The Rankine Cycle is one of many
thermodynamic cycles that is used to produce mechanical work. This cycle in particular is used to predict the work produced by a steam turbine system in a heat engine. Heat engines are commonly found in things like trains and air conditioners.

The Rankine cycle consists of four different components: a steam generator of some kind (for example, a boiler), a steam turbine, and condenser, and a pump. Each component changes the state and properties of the fluid that moves through it, by adding and taking away heat and work, in order to transfer energy from heat to work.

Steam generator: Turns the fluid at room temperature and pressure, to vapor or steam at an extremely high temperature, usually termed "superheated vapor.”

Turbine: The pressure is increased, resulting in a drop in temperature, and producing mechanical work that leaves the system to be used.

Condenser: Operates at a constant pressure where heat is released into the environment, compressing the state of the fluid back down to a liquid-gas mixture, that is mostly liquid. This state is otherwise known as “saturated liquid.”

Pump: Pressurizes the saturated liquid back up to the pressure of the turbine by bringing work back into the system, turning to a highly pressurized liquid which is usually referred to as a "compressed liquid.”

Finally, the fluid goes back through the steam generator at constant pressure to return to a superheated vapor.

Step 1: Understood Idealizations and Assumptions

In order to easily analyze the Rankine Cycle, there are many
idealizations that engineers commonly make, so that the system can be understood in ideal situations. Additionally, there are some things throughout the cycle analysis that we will consider to be 'ideal' meaning they work without any imperfections. Engineers tend to make these assumptions when analyzing systems, in order to have an accurate representation of what is going on, but the fact is that nothing ever works out exactly how we expect or calculate, and does not act the same way every time. This is especially true in the analysis of the Rankine Cycle. The reason that we say we are analyzing a "Simple, Ideal Rankine Cycle" is because it is exactly that; simplified and ideal. In reality there are dozens of things that would have to be accounted for to get an exact calculation, but they will be ignored for now in order to get a fundamental understanding of the cycle itself.

First, it is important to mention a few of the aspects of the cycle that make it less than 100% efficient, called irreversibilities. In each component there are characteristics that make the cycle as a whole less capable of creating energy. These include, friction created between the fluid and component, component structure imperfections, impure fluid, and heat loss/ fluid leakage to the environment. These, along with many other circumstances that just arise naturally take away from our ability to transfer all of the heat and work we put in to usable heat and work. After analyzing each state of the cycle, we will calculate the work we put in and work we got out, as well as the heat we put in, allowing us to calculate how efficient the cycle we used in the example was.

The concepts and vocabulary needed to consider the system ideal are:

Adiabatic: Does not lose heat.

Isentropic: Adiabatic and reversible (without any of the above irreversibilities) or no change in entropy. Used when analyzing the turbine and pump

Isobaric: Constant pressure; considered when analyzing the steam generator and condenser

Steady State: the state of the fluid does not change with time.

Steady Flow: that the mass of fluid flowing through the system is constant everywhere

Boundary Work: only consider changes that happen across the boundary of each component. There are other things that go on inside each component, but they are neglected for simplicity, since the final result would not be that different if they are included.

Step 2: Diagram

The First Step in the analysis is to
simply draw a diagram of the system.

Since we are doing the most fundamental of Rankine Cycles, the diagram is relatively straight forward. The Turbine is drawn as a sideways trapezoid to help depict the expansion of the fluid and which way it is flowing. Arrows are also used to help show direction, and the states (1-4) are labeled on the connections between each component because we are looking at the fluid before and after it undergoes each process; not while it is inside the component. Arrows are also used to show where heat and work are going in and out of the system. This will help in the calculation of the work at the end of the analysis.

Step 3: State Table

The setup of the State Table is one
the most crucial parts of analyzing a Rankine cycle, because without proper notation, and tracking of values it becomes extremely easy to lose or confuse values. The best way to set up a state table (as shown) is to have each state go down the right hand side, and each property across the top. Units are an important thing to keep track of, in thermodynamics in general, and are best shown at the top of the State Table, next to the property value. It is important to remember as you calculate, that each number should be considered in the specified units. In our example we will use consistent units in the English system.

One thing that is important to understand about filling out the State Table, is knowing what physical state the fluid is at the state of the cycle being analyzed. There are three states are:

Compressed Liquid: a liquid that has been extremely pressurized.

Liquid-Vapor Mix: the fluid is made up of both liquid and vapor and is further characterized. A liquid-vapor mix can either be a saturated liquid, saturated vapor, or have a quality. Saturated liquids are almost completely liquid, and if exposed to any more pressure will become a compressed liquid. Saturated vapors, which are at the other end of the spectrum, is almost all gas, and exposed to any

more pressure will become a
superheated vapor. Fluids characterized by quality are somewhere in the middle of the two, and are assigned a value based on the amount of gas in the mixture.

Each Property that we will be finding on the State Table serves a specific purpose:

Pressure (P): One of the most important properties, because it is usually given, allowing us to easily look up other properties in our State Tables, and gives us an easy gage for where we are as we travel through the cycle.

Temperature (T): Is not always necessary to record but is used to give you of an idea of where you are in the cycle, and a gage that is easily understood. It is also listed in the State Tables, so it is sometimes useful to look up other properties.

Entropy (s): An abstract concept used to represent the changes that occur to heat due to temperature.

Enthalpy(h): Measure of the internal energy within a system. Each component analyzed will be considered a 'system', allowing us to find an enthalpy value for each. This is essentially how the heat transfers through each of the components, and how we keep track of its increases and decreases. Enthalpy is the property we will use to calculate work at the end according to the First Law of Thermodynamics, to be introduced later.

Specific volume (v): Calculated based on the volume of fluid per unit mass (different from total volume, which is unnecessary) and is used in this cycle to calculate the enthalpy out of the pump.

Quality (x): Measure of a fluids liquid-gas mixture. If the fluid is in a super-heated vapor state, or compressed liquid state it will not have a quality. Otherwise, it is a mixture of liquid and gas and the value will range from 0 to 1, 0 being a saturated liquid (still unpressurized, but very close to the pressure limit) and 1 being a saturated vapor (steam, on the verge of being superheated).

Because calculating some properties can only be done via experimentation, the properties for liquid-gas mixtures, and superheated vapors are listed in tables that are used by engineers in analysis. There are different columns listed under the liquid-gas mixture table for saturated liquid and saturated vapor, which can be used along with the quality value to find the property for and quality fluid. Compressed liquid properties are not listed, because when taking their properties, it is accepted to assume the same properties of a saturated liquid because the values are extremely close.

Step 4: Given Values

In order to start any sort of
calculation a few things have to be given. It is first, important to know that to acquire all of the properties of any given state, you must first have two independent properties to calculate the rest. An independent property is one that does not change according to another property. For the sake of simplicity, the reasoning behind why properties are independent or dependent will be ignored, and the equations will just be used as they are given. Finally, in order to know anything about the cycle, we must be given a starting point. This usually consists of the Temperature and Pressure at State 1, and the Pressure at State 2. These values are marked on the State Table shown as P1, T1, and T2. This is how we will notate our values to keep track of them.

Step 5: Calculating State Property Values

Now that you have been introduced to
all of the states, properties, components and assumptions necessary to complete a basic analysis of the Rankine Cycle, it is time to begin our calculations. In order for everything to be easier to follow, the process will be explained, then written mathematically. Finally, for the purpose of making sense of all the equations and to better understand the calculations use the following example:

Water is the working fluid in an ideal Rankine Cycle, flowing at a rate of 1.8*10^6 lbm/hr. Steam enters the turbine at 1400 lbf/in.2 and 1459R. The condenser pressure is 2 lbf/in^2 and the mass flow rate is 20 lb/s Cooling water experiences a temperature increase from 608F to 768F, with negligible pressure drop, as it passes through the condenser.

State One: Turbine:

The initial pressure, P1 and temperature, T1 are given as 1400 lbf/in^2 and 1459 deg. R, respectively Therefore, we can easily record the remaining values x1, h1, and s1 from the superheated vapor table.

P1=1400 lbf/in^2 T1=1459 deg. R h1=1493.5 Btu/lbm s1=1.6094 Btu/lbmR

State Two: Condenser:

The condenser properties can now be found by identifying the two independent properties, pressure, P2 and entropy s2. Pressure is given as 2 lbf/in^2. You also know that the fluid goes isentropically (entropy, s, doesn’t change!) through the turbine from state 1 to state 2 so the values will be equal. After noting these values, turn to the saturated mixture table where at pressure 2 lbf/in^2 you will go across the table to the s values and notice that the value is between the saturated liquid value, s_f=0.1750 and saturated vapor value, s_g=1.9198. This is where the idea of quality comes into play.

When dealing with a saturated liquid there is an equation that allows you to use the saturated liquid value, s_f, the saturated vapor value, s_g, and the quality to find the value for the specific mixture in question. This equation is valid for specific volume, v, enthalpy, h, and entropy, s, and is as follows:

#mix = #_f + x(#_g - #_f), where # represents the property in question. (i.e. s3=sf+X(sg-sf))

In this case, you will use this equation with entropy s2, and the found s_f and s_g values observed earlier to find the quality.

1.6094=0.1750+x(1.9198-0.1750) giving a quality x2=.8221

From this quality we can find the enthalpy, h2 using the same equation:

h2=94.02+.8221(1116.1-94.02) giving enthalpy h2=934.02.

P2=2 lbf/in^2 x2=.8821 h2=934.02 Btu/lbm s2=1.6094Btu/lbmR

State Three: Pump:

The only thing that changes from state 2 to state 3 across the condenser is the quality of the fluid, which goes from a liquid-vapor mixture to a saturated liquid. This just means the quality, x3=0. The pressure across the condenser also stays constant, giving you two independent properties, P3=2 lbf/in^2 and x3=0, from which you can now find the remaining properties in the saturated liquid-vapor table. One thing that is unique about state 3 is that the specific volume is necessary for a later calculation.

P3=2 lbf/in^2 v3=.01623 ft^3/lbm h3=94.02 Btu/lbm s3=.1750 Btu/lbmR T3=585.04R

State Four: Steam Generator:

The two independent properties that are used here come simply from known characteristics of the cycle. First, the pressure increases across the pump back to the pressure that is needed to go through the turbine at state 1, because the fluid remains at a constant pressure across the steam generator. Because of this P4 must be equal to P1, 1400 lbf/lbm. Second, the pump works insentropically, just like the turbine, meaning that the entropy for state 4 will equal that at state 3; s3=s4=0.1750 Btu/lbmR. Now enthalpy at state 4 is the only other property needed. Given the properties we already have, the best way to find h4 is by using the definition of enthalpy; the change in enthalpy is equal to the specific volume multiplied by the change in pressure. Mathematically:

h4=h3+v3(P4-P3)

using this equation with your known values to solve for h4 gives:

h4=94.02+.01623(1400-2 =98.25

(there is a unit conversion here that for simplicity, can be disregarded without effected the final value)

P4= 1400 lbf/lbm h4=98.25Btu/lbm s4=.1750 Btu/lbmR

Step 6: Calculating Work, Heat and Efficiency

Now that all the calculations have
been done, the result is a fully populated State Table, making the rest of the calculations mere algebra.

After obtaining all of the necessary values, the First Law of Thermodynamics is used to calculate work, heat and efficiency of the cycle. In short, the First Law states that the difference between heat and work can be evaluated by the change in enthalpy multiplied by the mass. In otherwords:

Q-W=m(h2-h1)

Now, each component only utilizes one form of energy; heat or work. Therefore, the same equation can be used to calculate heat or work depending on the component. Another important note about heat and work is the sign convention. Heat that is put into the system is considered positive while heat out is negative, and contrarily, work put into the system is negative, while work out of the system is positive. This is because the heat into the system and the work out of the system are the properties of the cycle that are paid for and actively tracked in real world situations.

Net Work:

Calculate the total work into the system at the pump, then subtract the total work out of the system by the turbine:

Wnet=Win-Wout= m(h1-h2)-m(h4-h3)

Wnet=20lbm/s(1493.5-934.02)Btu/lbm-20(98.25-94.02)

Wnet=66572.4 KW

Heat In:

The heat is put into the system at the steam generator, thus we have:

Qin=m(h1-h4)

Qin=20(1493.5-98.25)

Qin=167370 KW

Thermal Efficiency:

Finally, the efficiency of the system, also called the first law efficiency because the values are based off of the Frist Law of Thermodynamics, is the ratio of the amount of energy that comes out of the system, to the amount that goes in giving and is represented by the variable nu, but for the sake of accessibility we will use the letter n:

n=Wnet/Qin =

n=66752.4/167370

n=.3977=39.77%

The efficiency of a basic Rankine Cycle is usually somewhere around the order of 40% so 39% is an acceptable value.

After completing all of the State Table values, and heat and work calculations, you have effectively analyzed the work produced and the efficiency of a basic vapor power cycle.