Thermodynamic of Steam and Boiler Processes

Thermodynamics of Steam Formation and SystemsBasic Concepts of ThermodynamicsThermodynamic System

A thermodynamic system, or simply system, is defined as a quantity of matter or a region in space chosen for study. The region outside the system is called the surroundings. The real or imaginary surface that separates the system from its surroundings is called the boundary. The boundary of a system may be fixed or movable.

Surroundings are physical space outside the system boundary.

Systems may be considered to be closed or open, depending on whether a fixed mass or a fixed volume in space is chosen for study.

Systems may be considered to be closed or open, depending on whether a fixed mass or a fixed volume in space is chosen for study.

An open system, or control volume, has mass as well as energy crossing the boundary, called a control surface. Examples of open systems are pumps, compressors, turbines, valves, and heat exchangers.

An isolated system is a general system of fixed mass where no heat or work may cross the boundaries. An isolated system is a closed system with no energy crossing the boundaries and is normally a collection of a main system and its surroundings that are exchanging mass and energy among themselves and no other system.

Since some of the thermodynamic relations that are applicable to closed and open systems are different, it is extremely important that we recognize the type of system we have before we start analyzing it.

Properties of a System

Any characteristic of a system in equilibrium is called a property. The property is independent of the path used to arrive at the system condition.

Some thermodynamic properties are pressure P, temperature T, volume V, and mass m.

Properties may be intensive or extensive.

Extensive properties are those that vary directly with size--or extent--of the system.

Some Extensive Properties

a. mass

b. volume

c. total energy

d. mass dependent property

Intensive properties are those that are independent of size.

Some Intensive Properties

a. temperature

b. pressure

c. age

d. color

e. any mass independent property

Extensive properties per unit mass are intensive properties. For example, the specific volume v, defined as

and density (, defined as

are intensive properties.

Units

An important component to the solution of any engineering thermodynamic problem requires the proper use of units. The unit check is the simplest of all engineering checks that can be made for a given solution. Since units present a major hindrance to the correct solution of thermodynamic problems, we must learn to use units carefully and properly. The system of units selected for this course is the SI System, also known as the International System (sometimes called the metric system). In SI, the units of mass, length, and time are the kilogram (kg), meter (m), and second (s), respectively. We consider force to be a derived unit from Newton's second law, i.e.,

In SI, the force unit is the newton (N), and it is defined as the force required to accelerate a mass of 1 kg at a rate of 1 m/s2. That is,

This definition of the newton is used as the basis of the conversion factor to convert mass-acceleration units to force units.

The term weight is often misused to express mass. Unlike mass, weight Wt is a force. Weight is the gravitational force applied to a body, and its magnitude is determined from Newton's second law,

where m is the mass of the body and g is the local gravitational acceleration (g is 9.807 m/s2 at sea level and 45(latitude). The weight of a unit volume of a substance is called the specific weight w and is determined from w = ( g, where ( is density.

Oftentimes, the engineer must work in other systems of units. Comparison of the United States Customary System (USCS), or English System, and the slug system of units with the SI system is shown below.

Sometimes we use the mole number in place of the mass. In SI units the mole number is in kilogram-moles, or kmol.

Newtons second law is often written as

where gc is called the gravitational constant and is obtained from the force definition. In the SI System 1 newton is that force required to accelerate 1 kg mass 1 m/s2. The gravitational constant in the SI System is

In the USCS 1 pound-force is that force required to accelerate 1 pound-mass 32.176 ft/s2. The gravitational constant in the USCS is

In the slug system, the gravitational constant is

Example 1-1

An object at sea level has a mass of 400 kg.

a) Find the weight of this object on earth.

b) Find the weight of this object on the moon where the local gravitational acceleration is one-sixth that of earth.

Solution(a)

Note the use of the conversion factor to convert mass-acceleration units into force units.

Process

Any change from one state to another is called a process. During a quasi-equilibrium or quasi-static process the system remains practically in equilibrium at all times. We study quasi-equilibrium processes because they are easy to analyze (equations of state apply) and work-producing devices deliver the most work when they operate on the quasi-equilibrium process.

ProcessProperty held constant

isobaric pressure

isothermal temperature

isochoric volume

isentropic entropy (see Chapter 7)

We can understand the concept of a constant pressure process by considering the above figure. The force exerted by the water on the face of the piston has to equal the force due to the combined weight of the piston and the bricks. If the combined weight of the piston and bricks is constant, then F is constant and the pressure is constant even when the water is heated.

We often show the process on a P-V diagram as shown below.

Steady-Flow Process

Consider a fluid flowing through an open system or control volume such as a water heater. The flow is often defined by the terms steady and uniform. The term steady implies that there are no changes with time. The term uniform implies no change with location over a specified region. Engineering flow devices that operate for long periods of time under the same conditions are classified as steady-flow devices. The processes for these devices is called the steady-flow process. The fluid properties can change from point to point with in the control volume, but at any fixed point the properties remain the same during the entire process.State Postulate

As noted earlier, the state of a system is described by its properties. But by experience not all properties must be known before the state is specified. Once a sufficient number of properties are known, the state is specified and all other properties are known. The number of properties required to fix the state of a simple, homogeneous system is given by the state postulate:

The thermodynamic state of a simple compressible system is completely specified by two independent, intensive propertiesCycle

A process (or a series of connected processes) with identical end states is called a cycle. Below is a cycle composed of two processes, A and B. Along process A, the pressure and volume change from state 1 to state 2. Then to complete the cycle, the pressure and volume change from state 2 back to the initial state 1 along process B. Keep in mind that all other thermodynamic properties must also change so that the pressure is a function of volume as described by these two processes.

Pressure

Force per unit area is called pressure, and its unit is the pascal, N/m2, in the SI system and psia, lbf/in2 absolute, in the English system.

The pressure used in all calculations of state is the absolute pressure measured relative to absolute zero pressure. However, pressures are often measured relative to atmospheric pressure, called gage or vacuum pressures. In the English system the absolute pressure and gage pressures are distinguished by their units, psia (pounds force per square inch absolute) and psig (pounds force per square inch gage), respectively; however, the SI system makes no distinction between absolute and gage pressures.

These pressures are related by

Or these last two results may be written as

Where the +Pgage is used when Pabs > Patm and Pgage is used for a vacuum gage.

The relation among atmospheric, gage, and vacuum pressures is shown below.

Example 1-4

A pressure gage connected to a valve stem of a truck tire reads 240 kPa at a location where the atmospheric pressure is 100 kPa. What is the absolute pressure in the tire, in kPa?

Temperature

Although we are familiar with temperature as a measure of hotness or coldness, it is not easy to give an exact definition of it. However, temperature is considered as a thermodynamic property that is the measure of the energy content of a mass. When heat energy is transferred to a body, the body's energy content increases and so does its temperature. In fact it is the difference in temperature that causes energy, called heat transfer, to flow from a hot body to a cold body. Two bodies are in thermal equilibrium when they have reached the same temperature. If two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. This simple fact is known as the zeroth law of thermodynamics.

The temperature scales used in the SI and the English systems today are the Celsius scale and Fahrenheit scale, respectively. These two scales are based on a specified number of degrees between the freezing point of water ( 0(C or 32(F) and the boiling point of water (100(C or 212(F) and are related by

Example 1-6

Water boils at 212 (F at one atmosphere pressure. At what temperature does water boil in (C.

Like pressure, the temperature used in thermodynamic calculations must be in absolute units. The absolute scale in the SI system is the Kelvin scale, which is related to the Celsius scale by

Steam Formation and Properties.

We now turn our attention to the concept of pure substances and the presentation of their data.

Simple System

A simple system is one in which the effects of motion, viscosity, fluid shear, capillarity, anisotropic stress, and external force fields are absent.

Homogeneous Substance

A substance that has uniform thermodynamic properties throughout is said to be homogeneous.

Pure SubstanceA pure substance has a homogeneous and invariable chemical composition and may exist in more than one phase.

Examples:

1. Water (solid, liquid, and vapor phases)

2. Mixture of liquid water and water vapor

3. Carbon dioxide, CO2

4. Nitrogen, N2

5. Mixtures of gases, such as air, as long as there is no change of phase.

and density (, defined as

P-V-T Surface for a Substance that expands upon freezing

Real substances that readily change phase from solid to liquid to gas such as water, refrigerant-134a, and ammonia cannot be treated as ideal gases in general. The pressure, volume, temperature relation, or equation of state for these substances is generally very complicated, and the thermodynamic properties are given in table form. The properties of these substances may be illustrated by the functional relation F(P,v,T)=0, called an equation of state. The above two figures illustrate the function for a substance that contracts on freezing and a substance that expands on freezing. Constant pressure curves on a temperature-volume diagram are shown in Figure 3-11.

These figures show three regions where a substance like water may exist as a solid, liquid or gas (or vapor). Also these figures show that a substance may exist as a mixture of two phases during phase change, solid-vapor, solid-liquid, and liquid-vapor.

Water may exist in the compressed liquid region, a region where saturated liquid water and saturated water vapor are in equilibrium (called the saturation region), and the superheated vapor region (the solid or ice region is not shown).

Let's consider the results of heating liquid water from 20(C, 1 atm while keeping the pressure constant. We will follow the constant pressure process shown in Figure 3-11. First place liquid water in a piston-cylinder device where a fixed weight is placed on the piston to keep the pressure of the water constant at all times. As liquid water is heated while the pressure is held constant, the following events occur.

Process 1-2:

The temperature and specific volume will increase from the compressed liquid, or subcooled liquid, state 1, to the saturated liquid state 2. In the compressed liquid region, the properties of the liquid are approximately equal to the properties of the saturated liquid state at the temperature.

Process 2-3:

At state 2 the liquid has reached the temperature at which it begins to boil, called the saturation temperature, and is said to exist as a saturated liquid. Properties at the saturated liquid state are noted by the subscript f and v2 = vf. During the phase change both the temperature and pressure remain constant (according to the International Temperature Scale of 1990, ITS-90, water boils at 99.975(C ( 100(C when the pressure is 1 atm or 101.325 kPa). At state 3 the liquid and vapor phase are in equilibrium and any point on the line between states 2 and 3 has the same temperature and pressure.

Process 3-4:

At state 4 a saturated vapor exists and vaporization is complete. The subscript g will always denote a saturated vapor state. Note v4 = vg.

Thermodynamic properties at the saturated liquid state and saturated vapor state are given in Table A-4 as the saturated temperature table and Table A-5 as the saturated pressure table. These tables contain the same information. In Table A-4 the saturation temperature is the independent property, and in Table A-5 the saturation pressure is the independent property. The saturation pressure is the pressure at which phase change will occur for a given temperature. In the saturation region the temperature and pressure are dependent properties; if one is known, then the other is automatically known.

Process 4-5:

If the constant pressure heating is continued, the temperature will begin to increase above the saturation temperature, 100 (C in this example, and the volume also increases. State 5 is called a superheated state because T5 is greater than the saturation temperature for the pressure and the vapor is not about to condense. Thermodynamic properties for water in the superheated region are found in the superheated steam tables, Table A-6.

This constant pressure heating process is illustrated in the following

Consider repeating this process for other constant pressure lines as shown below.

If all of the saturated liquid states are connected, the saturated liquid line is established. If all of the saturated vapor states are connected, the saturated vapor line is established. These two lines intersect at the critical point and form what is often called the steam dome. The region between the saturated liquid line and the saturated vapor line is called by these terms: saturated liquid-vapor mixture region, wet region (i.e., a mixture of saturated liquid and saturated vapor), two-phase region, and just the saturation region. Notice that the trend of the temperature following a constant pressure line is to increase with increasing volume and the trend of the pressure following a constant temperature line is to decrease with increasing volume.

The region to the left of the saturated liquid line and below the critical temperature is called the compressed liquid region. The region to the right of the saturated vapor line and above the critical temperature is called the superheated region. See Table A-1 for the critical point data for selected substances.

Review the P-v diagrams for substances that contract on freezing and those that expand on freezing given in Figure 3-21 and Figure 3-22.

At temperatures and pressures above the critical point, the phase transition from liquid to vapor is no longer discrete.

Figure 3-25 shows the P-T diagram, often called the phase diagram, for pure substances that contract and expand upon freezing.

The triple point of water is 0.01oC, 0.6117 kPa (See Table 3-3).

The critical point of water is 373.95oC, 22.064 MPa (See Table A-1).

Plot the following processes on the P-T diagram for water (expands on freezing)

and give examples of these processes from your personal experiences.

1. process a-b: liquid to vapor transition

2. process c-d: solid to liquid transition

3. process e-f: solid to vapor transition

Property Tables

In addition to the temperature, pressure, and volume data, Tables A-4 through A-8 contain the data for the specific internal energy u the specific enthalpy h and the specific entropy s. The enthalpy is a convenient grouping of the internal energy, pressure, and volume and is given by

The enthalpy per unit mass is

We will find that the enthalpy h is quite useful in calculating the energy of mass streams flowing into and out of control volumes. The enthalpy is also useful in the energy balance during a constant pressure process for a substance contained in a closed piston-cylinder device. The enthalpy has units of energy per unit mass, kJ/kg. The entropy s is a property defined by the second law of thermodynamics and is related to the heat transfer to a system divided by the system temperature; thus, the entropy has units of energy divided by temperature. The concept of entropy is explained in Chapters 6 and 7.

Temp., T (CSat. Press., Psat kPaSpecific volume,m3/kgInternal energy,kJ/kgEnthalpy,kJ/kgEntropy,kJ/kg(K

Sat. liquid,

vfSat.

vapor,

vgSat. liquid,

ufEvap.,

ufgSat. vapor, ugSat. liquid,

hfEvap., hfgSat. vapor, hgSat. liquid, sfEvap., sfgSat. vapor,

sg

0.010.61170.001000206.000.002374.92374.90.002500.92500.90.00009.15569.1556

50.87250.001000147.0321.022360.82381.821.022489.12510.10.07638.94879.0249

101.2280.001000106.3242.022346.62388.742.022477.22519.20.15118.74888.8999

151.7060.00100177.88562.982332.52395.562.982465.42528.30.22458.55598.7803

202.3390.00100257.76283.912318.42402.383.912453.52537.40.29658.36968.6661

253.1700.00100343.340104.832304.32409.1104.832441.72546.50.36728.18958.5567

304.2470.00100432.879125.732290.22415.9125.742429.82555.60.43688.01528.4520

355.6290.00100625.205146.632276.02422.7146.642417.92564.60.50517.84668.3517

407.3850.00100819.515167.532261.92429.4167.532406.02573.50.57247.68328.2556

459.5950.00101015.251188.432247.72436.1188.442394.02582.40.63867.52478.1633

5012.350.00101212.026209.332233.42442.7209.342382.02591.30.70387.37108.0748

5515.760.0010159.5639230.242219.12449.3230.262369.82600.10.76807.22187.9898

6019.950.0010177.6670251.162204.72455.9251.182357.72608.80.83137.07697.9082

6525.040.0010206.1935272.092190.32462.4272.122345.42617.50.89376.93607.8296

7031.200.0010235.0396293.042175.82468.9293.072333.02626.10.95516.79897.7540

7538.600.0010264.1291313.992161.32475.3314.032320.62634.61.01586.66557.6812

8047.420.0010293.4053334.972146.62481.6335.022308.02643.01.07566.53557.6111

8557.870.0010322.8261355.962131.92487.8356.022295.32651.41.13466.40897.5435

9070.180.0010362.3593376.972117.02494.0377.042282.52659.61.19296.28537.4782

9584.610.0010401.9808398.002102.02500.1398.092269.62667.61.25046.16477.4151

100101.420.0010431.6720419.062087.02506.0419.172256.42675.61.30726.04707.3542

360186660.0018950.0069501726.16625.72351.91761.53720.12481.63.91651.13735.0537

365198220.0020150.0060091777.22526.42303.61817.16605.52422.74.00040.94894.9493

370210440.0022170.0049531844.53385.62230.11891.19443.12334.34.11190.68904.8009

373.95220640.0031060.0031062015.802015.82084.302084.34.407004.4070

For the complete Table A-5, the last entry is the critical point at 22.064 MPa.

Saturation pressure is the pressure at which the liquid and vapor phases are in equilibrium at a given temperature.

Saturation temperature is the temperature at which the liquid and vapor phases are in equilibrium at a given pressure.

In Figure 3-11, states 2, 3, and 4 are saturation states.

The subscript fg used in Tables A-4 and A-5 refers to the difference between the saturated vapor value and the saturated liquid value region. That is,

The quantity hfg is called the enthalpy of vaporization (or latent heat of vaporization). It represents the amount of energy needed to vaporize a unit of mass of saturated liquid at a given temperature or pressure. It decreases as the temperature or pressure increases, and becomes zero at the critical point.

Quality and Saturated Liquid-Vapor Mixture

Now, lets review the constant pressure heat addition process for water shown in Figure 3-11. Since state 3 is a mixture of saturated liquid and saturated vapor, how do we locate it on the T-v diagram? To establish the location of state 3 a new parameter called the quality x is defined as

We note

Recall the definition of quality x

Note, quantity 1- x is often given the name moisture. The specific volume of the saturated mixture becomes

The form that we use most often is

It is noted that the value of any extensive property per unit mass in the saturation region is calculated from an equation having a form similar to that of the above equation. Let Y be any extensive property and let y be the corresponding intensive property, Y/m, then

The term yfg is the difference between the saturated vapor and the saturated liquid values of the property y; y may be replaced by any of the variables v, u, h, or s.

We often use the above equation to determine the quality x of a saturated liquid-vapor state.

The following application is called the Lever Rule:

The Lever Rule is illustrated in the following figures.

Superheated Water Table

A substance is said to be superheated if the given temperature is greater than the saturation temperature for the given pressure.

State 5 in Figure 3-11 is a superheated state.

In the superheated water Table A-6, T and P are the independent properties. The value of temperature to the right of the pressure is the saturation temperature for the pressure.

The first entry in the table is the saturated vapor state at the pressure.

Compressed Liquid Water Table

A substance is said to be a compressed liquid when the pressure is greater than the saturation pressure for the temperature.

It is now noted that state 1 in Figure 3-11 is called a compressed liquid state because the saturation pressure for the temperature T1 is less than P1.

Data for water compressed liquid states are found in the compressed liquid tables, Table A-7. Table A-7 is arranged like Table A-6, except the saturation states are the saturated liquid states. Note that the data in Table A-7 begins at 5 MPa or 50 times atmospheric pressure.

At pressures below 5 MPa for water, the data are approximately equal to the saturated liquid data at the given temperature. We approximate intensive parameter y, that is v, u, h, and s data as

The enthalpy is more sensitive to variations in pressure; therefore, at high pressures the enthalpy can be approximated by

For our work, the compressed liquid enthalpy may be approximated by

Saturated Ice-Water Vapor Table

When the temperature of a substance is below the triple point temperature, the saturated solid and liquid phases exist in equilibrium. Here we define the quality as the ratio of the mass that is vapor to the total mass of solid and vapor in the saturated solid-vapor mixture. The process of changing directly from the solid phase to the vapor phase is called sublimation. Data for saturated ice and water vapor are given in Table A-8. In Table A-8, the term Subl. refers to the difference between the saturated vapor value and the saturated solid value.

The specific volume, internal energy, enthalpy, and entropy for a mixture of saturated ice and saturated vapor are calculated similarly to that of saturated liquid-vapor mixtures.

where the quality x of a saturated ice-vapor state is

How to Choose the Right Table

The correct table to use to find the thermodynamic properties of a real substance can always be determined by comparing the known state properties to the properties in the saturation region. Given the temperature or pressure and one other property from the group v, u, h, and s, the following procedure is used. For example if the pressure and specific volume are specified, three questions are asked: For the given pressure,

The answer to one of these questions must be yes. If the answer to the first question is yes, the state is in the compressed liquid region, and the compressed liquid tables are used to find the properties of the state. If the answer to the second question is yes, the state is in the saturation region, and either the saturation temperature table or the saturation pressure table is used to find the properties. Then the quality is calculated and is used to calculate the other properties, u, h, and s. If the answer to the third question is yes, the state is in the superheated region and the superheated tables are used to find the other properties.

Some tables may not always give the internal energy. When it is not listed, the internal energy is calculated from the definition of the enthalpy as

Example 2-2Determine the enthalpy of 1.5 kg of water contained in a volume of 1.2 m3 at 200 kPa.

Recall we need two independent, intensive properties to specify the state of a simple substance. Pressure P is one intensive property and specific volume is another. Therefore, we calculate the specific volume.

Using Table A-5 at P = 200 kPa, vf = 0.001061 m3/kg , vg = 0.8858 m3/kg

Now,

We see that the state is in the two-phase or saturation region. So we must find the quality x first.

Example 2-3

Determine the internal energy of refrigerant-134a at a temperature of 0(C and a quality of 60%.

Using Table A-11, for T = 0(C, uf = 51.63 kJ/kg ug =230.16 kJ/kg

then,

Example 2-4

Consider the closed, rigid container of water shown below. The pressure is 700 kPa, the mass of the saturated liquid is 1.78 kg, and the mass of the saturated vapor is 0.22 kg. Heat is added to the water until the pressure increases to 8 MPa. Find the final temperature, enthalpy, and internal energy of the water. Does the liquid level rise or fall? Plot this process on a P-v diagram with respect to the saturation lines and the critical point.

Lets introduce a solution procedure that we will follow throughout the course. A similar solution technique is discussed in detail in Chapter 1.

System: A closed system composed of the water enclosed in the tank

Property Relation: Steam Tables

Process: Volume is constant (rigid container)

For the closed system the total mass is constant and since the process is one in which the volume is constant, the average specific volume of the saturated mixture during the process is given by

or

Now to find v1 recall that in the two-phase region at state 1

Then, at P = 700 kPa

State 2 is specified by: P2 = 8 MPa, v2 = 0.031 m3/kg

At 8 MPa = 8000 kPa,

vf = 0.001384 m3/kg vg = 0.02352 m3/kg

At 8 MPa, v2 = 0.031 m3/kg.

Therefore, State 2 is superheated.

Interpolating in the superheated tables at 8 MPa, v = 0.031 m3/kg gives,

T2 = 361 (Ch2 = 3024 kJ/kgu2 = 2776 kJ/kg

Since state 2 is superheated, the liquid level falls.

Thermodynamics of boiler steam formation.

The thermodynamic process of steam formation in boiler is isobaric process, with temperature increases because of the heat addition to the fluid. The figure 2.3 below shows boiler thermodynamics processes for two different pressures, on the coordinate of temperature (T) vs volume specific (v). The operating pressures are 100 kPa (1 bar) and 1000 kPa (10 bars), and it shown that the steam temperature of the fluid for 10 bars is always higher than for 1 bars during the process.Water was heated from the same temperature T1 and rises to the saturated liquid temperatures on the liquid saturation line, which are 99.61oC for 1 bar and 179.88oC for 10 bars. Further heat addition makes water start to evaporate, and the temperatures remain constant until it reaches the steam saturation line. More heat addition will increase the steam temperature, because fluid is already in superheated steam condition. Superheated steam is usually use in electricity power plants to drive steam turbine, where water is undesirable in this process. In water tube boilers the process of heating water up to liquid saturation line is done in economizer, while boiling water to produce steam is done in evaporator, and further heating steam in superheated condition is done in superheater. For boiler use in heating process, the steam is usually heated up to steam saturation line or less. The steam condition below the saturated line is usually called wet steam, where the steam quality x < 1.0. The value of x usually more than 0.90, since water is usually not releasing heat during heating process. The wet steam condition can not be avoided, because it is impossible to produce steam with exactly on the saturated condition during the evaporation process. Water will always entrained in the steam during steam separation process after the evaporation. Figure 2. shows the thermodynamic boiler process with produce wet steam. Figure 2. . Thermodynamics boiler process producing superheated steam Figure 2. . Thermodynamics boiler process producing wet steamProblems: 1. A boiler is used to produce steam for heating process in a food processing industry. The steam pressure is 0.6 MPa and the steam quality x = 0.95. Temperature of water enter the boiler is 50oC. Calculate the amount of heat required to form the steam per-kg of steam produced.2. A boiler is used to supply steam in large steam power plant with high pressure steam output of P = 16 MPa, and steam temperature T = 615oC. Calculate the amount of heat required to form the steam per-kg of steam produced.Thermodynamics of Steam Energy Systems.It has been discussed briefly in Chapter 1 that there four types of steam energy system depending on the different steam utilization, which are

1. Drive turbine for electric generating equipment, blowers, and pumps

2. Heating in industrial processes: chemical industries, agricultural, food processing industries, etc. These heating are usually done in non contact heat transfer.

3.Combination of the two utilizations, which combined heat and power (CHP). 4. Other purposes which usually require small steam capacity, such as cleaning, laundry, drying, sterilization, etc. These heating are usually done in direct contact heat transfer.Each system could be analyzed as a thermodynamics system, because the boundary of the system could be defined and there are exchange of heat and work between the system and the surrounding across the boundary. Thermodynamics analysis of systems related to boiler specification which are, steam power plant, combined heat and power system, and steam heating system are presented as follows. Steam Power Plant.

Carnot steam cycle.We consider power cycles where the working fluid undergoes a phase change. The best example of this cycle is the steam power cycle where water (steam) is the working fluid. The heat engine may be composed of the following components. The working fluid, steam (water), undergoes a thermodynamic cycle from 1-2-3-4-1. The cycle is shown on the following T-s diagram.

Ideal Carnot Steam Cycle Processes

Process Description

1-2

Isentropic compression in compressor

2-3

Constant pressure heat addition in boiler

3-4

Isentropic expansion in turbine

4-1

Constant pressure heat rejection in condenser

Figure 2. Diagram Instalation and Carnot Steam Cycle The thermal efficiency of this cycle is given as

Note the effect of TH and TL on (th, Carnot.

The larger the TH the larger the (th, Carnot

The smaller the TL the larger the (th, Carnot

To increase the thermal efficiency in any power cycle, we try to increase the maximum temperature at which heat is added.

Reasons why the Carnot cycle is not used:

Pumping process 1-2 requires the pumping of a mixture of saturated liquid and saturated vapor at state 1 and the delivery of a saturated liquid at state 2.

To superheat the steam to take advantage of a higher temperature, elaborate controls are required to keep TH constant while the steam expands and does work.

To resolve the difficulties associated with the Carnot cycle, the Rankine cycle was devised. Rankine Cycle

The simple Rankine cycle has the same component layout as the Carnot cycle shown above. The simple Rankine cycle continues the condensation process 4-1 until the saturated liquid line is reached.

Ideal Rankine Cycle Processes

Process Description

1-2

Isentropic compression in pump

2-3

Constant pressure heat addition in boiler

3-4

Isentropic expansion in turbine

4-1

Constant pressure heat rejection in condenser

The T-s diagram for the Rankine cycle is given below. Locate the processes for heat transfer and work on the diagram.

Example 10-1Compute the thermal efficiency of an ideal Rankine cycle for which steam leaves the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa.

We use the power system and T-s diagram shown above.

P2 = P3 = 6 MPa = 6000 kPa

T3 = 350oC

P1 = P4 = 10 kPa

Pump

The pump work is obtained from the conservation of mass and energy for steady-flow but neglecting potential and kinetic energy changes and assuming the pump is adiabatic and reversible.

Since the pumping process involves an incompressible liquid, state 2 is in the compressed liquid region, we use a second method to find the pump work or the (h across the pump.

Recall the property relation:

dh = T ds + v dP

Since the ideal pumping process 1-2 is isentropic, ds = 0.

The incompressible liquid assumption allows

The pump work is calculated from

Using the steam tables

Now, h2 is found from

BoilerTo find the heat supplied in the boiler, we apply the steady-flow conservation of mass and energy to the boiler. If we neglect the potential and kinetic energies, and note that no work is done on the steam in the boiler, then

We find the properties at state 3 from the superheated tables as

The heat transfer per unit mass is

Turbine

The turbine work is obtained from the application of the conservation of mass and energy for steady flow. We assume the process is adiabatic and reversible and neglect changes in kinetic and potential energies.

We find the properties at state 4 from the steam tables by noting s4 = s3 = 6.3357 kJ/kg-K and asking three questions.

The turbine work per unit mass is

The net work done by the cycle is

The thermal efficiency is

Based on Carnot efficiency principle, there are many methods to improve the simple Rankine cycle efficiency:

1. Superheat the steam, average temperature is higher during heat addition and moisture is reduced at turbine exit.2.Increase boiler pressure for fixed maximum temperature, availability of steam is higher at higher pressures, however moisture is increased at turbine exit.3.Reheat the steam, average temperature is higher during heat addition and moisture is reduced at turbine exit.4.Lower condenser pressure, less energy is lost to surroundings, however moisture is increased at turbine exit.5.Regenerative cycle, average temperature is higher during heat addition.

The first three methods for improving systems efficiency are related directly to the boiler, which basically related to the condition of steam output of the boiler.

1. The higher the steam output temperature the higher the system efficiency. 2.The higher the steam output pressure the higher the system efficiency.The following is an example to demonstrate the effect higher steam temperature using superheater and reheater, and higher steam pressure on steam power plants thermal efficiency.Example:Combined Heat and Power (CHP) Plant.In CHP plant the steam from boiler is utilized for generating power in turbine, and process heating. The CHP plant is usually used in an industry to supply electrical energy to drive the plant and heat energy taken from turbine outlet for process heating. The installation and thermodynamics diagram of CHP plant is shown below.

The thermodynamic diagram is the same as the steam power plant, since the condenser is replaced by the process heating which operate similarly to the condenser. The difference is the turbine outlet steam pressure is higher than the atmospheric, to provide sufficient temperature for the process and to overcome steam flow losses in pipes and process. The overall thermal efficiency of this system is very high, ideally equal to unity, since all the heat of the steam from boiler is utilized. Boiler specification for CHP system is determined based on the optimization calculation between two main factors, which are the amount of heat energy for process heating and electrical energy required by the plant. The output of the optimization is boiler capacity and steam pressure. Steam pressure of boiler for CHP is usually of moderate pressures, to achieve enough steam capacity for the process heating in the plant.Steam heating system.

In this system steam from boiler is directly used for process heating, and the requirement of electrical energy is usually supplied by the utility grid. Boiler steam pressure is usually of low pressure, and the steam is below saturated steam condition.

There are two types of process heating,1. Condensing steam or closed system, where heat transfer for process heating is non direct contact. The condensate is then used for boiler feedwater.

2. Non condensing steam or open system, where heat transfer for process heating is direct contact and the condensate mixed with the product being processed. The boiler teedwater is fully provided by water treatment facility.The installation and thermodynamic diagram of the two systems are shown below.

Figure 2.6. Closed system boiler steam heating a). Installation diagram

b). Thermodynamics diagram

Figure 2.6. Open system boiler steam heating

a). Installation diagram

b). Thermodynamics diagram

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b).

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b).

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V

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Process

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Process

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