EnergyFromThorium

The power conversion system (PCS) primarily exists to convert the heat generated by fission in the fuel salt of the halide reactor into useful work, most often electrical energy. The basic components of the power conversion system are: 1) a working fluid of some sort, 2) one or more heat exchangers where the fuel or an intermediate salt heats this working fluid, 3) one or more turbomachines using the working fluid to spin a turbine, 4) one or more electrical generators connected to the turbine(s), converting their rotational motion into electrical energy, 5) one or more heat exchangers cooling or condensing the working fluid before returning it to the turbomachinery, and 6) a heat rejection system releasing waste heat from the working fluid to the atmosphere, bodies of water, or some combination of both.

The principles of how a power conversion system works are typically covered in one or more semesters of an introduction to thermodynamics, often taught to mechanical engineers during their second or third years of an undergraduate education. Recognizing that many of the students using this text will not have had this introduction to thermodynamics, we will attempt to quickly cover some of the salient points of thermodynamics, allowing the student to analyze power conversion systems with a measure of success. Although a full treatment of thermodynamics will give greater insight into these principles than we will supply here, it is likely that many of the particular applications of power conversion systems, especially regarding their use in liquid-halide nuclear reactors, will be covered in far greater detail in this text than they would have been in a conventional thermodynamics course series.

In order to understand thermodynamic modeling, first consider the basic laws of thermodynamics, with which many people have a passing familiarity. The first law of thermodynamics states that energy is conserved. It can, and often does, change forms, but its total amount is always conserved. The second law of thermodynamics, at the risk of over-simplification, states that energy is always degraded. In other words, each time energy changes from one form to another, its usefulness always decreases, until finally the energy is not usefulness at all. The energy is still present, but in a form where no work can be extracted from it. An example might be a cup of hot chocolate left on a table. The hot chocolate will always cool until it reaches the same temperature as the room around it, but left alone a cup of chocolate at room temperature will never heat up on its own.

Another important thermodynamic principle is the concept of open and closed thermodynamic systems. A thermodynamic system, or simply a system, is defined as a quantity of matter or a region in space chosen for study. The mass or region outside of the system is called the surroundings. The real or imaginary surface separating the system from its surroundings is called the boundary. In a closed thermodynamic system, there is a fixed amount of mass, and no mass can cross the boundary. But energy, in the form of heat and work, can cross the boundary, and the volume of a closed system does not need to be fixed. Most of the power conversion systems that we will model for liquid-halide reactors will involve closed thermodynamic systems. On the other hand, an open thermodynamic system, often called a control volume, allows the passage of both mass and energy through a selected region of space. Some power conversion systems, like gas turbines that use atmospheric air for combustion, are modeled as open thermodynamic systems. These may find some applications in some liquid-halide reactor systems.

Next we introduce some of the different forms of energy. Some of these we are quite familiar with, such as the energy generated from the fission of a fissile nucleus, or the temperature of a flowing salt, or the velocity of salt in a channel. Thermodynamics gives us essentially no insight into the total energy of a system. Rather, it gives us insight into the change in energy. Thus, the total energy of a system can be assigned some base value of zero at some convenient reference point and all other energy states can be measured from this reference.

Any characteristic of a system is called a property. Some familiar examples of thermodynamic properties are the pressure, temperature, mass, or volume of some material within a system. Other properties of interest include density and specific volume. Specific volume is simply the reciprocal of density, or in other words, the volume per unit mass. Density and specific volume are also an excellent way to introduce the principle of an extensive or intensive property. Intensive properties are those which are independent of the size of the system. Examples include temperature, pressure, or density. If one had a volume of interest and measured its properties, and then decided to divide the volume in half and measure the properties again, one would find that the intensive properties would stay constant. Temperature, pressure, and density would measure the same as before. But the extensive properties such as mass, volume, and total energy would depend on the size of the volume. Cut the volume in half and these would get cut in half as well.

Generally, in thermodynamic equations, one finds uppercase letters used to denote extensive properties (with mass m being a notable exception) and lower-case letters used to denote intensive properties (with pressure P and temperature T being notable exceptions to this rule).

Consider a system which is not undergoing any change. At this point, all the properties can be measured or calculated throughout the entire system, which gives us a set of properties that completely describe the condition, or state, of the system. Thermodynamics deals with equilibrium states, and the word equilibrium implies a state of balance. Examples of equilibrium include temperature equilibrium, which is when a constant temperature exists throughout the system; mechanical equilibrium; chemical equilibrium; and phase equilibrium.

Any change that a system undergoes from one equilibrium state to another is called a process, and the series of states through which a system passes during a process is called the path of the process. To describe a process completely, one must specify the initial and final states of the process, as well as the path it follows, and its interactions with the surroundings. The prefix iso- is often used to describe a process for which some particular property remains constant. Important examples for our work include isothermal processes, which are processes where temperature remains constant, isobaric processes where pressure remains constant, isenthalpic processes where enthalpy remains constant, and isentropic processes where entropy remains constant. The properties of enthalpy and entropy may be unfamiliar to the reader, but will be introduced shortly.

An … process has constant…
isothermal temperature
isobaric pressure
isentropic entropy
isenthalpic enthalpy

A system undergoes a cycle if it returns to its initial state at the end of the process or series of processes. Thermodynamic cycle design will be a very important goal of this section. For instance, a helium gas turbine used as a power conversion system for a liquid-halide reactor can be modeled thermodynamically as a closed cycle. Helium would be compressed isentropically to a higher pressure, then heated isobarically to a maximum temperature using the salt from the reactor, then expanded isentropically back to the original pressure, and finally cooled isobarically back to the original temperature at which it started.

Designing thermodynamic cycles, using a series of processes, is the key objective of this section. The particular processes used to achieve a cycle, in combination with the working fluid chosen, are what define classes of cycles. For instance, the Rankine cycle is commonly used to generate electrical power using water as a working fluid. It is a closed cycle which consists of an isothermal compression of liquid water, followed by heating of the water to the point where it begins to change phases to steam, throughout that entire phase change process, and a degree of heating beyond that point. Then the hot, high-pressure gas consisting entirely of steam is expanded isentropically through a turbine, producing work which turns an electrical generator. The low-pressure steam is then cooled throughout its condensation process until only liquid water remains, at the same pressure at which it began. This series of processes, culminating in a fluid state identical to one in which it began, is a thermodynamic cycle.

In order to begin cycle analyses, a fairly rigorous understanding of each thermodynamic property is needed. Some will be fairly familiar, while others may be entirely new to the reader.

The property of pressure is usually one of the familiar ones. Pressure is the force exerted by a fluid per unit area. We speak of pressure only when we speak of a liquid or a gas. The counterpart of pressure in a solid is stress. For a fluid at rest, the pressure is the same in all directions. The unit of pressure is the pascal, which is defined as one Newton of force per square meter. The pascal is a small unit of pressure, so multiples of this unit are more commonly encountered, such as the kilopascal and the megapascal. Another variant is the bar, which is a value of pascals between the kilopascal and the megapascal, defined as 100,000 pascals. The bar is especially useful since it is within 1 percent of atmospheric pressure at sea level (101,325 pascals). Giving pressure readings in bars allows one to stay with metric or SI units while expressing pressure in a unit that is more familiar to non-engineers (an atmosphere of pressure).

Pressure measured against a vacuum reference is called absolute pressure, while pressure measured against the local atmospheric pressure is called gage pressure. Obviously, absolute pressure is the value needed when doing state calculations, whereas gage pressure is needed when computing the force on an area surrounded by atmosphere.

Just as pressure is measured against a vacuum reference for property calculations, in thermodynamics temperature is also measured against an absolute reference, in this case, absolute zero. The commonly used unit of absolute temperature is the Kelvin, named after Lord Kelvin (1824-1907). It has the same increment as a degree in the Celsius scale but is defined to begin at absolute zero (-273.15degC). Thus ice melts at 0 deg C and 273.15K, and water boils at 100 deg C and 373.15K. The Kelvin temperature scale is simply offset from the Celsius scale by the value of absolute zero. Because of this feature, when an equation or value has a temperature increment in Celsius degrees, it is synonymous with the same increment in Kelvin. For instance, if it takes one calorie of heat to raise one gram of water one degree Celsius, it will still take one calorie of heat to raise that water one Kelvin. However, actual values of Celsius temperatures and Kelvin temperatures are not synonymous. Zero degrees Celsius is very different than zero Kelvin. A similar relationship exists in the English/imperial system between the Fahrenheit and Rankine scales.

Density, as previously mentioned, is simply the mass per unit volume, and its reciprocal, specific volume, is simply the volume per unit mass. Often in thermodynamics, and especially in gas modeling, the density is given in molar density rather than in mass density. Molar density is the moles of a substance per unit volume rather than the mass. It can be found quite easily by multiplying the mass density by the molar mass. For example, if a gas had a mass density of 16 kg/m3, and had a molar mass of 1 mol/g or 1 kmol/kg, then the molar density would be 1 kmol/m3.

Internal energy of a gas or liquid is a thermodynamic property consisting of the various forms of molecular energy internal to the medium. There is rotational kinetic energy, translational kinetic energy, and vibrational kinetic energy. Of these, translational kinetic energy is the most common in gases that will be of interest to us, and is directly proportional to temperature. Therefore, internal energy can be considered to be, for the great majority of cases, a function only of temperature. In the more rigorous sense, internal energy will be a function of temperature and pressure for certain gases in certain regimes.

Enthalpy is a combination property that uses the symbol H and is defined as the addition of internal energy and the product of pressure and volume.

H = U + PV, or in unit mass form, h = u + Pv

Both the total enthalpy and the specific enthalpy are referred to as “enthalpy” because the context will make clear which one is meant. The widespread use of the term enthalpy came from Professor Richard Mollier, who recognized the importance of the group u + Pv in the analysis of steam turbines. The term enthalpy comes from the Greek work enthalpien which means “to heat”. Despite the similarity of the word enthalpy and entropy on first appearance, they are very, very different thermodynamic properties and should not be mixed up.

Defining entropy is something that often occupies chapters of a thermodynamics text and daunts even those who understand it. Some describe entropy simply as disorder, but that is too simplistic. Many have heard the word but only have a vague idea what it really means. To truly define entropy requires introducing the concepts of statistical thermodynamics and quantum mechanics, which even a mechanical engineering student might not encounter until postgraduate studies. But in an attempt to give a quick and fairly accurate depiction of what entropy is—entropy is the ability of a substance to occupy more and more discrete quantum states.

Consider a checkerboard filled with black pieces except for one empty square. Another black piece has only one location it can occupy on the checkerboard. Then consider an empty checkerboard and that same black piece. Now it could occupy any one of the 64 locations on the board. The black piece on the empty checkerboard has more entropy than the same black piece on the nearly-full checkerboard, because it could occupy any one of 64 states whereas the nearly-full checkerboard has only one state that could be occupied.

Consider another pair of checkerboards, one with 64 black pieces and another with 63 black pieces and one white piece. The latter checkerboard could have the white piece occupy any one of the 64 squares, with the black pieces occupying the others. But the former checkerboard has only one configuration—the board full of black pieces. The latter checkerboard has substantially more entropy than the former checkerboard, because of the greater number of states its pieces can occupy.

Consider another example closer to our personal experience—a messy bedroom. Perhaps when your bedroom is “clean” there is only one state for everything in the room. The bed is made, the clothes folded and packed in drawers, the floor vacuumed and the windows clean. But when the room is “messy” there are nearly an infinitude of different states that might exist—the clothes might be hanging on the end of the bed, the sheets might be down and rumpled, there might be clothes on the floor. You set about to reduce the entropy of your room by “cleaning” it. You move the items from a great variety of states back to the single state of “clean”. In doing so, you expend energy, sweat, digest food, and otherwise increase the overall entropy of the universe even as you reduce that of your room. No matter how hard you try, or how much you might wish you had a magic wand like a fairy godmother from a children’s story that could clean your room without effort, you cannot reduce the entropy of your messy room without increasing the overall entropy of the universe in some other way.

Although checkerboards and black and white pieces and dirty and clean rooms may seem to have only a tenuous connection to gases, liquids, states, properties, processes, cycles, and power conversion systems, the principles of entropy are absolutely central to the analysis of cycles for power conversion systems, because cycles that increase entropy the least have the highest efficiency and may offer the most economic forms of energy generation.

One of the basic principles of entropy is that the overall entropy of the universe is always increasing. The universe is always moving from an existence into another existence with more potential states. The entropy of a closed volume may be reduced, but somewhere else entropy is increasing to compensate for that local entropy reduction. The best that can be done in this situation is to attempt to define processes where entropy does not increase at all. Such processes are called isentropic processes, and although in the strictest sense they do not exist, there are many real processes that come quite close to being isentropic processes. At the very least, the definition of an isentropic process is one bounding the possible performance of a real process.

For instance, an isentropic compressor can compress a gas without increasing its entropy, but a real compressor may do nearly that well, achieving 90% of the performance of an isentropic compressor. But in no case will a real compressor achieve 120% of the performance of an isentropic compressor. It is an ideal that cannot be achieved, only aspired to. Ideal processes combined into ideal cycles are particularly useful in thermodynamics because they allow us to examine the “best-case” scenario for a given series of constraints. For example, if we are told that the maximum temperature of the coolant salt is 1000K, and that the seawater coolant around a reactor is at 300K, we can use an ideal cycle analysis to compute what the maximum efficiency of the thermodynamic cycle might be between those temperature limits (70%, if you’re really curious). A real cycle will never be more efficient than that.

This discussion of entropy has also given us no particular insights into the units used to express entropy, nor how entropy is to be depicted mathematically. The units of total entropy (depicted by a capital S) are energy per unit temperature, for instance, joules per Kelvin (J/K). In a similar manner, specific entropy (depicted by a lowercase s) is given in energy per unit mass and temperature, such as joules per kilogram-Kelvin (J/(kg*K)).

Calculating Processes

Earlier we discussed how a thermodynamic process involves moving from one thermodynamic equilibrium state to another, and how these processes often involve holding one thermodynamic property constant during the process. Processes that hold a property constant have convenient mathematical definitions, which will be exceptionally helpful to us as we calculate the different states of the system as it executes the processes of a cycle.

One of the first expressions we will use is the expression relating the pressure ratio across a process to the temperature ratio across the same process, assuming that the entropy is constant, or in other words, that it is an isentropic process.

(T2/T1) = (P2/P1)^(k-1)/k

In this equation, P1 represents the pressure of the gas at the beginning of the process, P2 represents the pressure at the end of the process, and in like manner, T2 and T1 represent the gas temperatures at the beginning and end of the processes, and k is the ratio of the gas’s specific heats. All of this is valid only under the assumption that the process does not change the entropy of the gas, which may seem to be a very esoteric concept at this stage of analysis, but don’t worry, it will become clearer through example. Undoubtedly, even the concept of what the entropy of a gas is right now is probably not very clear, but it will become clearer as we begin to use it in equations and processes.

Assembling Processes into Cycles

A cycle, as we previously mentioned, is an arrangement of processes that returns the system to its starting state. We will introduce the concept of a cycle with a particular cycle that will be very important to us in our ultimate objective, analyzing power conversion cycles for liquid-halide nuclear reactors.

We will learn cycles initially through the Brayton cycle. The Brayton cycle was developed by an American engineer, George Brayton, in the 1870s. The Brayton cycle was originally proposed for a reciprocating oil-burning engine, but now it is most commonly used in large aircraft gas-turbine engines, as well as stationary gas-turbine engines for power generation. The simple ideal Brayton cycle consists of four processes. Two of the processes are isentropic, and two are isobaric; the two isentropic processes involve machinery (a compressor and a turbine) and the two isobaric processes involve heat exchangers.

Starting at the temperature and pressure of state 1 (the initial state) the gas is compressed isentropically to a higher pressure at state 2.

From state 2 the gas is heated at constant pressure to state 3.

At state 3 the gas is expanded isentropically back to the initial pressure (albeit at an elevated pressure) at state 4.

From state 4 the gas is cooled at constant pressure back to the temperature of state 1.

A simple Brayton cycle has only two pressure levels, the initial pressure and the elevated pressure. If we make the assumption that the gas we are using as the working fluid is an ideal gas, and that it has constant specific heats (which are generally pretty good assumptions), then the actual values of initial pressure and final pressure don’t matter. What matters is their ratio, or what is simply called the pressure ratio. For instance, if the initial pressure of one simple Brayton cycle is 1 bar and the higher pressure is 10 bar, then the pressure ratio is 10. If another Brayton cycle has an initial pressure of 10 bar and an elevated pressure of 100 bar, then its pressure ratio is also 10 and we would see identical efficiency and net work from both Brayton cycles, given the same temperature limits. This independence from actual values of pressure is a very important feature of Brayton cycles and distinguishes them from other common power-generation cycles like Rankine cycles.

Now we need to apply some mathematical expressions to our simple ideal Brayton cycle so that we can calculate parameters that matter to us, like the efficiency of the cycle and how much net work it produces. We can use the expressions we previously developed to describe isentropic and isobaric processes, helping us describe the Brayton cycle.

Process 1-2: isentropic compression

From an initial pressure and temperature, we will compress the gas isentropically to a higher pressure. We assume that we know the initial pressure and the final pressure, as well as the initial temperature. What we desire to know is the final temperature at the end of the process. We also assume that because specific heats are constant, that knowing the temperature change tells us the change in enthalpy, and from the change in enthalpy, we know the amount of work required to execute the process.

T2 = T1*(P2/P1)^(k-1)/k

Let us assume that we have an initial pressure of 1 bar and a final pressure of 3 bar. This yields a pressure ratio (P2/P1) of 3. Let us assume that the working fluid is helium with a ratio of specific heat of 5/3. Let us also assume that the initial temperature is 300K. With these values inserted into the expression we get:

T2 = (300K)*(3 bar/1 bar)^(5/3 -1)/(5/3) = (300K)*(3)^(2/5) = 465.5K

Thus, after compressing helium isentropically from 1 bar to 3 bar, its temperature has been raised from 300K to 465.5K. How much work was consumed in compressing the helium? Now that we know the initial and final temperatures, we can compute the enthalpy change, and the amount of work consumed. The enthalpy is simply the specific heat at constant pressure multiplied by the temperature. The specific heat at constant pressure for helium is 5.1926 kJ/(kg*K).

Dh = h2 – h1 = cp(T2 – T1) = (5.1926 kJ/(kg*K))*(465.5K – 300K) = 859.6 kJ/kg

Therefore, the isentropic compression process consumed 859.6 kJ for each kilogram of helium it compressed from the lower pressure to the higher pressure. If this was expressed as a power and flowrate, then it would be equivalent to saying that it required 859.6 kW of power (kW = kJ/s) to compress each kg/s of helium flowrate from the lower pressure to the higher pressure.

Process 2-3: isobaric heating

Next we turn our attention to understanding the next step of the ideal Brayton cycle, the heating of the gas from its post-compression temperature to its maximum temperature of the cycle. We make the assumption that this heating takes place without any reduction in pressure or pressure loss. This is, of course, impossible in real practice, but we will not worry about this during our analysis of this ideal cycle. This maximum temperature that was just mentioned is something that we assume that we know in our analysis of the cycle, and is usually called the “turbine inlet temperature” since it is the temperature of the gas prior to entering the turbine that will execute process 3-4. Let us assume that the turbine inlet temperature of our cycle is 1000K. We desire to know that amount of heat that must be added to the gas in this stage. We know the temperature at the beginning of the process (465.5K) and we know the temperature at the end, so we can compute the enthalpy change during the process and the heat that must be added:

Dh = h3 – h2 = cp(T3 – T2) = (5.1926 kJ/(kg*K))*(1000K – 465.5K) = 2775 kJ/kg

Now it may seem somewhat strange that we are using enthalpy to describe the amount of work needed to compress the gas through our first process, and then turn around and use enthalpy again to describe the amount of heating required to heat the gas prior to its introduction into the turbine, but enthalpy can be used to describe both processes. One consumes energy in the form of work, the other consumes energy in the form of heat.

Process 3-4: isentropic expansion

After being heated to the maximum temperature of the cycle, the gas is expanded isentropically back to the lower temperature. Once again we can use the same expression we used previously to find the temperature of the gas at the end of this process, and in so doing, to find the enthalpy released as work by the expansion.

T4 = T3*(P4/P3)^(k-1)/k
T4 = (1000K)*(1 bar/3 bar)^(5/3 -1)/(5/3) = (1000K)*(3)^(2/5) = 644.4K

Dh = h3 – h4 = cp(T3 – T4) = (5.1926 kJ/(kg*K))*(1000K – 644.4K) = 1846 kJ/kg

We are nearly finished with our simple analysis of this simple cycle, but already we can notice several very important things. In our first process, it required 860 kJ to compress our gas from the initial pressure and temperature up to 3 bar and 465K. But now in our turbine we have generated 1846 kJ expanding that same gas from 3 bar back to 1 bar. If we assume that we need to “pay back” the work expended in the compression from the work generated by the expansion, then we still have (1846 – 860) = 986 kJ or work “left over” from the expansion. What does this 986 kJ of “net work” represent? It represents that actual net work of this simple ideal Brayton cycle. This is the amount of work we get from executing a simple Brayton cycle from helium, with a pressure ratio of 3 and a compressor inlet temperature of 300 K and a turbine inlet temperature of 1000 K.

A natural question to ask would be—is that “good”? And we can only answer that question by finishing our analysis of the cycle and comparing several of its factors against “ideal” cycles that represent the limits of thermodynamic performance.

Let us complete our analysis of this simple cycle by finishing the last step, and cooling the gas from the 644.4K of the turbine outlet to the 300K of the compressor inlet:

Dh = h4 – h1 = cp(T3 – T4) = (5.1926 kJ/(kg*K))*(644.4K – 300K) = 1788 kJ/kg

This last step will require removing 1788 kJ of enthalpy from each kilogram of working fluid. Now with each of our steps complete, we can summarize what we learned from the cycle:

Isentropic compression of helium from 1 to 3 bar, starting at 300K consumed 859.6 kJ/kg.

Isobaric heating of helium from the outlet temperature to 1000K consumed 2775 kJ/kg.

Isentropic expansion of helium from 3 to 1 bar, starting at 1000K, released 1846 kJ/kg.

Isobaric cooling of helium from the outlet temperature to 300K extracted 1788 kJ/kg.

Now we can quantify two important parameters of our cycle—its efficiency and its net work. The efficiency is defined as one minus the heat added divided by the heat extracted:

1 – (1788/2775) = 35.5%

The net work, as we previously computed, is simply the work released by the turbine less the work consumed by the compressor:

Net work = (1846 kJ/kg – 859.6 kJ/kg) = 986 kJ/kg.

Now we can state for this simple cycle that it is able to convert heat to work at an efficiency of 35.5% and a net work of 986 kJ/kg.

Now, having gone to all this effort to find the efficiency of a simple Brayton cycle between these temperatures and these pressures, it may come as somewhat of a disappointment to be told that there is a simple expression that can give the efficiency of such a cycle for any pressure ratio, but it is the case:

Efficiency = 1 – (P2/P1)^(1-k)/k

For our problem with a pressure ratio of 3 and a ratio of specific heats of 5/3, this becomes:

Efficiency = 1 – (3)^(1- 5/3)/(5/3) = 1 – (3)^(-0.4) = 35.5%

which is in exact agreement with our previous calculations. Sadly, there is no similar simple calculation for net work, so we must evaluate the processes of the cycle to compute that value.