Thermodynamics Notes

Introduction to Thermodynamics

Overview

• Thermodynamics studies the effects of work, heat, and energy on a system.
• It focuses on macroscopic changes and observations.
• All thermodynamics can be expressed using four quantities:
  • Temperature ($T$)
  • Internal Energy ($U$)
  • Entropy ($S$)
  • Heat ($Q$)

Classical vs. Statistical Thermodynamics

• Classical Thermodynamics:
  • Deals with relationships between bulk properties of matter.
  • Does not examine matter at the atomic or molecular level.
• Statistical Thermodynamics:
  • Explains bulk properties in terms of constituent atoms.
  • Treats the aggregation of atoms statistically, not individual atom behavior.

What Thermodynamics Deals With

• Energy
• The stability of molecules and direction of change.

Basic Terminology

• To understand the laws of thermodynamics, it's crucial to define terminology precisely.
• Some terms might seem familiar from everyday language, but they have technical meanings.

System and Surroundings

• System: The region where we focus our attention.
• Surroundings: The rest of the universe outside the system.
• Universe: System + Surroundings

Types of Exchange Between System and Surroundings

1. Energy exchange: Heat, radiation, etc.
2. Matter exchange: Movement of molecules across the boundary.

Types of Systems

1. Isolated systems: No exchange of matter or energy.
2. Open systems: Exchange of both matter and energy.
3. Closed systems: No exchange of matter, but exchange of energy.

System Boundaries and Interactions

• Open: All interactions possible (Mass, Work, Heat).
• Closed: Matter cannot enter or leave.
• Semi-permeable: Only certain species can enter or leave.
• Insulated: Heat cannot enter or leave.
• Rigid: Mechanical work cannot be done.
• Isolated: No interactions are possible.

Thermodynamic Process

• A passage of a thermodynamic system from an initial to a final state of thermodynamic equilibrium.
• Defined by initial and final states.

States of a System

• Initial state: Total energy of a system before a process.
• Final state: Total energy of a system after a process.
• Equilibrium state: Condition in which no further change occurs within the system or between the system and its surroundings.

Types of Thermodynamic Processes

• Isothermal process: Constant temperature (e.g., freezing water to ice at –10°C).
• Isobaric process: Constant pressure (e.g., heating water in open air under atmospheric pressure).
• Isochoric process: Constant volume (e.g., heating gas in a sealed metal container).
• Adiabatic process: No heat is added or removed from the system ($dq = 0$).
• Reversible process: System is close to equilibrium at all times; infinitesimal alteration can restore the universe to its original state.
• Cyclic process: Final and initial states are the same, but $q$ and $w$ need not be zero.

Heat and Work

• Work ($W$): Displacement ($d$) against a resisting force ($F$). $W = F \cdot d$
• Heat: Transfer of energy as a result of a temperature difference. Heat is NOT an entity or a form of energy; it's a mode of transfer of energy.

Heat vs. Work

• Work: Coordinated flow of matter.
  • Lowering a weight.
  • Motion of a piston.
  • Flow of electrons in a conductor.
• Heat: Random motion of matter.
  • Gas molecules in a cylinder.
  • Water molecules in a cup of water.
  • Atoms vibrating in a block of Cu.

Energy Transfer

• Energy may enter the system as heat or work.
• Once inside, it doesn't matter how the energy entered; it's stored as potential energy (PE) and kinetic energy (KE).
• This energy can be withdrawn as work or heat from the system.

Heat Capacity

• The amount of heat required to raise a certain mass of a material by a certain temperature.
• $Q = mc_x\Delta T$
• $c_x$ is the specific heat of substance x (SI units: J/kg·K).

Heat Capacity of Ideal Gas

• $C_V$ = heat capacity at constant volume = $\frac{3}{2} R$
• $C_P$ = heat capacity at constant pressure = $\frac{5}{2} R$
• For constant volume: $Q = nC_V\Delta T = \Delta U$
• The universal gas constant $R = 8.314 \text{ J/mol} \cdot \text{K}$

Laws of Thermodynamics (Summary)

According to C. P. Snow:

1. You can’t win.
2. You can’t even break even.
3. You can’t get out of the game.

First Law of Thermodynamics

• Energy can be exchanged between the system and its surroundings, but the total energy of the universe (system + surroundings) is constant.
• Energy is conserved; it can neither be created nor destroyed.
• Increase in internal energy = Heat added - Work done

Mathematical Representation of the First Law

• The change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
• $E<em>2 - E</em>1 = Q - W$

Process Terminology

• Adiabatic: No heat transferred ($Q = 0$).
• Isothermal: Constant temperature ($T = \text{constant}$).
• Isobaric: Constant pressure ($P = \text{constant}$).
• Isochoric (Isometric): Constant volume ($V = \text{constant}$).

Adiabatic Process Details

• An adiabatic process transfers no heat: $Q = 0$
• $\Delta U = Q – W$
• When a system expands adiabatically, $W$ is positive, so $\Delta U$ is negative.
• When a system compresses adiabatically, $W$ is negative, so $\Delta U$ is positive.
• Example: insulated system, quick changes like combustion.
• Specific heat ratio: $\gamma$

Adiabatic Equations for Ideal Gas

• Ideal gas, any process (Energy Added):
• Adiabatic process: Q = constant
• Adiabatic equation

Example of Adiabatic Process

Diesel Power: Fuel ignites in a diesel engine from the heat of compression (no spark plug needed). Compression is fast enough to be adiabatic. If the ignite temperature is 500°C, what compression ratio Vmax / Vmin is needed? Air’s specific heat ratio is g = 1.4, & before the compression the air is at 20 °C.

Isothermal Process Details

• An isothermal process is a constant temperature process.
• Any heat flow into or out of the system must be slow enough to maintain thermal equilibrium.
• For ideal gases, if $\Delta T = 0$, $\Delta U = 0$
• Therefore, $Q = W$
• Any energy entering the system ($Q$) must leave as work ($W$).

Isothermal Processes Equation

• Isothermal process: T = constant.
• Isothermal processes on ideal gas

Isobaric Process Details

• An isobaric process is a constant pressure process.
• $\Delta U$, $W$, and $Q$ are generally non-zero, but calculating the work done by an ideal gas is straightforward: $W = P\cdot\Delta V$
• Example: Water boiling in a saucepan.

Isobaric Processes & Specific Heat Equations

• Isobaric Process : constant P
• $C_P$ = molar specific heat at constant pressure
• Ideal gas, isobaric :

Isochoric Process Details

• An isochoric process is a constant volume process.
• When the volume of a system doesn’t change, it will do no work on its surroundings: $W = 0$
• $\Delta U = Q$
• Example: Heating gas in a closed container.

Constant-Volume Processes & Specific Heat Equations

• Constant-volume process ( isometric, isochoric, isovolumic ) : V = constant
• $C_V$ = molar specific heat at constant volume
• Ideal gas: U = U(T)
• Non-ideal gas:

Ideal Gas Processes Summary

Isothermal

• Defining characteristic: $T = \text{constant}$
• First law: Q = W
• Work done by gas: $W = nRT \ln{\frac{V<em>2}{V</em>1}}$
• Other relationships: $pV = \text{constant}$

Constant-Volume

• Defining characteristic: $V = \text{constant}$
• First law: Q = \Delta U
• Work done by gas: $W = 0$
• Other Relationships: $Q=nC_V\Delta T$

Isobaric

• Defining characteristic: $P = \text{constant}$
• First law: $Q=\Delta U + W$
• Work done by gas: $W = p(V<em>2 - V</em>1)$
• Other Relationships: $Q = nC<em>p\Delta T$, $C</em>p = C_V + R$

Adiabatic

• Defining characteristic: $Q=0$
• First law: $\Delta U = -W$
• Work done by gas: $W = \frac{P<em>1V</em>1 - P<em>2V</em>2}{1-\gamma}$
• Other Relationships: $PV^\gamma = \text{constant}$, $TV^{\gamma-1} = \text{constant}$

Cyclic Processes

• Cyclic Process: system returns to same thermodynamic state periodically.

Example: Work Done

An ideal gas with $\gamma = 1.4$ occupies 4.0 L at 300 K & 100 kPa pressure. It’s compressed adiabatically to ¼ of original volume, then cooled at constant V back to 300 K, & finally allowed to expand isothermally to its original V. How much work is done on the gas?

• AB (adiabatic)
• BC (isometric)
• CA (isothermal)

Second Law of Thermodynamics

• You can’t break even.
• Heat flows spontaneously from a hot object to a cold object without external work.

Entropy

• There exists a useful thermodynamic variable called entropy ($S$).
• A natural process that starts in one equilibrium state and ends in another will go in the direction that causes the entropy of the system plus the environment to increase for an irreversible process and to remain constant for a reversible process.
• $S<em>i = S</em>f \text{ (reversible)}$
• $S<em>f > S</em>i \text{ (irreversible)}$

Entropy and Disorder

• The second law of thermodynamics states that energy tends to become more evenly spread out across the universe.
• Entropy ($S$) is a measure of system disorder (messiness).
• $U$ is the quantity of a system’s energy, $S$ is the quality of a system’s energy.

Definition of Entropy

• Entropy measures the spontaneous dispersal of energy.

• $\text{Entropy} = \frac{\text{energy dispersed}}{T}$

• Entropy change $\Delta S$ shows us exactly how important to a system is a dispersion of a given amount of energy.

Entropy Examples

You can pump heat out of a refrigerator (to make ice cubes), but the heat is placed in the house and the entropy of the house increases, even though the local entropy of the ice cube tray decreases.

Entropy in Chemical Terms

• Related to the random movements of molecules and is measured by $T \Delta S$
• When a system is at equilibrium, no net reaction occurs and the system has no capacity to do work.
• $Q = T \Delta S$
• This is a condition of maximum entropy.
• Work can be done by system proceeding to equilibrium and measure of the maximum useful work is given by the following equation $W = - \Delta H + T \Delta S$

Heat Engines

• A device that converts heat into work while operating in a cycle.
• Heat engine $Q<em>H$, $Q</em>L$, $T<em>H$, $T</em>L$, $W_{net}$
• $\text{dQ} - W_{net} = dU \text{ (since dU=0 for a cycle)}$
• $\therefore W<em>{net} = Q</em>H - Q_L$
• Thermal efficiency, $\eta<em>{th} = \frac{W</em>{net}}{Q<em>H} = \frac{Q</em>H - Q<em>L}{Q</em>H} = 1 - \frac{Q<em>L}{Q</em>H}$

Heat Engines and the Carnot Cycle

All heat engines have:

• a high-temperature reservoir
• a low-temperature reservoir
• a cyclical engine

Steam Power Plant

A steam power plant is a good example of a heat engine where the working fluid, water, undergoes a thermodynamic cycle

• $W<em>{net} = W</em>{out} - W<em>{in} = Q</em>{in} - Q_{out}$

• $Q<em>{in}$ is the heat transferred from the high temp. reservoir, and is generally referred to as $Q</em>H$

• $Q<em>{out}$ is the heat transferred to the low temp. reservoir, and is generally referred to as $Q</em>L$

• Thermal efficiency $\eta<em>{th} = \frac{W</em>{net}}{Q<em>H} = \frac{Q</em>H - Q<em>L}{Q</em>H} = 1 - (\frac{Q<em>L}{Q</em>H})$

• Typical Efficiency of a large commercial steam power plant » 40%

• Thermal Reservoir: A hypothetical body with a very large thermal capacity to/from which heat can be transferred without changing its temperature. (E.g. the ocean, atmosphere, large lakes.)

Heat Pumps and Refrigerators

• A “heat pump” is defined as a device that transfers heat from a low-temperature source to a high-temperature one. E.g. a heat pump is used to extract energy from outside cold outdoor air into the warm indoors.
• A refrigerator performs the same function; the difference between the two is in the type of heat transfer that needs to be optimized.

Coefficient of Performance (COP)

For a Heat Pump: $COP<em>{HP} = \frac{Q</em>H}{W<em>{net}} = \frac{Q</em>H}{Q<em>H - Q</em>L} = \frac{1}{1 - \frac{Q<em>L}{Q</em>H}}$

For a Refrigerator: $COP<em>R = \frac{Q</em>L}{W<em>{net}} = \frac{Q</em>L}{Q<em>H - Q</em>L} = \frac{1}{\frac{Q<em>H}{Q</em>L} - 1}$

Note: $COP<em>{HP} = COP</em>R + 1$

Third Law of Thermodynamics

Absolute zero is a temperature that an object can get arbitrarily close to, but never attain. Temperatures as low as $2.0 \times 10^{-8} K$ have been achieved in the laboratory, but absolute zero will remain ever elusive – there is simply nowhere to “put” that last little bit of energy. This is the third law of thermodynamics: It is impossible to lower the temperature of an object to absolute zero in a finite number of steps.

Zeroth Law of Thermodynamics

If object A is in thermal equilibrium with object C, and object B is separately in thermal equilibrium with object C, then objects A and B will be in thermal equilibrium if they are placed in thermal contact.

Heat Engines and the Carnot Cycle

• A heat engine is a device that converts heat into work. A classic example is the steam engine. Fuel heats the water; the vapor expands and does work against the piston; the vapor condenses back into water again and the cycle repeats.

*All heat engines have:

• a high-temperature reservoir
• a low-temperature reservoir
• a cyclical engine

Heat engine efficiency

• An amount of heat Qh is supplied from the hot reservoir to the engine during each cycle. Of that heat, some appears as work, and the rest, Qc, is given off as waste heat to the cold reservoir.

• $\text{Efficiency} = \frac{W}{Q<em>h} = \frac{Q</em>h - Q<em>c}{Q</em>h}$

• $\text{Efficiency} = 1 - \frac{Q<em>c}{Q</em>h}$

• In order for the engine to run, there must be a temperature difference; otherwise heat will not be transferred.

Carnot's Theorem

• The maximum-efficiency heat engine is described in Carnot’s theorem: If an engine operating between two constant-temperature reservoirs is to have maximum efficiency, it must be an engine in which all processes are reversible. In addition, all reversible engines operating between the same two temperatures, Tc and Th, have the same efficiency. This is an idealization; no real engine can be perfectly reversible.

• If the efficiency depends only on the two temperatures, the ratio of the temperatures must be the same as the ratio of the transferred heats.

• $\frac{T<em>c}{T</em>h} = \frac{Q<em>c}{Q</em>h}$

• Therefore, the maximum efficiency of a heat engine can be written:

• $\eta = 1- \frac{T<em>c}{T</em>h}$

• The maximum work a heat engine can do is then:

• $W = Q<em>h(1- \frac{T</em>c}{T_h})$

• If the two reservoirs are at the same temperature, the efficiency is zero; the smaller the ratio of the cold temperature to the hot temperature, the closer the efficiency will be to 1.

Refrigerators, Air Conditioners, and Heat Pumps

• While heat will flow spontaneously only from a higher temperature to a lower one, it can be made to flow the other way if work is done on the system. Refrigerators, air conditioners, and heat pumps all use work to transfer heat from a cold object to a hot object.

• If we compare the heat engine and the refrigerator, we see that the refrigerator is basically a heat engine running backwards – It uses work to extract heat from the cold reservoir (the inside of the refrigerator) and exhausts to the kitchen. Note that more heat is exhausted to the kitchen than is removed from the refrigerator. An ideal refrigerator would remove the most heat from the interior while requiring the smallest amount of work.

• This ratio is called the coefficient of performance, COP:

• $COP = \frac{Q_c}{W}$

• Typical refrigerators have COP values between 2 and 6. Bigger is better!

• An air conditioner is essentially identical to a refrigerator; the cold reservoir is the interior of the house or other space being cooled, and the hot reservoir is outdoors. Exhausting an air conditioner within the house will result in the house becoming warmer, just as keeping the refrigerator door open will result in the kitchen becoming warmer.

• Finally, a heat pump is the same as an air conditioner, except with the reservoirs reversed. Heat is removed from the cold reservoir outside, and exhausted into the house, keeping it warm. Note that the work the pump does actually contributes to the desired result (a warmer house) in this case.

• In an ideal heat pump with two operating temperatures (cold and hot), the Carnot relationship holds; the work needed to add heat Qh to a room is:

• $W = Q<em>h(\frac{T</em>h}{T_c}-1)$

• The COP for a heat pump:

• $COP = \frac{Q<em>h}{W} = \frac{T</em>h}{T<em>h-T</em>c}$

Vapour Power Cycles

• Vapour power cycles are external combustion systems in which the working fluid is alternatively vaporized and condensed.

*Water/steam is easily available, is economical, chemically stable and physiologically harmless. Hence it is the most commonly employed working fluid.

*Due to its use as working substance in vapour power cycle, this cycle is often referred as steam power cycle.The vapour is generated in a steam boiler which then enters the steam turbine, a condenser and a feed pump. In a vapour power cycle, the main objectives are to convert the energy present in the fuels into mechanical energy and then to electrical energy.

*The fuel is burnt, hot flue gases are used to produce steam in the steam generator. This steam so produced is expanded in a steam turbine to do work.

*A power cycle continuously converts heat energy into work, in which a working fluid performs a succession of processes. In the vapor power cycle, the working fluid, which is water, undergoes a change of phase into steam, which may be in the form wet, dry saturated or super heated.

*A vapor power plant differs from a gas power plant in that, it’s working fluid may undergo a phase change during the working of the plant.

*Like in any other power cycle, the working fluid (steam/water) in a steam power plant undergoes four basic operations in a cyclic manner.
For each operation in a vapor power plant, we can think of a hypothetical or ideal process, which represents the basic intended operation. Since these operations are cyclic, the idealized processes representing these operations form an ideal cycle. That is known as vapor power cycle.

Carnot Vapor Cycle

• A Carnot cycle with two isothermal and two isentropic processes can be thought of as a vapor power cycle.

• However, in practice, it is almost impossible to design a vapor power plant, based on Carnot cycle.

• Carnot cycle is composed of four processes:

  • 1-2: isothermal heat addition
  • 2-3: isentropic (adiabatic + internally reversible) expansion
  • 3-4: isothermal heat rejection
  • 4-1: isentropic compression.

*Heat is transferred to water in the boiler from an external source to raise steam. The high pressure, high temperature steam leaving the boiler expands in the turbine to produce shaft work. The steam leaving the turbine condenses into water in the condenser, rejecting heat and then water is pumped back to the boiler.

Carnot Vapor Cycle Processes

• Process 1-2: The saturated water is isothermally converted into dry saturated steam in a boiler.

• Process 2-3: The dry steam expands isentropically in steam engine or steam turbine. The pressure and temperature falls from p2 to p3 and T2 to T3 respectively. No heat is supplied or rejected during the process.

• Process 3-4: The steam is now isothermally condensed in a condenser and heat is rejected at constant temperature T3 and pressure p3. Here T3 = T4.

• Process 4-1: The wet steam at point 4 is finally compressed isentropically in a compressor, till it returns back to initial state 1. The pressure and temperature rises from p4 to p1 and T4 to T1 respectively. Since no heat is absorbed or rejected during this process, therefore entropy remains constant.

• Work done: The work done during the cycle,

• Efficiency of the cycle: The efficiency of the Carnot cycle,

• Where, T1 = Highest temperature corresponding to the boiler.

• T2 = Lowest temperature corresponding to the condenser.

Carnot Cycle Drawbacks

• The isothermal processes 1-2 and 3-4 can be approached closely in actual boilers and condensers. Limiting the heat transfer processes to two-phase systems, however, severely limits the maximum temperature that can be used in the cycle (less than 374˚C for water).

Limiting the maximum temperature in the cycle also limits the thermal efficiency. Any attempt to raise the maximum temperature in the cycle involves heat transfer to the working fluid in a single phase, which is not easy to accomplish isothermally.

• During the isentropic expansion process 2-3 in the turbine the quality of the steam decreases. Thus the turbine has to handle steam with low quality, that is, steam with high moisture content. The impingement of liquid droplets on the turbine blades causes erosion and is a major source of wear. Thus steam with qualities less than about 90% cannot be tolerated in the operation of power plants.

• The isentropic compression process 4-1 involves the compression of a liquid-vapor mixture to a saturated liquid. There are two difficulties associated with this process. First, it is not easy to control the condensation process so precisely as to end up with the desired quality at state 4. Second, it is not practical to design a compressor that handles two phases.

*Steam condensation is not allowed to proceed to completion. The condensation process has to be stopped at state point 4.

*The working fluid at 4 is in both liquid and vapour state, it is difficult to compress two phase mixture isentropically.

*The vapour has larger specific volume; hence to accommodate greater volumes, the size of the compressor becomes quite big.

*For running a large sized compressor, more power is required; this results in poor plant efficiency.

*The steam at exhaust from the turbine is of low quality i.e. high moisture content. The liquid water droplets cause pitting and erosion of the turbine blades.

Rankine Vapour Cycle

• Many of the impractical things associated with the Carnot cycle can be eliminated by superheating the steam in the boiler and condensing it completely in the condenser.
• The cycle that results is the Rankine cycle, which is the ideal cycle for vapour power plants. The ideal Rankine cycle does not involve any internal irreversibilities and consists of the following four processes:
  • Process 1-2: Isentropic compression in a pump: Water enters the pump at state 1 as saturated liquid and is compressed isentropically to the operating pressure of the boiler. The water temperature increases somewhat during this isentropic compression process due to a slight decrease in the specific volume of water.
  • Process 2-3: Constant temperature heat addition in a boiler: Water enters the boiler as a compressed liquid at state 2 and leaves as a superheated vapor at state 3. The boiler is basically a large heat exchanger where the heat originating from the combustion gases, nuclear reactors or other sources is transferred to the water essentially at constant pressure. The boiler, together with the section where the steam is superheated, is often called as the steam generator.
  • Process 3-4: Isentropic expansion in a turbine: The superheated vapor at state 3 enters the turbine, where it expands isentropically and produces work by rotating the shaft connected to the electric-generator. The pressure and temperature of the steam drop during this process to the values at state 4, where steam enters the condenser.
  • Process 4-1: Constant pressure heat rejection in a condenser: At the state 4 the steam is usually a saturated liquid-vapor mixture with a high quality. Steam is condensed at constant pressure in the condenser, which is basically a large heat exchanger, by rejecting heat to a cooling medium such as lake, a river, or the atmosphere. Steam leaves the condenser as saturated liquid and enters the pump, completing the cycle.

Energy Analysis of the Ideal Rankine Cycle

• All four components associated with the Rankine cycle the pump, boiler, turbine and condenser are steady-flow devices, and thus all four processes that make up the Rankine cycle can be analyzed as steady-flow processes.

• The potential and kinetic energy changes of the steam are usually small relative to the work and heat transfer terms and are therefore neglected.

• The steady-flow energy equation per unit mass of the steam reduces to,

• The boiler and the condenser do not involve any work, and the pump and the turbine are assumed to be isentropic.

• Then the conservation of energy relation for each device can be expressed as follows:

• The pump work during process 1-2 is given by,

• Pump handles water which can be assumed to be incompressible.

• From property relationship for isentropic process, ds = 0

• therefore,

• For process 1-2,

• The turbine work during the process 3-4 is given by,

• The heat added per unit mass in the boiler during the process 2-3 is given by,

• The turbine work during the process 3-4 is given by,

• The heat rejected per unit mass in the condenser during the process 4-1 is given by,

• Where,

• Thermal efficiency: The thermal efficiency of the Rankine cycle is given by,

• The pump work is usually very small compared to turbine work. Hence, sometimes, it is neglected. In that case,

Rankine Cycle Calculations

• Work ratio:
• Steam flow rate:
  • It is defined as the rate of stem flow in kg/hr required to produce unit shaft power output (1kW).
  • It is a measure of the capacity of a steam power plant.
• Heat flow rate:
  • It is the rate of heat input Q1 in kJ/hr required to produce unit power output of 1kW.
  • Heat rate is an alternative to efficiency.

Comparison Between Rankine and Carnot Cycle

• For the same maximum and minimum temperatures Rankine cycle has lower efficiency than that of the Carnot cycle.
• For the same maximum and minimum temperatures Rankine cycle has the higher specific output than that of the Carnot cycle.
• Compression of wet vapor is difficult and involves large work in case of Carnot cycle when compared to the pumping work of feed water to the boiler in case of a Rankine cycle.

Effects of Pressure and Temperature on Rankine Cycle Performance

• Lowering the condenser pressure increases the thermal efficiency of the cycle:
  • The lowest pressure of condenser under ideal conditions is limited to the saturation temperature of the cooling water or air (cooling medium).
• Superheating the Steam to High temperatures increases thermal efficiency of the cycle:
  • It decreases the moisture content of the steam at the turbine exit
• Increasing the boiler pressure and temperature increases the efficiency of the cycle:
  • High moisture content results in erosion of blade surfaces, affecting their life. Normally the moisture content at the turbine exhaust should not exceed 15%.

Actual Vapor Power Cycles

• The actual vapor power cycle differs from the ideal Rankine cycle, as a result of irreversibilities in various components. Fluid friction and heat loss to the surroundings are the two common sources of the irreversibilities.

• Fluid friction causes pressure drops in the boiler, the condenser and the piping between various components. As a result, steam leaves the boiler at somewhat lower pressure.

*The other major source of irreversibility is the heat loss from the steam to the surroundings as the steam flows through various components. To maintain the same level of net work output, more heat needs be transferred to the steam in the boiler to compensate for these undesirable heat losses. As a result the cycle efficiency decreases.

*Under ideal conditions, the flow through pump and turbine is isentropic.

*The deviation of the actual pumps and turbines from the isentropic ones can be accounted for by utilizing isentropic efficiencies, defined as,

Problems on Vapour Power Cycle

• Ex-01 Dry saturated steam at 17.5 bar enters the turbine of a steam power plant and expands to the condenser pressure of 0.75 bar. Determine the Carnot and Rankine cycle efficiencies. Also find the work ratio of the Rankine cycle.

• Ex - 02 Steam enters the turbine of a steam power plant, operating on Rankine cycle, at 10 bar, 3000C. The condenser pressure is 0.1 bar. Steam leaving the turbine is 90% dry. Calculate the adiabatic efficiency of the turbine and also the cycle h, neglecting pump work.

• Ex-03 Dry and saturated steam at pressure 11 bar is supplied to a turbine and expanded isentropically to pressure 0.07 bar. Calculate the following (a) Heat rejected, (b) Heat supplied, (c) theoretical thermal efficiency.

• Ex-04 A steam turbine receives steam at pressure 20 bar and superheated to 88.6°C. The exhaust pressure is 0.07 bar and the expansion of steam takes place isentropically Using steam table, calculate the following. a Heat rejected b (b) Heat supplied, assuming that the feed pump supplies water to the boiler at 20 bar c Net work done d Work done by the turbine e Thermal efficiency f Theoretical steam consumption.