When I began my power industry career years ago (initially in the chemistry lab of a coal-fired power plant), the criticality of proper chemistry control of steam boilers, cooling water systems, and other equipment became quickly apparent. But to understand how the various steam generator sub-systems fit in with the process as a whole, a general knowledge of plant thermodynamics is quite valuable. I wrote about several of the most important Rankine cycle basics in 2008 for Power Engineering, but now, traditional fossil-fired plants are being replaced with combined cycle units that employee combustion turbines (Brayton cycle) for the bulk of power generation, but with Rankine cycle heat recovery steam generators (HRSGs) for additional power generation. A review of some of the basic steam thermodynamics from the original article seems warranted at this time, given the many new employees to the power industry. This revision omits the previous discussion of feedwater heaters, as most if not all HRSGs are not equipped with these devices, other than economizers. The focus this time is upon the steam turbine/condenser and superheater/reheater.

A Brief Review of Thermodynamic Definitions

The word “thermodynamics” conjures up visions of complex mathematics. Yet, relatively straightforward formulas can be used explain much about steam generator performance.

Thermodynamics is built around two laws. They are sometimes jokingly referred to as (first law), “You can’t get something for nothing,” and (second law), “You can’t break even.” In actuality, 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 – W s = ṁ 2 [V 2 2/2 + gz 2 + u 2 + P 2 υ 2 ] – ṁ 1 [V 1 2/2 + gz 1 + u 1 + P 1 υ 1 ] + dE c.v. /dt Eq. 1

Where,

Q = Heat input per unit time

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

ṁ 2 = Mass flow out of the system per unit time

ṁ 1 = Mass flow into the system per unit time

(V 2 2 – V 1 2)/2 = Change in kinetic energy

gz 2 – gz 1 = Change in potential energy

u 2 = Internal energy of the exiting fluid

u 1 = Internal energy of the entering fluid

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

P 1 υ 1 = Flow work of fluid as it enters the system

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

While this equation may look complicated, it can be readily 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 dE c.v. /dt is zero. Removal of these terms leaves the internal energy of the fluid (u) plus its flow work (Pυ) 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 a 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 0oC has been designated as having zero enthalpy. From these simplifications and definitions, the energy equation for steady state operation in a turbine reduces to:

Q – W s = ṁ(h 2 – h 1 ) Eq. 2

But this equation represents the ideal scenario, and here is where the second law steps in. Among other things, the second law describes process direction. A warm cup of coffee does not become hotter while the room grows colder. My office becomes more cluttered, not less, and so on.

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 (Q H ) at high temperature (T H ) and a heat discharge (Q L ) at low temperature (T L ), in which

Q H /T H – Q L /T L = 0 Eq. 3

This equation represents a theoretically ideal engine, where the heat utilized by the process is completely converted to work. For every real-world cycle or process, some energy losses occur, so in the real-world, Equation 3 becomes:

Q H /T H – Q L /T L < 0 Eq. 4

Losses 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 terms that apply to this discussion, is based on the ratios 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, as is reflected in Eq. 4.

While entropy may seem like a somewhat abstract concept, it is of great benefit in determining process efficiency. Like enthalpy, entropy values are included in the steam tables, and they can be used to determine process efficiency.

Two important points should be noted about the Carnot cycle, and by logical inference, other processes. First is that no process can be made to produce work without some extraction of heat, Q L in Equations 3 and 4. Q L in a steam power system is heat removed in the condenser (Q C ), as Figure 1 shortly illustrates.

Second, overall net efficiency (η) of a Carnot engine is defined as: η = 1 – T L /T H Eq. 5

So, as input temperature goes up and/or exhaust temperature goes down, efficiency increases. This aspect continues to drive research into advanced metals to handle higher steam temperatures.

The Turbine/Condenser

I have been asked on a number of occasions throughout the years why turbine exhaust steam must be condensed. Why not transport it directly back to the boiler? The primary answer is efficiency. Consider the basic system shown below with a turbine that has no frictional, heat or other losses, which means no entropy change (isentropic).

Fig. 1. A basic steam generation system with isentropic turbine.

In reality, 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 – 2000 psia

Main Steam Temperature – 1000oF

Turbine Outlet Steam Pressure – Atmospheric (14.7 psia)

Call this Example 1. The steam tables show that the enthalpy of the turbine inlet steam is 1474.1 Btu per pound of fluid (Btu/lbm). Thermodynamic calculations indicate that the exiting enthalpy from the turbine is 1018.2 Btu/lbm (steam quality is 86%). Equation 2 (the simplified first law, steady-state energy equation) becomes for the turbine, w T = ṁ(h 1 – h 2 ). Accordingly, the unit work available from this ideal turbine is (1474.1 Btu/lbm – 1018.2 Btu/lbm) = 455.9 Btu/lbm. To put this into practical perspective, assume steam flow (ṁ) to be 1,000,000 lb/hr. The overall work is then 455,900,000 Btu/hr = 133.6 megawatts (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 871.0 Btu/lbm. The unit work output equates to 1474.1 – 871.0 = 603.0 Btu/lbm. At 1,000,000 lb/hr steam flow, the total work is 603,100,000 Btu/hr = 176.7 MW. This represents a 32% 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. The unit volume of steam at the saturation pressure of the turbine exhaust (1 psia) is 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 by a factor of over 17,000. The condensing steam generates the strong vacuum in the condenser, which acts as a driving force to pull steam through the turbine. (It also generates the strong vacuum that can pull in air through minor openings, which in turn may cause efficiency losses.)

Let’s take this concept a step further. Suppose that waterside fouling or scaling (or excess air in-leakage on the steam-side) 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 603.1 to 568.5 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 34,600,000 Btu/hr or 10.1 MW of work. This is a primary, but not the only, reason why proper cooling water chemical treatment and condenser performance monitoring are very important. [3,4]

Superheating/Reheating

Ponder the common drum boiler, where the steam leaving the drum is saturated. If this steam were to be directly 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, and others, all fossil-fired utility steam generators include superheaters. The temperature to which the steam is raised above saturation represents the degree of superheat. A critical aspect to remember is that it takes nearly 1000 Btu to convert a pound of water to a pound of steam. As the examples above illustrated, the enthalpy of the superheated steam is only around 1500 Btu/lbm, so it should be recognizable that efficiencies of standard utility boilers are in the 30- to 35-percent range. The non-availability of the latent heat in conventional units is a primary factor why cogeneration energy production continues to increase in popularity, where the steam’s latent heat is utilized for process heating or other purposes. Overall net efficiencies of up to 80% have been reported for some co-generation applications. Net efficiencies of slightly over 60% are the maximum for modern combined cycle power generating units.

Ideally, superheat energy is just consumed at the last, low-pressure turbine blades. A delicate balance is necessary to extract all of the available energy from the steam but prevent excessive condensation in the low-pressure blades, as the water droplets will cause serious damage. The latter aspect is an important basis behind steam reheating, especially for high pressure units. In Example 2, the turbine exhaust steam quality is only 77%. This means that 23% of the fluid exits as condensed water droplets. (Had we accounted for entropy loss through the turbine, this value would have been a bit less but still excessive.) Such high moisture content is troublesome. A common rule of thumb suggests 10% moisture at the turbine exhaust as an upper limit.

Figure 2 shows a steam generator and turbine with a reheat system.

Fig. 2. Steam generator with reheater.



Main steam is at 2000 psia, 1000oF, and has an enthalpy of 1474.1 Btu/lbm. Condenser pressure is 1 psia. The steam extraction (cold reheat) pressure is 600 psia, which equates (isentropically) to a cold reheat temperature of 650oF and enthalpy of 1320.3 Btu/lbm. Assume negligible pressure drop through the reheater and a hot reheat temperature of 1000oF. This produces reheated steam with an enthalpy of 1517.8 Btu/lbm. Calculations from this basic example show that the reheating process improves the turbine exhaust steam quality from 77 percent to 86 percent. And, had typical values for entropy increase been factored into the calculations, the moisture content would probably be at or below the 10% recommended guideline.

The turbine exhaust enthalpy for this isentropic case is 958.7 Btu/lbm. Calculation of the work output 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: w t = (main steam enthalpy – cold reheat enthalpy) + (hot reheat enthalpy – turbine exhaust enthalpy).

In this instance, w t = (1474.1 – 1320.3) + (1517.8 – 958.7) = 712.9 Btu/lbm. Reheating considerably increases the work output as compared to the non-reheat example. Of course, reheating requires more energy input to the steam generator, but improved efficiency and higher steam quality at the turbine exhaust make reheating de rigueur for most high-pressure units.

Space limitations prevent a discussion of the influence of pressure on thermodynamic efficiency, but higher pressure typically results in higher efficiency. That is why in those countries that still rely heavily on coal-fired power, new units are often of the supercritical (>3208 psia) or ultra-supercritical design. Overall net efficiencies of some of these plants are in the mid-40 percent range.

Conclusion

The power production values outlined in the above examples are greater than normal because no accounting was made of heat losses and inefficiencies. Nonetheless, these examples illustrate the fundamental thermodynamic principles behind the operation of several important steam-generating components and subsystems. Much effort has been expended over the decades to optimize plant performance, with thermodynamics as the foundation.

Buecker

(Note: This discussion is based on the author’s personal investigations, and it does not include any input from his current employer.)

About the author: Brad Buecker is senior technical publicist with ChemTreat. He has 35 years of experience in or affiliated with the power industry, much of it in steam generation chemistry, water treatment, air quality control, and results engineering positions with City Water, Light & Power (Springfield, Illinois) and Kansas City Power & Light Company’s La Cygne, Kansas station. He also spent two years as acting water/wastewater supervisor at a chemical plant. Buecker has a B.S. in chemistry from Iowa State University with additional course work in fluid mechanics, energy and materials balances and advanced inorganic chemistry.

Brad Buecker is a regular contributor to Power Engineering. Click here to see many of his other PE stories.

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