# Performance Prediction for Industrial Boilers

By Viswanathan Ganapathy, boiler consultant |

*Understanding boiler performance calculations can allow engineers to improve the operation of their facility’s steam system and better engage with boiler-system vendors*

Chemical plants, petroleum refineries, power plants and cogeneration systems use oil-and-gas-fired steam generators, fluid heaters and waste-heat boilers of various types. Understanding how a boiler behaves with variations in parameters such as capacity, fluegas flow, inlet gas temperature or steam pressure can help plant engineers evaluate boilers better and understand why failures occur in crucial boiler components, such as superheaters, evaporators or economizers. Also, better understanding of boiler behavior can help plant engineers identify the reasons for decreases in efficiency.

Figures 1 and 2 show heat-recovery steam generators (HRSGs) of natural- and forced-circulation designs behind gas turbines. Figure 3 shows a typical fire tube boiler used in a chemical plant and Figure 4 shows an oil-and-gas fired steam generator.

In many of today’s cases, plant engineers rush to the vendor who supplied the equipment to ask opinions and get repair help, even if the boiler component is only experiencing minor performance problems. In this scenario, the vendor company is not likely to find faults with its own equipment design, even if there are flaws present. Rather, the vendor is likely to divert the attention of the plant engineers to the operations group or to the plant operating parameters. In other cases, the plant may not get a timely reply from the equipment vendor if the problem occurs several years after the equipment is installed.

Because boiler calculations are quite involved, many plant engineers are unable or unwilling to undertake them on their own. For many engineers, the basic heat-transfer principles covered in university degree programs are forgotten once they leave the institution behind for an industry job. This article is written to help process engineers perform basic boiler calculations and to help engineers apply heat-transfer and energy-engineering principles to understand and improve the operation of their plant’s steam system.

Using the simplified approach outlined in this article, process engineers can predict the thermal performance of boiler components, including superheaters, evaporators, economizers, fluid heaters or HRSGs of fire-tube or water-tube types, with ease and with reasonable accuracy. Plant engineers can then confirm their predictions using field measurements. These procedures may also be used to make minor modifications to existing equipment at a later date, when operating parameters are changed or the plant is upgraded.

While major modifications to the boiler may require help or advice from the equipment vendors, performing the calculations presented here will allow plant engineers to quickly get an idea of the performance or duty of the boiler and its components. Knowledge of these calculations will help plant engineers in the future as well.

During interactions with various facilities, I have suggested that plants develop a team of process engineers who specialize only in boiler-related calculations, in an effort to determine if the field data of steam-generating equipment are reasonable and whether or not they can be confirmed by thermal calculations. This type of specialized team can also develop calculation modules or computer codes to check if there are significant deviations from normal operating parameters for every boiler component in the plant. These calculation modules can also be used to determine whether fouling is present.

## Data for boiler purchases

Industrial boilers can cost millions of dollars, but often, facility personnel purchase a boiler without obtaining crucial data about it. Tables 1–6 show typical data sheets for a steam generator and a water-tube waste-heat boiler. Water-tube data are typically given in two parts: first, the tube’s geometric data, and second, its thermal performance. Plant engineers should demand similar data from boiler vendors before they purchase the boiler, and should ensure they have at least most of the information shown here in the tables.

The lack of such thorough data can often lead to confusion in evaluating boiler performance problems or tube failures and can lead to delays when the problem must be fixed immediately. A thermal process consultant tasked with analyzing boiler performance from a heat-transfer point of view must have such data for performing heat-transfer calculations to check whether the original proposal was reasonable or whether the field data correlate with the original design data. Many boiler vendors do not furnish these data as a matter of common practice, and plant engineers often do not realize that they need all this information to evaluate the boiler performance. They may not need the data at the time of the boiler purchase, but they likely will need it in the future.

I have experienced this problem when plants contact me to solve problems related to underperforming steam systems and lower-than-expected efficiency. The problems include boiler circulation issues, superheater or economizer tube failures, or underperforming steam generators and waste heat boilers. The tube geometry data are either unavailable or are not clear and have to be “made up” in order to evaluate the thermal performance. In some cases, I have also seen drawings that should show tube spacing or the number of streams, but nevertheless is not provided. In other cases, the thermal performance information given is sketchy and without details. This situation is like buying a gadget without an owner’s manual and trying to figure out a solution when the gadget fails.

There is a clear need for process and plant engineers to become more familiar with basic boiler calculations, as well as with the data that they should demand to obtain from boiler and HRSG suppliers.

# Nomenclature

## Basic energy-transfer equations

The basic equation for energy transfer in a boiler component is given by Equations (1) and (2).

*Q = UA∆T* **(1)**

∆*T* is log mean temperature difference (LMTD; nomenclature box shown above).

If *A* is based on the tube inner surface, then U should also be based on tube inner surface. Expressed as a equation, it would be the following:

*U _{i} A_{i} = U_{o} A_{o} = Q/∆T*

**(2)**

If a heat-transfer component receives external radiation from the furnace or cavity, then:

*Q – Q _{r} = Q_{c} + Q_{n} = UA∆T*

**(3)**

The energy lost by the hot fluegas stream is absorbed by the colder fluid (steam or water), as shown in the equation below:

*W _{h} ∆h_{h} = W_{c} ∆h_{c} = Q*

**(4)**

This equation neglects heat losses from the casing, which are typically 0.5 to 1.0%. Generally, the higher the boiler duty, the lower the heat loss percentage.

In all sizing or performance calculations, *U* (the overall heat-transfer coefficient) must be computed. For tubes with extended surfaces, the following equations apply:

1/*U _{o} = (A_{t} /A_{i})/h_{i} + ff_{i} (A_{t} /A_{i}) + ff_{o} + (A_{t} /A_{w}) (d/24K_{m}) ln(d/d_{i}) + *1

*/h*

_{o}η**(5)**

If plain tubes are used, Equation (5) may be simplified to the following Equation (6):

1/U_{o} = d/d_{i} /h_{i} + ff_{i} (d/d_{i}) + (d/24K_{m}) ln (d/d_{i}) + ff_{o} + 1/h_{o }** (6)**

The above equations may be used while sizing or evaluating the performance of water-tube boilers, superheaters, economizers and fire-tube boilers.

## Which coefficient governs *U*?

The gas-side heat-transfer coefficient governs the performance of boiler components, such as the evaporator, economizer, superheater and gas-fluid heaters. Why is this important to know? When simulating the performance of boiler components or of the boiler as a whole, the effect of gas-side parameters will govern the overall performance, while the steam-water-side parameters will have minimal impact. Therefore, concentrating on gas-side data will enable engineers to perform the analysis for off-design performance rather quickly, instead of having to evaluate both gas- and steam-side heat-transfer coefficients.

In a typical fire tube for example, the gas-side heat-transfer coefficient ranges from 10 to 20 Btu/ft^{2} h°F, while the boiling steam-side or water-side coefficient ranges from 1,500 to 3,000 Btu/ft^{2} h°F, depending on steam pressure and heat flux [1, 2]. If *d* = 2 in., *d _{i}* = 1.7 in.,

*ff*

_{i}=

*ff*= 0.001, and

_{o}*K*= 25 Btu/ft h°F, then the following calculation can be made:

_{m}1/*U _{o}* = (2/1.7)/15 + 0.001 × (2/1.7) + 0.001 + 1/1,500 + (2/24/25) ln (2/1.7) = 0.07843 + 0.001176 + 0.001 + 0.000667 + 0.00054 = 0.0818, or

*U*

_{o}= 12.22 or

*U*= 12.22 × 2/1.7 = 14.38 Btu/ft

_{i}^{2}h°F or

*U*= 0.958

_{i}*h*

_{i}Similarly, in a water-tube boiler, steam is boiling inside the tubes and hence, *h _{i}* = 1,500–4,000 Btu/ft

^{2}h°F, while fluegas flows outside and h

_{o}ranges from 10–20 Btu/ft

^{2}h°F. It can be shown from Equation (6) above that U

_{o}= 14.18 Btu/ft

^{2}h°F or U

_{o}= 0.945 h

_{o}.

Now, consider a finned tube evaporator with the following tube geometry: d = 2 in., d _{i} = 1.77 in., fins/in. = 2, fin height = 0.75 in., fin thickness = 0.049 in. The ratio of outside to inside area is 5.9.

Let us say h _{o} = 15 Btu/ft ^{2} h°F and h _{i} = 2,500 Btu/ft ^{2} h°F.

In this situation, the calculation goes as follows:

1/*U _{o}* = 5.9/2,500 + 0.001 × 5.9 + 0.001 + 5.9 × (2/24/25) ln(2/1.77) + 1/(0.73 × 15) = 0.00236 + 0.0059 +.001 + 0.0024 + 0.0913 = 0.109, or

*U*= 9.18 Btu/ft

_{o}^{2}h°F or

*U*= 9.18/0.73/15 = 0.84

_{o}*h*η (fin effectiveness is 0.73; Refs. 1 and 2 give details of their evaluation).

_{o}For a case in which there is a higher fin density and lower tube-side coefficient (such as a superheater), the ratio will be lower (around 0.75 *h _{o}* η. Hence, for finned tubes, depending on fin geometry,

*U*= 0.75 to 0.8 h

_{o}_{o}η. In this case,

*h*

_{o}still governs

*U*.

_{o}The gas-side heat-transfer coefficient, *h _{o}*, consists of convective and non-luminous coefficients. Procedures for estimating convective and non-luminous heat-transfer coefficients in boiler components are given in Ref. 1. Fluid pressure drops inside and outside tubes may also be obtained from Ref. 1. This article does not deal with the estimation of fluegas or steam-water-side pressure drops, only thermal performance.

*h _{o} = h_{c} + h_{n}*

**(7)**

The expanded online version of this article (at www.chemengonline.com/performance-prediction-industrial-boilers) describes the procedure for computing heat transfer coefficients for flow inside and outside plain tubes. The expanded version also provides an idea of how the fluegas transport properties can be obtained for a gas mixture. If one can obtain the steam-side heat-transfer coefficient *h _{i}* inside a superheater tube, it will help in evaluating tube wall temperatures, as shown in an example also included in the longer online version of this article. In a fire tube boiler,

*h*helps to determine the size of the boiler or predict its performance, as it governs

_{i}*U*, as shown above.

For gas temperatures below 1,500ºF, neglecting *h _{n}* may not lead to significant errors. In fire-tube boilers, since the beam length is tube inner diameter,

*h*, will be small and can be neglected. Then

_{n}*h*.

_{i}= h_{c}

## Two calculation categories

In general, calculations carried out in boiler practice can be divided into two categories: design calculations and off-design calculations (Figure 5).

Design, or sizing, calculations. Design calculations are used for sizing a new heat-transfer component, such as superheater, evaporator or economizer, steam generator or a waste heat boiler. Tube size, number of tubes wide and deep, tube and fin geometry and tube spacing are assumed for each component. The hot- and cold-fluid side flows and their inlet and exit temperatures are known from the energy balance. In other words, the energy required to be absorbed by each component is known, and also termed the LMTD (logarithmic mean temperature difference). *U* values are then computed based on the tube geometry and the surface area *A* is then obtained. Then, process designers check the gas- and fluid-side pressure drops and tube-wall temperatures to make sure they are reasonable. If not, the exercise is repeated. All boiler suppliers or anyone designing a new heating surface for a given duty would perform this type of calculation.

Off-design, or performance, calculations. In this type of calculation, the tube geometry and surface area of the boiler (or superheater or economizer, as the case may be) are known, as are the fluid flows and inlet temperatures of both hot and cold fluids. It is possible to arrive at the duty, as well as the gas- and fluid-side exit temperatures within a few iterations. Though boiler suppliers have to perform this calculation in part to check the load performance of their components, plant engineers invariably perform this type of calculation when the surface area and tube geometry of the heat-transfer equipment are already available, and the only variables are the hot- and cold-side fluid flows, analysis, fluid temperatures and steam pressure.

While there can be only a single design calculation, there can be numerous off-design calculations. Although boilers are designed for one set of parameters, any given boiler will likely operate under variable loads and parameters. Becoming familiar with off-design calculations will help plant engineers evaluate “what if” scenarios on boiler components and enable them to notify plant personnel of potential problems resulting from changing process parameters.

## Off-design calculation methods

There are two approaches to performing off-design calculations for equipment, such as a superheater or economizer or a fluid heater. Note that evaporator calculations are simpler due to the fact that the boiling water temperature is constant throughout the exchanger. The hot- and cold-side fluegas flows and inlet temperature and analysis are known. The objective is to obtain the duty and the exit hot- and cold-side fluid temperatures. These calculations are applicable for single-phase fluids only and no external radiation is assumed to be present.

* Conventional method.* The procedure for the conventional method is as follows:

- Assume the exit temperatures of the hot gas stream (
*T*) are known_{2}, T_{1}, t_{1} - Compute the assumed duty
*Q*=_{a}*W*(_{h}Cp_{h}*T*) (_{1}– T_{2}*W*are known)_{h}, W_{c} - Compute the exit temperature of the cold fluid
*t*_{2}using*t*=_{2}*t*_{1}+ Q_{a}/W_{c}/Cp_{c} - Compute the LMTD, since all four temperatures are known
- Compute
*U*using equations discussed in the online section of this article and in Refs. 1 and 2. Compute the transferred duty*Q*=_{t}*UA∆T* - If both
*Q*and_{a}*Q*are within a small range (~0.5%), then we may stop the iteration. If not, change_{t}*T*in Step 1 and repeat the protocol until the difference is 0.2% or less_{2}

Quick convergence logic may be used to speed up the calculations. Each time the gas or steam/water temperature is corrected, the corresponding gas and fluid properties, and also the heat-transfer coefficients, will change. A computer program is ideal for such an exercise, because the manual calculations become tedious. However, engineers should know how to do these calculations manually in order to develop a computer program. Ref. 1 also contains numerous manually calculated examples for various types of boiler equipment.

NTU method of performance evaluation. The Number of Transfer Units (NTU) method is the most widely used method in the chemical process industries (CPI), as it “directly” solves for the duty, although a few iterations will help to improve the accuracy. *U* is dependent on the average gas temperature or film temperature and on the fluid properties, which impact the heat-transfer coefficients with changing temperatures. A new equipment design also can also be evaluated by the NTU method. Simply change the tube geometry of the heat exchanger to match the desired performance, and then freeze the design. Following that, the off-design performance is evaluated for several cases. Hence, the NTU method can be used for a design calculation or an off-design calculation. Effectiveness, ε, is the most important factor in these calculations and once it is arrived at, the duty of the exchanger can be obtained.

Table 8 shows the effectiveness factor for various arrangements. It is sufficient if the fluid flows and the hot and cold fluid inlet temperatures are known. In that case, the following equations apply.

*Q = εC _{min} (T_{1} – t_{1})*

**(8)**

*C _{min} = (WCp)_{min} *

**(9)**

*C _{max} = (WCp)_{max}*

**(10)**

C = *C _{min} / C_{max}*

**(11)**

*NTU = UA/(WCp) _{min}*

**(12)**

These terms may be used in the equations shown in Table 8 to estimate ε for the exchanger configuration in question. Once ε is obtained, the duty may be obtained. Thereafter, the exit fluid temperatures are obtained. Using multiple iterations will improve the accuracy of the fluid properties and specific heats. Examples shown in the online version of this article illustrate this procedure for evaluating off-design performance of superheater, economizer, fluid heater, such as glycol heated by exhaust gases from a gas turbine.

## Evaporator performance

Fire-tube boiler or water-tube evaporator performance may be obtained in a far easier manner compared to superheaters or economizers. Since the cold-side fluid temperature is constant at saturation temperature, this evaluation is rather simple. *T _{1}* and

*T*refer to hot gas inlet and exit temperatures, respectively, and

_{2}*t*is the saturation temperature. Neglecting casing heat losses, the following equation can be written:

_{s}

*W _{g}Cp_{g} (T_{1} – T_{2})* = Q =

*UA∆T*=

*UA [(T*/ ln[(

_{1}– t_{s}) – (T_{2}– t_{s})]*T*) / (

_{1}– t_{s}*T*)]

_{2}– t_{s}**(13)**

Simplifying Equation (13) yields:

ln [(*T _{1} – t_{s}) / (T_{2} – t_{s})] = UA / (W_{g} Cp_{g})*

**(14)**

This equation may be used either for designing an evaporator or for checking its off-design performance.

If *T _{1}* and

*t*are known, Equation (14) may be used to obtain

_{s}*T*, and from that, duty (Q), as well as the steam generation in the evaporator, may be estimated, if the enthalpy of the feed water entering the evaporator and the enthalpy of saturated steam at operating pressure are both known. The value for

_{2}*T*obtained from the previous calculations may be compared with the field data to check whether the evaporator is performing well. This equation may also be used to arrive at the surface area

_{2}*A*, if

*T*and tube geometry and gas flow are known. The important Equation (14) also allows users to see what happens to the boiler duty with steam pressure changes.

_{1}, T_{2}, t_{s}## Challenging the HRSG vendor

Savvy process engineers familiar with such calculations can use the data provided by the HRSG supplier, along with field data, to evaluate whether or not the boiler supplier has overstated claims regarding steam generation and temperature profiles. Using the data provided by the HRSG supplier (Table 5) we see that the duty shown by the supplier for the evaporator is 60.37 million Btu/h, while our estimate (using the field data at a lower load) for the same guarantee case is only 58.7 million Btu/h.

Equation (14) comes in handy to see what is really happening here. Our estimate of UA for the evaporator from above is 4.92 × 86,379 = 424,985. Now using vendor data, we see from Equation (14) that: ln[(950–492) / (500–492)] = UA/(500,000 × 0.27 × 0.995) = 4.0474 from which *UA _{vendor} *= 543,670 Btu/h˚F], or a 28% higher value of

*UA*has been used by a vendor for the same cross-section, tube spacing, surface area. This means that the supplier has overestimated the

*U*value by 28% and submitted guaranteed results.

Plant engineers can perform such analyses for each section once they become familiar with these calculations and challenge the boiler supplier if performance is significantly different from predicted. However, in order to do this as mentioned earlier, good field data must be available. They can measure the gas temperatures at the evaporator inlet and exit, and prove that the performance is close to what is predicted by the vendor.

For example, in the actual operation case, if the steam temperature of 863°F, gas temperature at exit of evaporator and economizer of 485°F and 350°F and steam/water flow of 55,600 lb/h are accurately measured and shown to be close, then they have established their energy balance by calculation, as well as by measurements, and the HRSG vendor will have a difficult time justifying the design and guarantees.

Plant owners spend millions of dollars on HRSGs. They should also ensure the HRSG is well-instrumented with numerous gas and water/steam measurement points. Presently, many HRSG vendors provide a tapping point at the HRSG exit alone and plant owners do not generally think much about HRSG instrumentation at multiple locations of the HRSG. This is not a good practice. There should be multiple tapping points ahead and after each heating surface (particularly if the cross-section of HRSG is large), well-calibrated instruments and frequent checks on these readings by qualified instrumentation engineers to ensure the thermocouples are not plugged or damaged. Armed with such field data, it will be easier for the plant to simulate the HRSG performance at any operating point and extrapolate the results to guarantee conditions, or vice versa, and challenge the HRSG supplier if large deviations in steam generation or gas/steam temperatures are observed (Figure 6). Such independent analysis capability goes a long way toward ensuring effective HRSG operation and thorough examination of proposals from HRSG suppliers who do not “wing it” when it comes to stating thermal performance just to sell the equipment.

## Concluding remarks

The examples shown in the extended online-only version of this article illustrate the point that a simplified approach could be used to evaluate the thermal performance of boilers and their components, such as superheater, economizer fire tube or water tube evaporator or fluid heaters. *U* or (*UA*) could be obtained for the field data case and extrapolated for another set of operating conditions to determine the thermal performance. It is important that plant engineers obtain as much tube geometry data and thermal performance information from boiler suppliers before ordering the boilers, which will help them in their calculations at a later date. Using these, they can evaluate the performance at off-design conditions.

Ref. 1 provides more information on calculation procedures considering direct radiation, non-luminous radiation and hence more accurate procedures can be set up by plant engineers for each boiler in their plant, including steam generators. They can use the results to monitor the performance of various components and large deviations from “predicted” values can be taken up for more detailed analysis.

Though these procedures may not be as accurate as predicted by a complete boiler performance program, these calculations give an idea how process engineers can use field data to predict performance at any other operating point. The calculations should be backed up by proven instrumentation, which will make their performance predictions more reliable. Knowledge of these procedures will help plant engineers to become more informed and challenge boiler vendors in case they offer “inadequate” or inaccurate process thermal performance data while selling their boilers. Computer code can be developed by process engineers to model each boiler performance in their plant, which will help them to assess whether or not changes in the operating conditions are likely to cause unfavorable conditions in their boiler equipment.

*Edited by Scott Jenkins*

## References

1. Ganapathy, V., “Steam Generators and Waste Heat Boilers for Process and Power Plant Engineers”, CRC press, Florida, 2014.

2. Ganapathy, V., “Industrial Boilers and Heat Recovery Steam Generators”, CRC Press, Florida, USA, 2003.

3. Ganapathy, V., Simplify Heat Recovery Steam Generator Evaluation, Hydrocarbon Process., March 1990.

4. Ganapathy, V., Understanding Finned Heat Exchangers,

Chem. Eng., Sept 2013, pp. 62–65.

5. Ganapathy, V., Evaluate Extended Surface Exchangers Carefully,

Hydrocarbon Process., October 1990.

6. Ganapathy, V., Superheater Problems in Steam Generators,

Chem. Eng., February 2016, pp. 38–47.

## Author

**Viswanathan Ganapathy** is an international boiler consultant from Chennai, India (Email: v_ganapathy@yahoo.com). He has over 45 years of experience in the area of design and thermal performance aspects of steam generators, HRSGs and waste-heat boilers. He has authored over 250 boiler-related articles, which have been published in several U.S., U.K. and Indian publications. He has authored six books on boilers, the most recent being, “Steam Generators and Waste Heat Boilers for process and plant engineers,” published in 2015 by CRC Press, Florida.

## Online Extras

## HRSG performance calculations

Often, heat-recovery steam generators (HRSGs) are designed for a particular set of gas inlet conditions. However, once installed, due to various plant limitations, such as gas turbine load, steam or feed water parameters, the HRSG may not operate at the designed gas-inlet conditions. Then how does one know that the HRSG will perform as guaranteed once the appropriate gas inlet parameters are provided? This example shows what can be done.

HRSG performance calculation Example 1: An unfired HRSG is designed with the parameters shown in Tables 4 and 5, and the supplier has guaranteed performance as shown at 500,000 lb/h exhaust gas flow and at 1,100°F inlet gas temperature with a steam flow of 81,000 lb/h of 600 psig steam at 892°F. In operation however, the HRSG generates only 55,600 lb/h steam at 500 psig at 863°F. Inlet and exit gas temperatures are 1,000°F and 350°F, as shown in Table 6. Plant engineers want to know if this performance is acceptable. The boiler supplier says that, due to inadequate exhaust gas flow and temperature, the HRSG is making less steam than guaranteed, and that the vendor’s design is acceptable. Is the vendor correct?

Solution: Tables 4, 5 and 6 show the information given by the HRSG supplier and the field data. This is a problem commonly faced by plant engineers. The newly purchased HRSG operates at some gas/steam parameters that are not close to the design parameters. How can plant engineers determine if the HRSG was sized properly and if it will meet guaranteed performance after a few years, when gas inlet conditions for the guarantee case are provided? Suppliers’ warranties typically last for one year and any performance problems must be corrected by the vendor within that period. However, now since the plant is unable to provide the design gas inlet conditions within a year, the plant engineers are in a difficult position — that is, they are not sure if the HRSG has sufficient surface or not. However, using the simplified approach, the plant engineers can get a good estimate of how the HRSG will perform in the guarantee case or whether it will meet or not meet the guaranteed steam production.

Let us first establish the complete field data. It will help if we have accurately measured data, such as the water temperature entering and leaving the economizer, the drum temperature, the gas temperature at inlet and exit of the HRSG, the water and steam flows and steam temperatures. Using these data, it is possible for a process engineer to get a good idea of what the HRSG will do when the exhaust gas inlet is at “design” (guarantee case) conditions (namely 1,100°F and flow 500,000 lb/h). Presently, the gas flow is much lower. Exhaust gas-flow measurement is generally not accurate and hence it is obtained by energy balance based on energy absorbed by steam.

The data shown in Table 6 from the field should be used to generate the complete HRSG gas/steam temperature profiles. Since exhaust gas, steam and water temperatures and water flow are accurately measured, let us use these to obtain the exhaust gas flow.

From steam tables, we can obtain the following: enthalpy of final steam at 863°F, 515 psia is 1,446.2 Btu/lb. Enthalpy of feed water at say 550 psia, 230°F is 199.5 Btu/lb. Energy absorbed by steam (no blowdown used to simplify the calculations) = 55,600 × (1,446.2 – 199.5) = 69.32 million Btu/h.

Heat loss on the gas side is taken as 0.5%. Exhaust gas flow = 69.32 × 10 ^{6} / [0.995 × 0.2667 × (1,000 – 350)] = 401,880 lb/h (0.2667 is the gas specific heat at the average gas temperature of 675°F).

Let us do a similar energy balance at the superheater to determine the gas temperature leaving the superheater. Energy to steam = 55,600 × (1,446.2 – 1,204.8) = 13.42 million Btu/h (saturated steam enthalpy being 1,204.8 Btu/lb). At the average gas temperature of say 940°F, C _{p} = 0.275. Hence, gas temperature leaving superheater = 10,00 – 13.42 × 10 ^{6} / (401,882 × 0.995 × 0.275) = 878°F.

Energy absorbed by the economizer = 55,600 × (451.8 – 199.5) = 14.03 million Btu/h

Gas temperature entering economizer = 350 + 14.03 × 10 ^{6} / (401,882 × 0.995 × 0.2591) = 485°F.

Evaporator duty by difference = 69.32 – 14.03-13.42 = 41.87 million Btu/h.

Let us compute the U values for each section from field data, as now we can get LMTD for each section and we also know the surface area and duty of each section.

LMTD SH = [(1,000–863) – (878–472)] / ln[(1,000–863)/(878–472)] = 248°F.

LMTD evaporator = [(878–485) / ln [(878–472) / (485–472] = 115°F

LMTD economizer = [(120–16) / ln(120/16)] = 52°F

U _{1} = 13.42 × 106 / (8,941 × 248) = 6.06 Btu/ft ^{2} h°F (superheater)

U _{2} = 41.87 × 10 ^{6} / (86,379 × 115) = 4.21 Btu/ft ^{2} h°F (evaporator)

U _{3} = 14.03 × 10 ^{6} / (38,698 × 52) = 6.97 Btu/ft ^{2} h°F (economizer)

Using these computed values of U from field data, let us verify the steam generation in the guarantee case using the NTU method and the flow logic for the off-design performance calculations shown in Figure 6. We need to confirm if 81,000 lb/h at 600 psig and 892°F as guaranteed by the HRSG supplier is, in fact, feasible. First, let us correct U _{1}, U _{2}, U _{3} for the guarantee case as follows:

For the superheater:

(UA) _{p} = (UA) _{d} (Wg _{p} /Wg _{d}) ^{0.65} (Fg _{p} /Fg _{d}) (Ws _{p} /ws _{d}) ^{0.15} (15)

The steam flow term may be deleted for evaporator and economizer (UA) _{p} calculations as the effect of steam flow (evaporator) or water flow (economizer) has minimal impact on U. The correction factor F _{g} for fluegas properties is discussed in Appendix 2 of this online section.

Superheater

From Refs. 1, 2 and 3, the following can be observed:

U _{1c} = 6.06 × (500,000 / 401,882) ^{0.65} × (0.1438 / 0.1407) × (81,000 / 55,600) ^{0.15} = 7.55 Btu/ft ^{2} h°F. (0.1438 and 0.1407 are F _{g} factors based on gas properties shown in Table 7. F _{g} of 0.1438 is at the average gas temperature in superheater of 1,025°F in the guarantee case, while 0.1407 is F _{g} at the average gas temperature of 939°F in the field data case).

Cp _{s} = (1,458.28-1203.3)/(892-492) = 0.6375. Cp _{g} = 0.275

C _{min} = 81,000 × 0.6375 = 51637; C _{max} = 500,000 × 0.995 × 0.275 = 136,812

C = (51,637 / 136,812) = 0.3774. NTU = 7.55 × 8941 / 51,637 = 1.3081 (8,941 is the surface area of the superheater)

ε = [1-exp{–1.3081 × (1–0.3774)}]/[1-0.3774x exp{–1.3081 × (1–0.3774)}] = 0.6685

Q _{1} = 0.6685 × 51,637 × (1,100-492) = 20.99 million Btu/h

Gas temperature leaving superheater = 1,100 – 20.99 × 10 ^{6} / (500,000 × 0.995 × 0.275) = 947°F

Steam temperature leaving the superheater = 492 + 20.99 × 10 ^{6} / (81,000 × 0.6375) = 898°F

Enthalpy of superheated steam = 1,461.58 Btu/lb

Evaporator

U _{2c} = 4.21 × (500,000 / 401,882) ^{0.65} × (0.1329/0.131) = 4.92 (from Equation (14))

ln [(947-492)/(T _{2} –492)] = 4.92 x 86,379 / (500,000 × 0.995 × 0.27) or exit gas temperature from the evaporator T _{2} = 511°F

Duty of evaporator = 500,000 × 0.995 × (948 – 511) × 0.27 = 58.57 MM Btu/h

Economizer

U _{3c} = 6.97 × (500,000/401,882) 0.65 = 8.03. F _{g} correction is not applied since the average gas temperature in both the guarantee case and the field data case are nearly same.

Specific heat gas = 0.259 and that of water = 1.057 (from water enthalpy data)

C _{min} = 81,000 × 1.057 = 85,617. C _{max} = 500,000 × 0.995 × 0.259 = 128,825

C = 0.6646. NTU = 8.03 × 38,698 / 85,617 = 3.629

∈ = [1-exp(–3.629 × 0.3354)]/[1–0.6646 × exp{–3.629 × 0.3354] = 0.8766

Q _{3} = 0.8766 × 85,617 × (511–230) = 21.08 million Btu/h

Water temperature leaving economizer = 230 + 21,080,000 / 85,617 = 476°F

Exit gas temp = 511 – 21.08 × 10 ^{6} /(500,000 × 0.995 × 0.259) = 347°F

Total energy absorbed = 20.99 + 58.57 + 21.08 = 100.64 million Btu/h

From transferred duty, steam flow = 100.64 × 10 ^{6} /(1461.58 – 199.7) = 79,750 lb/h

Second iteration: superheater

Using the same Cp _{s} and Cp _{g}, C _{min} = 79,750 × 0.6375 = 50,841 and C _{max} = 500,000 × 0.995 × 0.275 = 136,812. C = 0.3716. NTU = 7.55 x 8,941 / 50,841 = 1.3277

∈ = [1–exp{–1.3277 × (1–0.3716)}] / [1–0.3718 x exp{–1.3277 × (1–0.3716)}] = 0.5658/0.8387= 0.6746

Q _{1} = 0.6746 × 50,841 × (1,100-492) = 20.85 MM Btu/h. t _{s2} = 492 + 20.85 × 10 ^{6} / 50,841 = 902°F

T _{g2} = 1,100 – 20.85 × 10 ^{6} / 136,812 = 948°F

Second iteration: evaporator

Using the same U _{2c}, ln[(948–492)/(T _{2} – 492)] = 4.92 × 86,379/(500,000 × 0.995 × 0.27) or exit gas temperature from evaporator = 511˚F. Q _{2} = 500,000 × 0.995 × 0.27 x (948–511) = 58.7 million Btu/h

Second iteration: economizer

Using the same specific heats for water and fluegas and U _{3c}, C _{min} = 79,750 × 1.057 = 84,296

C _{max} = 500,000 × 0.995 × 0.259 = 128,825.

C = 0.6543 NTU= 8.03 × 38,698 / 84,296 = 3.6863

ε = [1–exp{–3.6863 × (1 – 0.6543)}] / [1–0.6543 × exp{–3.6863 × (1 – 0.6543)}] = 0.8818

Q _{3} = 0.8818 × 84296 × (511 – 230) = 20.89 million Btu/h

t _{w2} = 230 + 20.89 × 10 ^{6} / 84296 = 478°F, T _{g2} = 511 – 20.89 × 10 ^{6} / 128,825 = 349°F

Total energy transferred = 20.99 + 58.7 + 20.89 = 100.58 million Btu/h

Corrected steam flow = 100.58 × 10 ^{6} / (1,463.78 – 199.7) = 79,560 lb/h

Let us go with 79,560 lb/h as the expected steam from this HRSG in guarantee case. Since the estimated value is far lower than that guaranteed, the plant has a strong case to inform the HRSG supplier that there is a need to verify the design and correct it. The exit gas temperature guaranteed by them is also on the low side by a significant margin (349°F versus 337°F). The energy absorbed 102 million Btu/h as guaranteed by them also cannot be met with the existing surface areas. The plant can show that the duty of the HRSG in the guaranteed case is only about 100.5 million Btu/h. The HRSG supplier will argue about measurement errors, calibration of instruments and so on. But there is a strong case that the HRSG will generate only about 79,500 lb/h steam at 900°F based on how it is presently performing. Plant engineers should familiarize themselves with such calculation processes. An excel work sheet or VBA coded excel can help in more accurate results instead of manual calculations.

## Superheater Example 2

A finned-tube counterflow superheater in a petroleum refinery is experiencing high tube-wall temperatures as measured at the steam outlet end. See data in Table 9. The operator is thinking of making it a parallel flow unit to lower the tube temperatures. What will be the performance and tube wall temperature if it is made a parallel flow unit? How much should the steam flow be reduced to get the same steam temperature as before for process?

Fluegas analysis: vol.% CO _{2} = 8 vol.%, H _{2} O = 18 vol.%,N _{2} = 72 vol.%, O _{2} = 3 vol.%.

The plant is experiencing high tube-wall temperatures and failures in its superheater. and wants to understand why this is happening. It wants to know if parallel flow design will help as a temporary measure and what will be its performance? If the same steam temperature of about 880˚F is required in parallel flow arrangement for process reasons, can they reduce the steam flow and by how much? What will be the maximum tube wall temperature in both cases?

Solution: Let us first understand why they have a problem. High finned superheaters increase the heat flux inside the tubes as discussed in several articles (Refs. 1 and 4) and hence should be designed with caution.

The ratio of external to internal surface areas = 10,195 × 12/(3.14 × 1.75 × 24 × 8 × 10.2) = 11.37

Let us compute U _{o} using the present set of data. The duty is 132,000 × (h _{o} – h _{i}).

From steam tables, h _{o} = enthalpy of steam at 894˚F and 715 psia is 1,455 Btu/lb. At inlet, saturated steam enthalpy at 730 psia is 1,201.2 Btu/lb and saturation temperature is 508˚F. Duty = 132,000 × (1,455–1,201.2) = 33.5 million Btu/h.

The average gas specific heat at 1155˚F is 0.3035 Btu/lbF. Assume heat loss is 1%.

Then the exit gas temperature = 1472-33.5 × 10 ^{6} /(176,000 × 0.99 × 0.3035) = 839˚F

LMTD = [(1,472 – 894) – (839–508)] / ln [(1,472 – 894) / (839 – 508)] = 443˚F

U _{o} = 33.5 × 10 ^{6} / 10,195 / 443 = 7.42 Btu/ft ^{2} h˚F.

This is the average heat transfer coefficient at the average gas temperature. The maximum U _{o} will be that at 1,472˚F. Using the factor F _{g} = Cp ^{0.33} k ^{0.67} /μ ^{0.32}, it can be shown using gas properties at 1,472˚F and 1,161˚F that F _{g} at 1,472˚F = 0.168, and that at 1,161˚F = 0.156.

Hence maximum U _{o} = 0.168 × 7.42 / 0.156 = 7.99 Btu/ft ^{2} h˚F. The maximum heat flux inside the tubes is = 7.99 × 11.37 × (1,472–894) = 52,509 Btu/ft ^{2} h.

The tube-side heat-transfer coefficient can be shown to be 275 Btu/ft ^{2} h˚F at 715 psia and 894˚F from Figure 10. Flow per tube = 132,000 / 24 = 5,500 lb/h (One can appreciate the importance of streams here as discussed in Appendix 1). From Figure 10, h _{i} = 243 × (5,500/3,300) ^{0.8} × (1.5/1.75) ^{1.8} = 277 Btu/ft ^{2} h˚F

The steam film temperature drop = 52,509 / 277 = 190˚F. If a fouling factor of 0.0005 is used on steam side, then inner tube-wall temperature will be 894 + 190 + 0.0005 × 52,509 = 1,110˚F. Wall resistance drop = 0.00052 × 52,509 = 26˚F. Hence, the maximum mid-wall temperature is 1,110 + 13 = 1,123˚F.

This is rather high, as thickness required by ASME code for T22 material exceeds 0.22 in for a design pressure of 800 psig, while only 0.125 in. has been used. The measured tube wall temperature also is around 1,120˚F. Hence the design was ill conceived in the first place. For superheaters, a small amount of finning only should be used, while in evaporators and economizers, we can use higher ratio of external to internal surface area as explained in Refs. 1 and 4.

## Parallel flow superheater

Let us use the NTU method to determine the performance if the superheater arrangement were made parallel flow with the same gas and steam flows as operating case. From the operating case earlier, the average gas side specific heat is 0.3035 and steam side specific heat is 0.657. F _{g} can be shown to be 0.156 and 0.168 at 1,155˚F and at 1,472˚F, respectively.

C _{min} = 176,000 × 0.99 × 0.3035 = 52,882. C _{max} =132,000 × 0.657 = 86724. C = 52,882 / 86,724 = 0.61

Use the same U _{o} as before for the first iteration. Then NTU = 7.42 × 10,195 / 52,882 = 1.43

ε = 1–exp[–1.43 × (1 + 0.61)]/(1+0.61) = 0.559

Hence, duty Q = 0.559 × 52, 882 × (1,472 – 508) = 28.5 million Btu/h

Exit steam temperature = 508 + 28.5 × 10 ^{6} / 132,000 / 0.657 = 837˚F.

Let us do another round of iteration to get more accurate results.

Exit gas temperature = 1,472–28.5 × 10 ^{6} / (176,000 × 0.99 × 0.3035) = 933˚F

Average gas temperature is 1,203˚F and average steam temperature is 673˚F. From the enthalpy of steam, Cp _{s} = (1,423.7–1,201.2)/(837–508) = 0.676 and Cp _{g} = 0.3051 from Table C2 in Appendix 3.

Then C _{min} = 176,000 × 0.99 × 0.3051 = 53,643 and C _{max} = 132,000 × 0.676 = 89,232

C = 53,643 / 89232 = 0.601. Average U = 7.42 × (0.1576/0.1558) = 7.51 Btu/ft ^{2} h˚F.

NTU = 7.51 × 10,195 / 53,643 = 1.427

ε = 1–exp[1.4 43 × (1 + 0.61)] / (1 + 0.61) = 0.559

Hence duty Q = 0.559 × 52,882 × (1,472–508) = 28.5 million Btu/h

Exit steam temperature 508 + 28.5 × 10 ^{6} /132,000/0.657 = 837˚F.

Let us do another round of iteration to get more accurate results:

Exit gas temperature = 1,472 – 28.5 × 10 ^{6} / (176,000 × 0.99 × 0.3035) = 933˚F

Average gas temperature is 1,203˚F and average steam temp is 673˚F.

From enthalpy of steam, Cp _{s} = (1,423.7 – 1,201.2) / (837 – 508) = 0.676 and Cp _{g} = 0.3051 from Table C2 in Appendix 3.

Then:

C _{min} = 176,000 × 0.99 × 0.3051 = 53,643 and C _{max} = 132,000 × 0.676 = 89,232

C = 53,643/89,232 = 0.601

Average U = 7.42 × (0.1576 / 0.1558) = 7.51 Btu/ft ^{2} h˚F

NTU = 7.51 × 10,195 / 53, 643 = 1.427

ε = 1- exp[127 × (1 + 601)] / 1.601 = 0.898 / 1.601 = 0.561 or

Q = 0.561 × 53,643 × (1,472-508) = 29 × 10 ^{6} Btu/h

Exit gas temperature = 29 × 10 ^{6} / (176,000 × 0.99 × 0.3051) = 926˚F and exit steam temperature is 29 x 10 ^{6} /132,000 / 0.676 = 833˚F.

It can be shown that the maximum tube wall temperature is far less than 900˚F and hence, tubes will not fail. However, the duty is lower. This is not a bad idea if a plant is willing to accept the lower duty of superheater as a temporary measure.

It can be shown that if the steam flow was reduced to 105,000 lb/h, then in parallel flow, one can achieve an exit steam temperature of 890˚F. The maximum tube wall temperature will be less than 950˚F and still acceptable. This exercise is left to the reader. Table 9 shows the results.

## Example 3

A petroleum refinery is planning to install a thermal fluid heater (paratherm1) utilizing energy from a gas turbine exhaust to preheat the fluid from 250 to 550˚F. It has obtained quotes from two suppliers and the data are shown here. Both offer the same duty and same exit fluid temperature. Both vendors use T11 tubes. Cross-section is the same. Plant wants to evaluate the two designs. The vendors have not offered more information. Prices were comparable. In fact, vendor 1 was 5% less in price.

Solution: The first thing that struck the plant engineers was the vast difference in surface areas. Vendor 1 appeared to be attractive as the price was also slightly lower. However, some process engineers felt that more surface area in finned tubes can be misleading as explained in references 1 and 4. Higher ratio of fin surface to tube surface lowers the gas-side heat-transfer coefficient. The product of (UA) will be the same if the duty and LMTD are same. Some plant engineers referred to the literature on the fluid and found that the maximum film temperature allowable was 650˚F. Hence a calculation was done to check the film temperature in both cases. Let us assume both vendors have sized the exchanger with proven correlations. In other words, let us assume that the surface areas given by them will perform the duty.

In order to compute the maximum film temperature, the maximum heat flux has to be computed. This required computation of tube side heat transfer coefficient at the fluid exit as also the U value.

The average U = Q/A/LMTD. LMTD in both cases is 350˚F. Hence U _{1} = 15.94 × 10 ^{6} / 7,740 / 350 = 5.89 Btu/ft ^{2} h˚F and U _{2} = 15.94 × 10 ^{6} / 350 / 4,895 = 9.30 Btu/ft ^{2} h˚F.

Fluegas properties at 950˚F and 750˚F (maximum and average fluegas temperatures are shown below. Tube-side properties at average fluid temperature and maximum fluid temperature are also shown below.

At average paratherm 1 temperature of 400˚F, Cp = 0.67,µ = 1.9, k = 0.055 and at 550˚F, Cp = 0.77, μ = 0.58, k = 0.055.

Hence, average fluid-transfer coefficient = 2.44 × (80,000/5) ^{0.8} × (0.67/1.9) ^{0.4} × 0.055 ^{0.6} /1.77 ^{1.8} = 233 Btu/ft ^{2} h˚F

And maximum fluid heat transfer coefficient at 550˚F is: 2.44 × (80,000/5) ^{0.8} × (0.77/0.58) ^{0.4} × 0.053 ^{0.6} / 1.77 ^{1.8} = 387 Btu/ft ^{2} h˚F

Maximum heat flux inside tubes: vendor 1:

6.21 × (950–550) × 7,740 × 12 / (3.14 × 1.77 × 15 × 10 × 8.5) = 32,550 Btu/ft ^{2} h and for vendor 2, it is 23,257 Btu/ft ^{2} h (average gas-side heat-transfer coefficients were corrected for gas temperature at inlet)

One has to be careful using vendor 1, because the film temperature is close to the limit of 650˚F. Small variations in U around tube periphery, non-uniformity in fluegas flow or temperatures can cause film temperature to be easily exceeded. Hence, the plant can consider vendor 2 a better and safer design. Vendor 1 may be asked to use smaller fin density and quote again. It may be noted that film temperatures cannot be measured. More often, they are calculated as shown. If the issue of film temperature had not been addressed, plant would have purchased vendor 1’s design and would have run into problems in operation later. Plants will do better to ask fluid heater vendors to provide film temperature calculations.

## Example of a fire tube boiler performance

In order to obtain the performance of any boiler components, the U value must be obtained first. This is a rather tedious calculation, if done manually, because fluid properties vary with temperature and analysis. Hence, a simplified approach is in order. Plant engineers are advised to start with these simplified calculations and with programming skills, develop calculation modules for boilers in their plants. This article gives several examples of how using the simplified approach plant engineers can check the accuracy of any water tube or fire tube boiler design or components such as superheater and economizer or a fluid heater.

One of the reasons the simplified approach works well in many boiler applications is that the gas-side heat-transfer coefficient governs the overall heat transfer coefficient U, as shown earlier, and hence, manipulating the variables connected with gas-side variations alone will help evaluate the performance with acceptable accuracy of any evaporator, superheater or economizer or fluid heater. Sometimes calculation of the tube inside or outside coefficient becomes a necessity and hence, Appendices 1, 2 and 3 are provided. The simplified approach for fire-tube waste-heat boiler evaluation will be of interest to plant engineers.

ln [(T _{g1} –t _{s})/(T _{g2} –t _{s})] = UA/(W _{g} Cp _{g}) (14)

As gas side governs U, for a given tube geometry (length, number, tube sizes) we can write:

UA = K _{1} (W _{g} /n) ^{0.8}/ (W _{g} Cp _{g}) (14a)

Where K _{1} is a constant for the given geometry, considering the tube length and diameter and gas properties

Equation (14) becomes:

ln[(T _{g1} –t _{s}) / (T _{g2} –t _{s})] = K _{1} ^{0.2} /W _{g} ^{0.2} = K / W _{g} ^{0.2} (16)

Where n is the tube count. The effect of gas properties is neglected in this simplified approach. If tube count does not change, then K _{1} ^{0.2} can be replaced by K, another constant. These simplifications work well for clean gases and when direct radiation from flame or high non-luminous heat transfer coefficients are absent.

Using the above equation, one can predict several aspects of a fire-tube waste-heat boiler performance:

1. Effect of steam pressure

2. Effect of plugging of tubes

3. Effect of change in gas flow

4. Effect of gas inlet temperature.

## Example 4

A fire-tube waste-heat boiler generating steam from hot air has been purchased with the parameters as shown in Table 12, Column 1. However, the plant is operating as shown in Column 2 at a lower steam pressure and at lower capacity due to various reasons. Tube geometry data are available as shown. How can we find out if the boiler is performing well or sized properly using the simplified approach? In Case 3, 15% of the tubes were plugged for leaks. What will be the performance then with the design gas flow?

Solution: Typically steam side flow measurements and low temperature gas, fluid temperatures are more reliable than gas flow or high temperature measurements. Hence, in such simulation studies, the fluegas flow is typically worked out from energy balance. This example also shows how one can find out the effect of steam pressure on an evaporator performance.

Case 2: The values shown are as measured in the field. Using these we have to check if the design case performance can be achieved. 4.93 million Btu/h is the energy absorbed by steam using the steam flow and enthalpy data with zero blowdown.

For a quick estimate of airflow in operating case, from Appendix 3, we can estimate the gas specific heat at the average gas temperature of (1,000 + 525) / 2 = 763˚F as 0.2592 Btu/lb˚F.

Then gas flow will be: 4.93 × 10 ^{6} /[0.2592 / (1,000-525)] = 40,040 lb/h

Let us first compute K.

ln[(1,000-448) / (525–448)] = K/40,042 ^{0.2} or K = 16.4.

Use this to check if design performance can be achieved. Let us see if exit gas temperature of 615˚F can be obtained with the stated gas flow of 70,000 lb/h at 1,200˚F and at 615 psia.

Using Equation (16) again, ln[(1,200-489 / (T _{g2} –489)] = 16.4 / 700,00 ^{0.2} or T _{g2} = 611˚F

The supplier has given 615˚F, which is close. The measured gas pressure drop between the boiler inlet and exit is 1.0 in. wc. The design pressure drop will be approximately (70,000/40,000) ^{2} × 1 = 3.1 in. wc across the tubes. Considering the higher average gas temperature the gas pressure will be higher, close to the 3.4 in wc, the design value stated. Hence the design/proposal data are reasonable, assuming the measurements in case 2 are good.

Case 3: If 15% of tubes are plugged, what will happen to the duty and gas pressure drop with the design gas flow and inlet gas temperature?

Let us check the revised K value for the original design. K = 16.4 × (361/425) ^{0.2} = 15.88

Using Equation (16), ln[(1,200-489) / (T _{g2} –489)] = 15.88/70,000 ^{0.2} = 1.705 or T _{g2} = 619˚F

The duty or steam flow after plugging will be (1,200-619) / (1200-611) = 0.986 × 100 = 98.6% of design value. The gas pressure drop will be about (425/361) ^{2} × 3.4 = 4.6 in wc. Gas pressure drop will be a major concern rather than decrease in duty with 15% plugging.

Hence using such simplified estimates backed up by one set of good field data, one can get a good idea of off-design performance of fire tube or water tube boilers.

## APPENDIX 1

Tube-side heat-transfer coefficient. Simplified estimates for heat transfer coefficients of single-phase fluids inside tubes are presented in Ref. 1. In fire tube boilers, the fluegas-side heat-transfer coefficient governs the overall heat transfer coefficient U. An idea of the tube-side coefficient h _{i} in a superheater will help determine its tube wall temperature and pressure drop. In a fluid heater, it helps to determine the film temperature of the thermic fluid.

Tube side coefficient may be estimated by the following:

Nu = 0.023 Re ^{0.8} Pr ^{0.4}

Nu = h _{i} d _{i} /12k

Re = 15.2wd _{i} /µ

[Re = 3,600ρ Vd _{i} /µ/12. V = 576 w/[3,600π d _{i} 2ρ.

Simplifying Re = 15.2 w/d _{i} /µ ]

Pr =µCp/k

All properties estimated at fluid bulk temperature.

Simplifying further, h _{i} = 2.44 w ^{0.8} C/d _{i} ^{1.8}

where factor C = (Cp/µ) ^{0.4} k ^{0.6}

C is available in tables and charts in order to quickly estimate hi for various fluids (see Refs. 1 and 2).

a: For wet air (% volume H _{2} O = 1, N _{2} = 78, O _{2} = 21)

Importance of streams in economizer and superheater

One of the important pieces of data to be furnished by boiler vendors in their description of a superheater, economizer or a fluid heater is streams, or the number of tubes carrying the entire steam or water flow of a boiler. Often one is required to estimate the steam or water velocity inside the tubes, heat transfer coefficient, pressure drop or tube wall temperatures. Many boiler vendors do not provide this crucial data or drawings which show baffle plates in the header, if used. Hence when confronted with solving a tube failure problem in a superheater, one gets stuck. Partitions in headers can mask this data and hence plant engineers should get this data from the boiler vendor for future analysis. The Figure 10 shows a few examples of what streams are.

## APPENDIX 2

Heat-transfer coefficients outside plain tubes

Fishenden and Saunders equation for convective heat transfer coefficient hc outside plain tubes takes the form:

Nu = 0.35FhRe ^{0.6} Pr ^{0.3}

The Fh factor depends on the tube arrangement, whether inline or staggered, and is shown in Table B1. It can be seen that the correction factor is nearly the same for both inline and staggered when Sl/d is above 2, which is the arrangement typically used in boiler practice. It also is used to improve ligament efficiency in drums. However if fluegas pressure drop is considered, it is far higher for staggered compared to inline arrangement. Hence, rarely will one see a staggered arrangement for plain tubes, because from a heat-transfer perspective, there is no significant benefit while operating costs in the form of fan power consumption will be much higher [1,2]

A conservative correlation used for both inline and staggered arrangements is:

Nu = 0.33 Re ^{0.6} Pr ^{0.33}

Nu = h _{c} d/12k

Re = Gd/12µ

Pr =µCp/k

All gas properties for heat transfer are evaluated at gas film temperature which is the average of gas and tube wall temperature and lower than the average gas temperature.

Simplifying the above,

h _{c} = 0.9G ^{0.6} F _{g} /d ^{0.4}

where

F _{g} = k ^{0.67} Cp ^{0.33} /µ ^{0.27}

Another correlation widely used is that of Grimson’s [1]

Nu = BRe ^{N}

where B and N are shown in Table B1.

Examples of using these correlations are given in Refs. 1 and 2.

Finned tube bundles

Correlations for finned tube bundles are more involved and shown in Refs. 1 and 2 with worked out examples.

The gas property factor affecting gas side heat transfer is given by [1,2,4]

F _{g} = k ^{0.67} Cp ^{0.33} /µ ^{0.32}

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