*(Editor’s Note: This story originally ran in November 2008 and has proved a perennial favorite of Power Engineering readers. With the New Year we thought we’d share it again. Click here to see the original post).*

**By Brad Buecker, ChemTreat**

Everyone knows baby boomers in the power sector have reached retirement age, meaning many new hires will be thrust into technical positions where fundamental information is valuable. One such area is steam generating efficiency. It’s therefore worthwhile to address a few fundamentals of steam production.

For many people, even some with technical backgrounds, the word “thermodynamics” conjures up visions of complex mathematics. Yet relatively simple thermodynamic formulas explain much about the fundamentals of steam generation.

To begin, thermodynamics is built around two laws, sometimes jokingly described thusly…

First Law: You can’t get something for nothing

Second Law: You can’t break even.

In reality, 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) is:

## Equation 1:

Q – WS = m2[V_{2}^{2}/2 + gz_{2} + u_{2} + P_{2}Ï…_{2}] – m_{1}[V_{1}^{2}/2 + gz_{1} + u_{1} + P_{1}Ï…_{1}] + dE_{C.V.}/^{dt}

In this equation,Q = Heat input per unit timeW _{S} = Shaft work such as that done by a turbine per unit timem _{2} = Mass flow out of the system per unit timem _{1} = Mass flow into the system per unit time(V _{2}^{2} – V _{1}^{2})/2 = Change in kinetic energygz _{2} – gz _{1} = Change in potential energyu _{2} = Internal energy of the exiting fluidu _{1} = Internal energy of the entering fluidP _{2}Ï… _{2} = Flow work of fluid as it exits the system (P = pressure, Ï… = specific volume)P _{1}Ï… _{1} = Flow work of fluid as it enters the systemdE _{C.V.}/dt = Change in energy within the system per unit time

While this equation appears complicated, it is easily understood through a few definitions and simplifications. First, in many systems (especially steam generators) potential and kinetic energies are very minor compared to other energy changes and can be overlooked. Second, in a steady flow process such as in a steam generator, the system does not accumulate energy, so dE_{c.v.}/dt = 0.

Removing these terms leaves the internal energy of the fluid (u) plus its flow work (PÏ…) 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. 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 0 C has been designated as having zero enthalpy. Using these simplifications and definitions, the energy equation for steady state operation in a turbine reduces to:

## Equation 2:

Q – W_{S = m(h2 – h1)}

But this represents the ideal scenario and here is where the second law steps in. Among other things, the second law describes process direction. What that means is that a warm cup of coffee placed on a kitchen table does not become hotter while the room grows colder. Human beings grow old and so on. The second law has as a foundation the concept of the Carnot Cycle, which says the most efficient engine that can be constructed operates with a heat input (Q_{H}) at high temperature (T_{H}) and a heat discharge (Q_{L}) at low temperature (T_{L}), in which

## Equation 3:

Q_{H}/T_{H} – Q_{L}/T_{L} = 0

This equation represents a theoretically ideal engine. In every process known to humans, some energy losses occur. These may be due to friction, heat escaping from the system, flow disturbances or a variety of other factors. 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

## Equation 4:

Q_{H}/T_{H} – Q_{L}/T_{L} < 0

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 by logical inference 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

## Equation 5:

Î· = 1 – T_{L}/T_{H}

So, as input temperature goes up and/or exhaust temperature goes down, efficiency increases.

## The Condenser, Reheater and Feedwater Heaters

So how does all of this apply to normal steam plant operation? We’ll next build upon a simple steam generator design to show the effects of common thermodynamic principles. First let’s examine condenser performance.

Many people wonder why steam exiting a turbine must be condensed back to water instead of being transported directly back to the boiler, thus creating the need for cooling towers or other means of cooling. Although doing so would create piping difficulties, the main reason relates to efficiency. Consider the simple system shown in Figure 1 on parge 106 with a turbine that has no frictional, heat or other losses, meaning no entropy change (isentropic).

In reality, turbines are typically 80 to 90 percent efficient. 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)

Consider 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 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 (1505.9 Btu/lbm – 1080.9 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.

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Now consider Example 2, where the system has a condenser that reduces the turbine exhaust pressure to 1 psia (approximately two 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 1505.9 – 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. Remember, Equation 5? This is a practical illustration of how the condenser lowers T_{L}.

Now think about 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 18 percent of the steam has condensed to water. However, the remaining steam takes up a specific volume of 274.9 ft^{3}/lbm. The corresponding volume of water in the condenser hotwell is 0.016136 ft^{3}/lbm. Thus, the condensation process reduces the fluid volume more than 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. 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 a primary reason why proper cooling water chemical treatment and condenser performance monitoring are important.

Next we turn to the superheater/reheater. Although the mechanics of superheating are well known, it is useful to discuss them briefly before discussing the reheater.

Consider the common drum boiler, where steam leaving the drum is saturated. If this steam were to be immediately injected into a turbine, very little work would occur because the steam would immediately begin condensing to water upon passage through the blades. That’s why all utility steam generators include superheaters. The temperature to which steam is raised above saturation represents the degree of superheat. An important point to remember is that it takes nearly 1,000 Btu to convert a pound of water to a pound of steam.

As the examples in the previous section illustrated, the enthalpy of superheated steam at 1,000 degrees F is only around 1,500 Btu/lbm. It should come as no surprise that efficiencies of standard utility boilers are in the 30 percent to 35 percent range. Research into more temperature- resistant superheater and reheater tube materials continues, in direct application of the equation, Î· = 1 – T_{L}/T_{H}. A recent article in *Power Engineering* (“Latest Steam Turbine Offerings Enhance Plant Performance,” May 2008, pp. 32-44) surveyed four steam turbine manufacturers who indicated that new materials will allow steam temperatures of 1,150 F or perhaps even higher.

Ideally, superheat energy is just completely consumed at the last, low-pressure turbine blades. A delicate balance must be maintained to extract all of the available energy from the steam but prevent excessive condensation in the low-pressure turbine blades. That’s because water droplets will cause serious blade damage. The latter aspect is an important basis behind steam reheating and operation at high pressures.

Thermodynamics show that work and efficiency of a steam generator improve with increased pressure. But consider a situation where steam pressure in increased to 2,000 psia from Example 2, in which 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. (The primary reason why supercritical [>3208 psia steam pressures] have become popular for modern coal-fired boilers is to achieve gains in efficiency through higher pressure.)

But at 2,000 psia, the turbine exit steam quality is only 77 percent. This means 23 percent of the fluid exits as condensed water droplets. Such high moisture content can damage low-pressure turbine blades. A rule of thumb suggests 10 percent moisture at the turbine exhaust as an upper limit. Reheating the steam helps alleviate this difficulty. Figure 2 on page 110 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 1474.1 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 1248.1 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 1526.5 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 where 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:

w_{t} = (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, q_{b} = (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 Boosts Efficiency

Feedwater heating provides regenerative effects to the power plant steam cycle. This illustration also builds on Example 4, and includes a single feedwater heater.

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. For multiple heaters, the temperature increase should be divided as equally as possible.

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 answer is that it improves efficiency. The heat input required by the boiler to produce the required steam is only 1,149.1 Btu/lbm (1,474.1 – 325.0), while 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.

Multiple feedwater heaters, especially in larger systems, increase efficiency even further. But at some point equipment costs offset efficiency gain. Six heaters are common in large, sub-critical systems, where five might be closed heaters with one deaerator. Interestingly, combined cycle power plants may have just one heater, often in the form of a deaerator integrally tied in with the low-pressure steam generator.

The values outlined in these examples are greater than normal because allowances were not made for heat losses in the boiler, inefficiencies in the turbine, frictional losses in the piping and other entropy-related factors. Nonetheless, these examples illustrate the fundamental principles and importance behind the operation of several important steam-generating components or subsystems. Numerous instances exist where power engineers have been involved with condenser performance improvement projects that have resulted in net savings of $500,000 to $1.5 million annually at just one plant.