 * lf * tf * nf) / 12

tf = fin thickness, in

nf = number of fins, fins/in

Reynolds correction factor, C2:

C2 = 0.07 + 8 * Re

-0.45

And,

Re = Gn * do/(12*µb)

Where,

µb = Gas dynamic viscosity, lb/ft-hr

Geometry correction, C4: For segmented fin tubes arranged in,


C4 = 0.11*(0.0 5*Pt/do)(-0.7*(lf/sf)^0.23)

an inline pattern,

C4 = 0.08*(0. 15*Pt/do)(-1.1*(lf/sf)^0.20)

For solid fin tubes arranged in,


C4 = 0.11*(0.0 5*Pt/do)(-0.7*(lf/sf)^0.20)

an inline pattern,

C4 = 0.08*(0. 15*Pt/do)(-1.1*(lf/sf)^0.15)

Where,

lf = Fin height, in

sf = Fin spacing, in

51

Non-equilateral & row correction, C6: For fin tubes arranged in,


C6 = 1.1+(1.8-2.1*e(-0.15*Nr^2)

)*e(-2.0*Pl/Pt)

- (0.7*e(-0.15*Nr^2)

)*e(-0.6*Pl/Pt)

an inline pattern,

C6 = 1.6+(0.75-1.5*e(-0.70*Nr))*e(-2.0*(Pl/Pt)^2)

Where,


Pl = Longitudinal tube pitch, in


Stud Tube Pressure Loss: For the stud tube pressure loss we will use the Muhlenforth method, The general equation for staggered or inline tubes,

∆p = Nr*0.0514*ns((Cmin-d0-0.8*ls)/((ns*(Cmin-do-1.2*ls)

2)0.555))1.8*G2*((Tg+460)/1460) Where,

∆p = Pressure drop across tubes, inH2O


Cmin = Min. tube space, diagonal or transverse, in


ls = Length of stud, in

G = Mass gass velocity, lb/sec-ft2

Tg = Average gas Temperature, °F

Correction for inline tubes,

∆p = ∆p*((do/Cmin)

0.333)2 And,

G = Wg/(An*3600)

An = Le*Nt*(Pt-do-(ls*ts*rs)/12)/12

52

Where,

Wg = Mass flow of gas, lb/hr

An = Net free area of tubes, ft2

Le = Length of tubes, ft

Nt = Number of tubes wide


ls = Length of stud, in

ts = Diameter of stud, in

rs = Rows of studs per foot

Ducting Pressure Losses

HRSG designers utilize ducting for many purposes in a system design. They are used for connecting flue gas plenums to stacks, distributing combustion air to burners, transfering flue gas to the HRSG or for bypassing the HRSG. The pressure losses through ducting pieces may be individually analyzed or the may be analyzed as a system.

We will first explore ducting losses by looking at the individual pieces. The following formulas and coefficients are from the American Petroleum Institute Practice API RP533.

Straight duct run friction loss:

∆p = (0.002989 * Fr * ρg * Vg2)*Le/De

Where,


Fr = Moody friction factor

ρg = Average gas density, lb/ft3

Vg = Velocity of gas, ft/sec

Le = Equivalent length of piece, ft

De = Equivalent diameter of piece, ft

And for round duct,

De = Diameter

And for rectangular duct,

De = (2 * Width * Height)/(Width + Height)

53

Moody friction factor, Fr

We can use the Colebrook equation to solve for the friction factor, with the roughness factor selected from the following:

Duct surface Roughness

Very rough 0.01

Medium rough 0.003

Smooth 0.0005

90° Round section elbow loss:

p = Vh * Cl

Where,

Vh = Velocity head of gas, inH2O

Cl = Loss Coefficient From Table

Radius/Diameter(R/D) Coefficient(Cl)

0.5 0.90

1.0 0.33

1.5 0.24

2.0 0.19

90° Rectangular section elbow loss:

p = Vh * Cl

Where,



Height/Width(H/W) Radius/Width(R/W) Coefficient(Cl)

0.25 0.5 1.25

1.0 0.37

1.5 0.19

0.50 0.5 1.10

1.0 0.28

1.5 0.13

54

1.00 0.5 1.00

1.0 0.22

1.5 0.09

4.0 0.5 0.96

1.0 0.19

1.5 0.07

Elbow of any degree turn loss:

This may be used for a rectangular or round duct elbow of N ° turn.

p = Vh * C90 * N/90

Where,


C90 = Loss coefficient from above for 90° turn

N = Number of degrees of turn

Sudden contraction loss:

p = Vh * Cl

Where,



Area2/Area1(A2/A1) Coefficient(Cl)

< 0.2 0.34

0.2 0.32

0.4 0.25

0.6 0.16

0.8 0.06

55

Gradual contraction loss:

p = Vh * Cl

Where,



Included Angle(N°) Coefficient(Cl)

30 0.02

45 0.04

60 0.07

No contraction change of axis loss:

p = Vh * Cl

Where,




<=14 0.15

Sudden enlargement loss:

p = Vh * Cl

Where,




0.1 0.81

0.3 0.49

0.6 0.16

0.9 0.01

56

Gradual enlargement loss:

p = Vh * Cl

Where,




5 0.17

10 0.28

20 0.45

30 0.59

40 0.73

Sudden exit loss:

p = Vh * Cl

Where,




0 1.0

90° Round miter elbow loss:

p = Vh * Cl

Where,


Cl = Loss Coefficient From Curve

57

90° Rectangular miter elbow loss:

p = Vh * Cl

Where,



Height/Width(H/W) Coefficient(Cl)

0.25 1.25

0.5 1.47

1.0 1.50

4.0 1.35

Pressure loss across stack damper:

This pressure loss is normally accounted for by rule of thumb. This may be 0.5 or 0.25 velocity head. We will use 0.25.

∆p = 0.25 * Vh Where,


Vh = Average velocity head of stack, inH2O

Draft gain or loss:

The draft gain or loss will be taken based on the height of the upward or downward flow of the flue gas. If the flow is upward, the pressure loss is negative.

58

∆p = (ρa - ρg)/5.2 * A Where,

∆p = Draft gain or loss, inH2O

ρg = Density of flue gas, lb/ft3

ρa = Density of ambient air, lb/ft3

A = Height of gas path, ft

Velocity head of gas:

Vh = Vg2 * ρg / 2 / 32.2 / 144 * 27.67783

Where,

Vg = Velocity of flue gas, ft/sec

ρg = Density of flue gas, lb/ft3

Now that we have some procedures for calculating the pressure loss for the various components that we might find in a duct system, how do we use them? The easiest way to analyze a ducting system and keep the pressure points straight, is to organize the system starting from the outlet and proceeding to the inlet. This may seem backwards at first, but when you examine the pressure at a given point in the system, you find that the pressure is always dependent on the downstream pressure. So, it makes sense to always work from outlet to inlet, then you always know the pressure of the outlet of the point you are at. To try this out, we will run the calculations for the simple example shown below.

For this example, we will assume we are picking up the flow at some point in a system, so we will calculate from the outlet(assumed to be to atmosphere) to the inlet without considering any condition at the inlet. We will assume the process conditions are those in above table.

So assuming,

D1 = 4 ft dia. D2 = 3 ft dia.

L1 = 10 ft L2 = 7 ft

L3 = 3 ft L4 = 7 ft

L5 = 3 ft

Desctription DeltaP, inH2O

Static, inH2O

Dynamic, inH2O

Total, inH2O

Outlet Condition 0 0 0 0

Sudden Exit 1.1242 0 1.1242 1.1242

Straight Duct Run, 2 ft 0.0136 0.0136 1.1242 1.1378

59

Miter Elbow, 2 Pc 1.4614 1.4750 1.1242 2.5992


Gradual Contraction, 45°

0.0450 2.3089 0.3557 2.6646

Miter Elbow, 2 Pc 0.4624 2.7713 0.3557 3.1270


Acid Dew Point Of Flue Gas

To improve the thermal efficiency of combustion equipment it is necessary to cool the flue gas to a low outlet temperature, to recover as much heat as possible. But the temperature must not be so low as to allow sulphur from the fuel to condense as sulphuric acid, resulting in a very corrosive flue gas.

The two graphs below indicate the maximum flue gas dew point versus sulphur content in typical fuels.

Maximum flue gas dew point versus percent sulphur in typical oil fuels:

Maximum flue gas dew point versus percent H2S in typical gas fuels:

60

NOx and other considerations Heat Recovery Steam Generators (HRSGs), do not form nitrogen oxides or other contaminants. However, the supplementary burners associated with this equipment, or the source of the flue gasses, such Gas Turbines or FCC Units, do create the conditions for the formation of these undesirable pollutants. Even though, NOx is not the product of combustion such as CO, CO2, and SO2, they are a side effect. When the fuel and air are mixed, as they are in the combustion process, the nitrogen, which is normally very stable, can react with other substances to form Nox. This reaction at higher temperatures causes the nitrogen to react with oxygen to form NO, and N2O referred to as NOx. Even with the available control methods such as steam/water injection in Gas Turbines and staged burner combustion in other equipment, there is sometimes the need to further reduce the NOx in the gas stream to the HRSG. Because of the low temperatures normally involved, this is normally accomplished with a Selective Catalytic Reduction (SCR) technology. SCR is the most effective and proven technology to reduce NOx emissions. With guaranteed reductions greater than 90%, SCR is adaptable to a wide range of temperature and service conditions. The SCR systems reduce NOx by optimally distributing ammonia over the catalyst bed, reducing NOx to harmless nitrogen and water. SCR, however, is not without its problems. The first is that the excess ammonia from the process is exhausted into the air. Both NOx and ammonia are acutely toxic, both contribute to fine particle formation, acidifying deposition, eutrophication and enrichment of terrestrial soils; and both may be converted to nitrous oxide (N20). Anhydrous ammonia (NH3) is the most economical reagent to use, however, the growing trend by end-users and operators is the use of aqueous ammonia (NH4OH). Aqueous ammonia is a safer reagent to transport, handle and store

61

than anhydrous ammonia. Vaporization of the aqueous ammonia solution is required to reach the base ammonia yield needed for the reaction. Typically, this vaporization is accomplished by using a hot air source. Exhaust gas can be withdrawn from the, or ambient air can be heated.

Circulation

Steam generation in boiler tubes is based on nucleate boiling. This means that a multiple of small steam bubbles are generated on the inside wall of the tube which remains "wet". If steam generation in any portion of the tube becomes excessive, film boiling results, which increases resistance to heat transfer and elevates the tube temperature. Steam blanketing can occur at low heat fluxes, in the region of nucleate boiling, if the steam bubbles formed are not continuously removed. At heat flues of 400,000 Btu/hr/ft2 and above, nucleate boiling changes to film boiling. At this point, even the most vigorous circulation cannot prevent the formation of an insulating steam film on the heating surface. To provide a reasonable factor of safety, HRSG's should be designed for a maximum heat flux of 100,000 Btu/hr/ft2.

The operating pressure of the evaporator also has a direct relation to the required circulation to achieve nucleate boiling. Since at a low pressure, the specific volume of the vapor is greater than at higher pressures, the circulation ratio must be greater at the lower pressure than at the higher. So, where a 10 :

ratio may be satisfactory for a system operating at 600 psia, you may need a 25 : 1 ratio for a 100 psia system.

For more specific information on circulation ratios and design of different configurations, see the Forced Circulation and Thermosyphon Circulation sections.

62

Forced Circulation

A forced circulation system uses a pump to maintain circulation in all the tubes of the evaporator tube bundle. A typical forced circulation system is shown at the right. The water is distributed within an inlet header to the various parallel circuits and the steam and water mixture is collected in an outlet header. The mixture is returned to a drum for separation of the steam and water.

The tubes may have any orientation but are generally horizontal with an upward water flow to improve stability between the parallel channels. The forced circulation evaporator will generally have more flow resistance than a natural circulation evaporator. In low pressure drop tube bundles, inlet orifices may be required to provide for better distribution to the parallel circuits.

When designing the circultion rate for these HRSG's these factors must be weighed agaist the cost of the pump and the pumping energy. Experience in steam boilers has shown that the "danger" zone in which film boiling can be expected to occur is between 20 and 25 percent steam generation by weight. This has led to the rule of thumb for a 5 : 1 circulation ratio. Unfortunately, this is in fact borderline and does not allow sufficient conservatism for unusual or extreme cases. For instance, an HRSG with a very long horizontal tube length(some units hav been manufactured with tubes as long as a hundred feet).

Also, the relation of the drum to the evaporator coil needs to be considered. As an example a steam coil in a fired heater may have the steam separation drum located a hundred feet away and at the geround, where the coil may be fifty feet up on top of the heater. In this case, the water pressure in the evaporator may be such that instead of total vaporization occuring in the coil, the temperature rises instead, and the major vaporization occurs at the inlet separators in the steam drum. In these cases, a 3 : 1 circulation ratio may be sufficient.

Thermosyphon(Natural) Circulation

Conventional, vertical tube boilers are generally designed for natural, "thermosyphon", circulation. This means the water and steam/water circuits are arranged so that the two phase mixture in the steam generating tubes rises to the steam drum by thermal lift of differential density and is replaced by water

63

from the drum by gravity flow. However, it should be noted that many horizontal tube natural circulation boilers have been sucessfully used for years.

Although the nature of a thermosyphon circulation boiler is quite simple, the calculation procedure to check the design to see what the circulation ratio is, can become very complicated. Most designs are such that each row of tubes have a different circulation than all the other rows, but some actually have a different circultion ratio for every tube in the evaporator, which would be difficult to determine, and probably not worth while. That is, of course, that an over all conservative check can be made that is reliable.

To take a look at the details of a thermo syphon evaporator, we will use the ever popular O-Frame design.

We will assume that our evaporator has the following design : Mechanical : SA178 Gr. A, 2.0" OD tubes with 0.120" wall, 28'-0" effective length, 28 tubes per row. 12 rows of finned tubes with 0.049" thick x 0.75" high x 6 fins per in. 2 bare downcomer rows located in the middle of the 12 finned rows. All tubes set on a triangular pitch. 48" ID steam drum and 24" ID mud drum located 33'-0" centerline to centerline. Process Conditions : Gas Side : 800,000 lbs/hr of Gas Turbine Exhaust at 898 °F


Nitrogen, N2 72.55 Oxygen, O2

12.34

Carbon Dioxide, CO2

3.72 Water, H2O

10.52

Argon, Ar 0.87 Sulphur Dioxide, SO2

0.0

Carbon Monoxide, CO

0.0

Tube Side : Steam is saturated at 630 psia.

64

So what we want to do first is select a single circuit in the system to review. We can do that with the sketch shown to the right. The 28 riser tubes would all be the same on this row. The 28 downcomer tubes would likewise be the same. The downcomers , both rows, 56 tubes, feed all the 336 of the riser tubes, so the circuit cannot be isolated from the rest of the circuits. If the heat absorbed by the downcomer tubes is so little that it will not cause a significant difference in the water density of that column, then we will be able to ignore the heat absorbed, for the purposes of checking the circulation. If this is true, then we would have one loss from the drum water down and through the mud drum up to the start of the heating zone of the riser. Another loss could be associated with the heating zone and a third loss covering from the end of the heating zone, up and through the primary separator.

The next step will be to determine the flux rate for each row of tubes through the unit so we can first see if the downcomer is going to be a significant heating tube. Also, we will need the flux rate on a row by row bases to determine the individual tube steam generation.

Row No. Flux Rate, Btu/hr/ft2 Tube Surf., ft2 Vapor Flow, lbs/hr

1 3,232 176.9 788

2 2,619 176.9 639

3 2,127 176.9 519

4 1,730 176.9 422

5 1,411 176.9 344

6 1,152 176.9 281

7 1,781 14.7 N/A

8 1,730 14.7 N/A

9 889 176.9 217

10 728 176.9 178

11 597 176.9 146

12 490 176.9 120

13 403 176.9 98

14 331 176.9 81

The total absorbed duty for the evaporator is 79,268,244 Btu/hr, or 109,288 lbs/hr of steam generated. If we assume, for now, that the circulation ratio is

65

10:1, then the water flow through the downcomer tubes will be about 1,092,880 lbs/hr, or 19,516 lbs/hr/tube. The pressure losses in the downcomers are, the sudden entry loss, the friction loss, and the sudden exit loss. This results in a pressure loss of 0.225 + 0.927 + 0.113 = 1.265 psi. But, since the head of water in the system is 11.322 psi, no vaporization will occur in the downcomers because the water in the mud drum is well above saturation for the system. The head of water/vapor in the riser leg will be less than that in the downcomer which will cause the water to circulate, thus, a thermosyphon system. How much circulation is what we need to determine.

We can break down the downcomer portion of the circuit as follows:

Sudden entry into downcomer tube : 0.5 velocity heads

Friction loss in downcomer tube : friction loss for 30'-0" of tube

Sudden exit into mud drum : 1.0 velocity heads

These losses are the same for all 56 downcomer tubes.

For the riser tube, up to the start of the heating zone:

Sudden entry into riser tube : 0.5 velocity heads

Friction loss in riser tube : friction loss for varing lengths of tube

Bend loss in riser tube : bend loss for varing angle of bend

These losses would be the same for each tube in a row, but different between each row.

For the riser tube heating zone:

Friction loss in riser tube : friction loss for 28'-0" of tube


For the riser tube, from end of the heating zone to steam drum:

Friction loss in riser tube : friction loss for varing lengths of tube

Bend loss in riser tube : bend loss for varing angle of bend

Sudden exit into steam drum baffle : 1.0 velocity heads

Primary separator loss : See Centrifugal Pressure Drop below


As you can see, we have a major problem in calculating the pressure loss for the primary separators since we do not have an actual design. So we will establish a proceedure to estimate the number of centrifugal separators that we

66

need for this sample. Then we will establish a formula for the pressure loss if we know the number of centrifugals.

Of course, if you are checking an existing design, you would know the number of centrifugal separators it has. It should also be noted that many steam drums don't use centrifugal separators, but rather a system of baffles, especially in low pressure systems.

Number of centrifugal separators required :

Ncent = {Ws(Vv + Vl[Cratio-1])}/(1080[(Vv-Vl)/Vl]0.5)

Pressure Drop through centrifugal separators, psi :

∆p = {2.28E-9(Vv + Vl[Cratio-1])/Cratio}{(Ws*Cratio)/Ncent)}

2 Where,

Ncent = Number of centrifugals

Ws = Steam Make, lbs/hr

Vv = Specific volume of vapor, ft3/lb

Vl = Specific volume of liquid, ft3/lb

Cratio = Design Circulation ratio

So for our sample HRSG design, the number of centrifugals required would be,

Ncent = {Ws(Vv + Vl[Cratio-1])}/(1080[(Vv-Vl)/Vl]0.5)

= {109288(0.73206 + 0.02024[10-1])}/(1080[(0.73206-0.02024)/0.02024]0.5) = 15.59 = 16 centrifugals

So now, assuming our actual circulation turns out to be 10:1, then the p would be,

p = {2.28E-9(Vv + Vl[Cratio-1])/Cratio}{(Ws*Cratio)/Ncent)}2

= {2.28E-9(0.73206 + 0.02024[10 - 1])/10}{(109288*10)/16)}2 = 0.9725 psi

If you are wondering why we didn't just say up front that, normally, these separators are sized at approximately 1.0 psi drop at design load, the answer is, we needed to get these complicated looking formulas established for our circulation calculations.

The method we have outlined above will give us a conservative answer since we will assume vaporization begins at the start of the heating zone, where it actually starts up the tube aways, and the final vaporization doesn't occur until separation in the centrifugals since it is only here that it is fully back to saturation temperature and pressure.

Since the flux rate on the first row of tubes is greatest, it will have the least circulation ratio, and the last row, where the flux is the least, will have the greatest circulation ratio. With our assumtion of a 10:1 circulation ratio, we will determine what the available loss for the rest of the circuit is,

67

Head in downcomer = l * hhead= 49.407 * 33 = 1630.43 lb/ft2 * 1/144 = 11.322 psi Head in riser up to heating zone = l * hhead = 49.407 * 2 / 144 = 0.686 psi Head in riser heating zone = m * hhead = 1/{[(0.02024 * 9.5) + (0.73206 * 0.5)]/10} * 28 / 144 = 3.483 psi Head in riser from heating zone = m * hhead = 1/{[(0.02024 * 9.0) + (0.73206 * 1.0)]/10} * 3 / 144 = 0.228 psi Head in riser total = 0.686 + 3.483 + 0.228 = 4.397 psi Left for riser losses = 11.322 - 4.397 - pdc= 11.322 - 4.397 - 1.265 = 5.660 psi

The problem with trying to analyze the circulation as we are here, is that the riser circulation is not 10:1, it is actually different for each of the 12 different sets of risers, and the downcomer data is not corrected for the combination of the 12 risers sets feeding the 2 downcomer sets. It is easy to see then, that any analysis must compute all 12 riser sets and the 2 downcomer sets at the same time, or the analysis is incorrect.

You will notice that when we run the whole evaporator as a single unit, we get a circulation ratio of 18.31:1, which we know from above is not correct. But, we can use our calculator and perform other checks as we would do if we were checking the circulation by hand. If you change the number of riser rows to 1, the lower straight length to 3', the upper to 2', the angle of the upper turn to 45 °, and the lower to 90°. Now for the downcomers, if we divide the number of downcomers by the number of riser rows, we get 6 downcomers to feed each row of risers. So change the downcomers to 1 row with 6 tubes wide. Then change the flux to the first row flux, 3,232 Btu/hr-ft2

We now calculate 10.58:1 circulation ratio for first row. Now change the flux rate to the last row, 331 Btu/hr-ft2 and we get a rate of 56.93:1. These checks, at least give us an approximation of the actual circulations that we would get with a program that can balance all the rows at the same time.

Sizing the riser circuits

The risers which we will be dicussing on this page are those that are generally external to the heat transfer sections. Since when the riser and the heat transfer are the same, the selection and sizing are based on the heat transfer requirement, i.e., the steam rate, and the circulation design. The circulation for the HRSG is discussed in previous pages.

The risers which we are now concerned with would be those found on I-Frame evaporators, horizontal tube evaporators, etc. This type riser is normally constructed of standard pipe and is connected to a header at the evaporator outlet with the other end connected to the steam drum.

Minimum riser flow area should be based on the following:

Amin = 0.0401(Wg / (Volv+Voll(CR-1)(1500(Volv+Voll(CR-1))/CR)0.5)

68

Where,

Amin = Riser area, in

Wg = Gross water/steam flow, lb/hr

Volv = Vapor specific volume, ft3

Voll = Liquid specific volume, ft3

Amin = Circulation ratio of evaporator coil

The target velocity in an individual riser pipe should be less than 12 ft/sec. The minimum number of riser pipes per header should be two. If the header is more than six feet and less than twelve feet in length, the preferred number of risers is three. For longer lengths, a similar number should be provided to prevent distribution problems in the header or manifold.

Sizing the downcomer circuits

The downcomers which we will be dicussing on this page are those that are generally external to the heat transfer sections. The circulation for the HRSG is discussed in previous pages.

The downcomers which we are now concerned with would be those found on I-Frame evaporators, horizontal tube evaporators, O-Frame, A-Frame and D-Frame evaporators with external downcomers. This type downcomer is normally constructed of standard pipe and is connected to upper drum with the other end connected to the lower drum or header.

Minimum downcomer flow area should be based on the following:

Amin = 0.007996 * Wg * Voll * CR Where,

Amin = Downcomer area, in

Wg = Gross water flow, lb/hr

Voll = Liquid specific volume, ft3

Amin = Circulation ratio of evaporator coil

The target velocity in an individual downcomer pipe should be less than 6 ft/sec. The minimum number of downcomer pipes per header is one. If the header is more than six feet and less than twelve feet in length, the preferred number of downcomers is two. For longer lengths, a similar number should be provided to prevent distribution problems in the header or manifold.

69

Other Considerations Boiler Water Treatment

Typical boiler system problems involve corrosion of feedwater systems; scale, deposition, and corrosion in the boiler; corrosion of condensate systems; and problems due to impurities in produced steam. ProChemTech manufactures various chemical products and equipment, which when properly applied, can economically control all of these problems.

Section of corroded boiler pipe

Oxygen Corrosion Control Oxygen which enters a boiler system will contribute to accelerated corrosion of the feedwater system, boiler, and condensate return system. The primary method to control oxygen entry into the boiler system is to deaerate added makeup. Once the makeup is deaerated, various organic and inorganic oxygen scavenger compounds are added to complete oxygen removal. An excess residual of oxygen scavenger in the boiler water, for instance 40 to 60 mg/l in the case of sulfite, is recommended to protect the boiler at all times. Oxygen scavengers should be added to the boiler makeup water stream immediately after deaeration so as to protect all downstream components of the system. Oxygen enters the condensate system via passage through the boiler and introduction via condensate tank vents. Oxygen based corrosion is controlled in the condensate system by a combination of good oxygen removal upstream, use of volatile oxygen scavengers, and/ or use of filming amines. Proper use of neutralizing amines, to maintain the pH above 8.0 and 8.5 su, also aids in oxygen corrosion control. ProChemTech produces a wide variety of oxygen scavenger products for use in boiler systems, which permits our water treatment specialists to specify the exact chemistry needed for safe, economical operation of any boiler system. Products are available as concentrated powders, easy to use liquids, and as blended multi-component formulations.

Scale/Deposition Control Scale and deposition in a boiler system can be caused by precipitation of two classes of materials: scale forming minerals, introduced with the makeup water; and corrosion products, generated within the boiler system itself due to poor corrosion control. The primary means for control of scale caused by minerals in the makeup water is to treat the makeup water prior to use via cation exchange softening, reverse osmosis, or demineralization. This treatment should remove the majority of the scale causing minerals, usually salts of calcium and magnesium, so that boiler chemical use is minimized.

Boiler coated with scale.

70

Even with the best pretreatment, some form of scale and deposition control chemistry is needed in the boiler water to control residual traces of scale forming minerals as well as any products of corrosion formed within the boiler system. Scale control products are basically classed as precipitating or non-precipitating based upon their action in preventing scale formation. Precipitating products, usually based on carbonate or phosphate chemistry, actually precipitate the hardness minerals within the bulk boiler water to form a fine "mud". This mud is maintained in suspension via use of polymer dispersants and removed in the routine boiler blowdown. Non-precipitating products, usually based upon sequestrant and chelant chemistries, function by chemically complexing the scale forming minerals to form a soluble compound, thus preventing any scale or mud formation.

As with oxygen scavengers, a residual of scale control chemistry is maintained in the boiler water to provide protection 100% of the time. For instance, phosphate chemistry is usually controlled to maintain a free phosphate residual of 20 to 40 mg/l.

ProChemTech produces products based on all four major scale control chemistries, as well as all commonly used polymer dispersants, for use in boiler systems. This range of products allows our water treatment specialists to specify the exact chemistry needed for safe, economical operation of any boiler system. Products are available as easy to use single, dual, and multi-component liquid formulations. In many cases, appropriate scale and deposition control chemistry can be obtained in one blended product by judicious selection of the various active components. This benefits the user by reducing the number of individual chemical products required for the water treatment program.

Boiler Corrosion Control While often not discussed, boiler internals must also be protected against corrosion caused by contact with water. The best method of protection is to simply maintain the boiler water OH alkalinity above 250 mg/l, thus rendering it non-corrosive to steel boiler internals. Both the carbonate and phosphate scale control chemistries also require the presence of OH alkalinity to properly form the scale preventing precipitates.

This fact, that boiler water should be maintained above a specific OH alkalinity value for two different specific reasons, nicely demonstrates the multiple actions often encountered in water treatment chemistry. In addition, it should be noted that in many cases sufficient alkalinity is present in the makeup water so that no supplemental alkalinity needs to be added to the boiler system to maintain the cycled boiler water alkalinity at an appropriate level. We bring this point up as many "water treatment experts" routinely add alkalinity as part of their boiler water treatment programs when it is not needed. This addition, in many cases, actually results in increased blowdown where the boiler cycles are limited by alkalinity considerations. Such actions amount to nothing less than theft, as the customer is both purchasing chemicals that are not really needed and the use results in increased boiler blowdown, which is extremely costly.

71

ProChemTech has a complete selection of single and multi- component alkalinity adding products, as well as many products with no added alkalinity, so that a water treatment program can be tailored exactly to each customer's needs.

Condensate Corrosion Control

Corrosion in a condensate return system can result from two items, entry of oxygen or low pH in the condensate.

Boiler condensate has a naturally low pH due to the formation of carbon dioxide in the boiler, from breakdown of carbonates present in the makeup water, and its subsequent carryout and dissolution in the condensed steam. When carbon dioxide dissolves in water, it forms carbonic acid, which can easily corrode the materials from which most condensate systems are constructed. To maintain the condensate pH in a minimal corrosion range, usually between 8.0 and 8.5 su, various volatile amines are added to the boiler. These compounds carry out with the steam and condense, forming an alkaline solution which neutralizes the carbonic acid.

Deposition from condensate system corrosion

Another approach, which can be used in place of or in combination with neutralizing amines, is to feed a filming amine into the boiler. Filming amines, typically based on octadedylamine, also volatilize and carry out with the steam. When they condense, however, they preferentially form an organic film on exposed metal surfaces, the film protecting the metal from attack by acidic condensate and oxygen.

Selection of the proper condensate corrosion control program can be quite complex as the ability of the various amines to carry throughout a steam system changes with steam pressure and distance from the boiler. The specific amine(s) fed into each boiler should be based upon the steam pressures and geometry of the steam system so as to ensure that the pH of all return condensate is within the recommended range.

ProChemTech produces a wide variety of condensate products, based on all accepted amines, for use in boiler systems. This permits our water treatment specialists to specify the exact chemistry needed for safe, economical operation of any boiler system. Products are available as concentrated single component and blended multi-component formulations.

Steam Quality Poor steam quality can cause problems via deposition on heat exchangers, erosion of turbines, and product contamination. These problems can be avoided by close control of boiler water dissolved solids levels and use of antifoam

72

compounds to control boiler foaming, thus giving improved steam quality. ProChemTech incorporates an extremely effective anti-foam in all our basic boiler scale and deposition control products, and also makes the material available as a concentrated single component product.

Water Conservation The various boiler chemistries developed by the company permit the maximum boiler cycles possible to be obtained via use of effective anti-foams, no alkalinity products, organic oxygen scavengers, and selection of minimal solids scale and deposition control products.

The company is also a major supplier of wastewater treatment technology and has integrated this knowledge with its boiler water technology to provide clients with a unique ability to reuse treated wastewaters as boiler makeup. Being a provider of complete wastewater systems and specific wastewater treatment chemistries, while knowing the various inherent limitations of boiler water treatment chemistry, has made the company the world leader in this new, rapidly expanding field.

To date, several projects in the railcar cleaning and repair industry have been successfully completed where wastewaters have been treated and then reused as boiler makeup water.

Chemistry Program Control We have also found that lack of proper chemistry control is a major cause of boiler system water problems. To insure proper application of our chemical products, the company has designed proprietary control systems which utilize a blend of manufactured and purchased units to provide the most reliable standard, and customized, chemical feed and blowdown systems for control of boiler water chemistry available.

Water Quaility Limits for HRSG

The use of water as a working engineering material, as in a Heat Recovery Steam Generator, introduces certain operational problems. These problems have their origin in the variability of the contaminants found in industrial waters, since, if pure water were readily available, water treatment would be incomplicated.

The initial supply of water for heat recovery steam generation is raw water from rivers, lakes, etc. This water undergoes primary treatment to convert it to makeup quality. it is then combined with condensate, that is, the condensed steam returned from a turbine or process and after further treatment is supplied to the boiler as feedwater. The amount of makeup, that is the proportion of the feedwater not furnished by returning condesate varies greatly, from 1% to 70% and is a major economic factor in determining the method of water treatment. The feedwater enters the boiler and is evaporated, separated in the steam drum and leaves for use in a turbine or process. Some solid material, called carryover, is usually entrained with the steam, and, if excessive, can cause

73

problems in superheaters and on turbine blades. The level of carryover is a function of the total solids, that is the sum of the dissolved and suspended solids, in the drum. The concentration of total solids is controlled by two means: The levels of total solids in the feedwater, and blowdown. Blowdown is the continuous or periodic removal to drain of a portion of the water in the drum. Sludge, which forms in the boiler, due to concentration of solids, and contributes to the suspended solids, is also removed in the blowdown since it settles in the lower portions of the drum.

The water conditioning cycle must include treatment of raw water from rivers, lakes, wells, etc.; steam condensate from processes and other uses; water within the steam generating system.

Since all the impurities present in feedwater are retained in the boiler, with the exception of those eliminated by blowdown, feedwater should be free of all deposit and scale forming materials, corrodants, and materials which cna fead to foaming and carryover in the steam drum. Preventing the formation of scale and other deposits is of particular importance, since depositation on waterside heat transfer surfaces both impedes heat transfer and increase the susceptibility of the boiler to corrosion. Conditioning and treatment of the makeup water and condensate will largely accomplish these objectives. In addition, deaeration of the feedwater, to remove corrosive gases (oxygen and free carbon dioxide), is required and the feedwater pH should be higher than 7.

Both users and suppliers of steam generating equipment generally agree that feedwater quality conditions in the following range are appropriate.

Feedwater Quality Limits

Boiler Pressure

<= 900 psia > 900 psia

Hardness 0.1 ppm 0

Total Metals(Fe,Cu, etc.) 0.2 ppm 0.02 ppm

Hydroxide Alkalinity (ppm Ca Co3) 25 5

Oxygen 0.007 ppm 0.007 ppm

pH 8.0 - 9.0 8.5 - 9.5

Organics 0 0

Silica 5.0 ppm 0.1 ppm

Suspended Solids 0 0

74

Designing the steam separation drums

Minimum Drum Diameter Sizing of steam disengaging drums relys on experience as much as specifics, but the following guidelines should be considered.

Except for very special designs, the minimum inside diameter of steam drums for HRSG service should be a minimum of 48 inches.

Steam drum water surge volume should be determined for rate of change due to anticipated process operating conditions and for anticipated rate of change due to system pressure increase or decrease. If not known, a minimum rate of change to be considered should not be less than 20% per minute. The drum sizing must allow for the swell volume to be accomplished without actuating the high liquid level alarm or causing liquid carryover into the saturated steam piping. Likewise, the shrinkage should be accomodated without actuating the low level alarm.

The vapor velocities in the free space above the normal liquid level should not exceed the following velocities at design flow rate and normal operating pressure.

Vh = 0.65{(ρl-ρv)/ρv}0.5

Vv = 0.26{(ρl-ρv)/ρv}0.5

Where,

Vh = Horizontal velocity, ft/s

Vv = Vertical velocity, ft/s

ρl = Liquid density, lb/ft3

ρv = Vapor density, lb/ft3

Feedwater Holdup Time In addition to accomodating the shrinkage volume discussed above, the drum size should allow for a minimum of two minutes holdup time at design flowrate in the advent of a loss of feedwater.

Internals, Separators The steam drum must be sized to accomodate the internals necessary to meet the guaranteed steam purity requirements, as well as the riser and downcomer conections. The internals include the primary separators, baffles or centrifugal, the secondary separators or dry pipe, as well as the channels or plenums needed for collecting the steam and water mixture from the risers. All steam drum internals should be removable without cutting. All internals should be able to be removed through the drum manhole.

There are many types of drum internals available, so we are only suggesting what may be used. For low pressure drums, under 50 psia, simple primary baffles and secondary dry pipe should be sufficient for most services. For low pressure drums integrally connected to a deaerator, no dry pipe is required. For other pressures and services the use of centrifugal primary separators with

75

chevron secondary scrubbers are recommended. We aslo recommend the use of a dry pie on internal connection of steam outlets even when chevrons are used.

Chevron Scrubbers The minumum area of chevrons required may be calculated using the following:

. Amin =(Wn * Vv)/(1080((Vv-Vl)/Vl)

0.5) Where,

Amin = Minimum area of chevrons, ft2

Wn = Net steam flow, lb/hr

Vv = Specific volume of drum vapor, ft3/lb

Vl = Specific volume of drum liquid, ft3/lb

So, if you had a single row of chevrons 8" high, you would need 1.5*Amin in length or if you had a double row, i.e., one row on each side of dry pipe, you would need a length of 0.75*Amin. Of course, for larger diameter drums, you could use 12" high chevrons and reduce the length requirement accordingly. You should keep in mind, however, that increasing the height of chevrons reduces the working area in the drum which is needed for drum swell, etc.

Centrifugal Separators The minumum number of centrifugal separators may be calculated using the following:

. Ncent = {Ws(Vv + Vl[Cratio-1])}/(1080[(Vv-Vl)/Vl]

0.5) Where,

Ncent = Number of centrifugals

Ws = Steam Make, lbs/hr

Vv = Specific volume of vapor, ft3/lb

Vl = Specific volume of liquid, ft3/lb

Cratio = Design Circulation ratio

We are assuming 12" centrifugals for this case(remember, they must go through our 12"x16" manhole) so it is fairly easy to determine the length required for a single row. But to reduce the required length, you may stagger the cans and/or place them along both sides of the drum. Additionally, the formula is designed to provide a 1 psi pressure loss through the centrifugal separators at design load.

Feedwater Distribution Pipe The boiler feedwater distribution pipe should have the following minimum pipe inside diameter.

di = 0.0921(Wbfw * Volbfw)0.5

76

Where,

di = Inside pipe diameter, in

Wbfw = Feedwater flow, lb/hr

Volbfw = Specific volume of feedwater, ft3/lb

Length of the pipe should be approximately the full length of the drum. Pipe should be secured and supported approximately every 35 diameters. Pipe should be perforated approximately 12 inches on center. Total flow area of perforations should not be less than area of the pipe. Feed pipe should have at least one breakaway joint at connection to external nozzle inside drum. Above 3 inch size should be flanged and below 3" should be threaded. End of pipe should be capped with a vent hole at top. The perforations should be orientated downward and toward the chemical feed pipe. The pipe should be below low liquid level and, if possible, in the stream of effluent coming from the primary separators and going to the downcomers. The entry into the drum must be fitted with a thermal sleeve if required by ASME code. Preferred location is horizonal, on centerline of drum head. If drum has thermal sleeve, it may not be located below horizontal, which would allow debris to collect in sleeve. If the economizer is designed such that steaming can occur at any operating condition, this design is not adequate. Special design for steaming economizer is not covered in this discussion.

Continuous Blowdown Pipe The boiler continuous blowdown pipe should have the following minimum pipe inside diameter.

di = 0.0921(Wbldn * Volsatliq)0.5

Where,


Wbldn = Blowdown flow, lb/hr

Volsatliq = Specific volume of drum water, ft3/lb

To calculate the blowdown flow, you must know the solids in the feedwater. If known, you can base the blowdown flow on the following ABMA standards for drum water conditions, unless the specifications require a more conservative level.

77

American Boiler Manufacturers Association

Boiler Water Standards

Pressure at Drum Outlet, psig

Total Solids, ppm

Total Alkalinity, ppm

Suspended Solids, ppm

0-300 3500 700 300

301-450 3000 600 250

451-600 2500 500 150

601-750 2000 400 100

751-900 1500 300 60

901-1000 1250 250 40

1001-1500 1000 200 20

1501-2000 750 150 10

2001 & Higher 500 100 5

So if you had a feedwater with a total solids of 20 ppm and you were operating at 650 psia, then the blowdown based on percent of feedwater flow would equal 20/2000*100 = 1%, or based on feedwater flow it would equal 20/(2000-20)*100 = 1.01%.

The minimum pipe diameter to be used for this service is 3/4 inch IPS. Length of pipe should be full length of drum. Pipe should be perforated on approximately 12 inch centers. Perforations should be orientated in an upward direction. Blowdown pipe should have a threaded breakaway joint at connection to internal nozzle. The blowdown pipe should be located below the water discharge of the primary separators. If the separators are located on both sides of the drum, the blowdown pipe should be divided and ran on each side.

Intermittent Blowdown Pipe A rule of thumb to use is for steam flows up to 150,000 lbs/hr, use 1 1/2 inch; and for greater than 150,000 lbs/hr, use 2 inch size. Preferred location is dependent on type of HRSG. In double drum designs, the intermittent blowdown pipe should be located in the mud drum. For single drum designs, the blowdown pipe should be located in the bottom of the drum. The purpose of the intermittent blowdown is to make quick corrections to water level as well as sludge removal. It should be pointed out that there is not much "sludge" in modern boilers with sufficient water treatment. This blowdown pipe may also be used as part of the overall boiler draining procedure.

Chemical Feed Pipe Minimum internal pipe size should not be less than 1/2 inch IPS. There is no specific sizing rules since normally at time of design, the flow is not known. A rule of thumb to use is for steam flows up to 50,000 lb/hr, use 1/2 inch; for 50,000 to 150,000 lbs/hr, use 3/4 inch; and for greater than 150,000 lbs/hr, use 1 inch size. This pipe should be 304 SS material including the nozzle entering the drum. Good practice is to use a ss sleeved entry nozzle designed such the

78

the chemical feed pipe can be replaced easilly. Pipe should be supported every thirty-five diameters. Chemical feed pipe should be capped and vented at end. Preferred location is where the wash from the feedwater pipe will mix the chemicals well before they can enter the downcomers. Care should be taken so that chemicals cannot be drawn up by the contiuous blowdown.

Other Miscellaneous Nozzles The minimum connection size for any connection should not be less than 3/4 inch. All nozzle connections up through 2 inch shall be minimum schedule 160.

Manholes Manholes should be 12 inch by 16 inch or larger and must be equiped with yoke and hinged cover. Sizing must be able to facilitate removal of all internals without cutting.

Steam Outlets All drums should have a minimum of two steam outlets, manifolded together outside the drum. This reduces the demand on the dry pipe and chevrons, as well as, lowers the horizontal velocity in the vapor space. An exception to this rule may be made for units with a total steam flow of less than 50,000 lbs/hr and there is no superheater. The minimum piping area should not be less than:

di = ((Wn1.35 * Volsatvap * 0.0000105)0.201)2 * 0.7854

Where,


Wn = Net steam flow, lb/hr

Volsatvap = Specific volume of drum vapor, ft3/lb

Normal Steam Drum Connections

Connection Design Pressure, psig Type

Steam Outlets All Pressures Welded

Safety Valves Under 650 Flanged

650 and Over Welded

Chemical Feed w/Sleeve All Pressures Flanged

Feedwater Inlet w/Sleeve Under 650 Flanged

650 and Over Welded

Water Columns, Lower Under 650 Flanged

Conns w/Sleeves 650 and Over Welded

Test Connections Under 650 Flanged

650 and Over Welded

Pressure Gauges Under 650 Flanged

650 and Over Welded

Vents Under 650 Flanged

650 and Over Welded

79

Sampling Connections Under 650 Flanged

650 and Over Welded

Continuous Blowdown Under 650 Flanged

650 and Over Welded

Intermittent Blowdown Under 650 Flanged

650 and Over Welded

Risers All Pressures Rolled or Welded

Downcomers All Pressures Rolled or Welded

For our sample HRSG that we have frequently referred to in this discussion, our steam drum might look like the one shown below. When we reviewed the natural circulation that would be the case for this HRSG, since it is an O-Frame design, we concluded that we needed 16 centrifugals. Since we have risers entering the drum from both sides, it makes sense to use two rows of centrifugals, so without staggering, we would need at least 8 feet of length. Since we had 28 tubes wide at 4" spacing, the inside width of the evaporator would be 28 * 4/12 = 9.33'. If we add 18" to allow for the casing and structure, then the seam length of the drum would be 10'-8" which is plenty of room for our centrifugals.

Our guide for the chevrons indicates we need a minimum of 12 ft2. So if we assume 8" high with two rows, we would need 9 feet of length. With our seam length, we have plenty of room for our chevrons.

Following our rule of two outlet steam connections, the maximum vapor flow at any point in the upper part of the drum would be 104396/4 = 26,099 lb/hr. Using our formula for maximum velocity, Vh, from above, we are okay up to 3.85 ft/s. if we assume the normal water level, NWL,is at centerline of drum, our net flow area is 6.28 ft2, and our velocity would be only 1.15 ft/s without discounting the area blocked by chevron hangers, so we are okay.

80

The first thing we notice after checking the times is that we don't meet our requirement for 2 minutes between NWL and LLSD. We can achieve this by raising the NWL to 26 inches and lowering the LLSD to 6 inches. But, this shows how tight the design of this drum is due to the short length. We could of course increase the diameter, but this would affect the cost, or we could increase the length, but then we couldn't ship because of the width of the module. Remember, when we change these levels, we must recheck the vapor velocities.

Insulation & Heat Loss

The insulation in an HRSG is extremely important for a number of reasons. The insulation provides a means of keeping the heat contained in the HRSG where it can be absorbed by the heat exchanger tubes, resulting in higher overall efficiencies. The insulation also keeps the external shell cooler making it safe for operating and maintenance personnel to safely work around the equipment. This cooler casing temperature also results in the structural stability of the overall structure.

Twenty years ago, much of the insulation used in HRSGs was the gunned or cast refractory. This material often, was mixed on site at the HRSG manufacturer's shop, and thus frequently varied in insulating properties. The more popularly used mixes like 1:2:4 LHV and others became standard and over time the insulating properties became very predictable. This was improved upon by the offering of proprietary mixes, by a number of companies, which were packaged in controlled environments and were thus more predictable in their application.

During the early eighties, ceramic fibers became accepted in the industry and since they are a much better insulator, they quickly caused a decline in the use of refractory. In general, 3 inches of ceramic fiber blanket could do a better job than 6 inches of refractory and weighed much less. As an example, if we have a hot face temperature of 1200 °F and an air temperature of 70 °F on a vertical wall with no wind blowing, 6" 1:2:4 LHV gunned has a cold face temperature of 205.5 °F where 3" 8# 2300°F ceramic fiber would have a cold face temperature of 161.2 °F. And this is with a weight less than 10% of the gunned refractory, which reduces freight cost. Furthermore, the ceramic fiber blanket does not require "drying" in the field as would be required with the refractory.

Refractory is still used in special cases and in areas where it is more durable or easier to install. The floor of a unit which must be walked on during maintenance and inspection, may use castable refractory or brick, or both because it is more durable. End tube sheets, when they have multiple tube penetrations such as in an end supported tube convection may utilize gunned refractory because it is easier(less costly) to apply between the openings for the tubes then ceramic fiber blanket.

The use of fiber insulation in the high velocity ducting, normal in HRSG designs, quickly brought out a problem that didn't occur in other equipment such as Direct Fired Heaters or Boilers. The damage to the fiber material due to the high

81

velocities, over 50 ft/sec, encountered in the ducting requried the use of metal liners. These thin metal liners themslves also present a design problem which is discussed elsewhere.

Heat Loss Through Insulation:

The heat loss due to radiation may be calculated using the Stefan-Boltzman formula.

hr = 17.4*10-10

*e*(T14 - T2

4)

Where,

hr = Heat loss by radiation, Btu/hr-ft2

e = Emisivity of surface, assumed at 0.95

T1 = Temperature of surface, °R

T2 = Temperature of surroundings,°R

The heat loss due to free convection may be calculated using the following method.

hc = 0.53*C*(1/Tavg)0.18*(T1 - T2)1.27

Where,

hc = Heat loss by convection, Btu/hr-ft2

C = A constant, assumed at :

1.79 for an arch or roof

1.39 for a wall

0.92 for a floor

Tavg = Average temperature of wall and surroundings,°R

The heat loss due to forced convection, where the air velocity is greater than zero, may be calculated using the following method.

hfc = (1 + 0.225 * V)*(T1 - T2) Where,

hfc = Heat loss by forced convection, Btu/hr-ft2

V = Velocity of air across surface, ft/sec

Documents

HRSG Design