# Steam Generator Efficiency

The Impact of Condenser Performance, Feedwater Heating and Steam Reheat

## By Brad Buecker, Contributing Editor

The word “thermodynamics” conjures up visions of complex mathematics to many people, even some with technical backgrounds. Yet, relatively simple formulas from thermodynamics can be used explain much about steam generator fundamentals.

Thermodynamics is built around two laws. The first law is that of conservation of energy. It says that energy used within a system is neither created nor destroyed but only transferred. The classic energy equation for a simple system (defined as a control volume in textbooks)1,2 is:

Q – WS = m2[V22/2 + gz2 + u2 + P2v2] – m1[V12/2 + gz1 + u1 + P1v 1] + dEc.v./dt Eq. 1

Where,

Q = Heat input per unit time

WS = Shaft work such as that done by a turbine per unit time

m2 = Flow out of the system per unit time

m1 = Flow into the system per unit time

(V22 – V12)/2 = Change in kinetic energy

gz2 – gz1 = Change in potential energy

u2 = Internal energy of the exiting fluid

u1 = Internal energy of the entering fluid

P2v 2 = Flow work of fluid as it exits the system (P = pressure, v = specific volume)

P1v 1 = Flow work of fluid as it enters the system

dEc.v./dt = Change in energy within the system per unit time

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This equation can be easily understood through a few definitions and simplifications.

First, in many systems and especially steam generators, potential and kinetic energies are very minor compared to other energy changes and can be neglected.

Second, in a steady flow process such as a steam generator, the system does not accumulate energy, so dEc.v./dt is zero. Removal of these terms leaves the internal energy of the fluid (u) plus its flow work (Pv) capabilities. Scientists have combined these two terms into the very useful property known as enthalpy (h). Enthalpy is a measure of the available energy of the fluid, and enthalpies have been calculated for a wide range of steam and saturated liquid conditions. These values may be found in the standard ASME steam tables, where saturated water at 0C has been designated as having zero enthalpy. Using these simplifications and definitions, the energy equation for steady flow operation reduces to:

Q – WS = m(h2 – h1) Eq. 2

But this equation represents the ideal scenario and here is where the second law steps in. The second law describes process direction. For example, a warm cup of coffee placed on a kitchen table does not become hotter while the room grows colder. Many more examples are possible, but this one conveys the essence of the second law.

The second law has as a foundation the concept of the Carnot cycle, which says that the most efficient engine that can be constructed operates with a heat input (QH) at high temperature (TH) and a heat discharge (QL) at low temperature (TL), in which

QH/TH – QL/TL = 0 Eq. 3

This equation represents a theoretically ideal engine. In every process known to humans, some energy losses occur. Scientists have defined a property known as entropy (s), which, in its simplest terms, is based on the ratio of heat transfer in a process to the temperature (Q/T). In every process, the overall entropy change (of a system and its surroundings) increases. So, in the real world, Equation 3 becomes

QH/TH – QL/TL < 0 Eq. 4

While entropy may seem like a somewhat abstract term, it is of great benefit in determining process efficiency. Like enthalpy, entropy values are included in the steam tables.

Two important points should be noted about the Carnot cycle and all real world processes. First, no process can be made to produce work without some extraction of heat from the process (QL) in Equation 3. QL in a conventional steam generator is heat removed in the condenser.

Second, the efficiency (Î·) of a Carnot engine is defined as

Î· = 1 – TL/TH Eq. 5

So, as input temperature goes up and/or exhaust temperature goes down, efficiency increases. This concept has some limitations in complex systems and an example is outlined later.

So how does this discussion apply to normal steam plant operation? First, let us examine condenser performance.

## Condensers

Why must turbine exhaust steam be condensed? Why not transport it directly back to the boiler? The reason is one of efficiency. For simplicities’ sake, consider the system shown below with a turbine that has no frictional, heat or other losses, which means no entropy change (isentropic).

In actuality, turbines are typically 80 to 90 percent efficient, but this factor does not need to be included here to show the importance of condenser performance. Conditions are:

Main Steam (Turbine Inlet) Pressure – 1,000 psia
Main Steam Temperature – 1,000 F
Turbine Outlet Steam Pressure – Atmospheric (14.7 psia)

Let us call this Example 1. The steam tables show that the enthalpy of the turbine inlet steam is 1,505.9 Btu per pound of fluid (Btu/lbm). Thermodynamic calculations indicate that the exiting enthalpy from the turbine is 1,080.9 Btu/lbm (steam quality is 93 percent). Equation 2 (the first law, steady-state energy equation) becomes for the turbine, wT = m(h1 – h2). Accordingly, the unit work available from this ideal turbine is (1,505.9 Btu/lbm – 1,080.9 Btu/lbm) = 425.0 Btu/lbm. To put this into practical perspective, assume steam flow (m) to be 1,000,000 lb/hr. The overall work is then 425,000,000 Btu/hr = 124.5 MW.

Now consider Example 2, where the system has a condenser that reduces the turbine exhaust pressure to 1 psia (approximately 2 inches of mercury). Again assuming an ideal turbine, the enthalpy of the turbine exhaust is 923.4 Btu/lbm. The unit work output equates to 1,505.9 – 923.4 = 582.4 Btu/lbm. At 1,000,000 lb/hr steam flow, the total work is 582,400,000 Btu/hr = 170.6 MW. This represents a 37 percent increase from the previous example. Obviously, condensation of the steam has an enormous effect upon efficiency. This is a practical illustration of how the condenser lowers TL from equation 5.

Another point is that if the steam from Example 1 were rerouted directly to the boiler, the power requirements and size for the feed pump would be huge. Much less work is required to pressurize a liquid than a gas.

One can also look at this example from a physical perspective. Calculations indicate that the steam quality at the turbine exhaust (at 1 psia condenser pressure) is 82 percent. This means that 18 percent of the steam has condensed to water. However, the remaining steam takes up a specific volume of 274.9 ft3/lbm. The corresponding volume of water in the condenser hotwell is 0.016136 ft3/lbm. Thus, the condensation process reduces the fluid volume over 17,000 times. The condensing steam generates the strong vacuum in the condenser, which actually acts as a driving force to pull steam through the turbine.

Let’s take this concept a step further in Example 3. Consider if waterside fouling or scaling (or excess air in-leakage) causes the condenser pressure of the previous example to increase from 1 psia to 2 psia. Thermodynamic calculations show that the work output of the turbine drops from 582.4 to 546.1 Btu/lbm. So, at 1,000,000 lb/hr steam flow, a rise of 1 psia in the condenser backpressure equates to a loss of 36,300,000 Btu/hr or 10.6 MW of work. This is why proper cooling water chemical treatment and condenser performance monitoring are important.3

For simple steam generating systems, general efficiency is represented by this equation:

Î· = (wT – wP)/qB Eq. 6, where

wT = Work produced by the turbine

wP = Work needed by the feedwater pump

qB = Heat input to the boiler

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The energy required by the feedwater pump is much less than the work produced by the turbine, so it is often left out in basic energy calculations.

The heat input (qB) is equivalent to the difference in enthalpy of the condensate entering the boiler vs. that of the main steam exiting the boiler. For Examples 2 and 3 outlined above, qB calculates to 1,436.2 and 1,411.9 Btu/lbm, respectively. From the simplified efficiency equation (Î· = wT/qB) the respective efficiencies are 40.6 percent and 38.7 percent.

## Superheating Mechanics

Consider the common drum boiler, where the steam leaving the drum is saturated. If this steam were to be immediately injected into a turbine, very little work would occur, as the steam would immediately begin condensing to water upon passage through the blades. For this reason, all utility steam generators include several tube circuits, which reside in the boiler backpass, through which the saturated steam passes for additional heating. These tubes comprise the superheater. The temperature to which the steam is raised above saturation represents the degree of superheat. Remember that it takes nearly 1,000 Btu to convert a pound of water to a pound of steam in utility steam generators. As the examples in the previous section illustrated, it should come as no surprise that efficiencies of standard utility boilers are in the 30- to 35-percent range.

Research into more temperature resistant superheater and reheater tube materials continues, in direct application of the equation,Î· = 1 – TL/TH. Maximum steam temperatures in even the most advanced supercritical units have been limited to about 1,100 F due to materials performance issues. However, new alloys are being developed that may allow higher steam temperatures, especially for future supercritical boilers.

Ideally, superheat energy is almost completely consumed at the last, low-pressure turbine blades. A balance is necessary to extract all of the available energy from the steam but prevent excessive condensation in the turbine blades. Thermodynamics shows that work and efficiency of a steam generator will improve with increased pressure.

Let’s increase the steam pressure to 2,000 psia from Example 2, where the condenser pressure was 1 psia. Note this as Example 4. The main steam enthalpy becomes 1,474.1 Btu/lbm and the turbine exhaust enthalpy is 871.0 Btu/lbm. The turbine work output rises to 603.1 Btu/lbm (176.7 MW at 1,000,000 lb/hr steam flow), and the efficiency increases from 40.6 percent to 42.9 percent. (Efficiency gain through higher pressure is a primary reason why supercritical have become popular for coal-fired boilers.) But, at 2,000 psia the turbine exit steam quality is only 77 percent. This means that 23 percent of the fluid exits as condensed water droplets. Such high moisture content can damage low-pressure turbine blades. Ten percent moisture at the turbine exhaust is typically an upper limit.

Reheating the steam helps to alleviate this difficulty. Consider Figure 2, which shows a steam generator and turbine with a reheat system.

Main steam is at 2,000 psia, 1,000 F, and has an enthalpy of 1,474.1 Btu/lbm. The steam extraction (cold reheat) pressure is 300 psia, which equates (isentropically) to a cold reheat temperature of 485 F and enthalpy of 1,248.1 Btu/lbm. Assume no pressure drop through the reheater and a hot reheat temperature of 1,000 F, producing reheated steam with an enthalpy of 1,526.5 Btu/lbm. Calculations show that the reheating process improves the turbine exhaust steam quality from 77 percent to 90 percent. Because the steam quality increases, the turbine exhaust enthalpy increases slightly to 1,003.9 Btu/lbm.

Calculation of the work output, boiler heat input and efficiency of this example becomes slightly more complicated, as in this case work is done by two separate steam feeds to the turbine and heat is added to two separate steam systems in the boiler. The unit work equation is

wt = (inlet steam enthalpy – cold reheat enthalpy) + (hot reheat enthalpy – turbine exhaust enthalpy).

In this case, wt = (1,474.1 – 1,248.1) + (1,526.5 – 1,003.9) = 748.6 Btu/lbm. As can be seen, reheating considerably increases the work output as compared to the non-reheat example. The boiler heat input is defined as (main steam enthalpy – feedwater enthalpy) + (hot reheat enthalpy – cold reheat enthalpy). For this example, qb = (1,474.1 – 69.7) + (1,526.5 – 1,248.1) = 1,682.8 Btu/lbm. An obvious conclusion is that reheat increases the energy output but also the fuel requirements to the boiler. The efficiency calculates to 44.5 percent, which is 2 percent higher than the non-reheat example. The increased fuel requirement is counterbalanced by increased work output and better steam quality of the turbine exhaust. A well-designed reheat system can reduce moisture to low levels in the turbine exhaust steam. Supercritical units may have two reheaters to maximize turbine performance.

## Feedwater Heating

A general rule of thumb suggests that for a single heater, the extraction steam flow rate should be designed to raise the feedwater temperature to a point halfway between the condensate temperature and saturation temperature of the boiler.

In this scenario (condenser pressure of 1 psia and boiler pressure of 2,000 psia) the condensate temperature is 69.7 F and the boiler saturation temperature is 636.0 F. The midway point between these two temperatures is 353 F. If steam is extracted from the turbine at a pressure of 500 psia, energy/mass balance calculations show that the flow rate to the heater should be 20.8 percent of the total steam flow. The extraction steam enthalpy is 1,299.7 Btu/lbm. The heat exchange produces feedwater with an enthalpy of 325.0 Btu/lbm. The turbine work equals the enthalpy difference between the main steam and extraction point (1,474.1 – 1,299.7 Btu/lbm), plus the remaining steam (79.2 percent) that passes to the turbine exhaust (0.792*[1,299.7 – 871.0] Btu/lbm).

In this case, the turbine work equates to 513.8 Btu/lbm. This is less than the work obtained in Example 4 (603.1 Btu/lbm), which had no feedwater heater. One might logically ask how feedwater heating improves the process. The benefits are efficiency related. The heat input required by the boiler to produce the required steam is only 1,149.1 Btu/lbm (1,474.1 – 325.0), as the feedwater temperature is much warmer. Thus, the efficiency of this system is (513.8/1,149.1) * 100 = 44.7 percent. This represents an 11 percent increase from Example 4.

The principal concept behind the efficiency gain is that much of the heat re-used in the feedwater heater would have otherwise been exhausted in the condenser.

## Putting It All Together

A large utility steam generator usually has several feedwater heaters, at least a single reheater and condenser that has excess surface area for cooling. Proper monitoring and chemical treatment programs are vital in keeping these systems free from fouling or corrosion. My co-workers and I have personally been involved with condenser performance improvement projects that have resulted in net savings of \$500,000 to \$1,500,000 a year at just one plant. To borrow an old phrase, such savings “are not chicken feed.”

## References :

1. Van Wylen, G., and R. Sonntag, “Fundamentals of Classical Thermodynamics, 3rd Ed.”; John Wiley & Sons, 1986.

2. Potter, M., and C. Somerton, “Thermodynamics for Engineers”; Schaum’s Outline Series, McGraw-Hill, 1993.

3. B. Buecker, “Condenser Chemistry and Performance Monitoring: A Critical Necessity for Reliable Plant Operation”; from the Proceedings of the 60th International Water Conference, Pittsburgh, Pennsylvania, October 18-20, 1999.

Author: Brad Buecker is an air quality control specialist at a large Midwestern power plant. Buecker has a BS in Chemistry from Iowa State University. He is a member of the ACS, AIChE, ASME and NACE.