We consider the First Law of Thermodynamics applied to stationary closed systems as a conservation of energy principle. Thus energy is transferred between the system and the surroundings in the form of heat and work, resulting in a change of internal energy of the system. Internal energy change can be considered as a measure of molecular activity associated with change of phase or temperature of the system and the energy equation is represented as follows:

Energy transferred across the boundary of a system in the form of heat always results from a difference in temperature between the system and its immediate surroundings. We will not consider the mode of heat transfer, whether by conduction, convection or radiation, thus the quantity of heat transferred during any process will either be specified or evaluated as the unknown of the energy equation. By convention, positive heat is that transferred from the surroundings to the system, resulting in an increase in internal energy of the system.

In this course we consider three modes of work transfer across the boundary of a system, as shown in the following diagram:

Boundary work is evaluated by integrating the force F multiplied by the incremental distance moved dx between an initial state (1) to a final state (2). We normally deal with a piston-cylinder device, thus the force can be replaced by the piston area A multiplied by the pressure P, allowing us to replace Adx by the change in volume dV, as follows:

This is shown in the following schematic diagram, where we recall that integration can be represented by the area under the curve.

**Isothermal**(constant temperature process)**Isochoric**or**Isometric**(constant volume process)**Isobaric**(constant pressure process)**Adiabatic**(no heat flow to or from the system during the process)

It is sometimes convenient to evaluate the specific work done which can be represented by a *P-v* diagram thus if the mass of the system is m [kg] we have finally:

where: P is pressure [kPa], V is volume [m^{3}]

m is mass[kg], v is specific volume

W is work done [kJ], w is specific work done

We note that work done by the system on the surroundings (expansion process) is positive, and that done on the system by the surroundings (compression process) is negative.

Finally for a closed system **Shaft Work** (due to a paddle wheel) and **Electrical Work** (due to a voltage applied to an electrical resistor or motor driving a paddle wheel) will always be negative (work done on the system). Positive forms of shaft work, such as that due to a turbine, will be considered in **Chapter 4** when we discuss open systems.

The third component of our Closed System Energy Equation is the change of internal energy resulting from the transfer of heat or work. Since specific internal energy is a property of the system, it is usually presented in the Property Tables such as in the Steam Tables

.

In the case studies that follow we find that one of the major applications of the closed system energy equation is in heat engine processes in which the system is approximated by an ideal gas, thus we will develop relations to determine the internal energy for an ideal gas. We will find also that a new property called **Enthalpy** will be useful both for Closed Systems and in particular for Open Systems, such as the components of steam power plants or refrigeration systems. Enthalpy is not a fundamental property, however is a combination of properties and is defined as follows:

Enthalpy [kJ]:

Specific Enthalpy :

As an example of its usage in closed systems, consider the following constant pressure process:

However, since the pressure is constant throughout the process:

Substituting in the energy equation and simplifying:

(constant pressure process)

Values for specific internal energy (u) and specific enthalpy (h) are available from the Steam Tables, however for ideal gasses it is necessary to develop equations for Δu and Δh in terms of Specific Heat Capacities. We develop these equations in terms of the differential form of the energy equation in the next section of this chapter (Specific Heat Capacities of an Ideal Gas).

Two kilograms of water at 25°C are placed in a piston cylinder device under 3.2 MPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the temperature of the steam reaches 350°C (State (2)). Determine the work done by the fluid (W) and heat transferred to the fluid (Q) during this process using the energy equations, then compare this value of Q to that obtained from the change in enthalpy of the system.

We first draw the diagram of the process including all the relevant data as follows:

Since work involves the integral of Pdv we find it convenient to sketch the *P-v* diagram of the problem as follows:

Substituting equation (2) in the energy equation (1) and simplifying, we obtain:

Now for a constant volume process (dv = 0):

where: C_{v} is the specific constant volume heat capacity

That is, the specific constant volume heat capacity of a system is a function only of its internal energy and temperature. Now in his classic experiment of 1843 Joule showed that the internal energy of an ideal gas is a function of temperature only, and not of pressure or specific volume. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the change in internal energy can be expressed as:

Consider now the enthalpy. By definition h = u + Pv, thus differentiating we obtain:

Again for a simple system, enthalpy (h) is a function of two independent variables, thus we assume it to be a function of temperature T and pressure P, hence:

Substituting equation (6) in the energy equation (5), and simplifying:

Hence for a constant pressure process, since dP = 0:

where: C_{p} is the specific constant pressure heat capacity

That is, the specific constant pressure heat capacity of a system is a function only of its enthalpy and temperature. Now by definition:

Now since for an ideal gas Joule showed that internal energy is a function of temperature only, it follows from the above equation that enthalpy is a function of temperature only. Thus for an ideal gas the partial derivatives can be replaced by ordinary derivatives, and the differential changes in enthalpy can be expressed as:

Finally, from the definition of enthalpy for an ideal gas we have:

Define: (ratio of specific heat capacities)

Values of R, C_{P}, C_{v} and k for ideal gases are presented (at 300K) in the table on Properties of Various Ideal Gases. Note that the values of C_{P}, C_{v} and k are constant with temperature only for mon-atomic gases such as helium and argon. For all other gases their temperature dependence can be considerable and needs to be considered. We find it convenient to express this dependence in tabular form and have provided a table of Specific Heat Capacities of Air.

Conceptually the Stirling engine is the simplest of all heat engines. It has no valves, and includes an externally heated space and an externally cooled space. It was invented by Robert Stirling, and an interesting website by Bob Sier includes a photograph of Robert Stirling, his original patent drawing of 1816, and an animated model of Stirling’s original engine.

In its original single cylinder form the working gas (typically air or helium) is sealed within its cylinders by the piston and shuttled between the hot and cold spaces by a displacer. The linkage driving the piston and displacer will move them such that the gas will compress while it is mainly in the cool compression space and expand while in the hot expansion space. This is clearly illustrated in the adjacent animation which was produced by Richard Wheeler (Zephyris) of Wikipedia.

Refer also to the animation produced by Matt Keveney in his Stirling engine animation website. Since the gas is at a higher temperature, and therefore pressure, during its expansion than during its compression, more power is produced during expansion than is reabsorbed during compression, and this net excess power is the useful output of the engine. Note that there are no valves or intermittent combustion, which is the major source of noise in an internal combustion engine. The same working gas is used over and over again, making the Stirling engine a sealed, closed cycle system. All that is added to the system is steady high temperature heat, and all that is removed from the system is low temperature (waste) heat and mechanical power.

Athens, Ohio, is a hotbed of Stirling cycle machine activity, both engines and coolers, and includes R&D and manufacturing companies as well as internationally recognized consultants in the area of Stirling cycle computer analysis. The parent company of this activity is Sunpower, Inc. It was formed by William Beale in the early 1970’s, mainly based on his invention of the free-piston Stirling engine which we describe below. **Update (Jan. 2013)**: Sunpower was recently acquired by AMETEK, Inc. in Pensylvania, however continues doing Stirling cycle machine development in Athens, Ohio. **Update (Nov. 2013)**: Sunpower has recently introduced a 1 kW Stirling Developers Kit based on a free piston Stirling engine fired by Propane or natural gas.

**Some examples of single cylinder Stirling engines:** Stirling Technology Inc. is a spinoff of Sunpower, and was formed in order to continue the development and manufacture of the 5 kW ST-5 Air engine. This large single cylinder engine burns biomass fuel (such as sawdust pellets or rice husks) and can function as a cogeneration unit in rural areas. It is not a free-piston engine, and uses a bell crank mechanism to obtain the correct displacer phasing. Another important early Stirling engine is Lehmann’s machine on which Gusav Schmidt did the first reasonable analysis of Stirling engines in 1871. Andy Ross of Columbus, Ohio built a small working replica of the Lehmann machine, as well as a model air engine.

**Solar Heat and Power Cogeneration:** With the current energy and global warming crises, there is renewed interest in renewable energy systems, such as wind and solar energy, and distributed heat and power cogeneration systems. Cool Energy, Inc. of Boulder, Colorado, is currently in advanced stages of developing a complete solar heat and power cogeneration system for home usage incorporating Stirling engine technology for electricity generation. This unique application includes evacuated tube solar thermal collectors, thermal storage, hot water and space heaters, and a Stirling engine/generator.

**Ideal Analysis:** Please note that the following analysis of Stirling cycle engines is ideal, and is intended only as an example of **First Law Analysis** of closed systems. In the real world we cannot expect actual machines to perform any better than 40 – 50% of the ideal machine. The analysis of actual Stirling cycle machines is extremely complex and requires sophisticated computer analysis (see for example the course notes on: Stirling Cycle Machine Analysis.)

The free-piston Stirling engine developed by Sunpower, Inc is unique in that there is no mechanical connection between the piston and the displacer, thus the correct phasing between them occurs by use of gas pressure and spring forces. Electrical power is removed from the engine by permanent magnets attached to the piston driving a linear alternator. Basically the ideal Stirling engine undergoes 4 distinct processes, each one of which can be separately analysed, as shown in the *P-V* diagram below. We consider first the work done during all four processes.

- Process 1-2 is the compression process in which the gas is compressed by the piston while the displacer is at the top of the cylinder. Thus during this process the gas is cooled in order to maintain a constant temperature T
_{C}. Work W_{1-2}required to compress the gas is shown as the area under the*P-V*curve, and is evaluated as follows.

- Process 2-3 is a constant volume displacement process in which the gas is displaced from the cold space to the hot expansion space. No work is done, however as we shall see below, a significant amount of heat Q
_{R}is absorbed by the gas from the regenerator matrix. - Process 3-4 is the isothermal expansion process. Work W
_{3-4}is done by the system and is shown as the area under the*P-V*diagram, while heat Q_{3-4}is added to the system from the heat source, maintaining the gas at a constant temperature T_{H}.

- Finally, process 4-1 is a constant volume displacement process which completes the cycle. Once again we will see below that heat Q
_{R}is rejected by the working gas to the regenerator matrix.

The net work, W_{net}, done over the cycle is given by: W_{net} = (W_{3-4} + W_{1-2}), where the compression work W_{1-2} is negative (work done on the system).

We now consider the heat transferred during all four processes, which will allow us to evaluate the thermal efficiency of the ideal Stirling engine. Recall from the previous section that in order to do a First Law analysis of an ideal gas to determine the heat transferred we needed to develop equations to determine the internal energy change Δu in terms of the Specific Heat Capacities of an Ideal Gas.

The two constant volume processes are formed by holding the piston in a fixed position, and shuttling the gas between the hot and cold spaces by means of the displacer. During process 4-1 the hot gas gives up its heat Q_{R} by passing through a regenerator matrix, which is subsequently completely recovered during the process 2-3.

Thus thermal efficiency:

Note that: thus we find that:

We will find in **Chapter 5** that this is the maximum theoretical efficiency that is achievable from a heat engine, and usually referred to as the **Carnot** efficiency.

Note that if no regenerator is present the heat QR must be supplied by the heater. Thus the efficiency will be significantly reduced to . Furthermore the cooler will then have to reject the heat that is normally absorbed by the regenerator, thus the cooling load will be increased to Q_{out} + Q_{R}. Recall that Q_{2-3} = Q_{R} = -Q_{4-1}.

Note that the practical Stirling cycle has many losses associated with it and does not really involve isothermal processes, nor ideal regeneration. Furthermore since the Free-Piston Stirling cycle machines involve sinusoidal motion, the *P-V* diagram has an oval shape, rather than the sharp edges defined in the above diagrams. Nevertheless we use the ideal Stirling cycle to get an initial understanding and appreciation of the cycle performance.

One important aspect of Stirling cycle machines that we need to consider is that the cycle can be reversed – if we put net work into the cycle then it can be used to pump heat from a low temperature source to a high temperature sink. Sunpower, Inc. has been actively involved in the development of Stirling cycle refrigeration systems and produces Stirling cycle cryogenic coolers for liquefying oxygen. In 1984 Sunpower developed a free piston Duplex Stirling Machine having only three moving parts including one piston and two displacers, in which a gas fired Stirling cycle engine powered a Stirling cycle cooler. Global Cooling, Inc. was established in 1995 as a spinoff of Sunpower, and was formed mainly in order to develop free-piston Stirling cycle coolers for home refrigerator applications. These systems, apart from being significantly more efficient than regular vapor-compression refrigerators, have the added advantage of being compact, portable units using helium as the working fluid (and not the HFC refrigerants such as R134a, having a Global Warming Potential of 1,300). More recently Global Cooling decided to concentrate their development efforts on systems in which there are virtually no competitive systems – cooling between -40°C and -80°C, and they established a new company name: Stirling Ultracold.

We are fortunate to have obtained two original M100B coolers from Global Cooling. The one is used as a demonstrator unit, and is shown in operation in the following photograph. The second unit is set up as a ME Senior Lab project in which we evaluate the actual performance of the machine under various specified loads and temperatures.

The **Air Standard Diesel cycle** is the ideal cycle for **Compression-Ignition** (CI) reciprocating engines, first proposed by Rudolph Diesel over 100 years ago. The following link by the Kruse Technology Partnership describes the four-stroke diesel cycle operation including a short history of Rudolf Diesel. The four-stroke diesel engine is usually used in motor vehicle systems, whereas larger marine systems usually use the two-stroke diesel cycle. Once again we have an excellent animation produced by Matt Keveney presenting the operation of the four-stroke diesel cycle.

The actual CI cycle is extremely complex, thus in initial analysis we use an ideal “air-standard” assumption, in which the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state.

The ideal air-standard diesel engine undergoes 4 distinct processes, each one of which can be separately analyed, as shown in the *P-V* diagrams below. Two of the four processes of the cycle are **adiabatic** processes (adiabatic = no transfer of heat), thus before we can continue we need to develop equations for an ideal gas adiabatic process as follows:

The Adiabatic PRocess of an Ideal Gas (Q=0)

The analysis results in the following three general forms representing an adiabatic process:

where k is the ratio of heat capacities and has a nominal value of 1.4 at 300K for air.

Process 1-2 is the adiabatic compression process. Thus the temperature of the air increases during the compression process, and with a large compression ratio (usually > 16:1) it will reach the ignition temperature of the injected fuel. Thus given the conditions at state 1 and the compression ratio of the engine, in order to determine the pressure and temperature at state 2 (at the end of the adiabatic compression process) we have:

Work W_{1-2} required to compress the gas is shown as the area under the *P-V* curve, and is evaluated as follows.

Process 3-4 is thus the adiabatic expansion process. The total expansion work is W_{exp} = (W_{2-3} + W_{3-4}) and is shown as the area under the *P-V* diagram and is analyzed as follows:

The net work W_{net} done over the cycle is given by: W_{net} = (W_{exp} + W_{1-2}), whereas before the compression work W_{1-2} is negative (work done *on* the system).

In the Air-Standard Diesel cycle engine the heat input Q_{in} occurs by combusting the fuel which is injected in a controlled manner, ideally resulting in a constant pressure expansion process 2-3 as shown below. At maximum volume (bottom dead center) the burnt gasses are simply exhausted and replaced by a fresh charge of air. This is represented by the equivalent constant volume heat rejection process Q_{out} = -Q_{4-1}. Both processes are analyzed as follows:

Again from the First Law for a cycle:

Thus thermal efficiency:

The **Air Standard Otto cycle** is the ideal cycle for **Spark-Ignition** (SI) internal combustion engines, first proposed by Nikolaus Otto over 130 years ago, and which is currently used most motor vehicles. The following link by the Kruse Technology Partnership presents a description of the four-stroke Otto cycle operation including a short history of Nikolaus Otto. Once again we have excellent animations produced by Matt Keveney presenting both the four-stroke and the two-stroke spark-ignition internal combustion engine operation.

The analysis of the Otto cycle is very similar to that of the Diesel cycle which we analyzed in the previous section. We will use the ideal “air-standard” assumption in our analysis. Thus the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state.

The most significant difference between the ideal Otto cycle and the ideal Diesel cycle is the method of igniting the fuel-air mixture. Recall that in the ideal Diesel cycle the extremely high compression ratio (around 18:1) allows the air to reach the ignition temperature of the fuel. The fuel is then injected such that the ignition process occurs at a constant pressure. In the ideal Otto cycle the fuel-air mixture is introduced during the induction stroke and compressed to a much lower compression ratio (around 8:1) and is then ignited by a spark. The combustion results in a sudden jump in pressure while the volume remains essentially constant. The continuation of the cycle including the expansion and exhaust processes are essentially identical to that of the ideal Diesel cycle.

In this chapter we consider the property values and relationships of a pure substance (such as water) which can exist in three phases – solid, liquid and gas. We will not consider the solid phase in this course.In order to introduce the rather complex phase change interactions that occur in pure substances we consider an experiment in which we have liquid water in a piston-cylinder device at 20°C and 100kPa pressure. Heat is added to the cylinder while the pressure is maintained constant until the temperature reaches 300°C, as shown in the following *T-v* diagram (temperature vs specific volume):

Notice that during this entire process the specific volume of the water increased by more than three orders of magnitude, which made it necessary to use a logarithmic scale for the specific volume axis.

We can repeat this same experiment at different pressures to attain more curves as shown in the figure below.

Saturation lines can be drawn by connecting the loci of the saturated liquid and saturated vapor points as shown in the figure below.

interest as shown in the diagram, being the

the left, the

region (which also includes the

the saturated vapor line and above the critical point. We will use

Tables

properties. Notice that we have provided property tables of steam, Refrigerant

R134a, and Carbon Dioxide, which due to environmental concerns involving R134a

is likely to become the refrigerant of common usage in the future.

The **Quality Region**, also referred to as

the **Saturated Liquid-Vapor Mixture Region**, is the area enclosed between

the saturated liquid line and the saturated vapor line. At any point within

this region the quality of the mixture (sometimes referred to as the dryness

factor) is defined as the mass of vapor divided by the total mass of the fluid,

as shown in the following diagram:

The above discussion was done in terms of the

temperature (T) and specific volume (v). You may recall from Chapter 1 when we

defined the State Postulate however, that any two independent intensive

properties can be used to completely define all other intensive state

properties. This means we can also evaluate a substance in terms of Pressure

(P) and specific volume (v) as shown below:

Two kilograms of water at 25°C are placed in a piston cylinder device under 100 kPa pressure as shown in the diagram (State (1)). Heat is added to the water at constant pressure until the piston reaches the stops at a total volume of 0.4 m^{3} (State (2)). More heat is then added at constant volume until the temperature of the water reaches 300°C (State (3)). Determine (a) the quality of the fluid and the mass of the vapor at state (2), and (b) the pressure of the fluid at state (3).

** Step 2:** In the case of a closed system with a phase change fluid,

Thus v_{2} = = = 0.2

Quality x_{2} =

mass of water vapor at state 2:

Concerning state (3), the problem statement did not specify that it is in the superheat region. We needed to first determine the saturated vapor specific volume v_{g} at 300°C. This value is 0.0216 , which is much less than the specific volume v_{3} of 0.2 , thus placing state (3) well into the superheated region. Thus the two intensive properties which we use to determine the pressure at state (3) are T_{3} = 300°C, and v_{3} = 0.2 . On scanning the superheat tables we find that the closest values lie somewhere between 1.2 MPa and 1.4 MPa, thus we use linear interpolation techniques to determine the actual pressure P_{3} as shown below:

When boiling food the cooking temperature is limited to about 100° C (depending on elevation). Pressure cookers allow for faster cooking by increasing the boiling temperature of water using increased pressure. The pressure within the sealed cooker can be varied by using different weights to seal the vent. A pressure cooker containing 4kg of water is to be used to cook potatoes at 150° C determine (a) what pressure in necessary to maintain the desired boiling temperature and (b) the mass of the weight that must be used in order to maintain the pressure determined in part (a) given a round vent with a diameter of 1cm, (c) if the pressure cooker has a volume of 10L what is the quality of the steam within the pressure cooker the instant it reaches temperature, before any steam is vented (assuming all air has been purged from the cooker by this point) , and (d) how long can the potatoes be cooked at the desired temperature before all the water is boiled away if heat is being added to the system by a 7000 Watt burner.

In order to maintain a consistent temperature we know we want a pressure that will place the boiling point of the water at exactly 150° C. To find this pressure we look to the steam tables in order to find the pressure of saturated steam at the desired temperature, in this case it is 476.16 kPa.

To solve part (b) we need to find the weight needed to exactly counter the steam pressure trying to escape the vent. Since force is pressure multiplied by area we must determine the area of the vent:

Then we multiply this value by the pressure found in part (a) to find the weight required and divide by g to find the mass required to create that weight:

For part (c) we will use the formula for x as a ratio of specific volumes listed above:

We obtain the values of v_{f} and v_{g} from the steam tables and calculate v from the volume of the pressure cooker and the total mass of the water:

Plugging these values into our equation for x we obtain:

The solution for part (d) brings us back to the steam tables once again. This time we are looking at the specific enthalpy of saturated steam at our given temperature, which in this case is listed as 2113.7 . We then have to multiply this value by the total mass of water (4kg) to find the total required energy:

And finally we need to divide that energy by the heat input of the burner (7000W) to find the time required to transfer the required energy to the water:

Continuing on our discusstion of pure substances,

we find that for a pure substance in the superheated region at specific volumes

much higher than the critical point, the *P-v-T* relation can be

conveniently expressed very accurately by the **Ideal Gas Equation of State**

as follows:

where: R is constant for a particular substance and is called the **Gas Constant**.

Note that for the ideal gas equation both the pressure P and the temperature T must be expressed in absolute quantities.

Consider for example the *T-v* diagram for water as shown below:

where: is the Universal Gas Constant and

is the molar mass of the substance.

For Air:

Steam:

The three commonly used formats to express the Ideal Gas Equation of State are:

A piston-cylinder device contains 0.5 kg saturated liquid water at a pressure of 200 kPa. Heat is added and the steam expands at constant pressure until it reaches 300°C.

- Draw a diagram representing the process showing the initial and final states of the system.
- Sketch this process on a
*T-v*(temperature-specific volume) diagram with respect to the saturation lines, critical point, and relevant constant pressure lines, clearly indicating the initial and final states. - Using steam tables determine the initial temperature of the steam prior to heating.
- Using steam tables determine the final volume of the steam after heating
- Using the ideal gas equation of state determine the final volume of the steam after heating. Determine the percentage error of using this method compared to that of using the steam tables.

Note: The critical point data and the ideal gas constant for steam can be found on the first page of the steam tables.

Even if questions a) and b) were not required, this should always be the first priority item in solving a thermodynamic problem.

d) From the *T-v* diagram we determine that state (2) is in the superheated region, thus we use the superheated steam tables to determine that v_{2} = v_{200kPa,300°C} = 1.3162 . Thus V_{2} = mv_{2}= (0.5kg)(1.3162 ) = 0.658 m^{3} (658 liters).

e) Determine V_{2} from the Ideal Gas Equation of State

T **MUST** be absolute!

Note that in doing a units check, we find that the following conversion appears so often that we feel it should be added to our Units Conversion Survival Kit (recall Chapter 1):

Finally we determine the percentage error of using the ideal gas equation at state (2):

We noticed in the above *T-v* diagram for water that the gasses can deviate significantly from the ideal gas equation of state in regions nearby the critical point and there have been many equations of state recommended for use to account for this non-ideal behaviour. However, this non-ideal behaviour can be accounted for by a correction factor called the **Compressibility Factor** Z defined as follows:

thus when the compressibility factor Z approaches 1 the gas behaves as an ideal gas. Note that under the same conditions of temperature and pressure, the compressibility factor can be expressed as:

Different fluids have different values of critical point pressure and temperature data PCR and TCR, and these can be determined from the Table of Critical Point Data of Various Substances. Fortunately the **Principle of Corresponding States** shows that we can normalize the pressure and temperature values with the critical values as follows:

All fluids normalized in this manner exhibit similar non-ideal gas behaviour within a few percent, thus they can all be plotted on a Generalized Compressibility Chart. A number of these charts are available, however we prefer to use the Lee-Kesler (logarithmic) Compressibility Chart. The use of the compressibility chart is shown in the following example.

Carbon Dioxide gas is stored in a 100 liter tank at 6 MPa and 30°C. Determine the mass of CO_{2} in the tank based on (a) values obtained from the CO_{2} tables of data, (b) the ideal gas equation of state, and (c) the generalized compressibility chart. Compare (b) and (c) to (a) and determine the percentage error in each case.

We first determine the Critical Point data for CO_{2} from the Table of Critical Point Data of Various Substances

In this course we consider three types of Control Volume Systems – Steam Power Plants, Refrigeration Systems, and Aircraft Jet Engines. Fortunately we will be able to separately analyze each component of the system independent of the entire system, which is typically represented as follows:

Consider an elemental mass **d**m flowing through an inlet or outlet port of a control volume, having an area A, volume **d**V, length **d**x, and an average steady velocity , as follows.

where: is the mass flow rate

is the volumetric flow rate

is the density , is the specific volume

is the velocity , is the flow area []

The fluid mass flows through the inlet and exit ports of the control volume accompanied by its energy. These include four types of energy – internal energy (**u**), kinetic energy (**ke**), potential energy (**pe**), and flow work (**w _{flow}**). In order to evaluate the flow work consider the following exit port schematic showing the fluid doing work against the surroundings through an imaginary piston:

Consider the control volume shown in the following figure. Under steady flow conditions there is no mass or energy accumulation in the control volume thus the mass flow rate applies both to the inlet and outlet ports. Furthermore with a constant mass flow rate, it is more convenient to develop the energy equation in terms of power [kW] rather than energy [kJ] as was done previously.

The specific energy e can include kinetic and potential energy, however will always include the combination of internal energy (**u**) and flow work (**Pv**), thus we conveniently combine these properties in terms of the property enthalpy (as was done in **Chapter 3a**), as follows:

Notice that enthalpy h is fundamental to the energy equation for a control volume.

When dealing with closed systems we found that sketching *T-v* or *P-v* diagrams was a significant aid in describing and understanding the various processes. In steady flow systems we find that the Pressure-Enthalpy (*P-h*) diagrams serve a similar purpose, and we will use them extensively. In this course we consider three pure fluids – water, refrigerant R134a, and carbon dioxide, and we have provided *P-h* diagrams for all three in the Property Tables section. We will illustrate their use in the following examples. The *P-h* diagram for water is shown below. Study it carefully and try to understand the significance of the distinctive shapes of the constant temperature curves in the compressed liquid, saturated mixture (quality region) and superheated vapor regions.

A basic steam power plant consists of four interconnected components, typically as shown in the figure below. These include a steam turbine to produce mechanical shaft power, a condenser which uses external cooling water to condense the steam to liquid water, a feedwater pump to pump the liquid to a high pressure, and a boiler which is externally heated to boil the water to superheated steam. Unless otherwise specified we assume that the turbine and the pump (as well as all the interconnecting tubing) are adiabatic, and that the condenser exchanges all of its heat with the cooling water.

**A Simple Steam Power Plant Example** – In this example we wish to determine the performance of this basic steam power plant under the conditions shown in the diagram, including the power of the turbine and feedwater pump, heat transfer rates of the boiler and condenser, and thermal efficiency of the system.

- Turbine output power and the power required to drive the feedwater pump.
- Heat power supplied to the boiler and that rejected in the condenser to the cooling water.
- The thermal efficiency of the power plant η
_{th}, defined as the net work done by the system divided by the heat supplied to the boiler. - The minimum mass flow rate of the cooling water in the condenser required for a specific temperature rise.

Do not be intimidated by the complexity of this system. We will find that we can solve each component of this system separately and independently of all the other components, always using the same approach and the same basic equations. We first use the information given in the above schematic to plot the four processes (1)-(2)-(3)-(4)-(1) on the *P-h* diagram. Notice that the fluid entering and exiting the boiler is at the high pressure 10 MPa, and similarly that entering and exiting the condenser is at the low pressure 20 kPa. State (1) is given by the intersection of 10 MPa and 500°C, and state (2) is given as 20 kPa at 90% quality, State (3) is given by the intersection of 20 kPa and 40°C, and the feedwater pumping process (3)-(4) follows the constant temperature line, since T4 = T3 = 40°C.

(Note: We find it strange that the only thermodynamics text that we know of that even considered the use of the *P-h* diagram for steam power plants is Engineering Thermodynamics – Jones and Dugan (1995). It is widely used for refrigeration systems, however not for steam power plants.)

We now consider each component as a separate control volume and apply the energy equation, starting with the steam turbine. The steam turbine uses the high-pressure – high-temperature steam at the inlet port (1) to produce shaft power by expanding the steam through the turbine blades, and the resulting low-pressure – low-temperature steam is rejected to the condenser at port (2). Notice that we have assumed that the kinetic and potential energy change of the fluid is negligible, and that the turbine is adiabatic. In fact any heat loss to the surroundings or kinetic energy increase would be at the expense of output power, thus practical systems are designed to minimize these loss effects. The required values of enthalpy for the inlet and outlet ports are determined from the steam tables.

The very low-pressure steam at port (2) is now directed to a condenser in which heat is extracted by cooling water from a nearby river (or a cooling tower) and the steam is condensed into the subcooled liquid region. The analysis of the condenser may require determining the mass flow rate of the cooling water needed to limit the temperature rise to a certain amount – in this example to 10°C. This is shown on the following diagram of the condenser:

Thus we see that under the conditions shown, 17.6 MW of heat is extracted from the steam in the condenser.

I have often been queried by students as to why we have to reject such a large amount of heat in the condenser causing such a large decrease in thermal efficiency of the power plant. Without going into the philosophical aspects of the Second Law (which we cover later in **Chapter 5**, my best reply was provided to me by Randy Sheidler, a senior engineer at the Gavin Power Plant. He stated that the **Fourth Law of Thermodynamics** states: **“You can’t pump steam!”**, so until we condense all the steam into liquid water by extracting 17.6 MW of heat, we cannot pump it to the high pressure to complete the cycle. (Randy could not give me a reference to the source of this amazing observation).

In order to determine the enthalpy change Δh of the cooling water (or in the feedwater pump which follows), we consider the water to be an **Incompressible Liquid**, and evaluate Δh as follows:

Tnthalpy h is defined as:

On differentiating:

For an incompressible liquid: and

where is the Specific Heat Capacity of the liquid

Thus:

Integrating:

From the steam tables we find that the specific heat capacity for liquid water C_{H2O} = 4.18 . Using this analysis we found on the condenser diagram above that the required mass flow rate of the cooling water is 421 . If this flow rate cannot be supported by a nearby river then a cooling tower must be included in the power plant design.

We now consider the feedwater pump as follows:

The final component that we consider is the boiler, as follows:

Thermal Efficiency:

Note that the feedwater pump work can normally be neglected.

In an effort to decentralize the national power distribution grid, the following supercritical (25 MPa), coal fired steam power plant (modeled after the Gavin Power Plant in Cheshire, Ohio) has been proposed to service about 10,000 households in Athens, Ohio. It is to be placed close to the sewage plant on the east side of Athens and cooled by water from the Hocking River. We consider first a simplified system as shown below. Notice that we have replaced the “Boiler” with a “Steam Generator”, since at supercritical pressures the concept of boiling water is undefined. Furthermore we have specifically split the turbine into a High Pressure (HP) turbine and a Low Pressure (LP) turbine since we will find that having a single turbine to expand from 25MPa to 10kPa is totally impractical. Thus for example the Gavin Power Plant has a turbine set consisting of 6 turbines – a High Pressure Turbine, an Intermediate Pressure (Reheat) turbine, and 4 large Low Pressure turbines operating in parallel.

- The outlet pressure of the LP turbine at port (3) is 10 kPa, which is well below atmospheric pressure. The extremely low pressure in the condenser will allow air to leak into the system and ultimately lead to a deteriorated performance.
- The quality of the steam at port (3) is 80%. This is unacceptable. The condensed water will cause erosion of the turbine blades, and we should always try to maintain a quality of above 90%. One example of the effects of this erosion can be seen on the blade tips of the final stage of the Gavin LP turbine. During 2000, all four LP turbines needed to be replaced because of the reduced performance resulting from this erosion. (Refer: Tour of the Gavin Power Plant – Feb. 2000)

The following revised system diagram corrects both flaws. The steam at the outlet of the HP turbine (port (2)) is reheated to 550 C before entering the LP turbine at port (3). Also the low pressure liquid condensate at port (5) is pumped to a pressure of 800 kPa and passed through a de-aerator prior to being pumped by the feedwater pump to the high pressure of 25 MPa.

- Assuming that both turbines are adiabatic and neglecting kinetic energy effects determine the combined output power of both turbines [10.6 MW].
- Assuming that both the condensate pump and the feedwater pump are adiabatic, determine the power required to drive the two pumps [-204 kW].
- Determine the total heat transfer to the steam generator, including the reheat system [26.1 MW].
- Determine the overall thermal efficiency of this power plant. (Thermal efficiency (η
_{th}) is defined as the net work done (turbines, pumps) divided by the total heat supplied externally to the steam generator and reheat system) [40 %]. - Determine the heat rejected to the cooling water in the condenser [-15.7 MW].
- Assume that all the heat rejected in the condenser is absorbed by cooling water from the Hocking River. To prevent thermal pollution the cooling water is not allowed to experience a temperature rise above 10°C. If the steam leaves the condenser as saturated liquid at 40°C, determine the required minimum volumetric flow rate of the cooling water [22.6 cubic meters/minute].
- Discuss whether you think that the proposed system can be cooled by the Hocking river. You will need to do some research to determine the minimal seasonal flow in the river in order to validate your decision. (Hint- Google: Hocking River Flow)

Thanks to Kris Dambrink from Imtech.nl for making me aware of this alternative approach to adapting an Open Feedwater Heater to a steam power plant (4 Feb 2010)

This Solved Problem is an alternative extension of Solved Proble 4.1 in which we extend the de-aerator by tapping steam from the outlet of the High Pressure turbine and reduce the pressure to 800 kPa by means of a **Throttling Control Valve** before feeding it into the de-aerator. This allows one to conveniently convert the de-aerator into an **Open Feedwater Heater** without requiring a bleed tap from the Low Pressure turbine at exactly the de-aerator pressure, as shown in the following diagram:

- A mass fraction of the steam y is tapped from the outlet of the HP turbine (2) and passed through the throttle such that its pressure is reduced to that of the de-aerator (9). It is then mixed with a mass fraction (1-y) of the liquid water at station (6). The mass fraction y is chosen to enable the fluid to reach a saturated liquid state at station (7).
- The feedwater pump then pumps the liquid to station (8), thus saving a significant amount of heat from the steam generator in heating the fluid from station (8) to the turbine inlet at station (1). It is true that with a mass fraction of (1-y) there is less power output due to a reduced mass flow rate in the LP turbine, however the net result is normally an increase in thermal efficiency.

Thus once more we see that in spite of the complexity of the system, the *P-h* diagram plot enables an intuitive and qualitative initial understanding of the system. Using the methods described earlier in this chapter for analysis of each component, as well as the steam tables for evaluating the enthalpy at the various stations (shown in red), and neglecting kinetic and potential energy effects, determine the following:

- Assuming that the open feedwater heater is adiabatic, determine the mass fraction of steam y required to be bled off the outlet of the HP turbine which will bring the fluid from station (6) to a saturated liquid state in the de-aerator. [y = 0.20]We first need to evaluate the enthalpy of the fluid at station (9) after passing through the throttling control valve:

- Assuming that both the condensate pump and the feedwater pump are adiabatic, determine the power required to drive the two pumps [235 kW].On examining the system diagram above we noticed something very strange about the feedwater pump. Until now we considered liquid water to be incompressible, thus pumping it to a higher pressure did not result in an increase of its temperature. However on a recent visit to the Gavin Power Plant we discovered that at 25MPa pressure and more than 100°C water is no longer incompressible, and compression will always result in a temperature increase. We cannot use the simple incompressible liquid formula to determine pump work, however need to evaluate the difference in enthalpy from the Compressed Liquid Water tables, leading to the following results:

- Assuming that both turbines are adiabatic, determine the new (reduced) combined power output of both turbines. Recall from Solved Problem 4.1 that the power output of the turbines was found to be 10.6 MW if no steam is bled from the LP turbine [8.98 MW]

- Determine the total heat transfer to the steam generator, including the reheat system [21.4 MW].

- Determine the overall thermal efficiency of this power plant. (Thermal efficiency (η
_{th}) is defined as the net work done (turbines, pumps) divided by the total heat supplied externally to the steam generator and reheat system) [41 %].

- Determine the heat rejected to the cooling water in the condenser [-12.6 MW].
- Assume that all the heat rejected in the condenser is absorbed by cooling water from the Hocking River. To prevent thermal pollution the cooling water is not allowed to experience a temperature rise above 10°C. If the steam leaves the condenser as saturated liquid at 40°C, determine the required minimum volumetric flow rate of the cooling water [18.1 cubic meters/minute].

**Discussion:** Thus we find that the open feedwater heater did in fact raise the efficiency from 40% to 41%. This may not seem like a significant amount, however all the basic components were already in place, since without a de-aerator the steam power plant will deteriorate and become non-functional within a very short time due to leakage of air into the system. Furthermore, if the reduction in power output is not acceptable, then it can be easily remedied by increasing the mass flow rate in the system design. Note that this is a contrived example in order to demonstrate that no matter how complex the system is, we can easily plot the entire system on a *P-h* diagram and obtain an immediate intuitive understanding and evaluation of the system performance. It is helpful to check each value of enthalpy read or evaluated from the steam tables and compare them to the values on the enthalpy axis of the *P-h* diagram.

In the early days of refrigeration the two refrigerants in common use were ammonia and carbon dioxide. Both were problematic – ammonia is toxic and carbon dioxide requires extremely high pressures (from around 30 to 200 atmospheres!) to operate in a refrigeration cycle, and since it operates on a transcritical cycle the compressor outlet temperature is extremely high (around 160°C). When Freon 12 (dichloro-diflouro-methane) was discovered it totally took over as the refrigerant of choice. It is an extremely stable, non toxic fluid, which does not interact with the compressor lubricant, and operates at pressures always somewhat higher than atmospheric, so that if any leakage occurred, air would not leak into the system, thus one could recharge without having to apply vacuum.

Unfortunately when the refrigerant does ultimately leak and make its way up to the ozone layer the ultraviolet radiation breaks up the molecule releasing the highly active chlorine radicals, which help to deplete the ozone layer. Freon 12 has since been banned from usage on a global scale, and has been essentially replaced by chlorine free R134a (tetraflouro-ethane) – not as stable as Freon 12, however it does not have ozone depletion characteristics.

Recently, however, the international scientific consensus is that Global Warming is caused by human energy related activity, and various man made substances are defined on the basis of a Global Warming Potential (GWP) with reference to carbon dioxide (GWP=1). R134a has been found to have a GWP of 1300 and in Europe, within a few years, automobile air conditioning systems will be barred from using R134a as a refrigerant.

The new hot topic is a return to carbon dioxide (R744) as a refrigerant (refer for example to the website: R744.com). The previous two major problems of high pressure and high compressor temperature are found in fact to be advantageous. The very high cycle pressure results in a high fluid density throughout the cycle, allowing miniaturization of the systems for the same heat pumping power requirements. Furthermore the high outlet temperature will allow instant defrosting of automobile windshields (we don’t have to wait until the car engine warms up) and can be used for combined space heating and hot water heating in home usage (refer for example: Norwegian IEA Heatpump Program Annex28.

In this chapter we cover the vapor-compression refrigeration cycle using refrigerant R134a, and will defer coverage of the carbon dioxide cycle to **Chapter 9**.

Unlike the situation with steam power plants it is common practice to begin the design and analysis of refrigeration and heat pump systems by first plotting the cycle on the *P-h* diagram.

The following schematic shows a basic refrigeration or heat pump system with typical property values. Since no mass flow rate of the refrigerant has been provided, the entire analysis is done in terms of specific energy values. Notice that the same system can be used either for a refrigerator or air conditioner, in which the heat absorbed in the evaporator (q_{evap}) is the desired output, or for a heat pump, in which the heat rejected in the condenser (q_{cond}) is the desired output.

- Heat absorbed by the evaporator (q
_{evap}) - Heat rejected by the condenser (q
_{cond}) - Work done to drive the compressor (w
_{comp}) - Coefficient of Performance (COP) of the system, either as a refrigerator or as a heat pump.

As with the Steam Power Plant, we find that we can solve each component of this system separately and independently of all the other components, always using the same approach and the same basic equations. We first use the information given in the above schematic to plot the four processes (1)-(2)-(3)-(4)-(1) on the *P-h* diagram. Notice that the fluid entering and exiting the condenser (State (2) to State (3)) is at the high pressure 1 MPa. The fluid enters the evaporator at State (4) as a saturated mixture at -20°C and exits the evaporator at State (1) as a saturated vapor. State (2) is given by the intersection of 1 MPa and 70°C in the superheated region. State (3) is seen to be in the subcooled liquid region at 30°C, since the saturation temperature at 1 MPa is about 40°C. The process (3)-(4) is a vertical line (h3 = h4) as is discussed below.

In the following section we develop the methods of evaluating the solution of this example using the R134a refrigerant tables. Notice that the refrigerant tables do not include the subcooled region, however since the constant temperature line in this region is essentially vertical, we use the saturated liquid value of enthalpy at that temperature.

We now consider each component as a separate control volume and apply the energy equation, starting with the compressor. Notice that we have assumed that the kinetic and potential energy change of the fluid is negligible, and that the compressor is adiabatic. The required values of enthalpy for the inlet and outlet ports are determined from the R134a refrigerant tables.

Notice that for the same system we always find that COP_{HP} = COP_{R} + 1.

Notice also that the COP values are usually greater than 1, which is the reason why they are never referred to as “Efficiency” values, which always have a maximum of 100%.

Thus the *P-h* diagram is a widely used and very useful tool for doing an approximate evaluation of a refrigerator or heat pump system. In fact, in the official Reference Handbook supplied by the NCEES to be used in the Fundamentals of Engineering exam, only the *P-h* diagram is presented for R134a. You are expected to answer all the questions on this subject based on plotting the cycle on this diagram as shown above.

With the global quest for energy efficiency, there is renewed interest in Geothermal Heat Pumps which were have been in limited use for more than 60 years. Essentially this technology relies on the fact that a few meters below the surface of the earth the temperature remains relatively constant throughout the year (around 55°F (13°C)), warmer than the air above it during winter, and cooler during summer. This means that we can design a heat pump which can combine hot water and space heating in winter in which the earth is used as a heat source (rather than the outside air) at a considerable increase in coefficient of performance COP. Similarly, with suitable valving, we can use the same system in summer for hot water heating and air conditioning in which the earth is used as a heat sink, rather than the outside air. This is achieved by using a **Ground** Loop in order to enable heat transfer with the earth, as described in the USDOE website: Geothermal Heat Pumps. Another description of geothermal heat pumps has been provided by David White Services of Southeastern Ohio and includes a YouTube video by WaterFurnace: GEOTHERMAL How does it work.

We wish to do a preliminary thermodynamic analysis of the following home geothermal heat pump system designed for wintertime hot water and space heating. Notice that with suitable valving this system can be used both in winter for space heating and in summer for air conditioning, with hot water heating throughout the year.

- Using the R134a property tables determine the enthalpies at all five stations and verify and indicate their values on the
*P-h*diagram.- Station 1 we get from the saturated vapor pressure table: hg = 253.8
- Station 2 we get from the superheated vapor table : h = 293.3
- Station 3 we get from the saturated liquid pressure table: hf = 136
- Station 4 we approximate as a saturated liquid at 20°C: h= 79.32
- Station 5 is the same as station 4 since the only part in between is an adiabatic throttle: h= 79.32

- Determine the mass flow rate of the refrigerant R134a.
- The mass flow rate is determined by the compressor which puts 500W of power into the system. We can find the mass flow rate by using the power from the pump and dividing it by the difference in enthalpy across the compressor (between Stations 1 & 2):

- The mass flow rate is determined by the compressor which puts 500W of power into the system. We can find the mass flow rate by using the power from the pump and dividing it by the difference in enthalpy across the compressor (between Stations 1 & 2):
- Determine the power absorbed by the hot water heater and that absorbed by the space heater.
- At this point we know the mass flow rate and the enthalpies at each station so finding the power absorbed by the water heater and space heater is as simple as finding the difference in enthalpies across each component and multiplying by the mass flow rate:

- At this point we know the mass flow rate and the enthalpies at each station so finding the power absorbed by the water heater and space heater is as simple as finding the difference in enthalpies across each component and multiplying by the mass flow rate:
- Determine the time taken for 100 liters of water at an initial temperature of 20°C to reach the required hot water temperature of 50°C.
- We know one liter of water has a mass of 1kg so total mass of the 100 L is 100 kg and from the steam tables we can approximate the change in enthalpy of water over this 30 degree change in temperature to be 125.4 so we can use these numbers along with our power from the previous step:

- We know one liter of water has a mass of 1kg so total mass of the 100 L is 100 kg and from the steam tables we can approximate the change in enthalpy of water over this 30 degree change in temperature to be 125.4 so we can use these numbers along with our power from the previous step:
- Determine the Coefficient of Performance of the hot water heater (defined as the heat absorbed by the hot water divided by the work done on the compressor)
- Determine the Coefficient of Performance of the heat pump (defined as the total heat rejected by the refrigerant in the hot water and space heaters divided by the work done on the compressor)
- Think about what changes would be required of the system parameters if no geothermal water loop was used, and the evaporator was required to absorb its heat from the outside air at -10°C. Discuss the advantages of the geothermal heat pump system over other means of space and water heating.

In this chapter we consider a more abstract approach to heat engine, refrigerator and heat pump cycles, in an attempt to determine if they are feasible, and to obtain the limiting maximum performance available for these cycles. The concept of mechanical and thermal reversibility is central to the analysis, leading to the ideal Carnot cycles. (Refer to Wikipedia: Sadi Carnot a French physicist, mathematician and engineer who gave the first successful account of heat engines, the Carnot cycle, and laid the foundations of the second law of thermodynamics).

We represent a heat engine and a heat pump cycle in a minimalist abstract format as in the following diagrams. In both cases there are two temperature reservoirs T_{H} and T_{L}, with T_{H} > T_{L}.

We now present two statements of the Second Law of Thermodynamics, the first regarding a heat engine, and the second regarding a heat pump. Neither of these statements can be proved, however have never been observed to be violated.

**The Kelvin-Planck Statement:** It is impossible to construct a device which operates on a cycle and produces no other effect than the transfer of heat from a single body in order to produce work.

We prefer a less formal description of this statement in terms of a boat extracting heat from the ocean in order to produce its required propulsion work:

It is remarkable that the two above statements of the Second Law are in fact equivalent. In order to demonstrate their equivalence consider the following diagram. On the left we see a heat pump which violates the Clausius statement by pumping heat Q_{L} from the low temperature reservoir to the high temperature reservoir without any work input. On the right we see a heat engine rejecting heat Q_{L} to the low temperature reservoir.

Notice that the statements on the Second Law are negative statements in that they only describe what is impossible to achieve. In order to determine the maximum performance available from a heat engine or a heat pump we need to introduce the concept of **Reversibility**, including both mechanical and thermal reversibility. We will attempt to clarify these concepts in terms of the following example of a reversible piston cylinder device in thermal equilibrium with the surroundings at temperature T_{0}, and undergoing a cyclic compression/expansion process.

In summary, there are three conditions required for reversible operation:

- All mechanical processes are frictionless.
- At each incremental step in the process thermal and pressure equilibrium conditions are established.
- All heat transfer processes are isothermal.

Carnot’s theorem, also known as Carnot’s rule, or the Carnot principle, can be stated as follows:

No heat engine operating between two heat reservoirs can be more efficient than a reversible heat engine operating between the same two reservoirs.

The simplest way to prove this theorem is to consider the scenario shown below, in which we have an irreversible engine as well as a reversible engine operating between the reservoirs T_{H} and T_{LH} from the high temperature reservoir, however the irreversible engine produces more work W_{I} than that of the reversible engine W_{R}.

The first Corollary of Carnot’s theorem can be stated as follows:

All reversible heat engines operating between the same two heat reservoirs must have the same efficiency.

Thus regardless of the type of heat engine, the working fluid, or any other factor if the heat engine is reversible, then it must have the same maximum efficiency. If this is not the case then we can drive the reversible engine with the lower efficiency as a heat pump and produce a Kelvin-Planck violator as above.

The second Corollary of Carnot’s theorem can be stated as follows:

The efficiency of a reversible heat engine is a function only of the respective temperatures of the hot and cold reservoirs. It can be evaluated by replacing the ratio of heat transfers Q_{L}and Q_{H}by the ratio of temperatures T_{L}and T_{H}of the respective heat reservoirs.

Thus using this corollary we can evaluate the thermal efficiency of a reversible heat engine as follows:

- Draw a diagram representing the heat pump system showing the flow of energy and source and sink temperatures and determine the following:
- The maximum possible Coefficient of Performance of a hot water heater (COP
_{HW}) that could be obtained by a reversible heat pump. - The maximum possible Coefficient of Performance of a space cooling air conditioner (COP
_{AC}) that could be obtained by a reversible heat pump.

- The maximum possible Coefficient of Performance of a hot water heater (COP

- Comparing the actual Coefficients of Performance shown above to those of the reversible heat pump determine if the actual heat pump shown above is feasible. State the reasons for your conclusion.

Derive all equations used starting from the basic definition of COP_{HW} and COP_{AC} and the Carnot relations for the ratio of heats of a heat pump.

**Solution:** For 1), 2), and 3) we need to reduce the system complexity shown above to an energy flow diagram showing only the basic requirements – cool the air to 20°C and heat the water to 50°C. Given the temperature if the heat source (20°C) and the heat sink (50°C) we can evaluate the respective reversible Coefficients of Performance.

**Defining Entropy (S) through the Clausius Inequality**

Consider two heat engines, one a reversible (Carnot) engine and the other an irreversible heat engine. For purposes of developing the Clausius Inequality we assume that both engines are sized to accept the same amount of heat Q_{H} from the thermal source. Thus since the irreversible engine must be less efficient than the Carnot engine, it must reject more heat Q_{L,irrev} to the thermal sink than that rejected by the Carnot engine Q_{L,rev}, as shown:

Recall from Chapter 5 that whenever we considered the efficiency of a reversible heat engine, we went into “meditation mode”, replacing the ratio of heat flows with the ratio of temperatures:

Let Q_{diff} = (Q_{L,irrev} – Q_{L,rev}), then the cyclic integral for an irreversible heat engine becomes:

Thus finally, for any reversible or irreversible heat engine we obtain the Clausius Inequality:

**Defining the Property Entropy – S**

All properties (such as pressure P, volume V, etc.) have a cyclic integral equal to zero.

A very strange definition indeed, and difficult to comprehend. It is defined in differential format as the reversible heat transfer divided by the temperature. In an attempt to try and understand it we rewrite the definition as follows:

Heat:

It is advantageous to compare this definition with the equivalent definition of work as follows:

Work:

Thus it begins to make sense. Work done requires both a driving force (pressure P) *and* movement (volume change dV). We implicitly evaluated the work done for reversible processes – always neglecting friction or any other irreversibility. Similarly we can state that heat transfer requires both a driving force (temperature T) and some equivalent form of “movement” (entropy change dS). Since temperature can be considered as represented by the vibration of the molecules, it is this transfer of vibrational energy that we define as entropy.

We now continue with the **Increase in Entropy Principle** which is also derived from the Clausius Inequality, and states that for any process, the total change in entropy of a system or control volume together with its enclosing adiabatic surroundings is always greater than or equal to zero. This total change of entropy is denoted the **Entropy Generated** during the process (S_{gen} or s_{gen} ). For reversible processes the entropy generated will always be zero.

**1. Non-flow Process**

We previously found from considerations of the Clausius Inequality that the following cyclic integral is always less than or equal to zero, where the equality occurred for a reversible cycle.

This lead to the definition of the property Entropy (S). Consider now an irreversible cycle in which process (1) -> (2) follows an irreversible path, and process (2) -> (1) a reversible path, as shown:

**2. Flow Processes (Steady Flow)**

We now consider the entropy generated during a steady flow process through a single-input/single-output Control Volume (CV) enclosed in an adiabatic surroundings as shown:

Surprisingly the form of the specific entropy generated function s_{gen} for a control volume is identical to that for a system.

For multiple-input, multiple-output control volumes under steady flow conditions, the entropy generated function is extended to:

where the summations () are taken over all the exit ports (e) and inlet ports (i).

We use the differential form of the energy equation to derive the Tds relations which can be used to evaluate the change of entropy (Δs) for processes involving 2-phase fluids (Steam, R134a, CO_{2}), solids or liquids, or ideal gasses.

That is “Tedious”, but without the vowels, of course! Recall from consideration of the Clausius Inequality that we defined entropy as follows:

Define Entropy: thus:

Consider the differential form of the system energy equation:

Also, from the definition of enthalpy h:

**Tables of 2-phase fluid (Steam, R134a)**

Notice that all the properties on the right hand side of this equation T, P, v, u are measurable properties given in the tables, thus this differential equation can be solved numerically, leading to the table values of entropy s.

**Liquid (or solid-incompressible)**

Finally we present a convenient Entropy Equation Summary Sheet which summarizes the relevant relations concerning entropy generation and evaluation of entropy change Δs. The Isentropic Processes Summary Sheet extends the relations of entropy change to enable the evaluation of isentropic processes.

One of the important applications of isentropic processes is in determining the efficiency of various adiabatic components. These include turbines, compressors and aircraft jet nozzles. Thus we have made the statement that steam turbines are designed to be adiabatic, and that any heat loss from the turbine will result in a reduction in output power, however only now can we make the statement that the ideal turbine is isentropic. This enables us to evaluate the **Adiabatic Efficiency** (sometimes referred to as isentropic efficiency) of these components, and we extend the isentropic process sheet with an Adiabatic Efficiency Summary Sheet.

There are two property diagrams involving entropy in common usage, the temperature-entropy (*T-s*) and enthalpy-entropy (*h-s*) “Mollier” diagrams. We will find that the *h-s* diagram is extremely useful for evaluating adiabatic turbines and compressors, and complements the *P-h* diagram which we used in Chapter 4 to evaluate entire steam power plants or refrigerator systems. The *h-s* diagram for steam is presented below:

**An Adiabatic Steam Turbine Example**

Consider an adiabatic steam turbine having a turbine adiabatic efficiency η_{T} = 80%, operating under the conditions shown in the following diagram:

- Using steam tables, determine the enthalpy and entropy values at station (1) and station (2s) assuming that the turbine is isentropic. [h
_{1}= 3479 , s_{1}= 7.764 ; h_{2s}= 2461 , s_{2s}= s_{1}] - From the definition of turbine adiabatic efficiency (shown on the diagram), and given that η
_{T}= 80%, determine the actual enthalpy and entropy values as well as the temperature at station (2a). [h_{2a}= 2665 kJ/kg, s_{2a}= 8.38 kJ/kg.K, T_{2a}= 88°C] - Plot the actual and isentropic turbine processes (Stations (1)-(2a) and (1)-(2s)) on the enthalpy-entropy
*h-s*“Mollier” diagram, and indicate the actual turbine specific work (w_{a}) as well as the isentropic turbine specific work (w_{s}) on the diagram. - Determine the actual power output of the turbine (kW). [1629 kW]

The *h-s* diagram plot follows. Notice that we have indicated all the enthalpy and entropy values (which we determined from the steam tables) on the plot. This allows a check on the feasibility of our results.

**Aircraft Gas Turbine Engines**

There are many different forms and modifications of aircraft gas turbine engines, and in this course we discuss two variants – the ideal turbojet engine, and the gas turbine engine for usage in helicopters.

The gas turbine engine for usage in helicopters is shown below:

Consider a supercritical steam power plant with reheat for Athens, Ohio:

- Plot the two turbine processes (Stations (1)-(2) and (3)-(4)) on the enthalpy-entropy
*h-s*“Mollier” diagram. Plot also the equivalent isentropic turbine processes on the diagram, and indicate the actual turbine specific work as well as the isentropic turbine specific work for both turbines on the*h-s*diagram. - Using steam tables, determine the turbine adiabatic efficiency η
_{T}of both turbines. - Discuss your results as well as the feasibility of the turbine set.

*Justify* all values used and *derive* all equations used starting from the basic energy equation for a flow system, the basic definition of turbine adiabatic efficiency η_{T}.

**Solution Approach:**

- Plot the two turbine processes (Stations (1)-(2) and (3)-(4)) on the enthalpy-entropy
*h-s*“Mollier” diagram. Plot also the equivalent isentropic turbine processes on the diagram, and indicate the actual turbine specific work as well as the isentropic turbine specific work for both turbines on the*h-s*diagram. [refer*h-s*diagram below]

- Using steam tables, determine the turbine adiabatic efficiency η
_{T}of both turbines. - Discuss your results as well as the feasibility of the turbine set.

The University of Oklahoma catalogue describes AME 2213 this way: *“First and second law of thermodynamics are developed and applied to the solutions of problems from a variety of engineering fields. Extensive use is made of differential calculus to interrelate thermodynamics functions”*. My interpretation of this description is more specific. Thermodynamics, generally speaking, is the science of energy. The transformation of energy from one form to another, and in many cases thermodynamics is about transforming heat into work, such as in an automobile engine or at a power plant.

The application of Thermodynamics is almost everywhere in our daily life. Examples of some application areas of this subject are: Propulsion, Internal Combustion Engines, Power Plants, Refrigeration and Air Conditioning, Solar Heating, Interaction of Human Body with Surroundings, Biomedical devices, Human body, animal, plants, Ecological systems, etc. In this course on thermodynamics we will focus on the analysis of energy systems and the application of these systems to real world contexts.

There are two approaches of teaching Thermodynamics – microscopic and macroscopic. Classical Thermodynamics (macroscopic approach) does not require detailed knowledge of molecular motion to describe a *System*. We will mainly use this approach in AME 2213. Statistical Thermodynamics (macroscopic approach) considers quantum mechanical description of molecules. We will occasionally use this approach to improve our basic understanding.

To understand the fundamental concepts of thermodynamics, basic principles (conservation of mass, conservation of energy, concept of entropy) and basic thermodynamics terminology (System, Property, State, Equilibrium, Process, Pressure) should be learn. This first chapter is dedicated to these introductory concepts, as well as to the importance of dimensions and units. We will distinguish primary dimensions in Thermodynamics such as mass (**m**), length (**L**), time (**t**), temperature (**T**), electric current (**I**), luminous intensity (**lv**), amount of substance (**n**), and secondary dimensions (derived) such as: velocity (), volume (), etc. In this course we will use two important unit systems: English (**E**) and International System of Units (**SI**).

Thermodynamics is the science of energy, including energy storage and energy in transit. The Conservation of Energy Principle states that energy cannot be created or destroyed, but can only change its form. The three forms of energy storage of greatest interest to us are Potential Energy (**PE**), Kinetic Energy (**KE**), and Internal Energy (**U**), which we introduce below. The two forms of energy in transit that we consider are Work (**W**) and Heat (**Q**), and the interactions between these various forms of energy are defined in terms of the First Law of Thermodynamics.

Newton’s Second Law states:

where: F is the force in Newtons [N]

m is the mass in kilograms [kg]

a is the acceleration in meters per second squared

Here we define a pound mass (lbm) and a pound force (lbf), thus since the acceleration due to gravity g = 32.2 we have:

Unity Conservation Ratio:

We now consider the work done (W), the energy in transit requiring both the applied force (F) and movement (x). If the force (F) is constant over the distance moved (x) then the work done is given by:

where: W is the work done in Joules [J]

F is the force in Newtons [N]

x is the distance moved in meters [m]

However, in general the force (F) is not constant over the distance x, thus we need to sum all the incremental work processes taking into consideration the variation of the force (F). This leads to the equivalent integral form for determining work done (W) as follows:

where: W is the work done in Joules [J]

F is the force in Newtons [N]

x is the distance moved in meters [m]

- What is the accuracy of this conversion?
- Use this information to show that 9 mph is equivalent to 4 .

We find that with the above survival kit we can determine many unit conversions between SI & English units, typically as demonstrated in the following block:

The various forms of energy of interest to us are introduced in terms of a solid body having a mass m [kg]. These include potential, kinetic and internal energy. Potential energy (**PE**) is associated with the elevation of the body, and can be evaluated in terms of the work done to lift the body from one datum level to another under a constant acceleration due to gravity g, as follows:

Kinetic energy (**KE**) of a body is associated with its velocity and can be evaluated in terms of the work required to change the velocity of the body, as follows:

however, velocity , thus integrating from :

Internal energy (**U**) of a body is that associated with the molecular activity of the body as indicated by its temperature T [°C], and can be evaluated in terms of the heat required to change the temperature of the body having a specific heat capacity C, as follows:

In order to gain an intuitive appreciation for the relative magnitudes of the different forms of energy we consider the (tongue-in-cheek) example of an attempt to cook a turkey by potential energy. The turkey is brought to the top of a 100 m building (about 30 stories) and then dropped from the ledge. The potential energy is thus converted into kinetic energy, and finally on impact the kinetic energy is converted into internal energy. The increase in internal energy is represented by an increase in temperature, and hopefully, if this experiment is repeated enough times the temperature increase will allow the turkey to cook. This remarkable experiment was first reported by R.C.Gimmi and Gloria J Browne – “**Cooking with Potential Energy**“, published in the **Journal of Irreproducible Results** (Vol. 33, 1987, pp 21-22).

For purposes of analysis we consider two types of Thermodynamic Systems:

**Closed System**– usually referred to as a**System**or a**Control Mass**. This type of system is separated from its surroundings by a physical boundary. Energy in transit in the form of Work or Heat can flow across the system boundary, however there can be no mass flow across the boundary. One typical example of a system is a piston / cylinder device in which the system is defined as the fixed mass of fluid contained within the cylinder.

**Open System**– usually referred to as a Control Volume. In this case, in addition to work or heat, we have mass flow of the working fluid across the system boundaries through inlet and outlet ports. In this course we will be exclusively concerned with steady flow control volumes, in that the net mass of working fluid within the system boundaries remains constant (i.e. mass flow in ) = mass flow out ). The following sections refer mainly to systems – we will consider control volumes in more detail starting with**Chapter 4**.

The closed system shown above can be defined by its various Properties, such as its pressure (P), temperature (T), volume (V) and mass (m). We will introduce and define the various properties of thermodynamic interest as needed in context. Furthermore the properties can be either Extensive or Intensive (or Specific). An extensive property is one whose value depends on the mass of the system, as opposed to an intensive property (such as pressure or temperature) which is independent of the system mass. A specific property is an intensive property which has been obtained by dividing the extensive property by the mass of the system. Two examples follow – notice that specific properties will always have kilograms (kg) in the units’ denominator.

Specific volume:

Specific internal energy:

The State of a system is defined by the values of the various intensive properties of the system.

The State Postulate states that if two independent intensive property values are defined, then all the other intensive property values (and thus the state of the system) are also defined. This can significantly simplify the graphical representation of a system, since only two-dimensional plots are required. Note that pressure and temperature are not necessarily independent properties, thus a boiling liquid will change its state from liquid to vapor at a constant temperature and pressure.

We assume that throughout the system Equilibrium conditions prevail, thus there are no temperature or pressure gradients or transient effects. At any instant the entire system is under chemical and phase equilibrium.

A Process is a change of state of a system from an initial to a final state due to an energy interaction (work or heat) with its surroundings. For example in the following diagram the system has undergone a compression process in the piston-cylinder device.

- Isothermal (constant temperature process)
- Isochoric or Isometric (constant volume process)
- Isobaric (constant pressure process)
- Adiabatic (no heat flow to or from the system during the process)

We assume that all processes are Quasi-Static in that equilibrium is attained after each incremental step of the process.

A system undergoes a Cycle when it goes through a sequence of processes that leads the system back to its original state.

The basic unit of pressure is the Pascal [Pa], however practical units are kilopascal [kPa], bar [100 kPa] or atm (atmosphere) [101.32 kPa]. The Gage (or Vacuum) pressure is related to the Absolute pressure as shown in the diagram below:

In this problem we use the basic mercury barometer to determine the height of a building. Consider the case that the barometer reading at the top of the building is 751 mm Hg, and at the bottom of the building is 760 mm Hg. Assume the density of mercury ρ_{Hg} = 13,600 , and that the average density of the column of air ρ_{air} = 1.18 .

The approach to solution is illustrated in the following diagram. A free-body force diagram on the column of air allows us to determine the height as a function of the pressure difference ΔP from top to bottom.

In Engineering Thermodynamics we normally present results to within 3 or 4 significant digits. The question that one really should ask is “Is this a reasonable method to measure the height of a building?” and the answer is a resounding NO! In the following we do an uncertainty analysis and find that unless we also measure the air temperature during this experiment (why? temperature doesn’t even appear in the above equations!) then this method has an accuracy of:

height

Unacceptable by any standards.

A Throttling Valve is often used to control the downstream pressure of a high pressure fluid (such as steam or air) flowing in a pipe. In the following diagram we have a water manometer to measure the pressure drop ΔP caused by the throttling valve as well as a mercury manometer to measure the downstream pressure of the air.

Temperature is a measure of molecular activity, and a temperature difference between two bodies in contact (for example the immediate surroundings and the system) is the driving force leading to heat transfer between them.

Both the Fahrenheit and the Celsius scales are in common usage in the US, hence it is important to be able to convert between them. Furthermore we will find that in some cases we require the Absolute (Rankine and Kelvin) temperature scales (for example when using the Ideal Gas Equation of State), thus we find it convenient to plot all four scales as follows: