Thermodynamics

Introduction to Thermodynamics

Ability to acquire and explain the basic concepts in thermodynamics

The student should be able to explain:


System, boundary and surroundings.




Non-flow (control mass, closed) and flow (control volume, open) processes.




Intensive and extensive properties, zeroth law of thermodynamics





Thermodynamics state (equilibrium) Process (isobaric, isochoric, isothermal), cycles, steady flow process

1.1 System, boundary and surroundings 1.2 Non-flow and flow processes 1.3 Intensive and extensive properties 1.4 Thermodynamic states and equilibrium

What is Thermodynamics? Greek Words

Therme (heat)

Dynamis (Power)

The study of:
Energy
Transformation of useless energy (heat) to useful one (work or power)
Interaction between energy and matter (liquids and gases) 5

 HouseHouse-hold utensils appliances: Air-conditioner, heater, refrigerator  Engines: Engines: Automotive, aircraft, rocket  Plant/ Factory Refinery, power plants, nuclear power plant

Surroundings System Boundary

Real or imaginary surface that separates the system from its surroundings

region chosen to study the changes of a physical property

region outside the system

fixed Boundary movable

1.2 Non-flow and flow processes Types of systems: (a) isolated - no heat/ mass transfer across boundary (b) closed(control mass) - only heat transfer across boundary (c) open system(control volume) - heat & mass transfer across boundary

Non-flow processes

Flow processes

Forms of Energy Forms of energy - thermal, mechanical, chemical, kinetic, potential, electric, magnetic & nuclear E = total energy i.e sum of all energy in a system e = total energy = E (kJ/kg) mass m Forms of energy that make up the total energy of a system :

Energy form macroscopic microscopic

energy of a system as a whole with respect to some outside reference frames, e.g. KE, PE - related to molecular structure of a system and the degree of molecular activity - independent of outside reference frames

Sum of all microscopic forms of energy = Internal Energy (U) Macroscopic forms of energy

Kinetic energy (KE) - result of motion relative to some reference frame KE = mv2/2 (kJ) where v = velocity of the system relative to some fixed reference frame (m/s) m = mass of an object (kg)

Potential energy (PE) - due to elevation in a gravitational field PE = mgh (kJ) where g = gravitational acceleration, 9.81 m/s2 h = elevation of center of gravity of a system relative to some arbitrarily plane (m)

Therefore, E = U + KE + PE (kJ) 11

Internal energy - sum of all microscopic forms of energy of a system  related to - 1) molecular structure 2) degree of molecular activity I. E

KE

PE

molecular translation molecular rotation electron translation molecular vibration electron spin nuclear spin

sensible energy depend on the temperature

Latent heat - Internal energy associated to with the phase of a system - phase -change process can occur without a change in the chemical composition of a system 12

Property - any characteristic of a system that describes a system 

Some familiar properties are P, T, V and m. But can be extended to include less familiar ones such as viscosity, thermal conductivity, thermal expansion coefficient and etc



Density (mass per unit volume), ρ =



Specific gravity or relative density (ratio of the density of a substance to the density of some standard substance at a specified temperature) e.g. for water, ρ = ρ s ρH O V Specific volume, ν = (m3/kg)

m (kg/m3) depends on T & P V

2



m

Intensive Properties

independent of the size/extent of the system

T, P,ρ age, colour

Extensive dependent on the size/extent of the system

m V total E

Specific properties - extensive properties per unit mass E.g. specific volume (v = V/m) and specific total energy (e = E/m) 14



State

a set of properties that describe the condition of a system at certain time

At a given state, all the properties of a system have fixed values. If the value of one property changes, the state will change to a different one. 

Equilibrium state

steady state/ state of balance & no change with time



Thermal equilibrium

T is the same throughout the system



Mechanical equilibrium

P is the same throughout………



Phase equilibrium

m of each phase unchanged



Chemical equilibrium

chemical composition unchanged

Thermal equilibrium (uniform temperature) 16

Process one

Any change that a system undergoes from equilibrium state to another

Path

Series of states through which a system during a process

passes

 Need to specify the initial & final states of the process, as well as the path it follows, and the interactions with the surroundings.

 When a process proceeds in such a manner that the system remains infinitesimally close to equilibrium state at all times.  Sufficiently slow process that allows the system to adjust to itself internally so that properties in one part of the system do not change any faster than those at other parts. Slow compression (quasi-equilibrium)

Very fast compression (non-quasi equilibrium)

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 The prefix iso- is often used to designate a process for which a particular property remains constant. Isothermal Process a process when T remains constant Isobaric

P constant

Isochoric/ Isometric

specific volume v remains constant

P

2

Process B 1

Process A V

 A system is said to have undergone a cycle if it returns to its initial state at the end of the process.  For a cycle, the initial & final states are identical

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Pressure P =

Force Area

=

F Unit = N/m2 or Pa A

 Gas or liquid Pressure  Solids Stress  Common units 1 bar = 105 Pa 1 atm = 101,325 Pa = 1.01325 bars 1 kgf/ cm2 = 0.9807 bar = 0.96788 atm  English unit Ibf/in2 or psi Absolute pressure Gage pressure Vacuum pressure

Actual pressure at at given position & measured relative to absolute vacuum Difference between absolute pressure & local atmospheric pressure Difference between atmospheric pressure & absolute pressure

 Absolute, gage & vacuum pressures are all +ve quantities & related to each other by: Pgage = Pabs - Patm (for pressure above Patm) Pvac = Patm - Pabs (for pressure below Patm)

 In thermo, absolute pressure is always used unless stated. 21

Manometer  Small to moderate pressure difference are measured by a manometer and a differential fluid column of height h corresponds to a pressure difference between the system and the surrounding of the manometer.

∆P = ρ g h

( kPa )

22

Other Pressure Measurement Devices Bourdon Tube

Modern pressure sensors: 1) Pressure transducers 2) Piezoelectric material

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Example 1.1 A vacuum gage connected to a chamber reads 5.8 psi at a location where the atmospheric pressure is 14.5 psi. Determine the absolute pressure in the chamber. Solution: Using Pvac = Patm - Pabs = 14.5 - 5.8 = 8.7 psi 24

A vacuum gage connected to a tank reads 30 kPa at a location where the atmospheric pressure is 98 kPa. What is the absolute pressure in the tank? Solution: Pabs

= Patm - Pgage = 98 kPa - 30 kPa = 68 kPa

25

Example 1.3 A pressure gage connected to a valve stern 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 and in psia? Solution: Pabs

= Patm - Pgage = 100 kPa + 240 kPa = 340 kPa

The pressure in psia is Pabs = 340 kPa =

14.7 psia = 49.3 psia 101.3kPa

What is the gage pressure of the air in the tire, in psig? Pgage = Pabs - Patm = 49.3 psia - 14.7 psia = 34.6 psig 26

Example 1.4 Both a gage and a manometer are attached to a gas tank to measure its pressure. If the pressure gage reads 80 kPa, determine the distance between the two fluid levels of the manometer if the fluids is mercury whose density is 13,600 kg/m3.

∆P h= ρg 103 N / m3 80 kPa kPa h= kg m 1N 13600 3 9.807 2 m s kg m / s2 = 0.6 m 27

 Measure of hotness and coldness  Transfer of heat from higher to lower temp. until both bodies attain the same temp. At that point, heat transfer stops and the two bodies have reached thermal equilibrium requirement: equality of temperature  Zeroth Law of Thermodynamics: 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. Temperature scales:

Celcius (°C) Fahrenheit (°F) Kelvin (K) Rankine (R) 28

Conversion:

T (K) = T ( oC) + 273.15 T (R) = T (oF) + 459.67 T (R) = 1.8 T(K) T (oF) = 1.8 T(oC) + 32 29

Conversion: T(K) = T(°C) + 273.15 T(R) = T(°F) + 459.67 ∆T K = (T2°C +273.15) - (T1°C + 273.15) = T2°C - T1°C = ∆T°C ∆T R = ∆T°F

30

Example 1.5 Consider a system whose temperature is 18°C. Express this temperature in K, R and °F. Ans: Ans: 291 K, 523.8 R, 64.4 oF

Example 1.6 The temperature of a system drops by 27°F during a cooling process. Express this drop in temperature in °C, K, R

Ans: Ans: 15 oC, 15 K, 27 R

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Properties of Pure Substances

Ability to acquire and explain the basic concepts in thermodynamics

Course Learning Outcomes The student should be able to: •

Define saturated liquid, saturated vapor, saturated liquid-vapor mixture, compressed liquid, superheated vapor, critical points, and triple point.

•

Sketch a P-v, T-v, and P-T diagrams and identify the phase regions of pure substances on the diagrams.

•

Obtain thermodynamic properties of pure substances from property tables.

•

Show the state of pure substance on a P-v and T-v diagram with respect to saturation lines.

•

Show the isobaric, isochoric and isothermal processes on a P-v and T-v diagram with respect to saturation lines.

•

Solve problems related to properties and processes of pure substance

2.1 Pure substance 2.2 Equilibrium phases of pure substance 2.3 Phase change processes of pure substance 2.4 Property diagrams for phase change processes 2.5 Property tables 2.6 The ideal gas equation of state

2.1 Pure Substance








Pure substance - A substance that has a fixed chemical composition throughout Examples of pure substances: 1. Water (solid, liquid, and vapor phases) 2. Mixture of liquid water and water vapor 3. CO2 4. N2 5. Mixtures of gases, such as air, as long as there is no change of phase 6. He Air is a mixture of several gases, but it is considered to be a pure substance.

Nitrogen and gaseous air are pure substances.

6

A mixture of liquid and gaseous water is a pure substance, but a mixture of liquid and gaseous air is not.

2.2 Equilibrium phases of pure substance Phase is identified as having a distinct molecular arrangement that is homogeneous throughout and separated from the others by easily identifiable boundary surfaces.

The arrangement of atoms in different phases: (a) solid phase - molecules are at relatively fixed positions (b) liquid phase - groups of molecules move about each other in the molecules (c) gas phase - move about at random

Equilibrium phases of pure substance Phase equilibrium: If a system involves two phases and when the mass of each phase reaches an equilibrium level and stays there. State Postulate The state postulate for a simple, pure substance states that the equilibrium state can be determined by specifying any two independent intensive properties.

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2.3 Phase-change processes of pure substance Saturated vapor Saturated liquid

- Water exists in the liquid phase - compressed liquid or subcooled liquid: A substance that it is not about to vaporize.

- Water exists as a liquid that is ready to vaporize - saturated liquid: A liquid that is about to vaporize..

- As more heat is transferred, part of the saturated liquid vaporizes -saturated liquidvapor mixture

- At 1 atm pressure, the temperature remains constant at 100°C until the last drop of liquid is vaporized - saturated vapor

- As more heat is transferred, the temperature of the vapor starts to rise - superheated vapor

If the entire process between state 1 and 5 described in the figure is reversed by cooling the water (maintain the pressure at the same value), the water will go back to state 1, retracing the same path, and in so doing, the amount of heat released (during the cooling process) will exactly match the amount of heat added during the heating process.

T-v diagram for the heating process of water at constant pressure

• Saturation temperature Tsat: The temperature at which a pure substance changes phase at a given pressure. • Saturation pressure Psat: The pressure at which a pure substance changes phase at a given temperature. Example: For water (pure substance) At a pressure of 101.325 kPa, Tsat is 99.97°C. At a temperature of 99.97°C, Psat is 101.325 kPa.

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• The temperature at which water starts boiling depends on the pressure; therefore, if the pressure is fixed so is the boiling temperature • Water boils at 100°C at 1 atm pressure.

The liquid-vapor saturation curve of a pure substance (water). Cengel 6th Ed pg 116

12 12

Example 2.1 Determine the saturation pressure, Psat for water at temperature of i) 25°C ii) 225°C.

Determine the saturation temperature, Tsat for water at pressure of i) 1.23 kPa ii) 500 kPa.

13




Latent heat: heat The amount of energy absorbed or released during a phase-change process.




Latent heat of fusion: fusion The amount of energy absorbed during melting (equivalent to the amount of energy released during freezing).




Latent heat of vaporization: The amount of energy absorbed during vaporization (equivalent to the energy released during condensation)

14










The magnitudes of the latent heats depend on the temperature or pressure at which the phase change occurs. At 1 atm pressure, the latent heat of fusion of water is 333.7 kJ/kg and the latent heat of vaporization is 2256.5 kJ/kg. Atmospheric pressure and boiling temperature of water decrease with increases of elevation.

15

The variations of properties during phase-change processes are best studied using property diagrams for pure substances.

T-v

P-T

Diagram

Diagram

P-v Diagram

The T-v Diagram

T-v diagram of constant-pressure phase-change processes of water (pure substance) at various pressures. Cengel 6th Ed pg 119.

17

At supercritical pressures (P > Pcr), there is no a distinct phasechange process. Critical point: point The point at which the saturated liquid and saturated vapor states are identical

18

19

The P-v Diagram

The pressure in a piston-cylinder device can be reduced by reducing the weight of the piston.

P-v diagram of a pure substance.

20

For water, Ttp = 0.01°C Ptp = 0.6117 kPa At triple-point pressure and temperature, a substance exists in three phases in equilibrium.

P-v diagram of a substance that contracts on freezing.

P-v diagram of a substance that expands on freezing (such as water).

21

The P-T Diagram Phase Diagram

22

P > Ptp: Melting -> Evaporation P < Ptp: Sublimation (Evaporation directly without melting first)

P > Ptp

Sublimation P < Ptp

23

2.5 Property Tables • For most substances, the relationships among thermodynamic properties are too complex to be expressed by simple equations. • Therefore, properties are frequently presented in the form of tables. • Some thermodynamic properties can be measured easily, but others cannot and are calculated by using the relations between them and measurable properties. • The results of these measurements and calculations are presented in tables in a convenient format. Table A–4: Saturation properties of water under temperature. Pg 916-917 Table A–5: Saturation properties of water under pressure. Pg 918-919 Table A–6: Superheated properties of water. Pg 920-923 Table A–7: Compressed liquid water. Pg 924

Enthalpy - A Combination Property

or

The product pressure × volume has energy units.

The combination u + Pv is frequently encountered in the analysis of control volumes. 25

Saturated Liquid and Saturated Vapor States Table A–4: Saturation properties of water under temperature. Pg 916 Table A–5: Saturation properties of water under pressure. Pg 918 A partial list of Table A–4.

Enthalpy of vaporization, hfg (Latent heat of vaporization): The amount of energy needed to vaporize a unit mass of saturated liquid at a given temperature or pressure.

26

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,

u fg = ug − u f h fg = hg − h f s fg = sg − s f

27

Example 3.2 A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.

28

Example 3.2 A piston-cylinder device contains 0.06 m3 of saturated water vapor at 350 kPa pressure. Determine the temperature and the mass of the vapor inside the cylinder.

29

Example 3.3 A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine a) The volume change b) The amount of energy transferred to the water

30

• During vaporization process, a substance exists as part liquid and part vapor
mixture of saturated liquid and saturated vapor

The relative amounts of liquid and vapor phases in a saturated mixture are specified by the quality x.

31

• Quality, x : The ratio of the mass of vapor to the total mass of the mixture. • Quality is between 0 and 1
0: sat. liquid, 1: sat. vapor.

x=

mass vapor masstotal

=

mg mt

=

mg m f + mg

32

We note

V = V f + Vg mt = m f + mg V = mt vavg , V f = m f v f , Vg = mg vg mtvmv avg = m f v f + mg v g

(1)

Substituting mf = mt – mg, dividing (1) by mt and substituting mg/mt = x yields

A two-phase system can be treated as a homogeneous mixture for convenience.

vavg = (1 − x)v f + xvg Then

vavg = v f + xv fg x=

where vfg = vg - vf

vavg − v f v fg

33

• The previous relationships can be summarized in a single equation as:

y

v, u, or h.

• This application is called the Lever Rule

x=

vavg − v f v fg

The v value of a saturated liquid–vapor mixture lies between the vf and vg values at 34 the specified T or P.

Example 3.4 A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine: a) The pressure in the tank b) The volume of the tank

35

Example 3.5 An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa. Determine: a) The temperature b) The quality c) The enthalpy of the refrigerant d) The occupied by the vapor phase

36

Table A– A–6: Superheated properties of water. Pg 920 • In the region to the right of the saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor. • In this region, temperature and pressure are independent properties.

Exercise Identify: a) Saturated vapor line b) Critical point c) Superheated vapor region! What is the phase of this region?

P or T

v

Compared to saturated vapor, superheated vapor is characterized by

A partial listing of Table A– 6.

At a specified P, superheated vapor exists at a higher h than the saturated vapor.

Example 3.6 Determine the internal energy of water at 200 kPa and 300°C.

Example 3.7 Determine the temperature of water at a state of P=0.5 MPa and h=2890 kJ/kg.

Table A– A–7: Compressed liquid water. Pg 924 Compressed liquid is characterized by

At a given P and T, a pure substance will exist as a compressed liquid if

• The compressed liquid properties depend on temperature much more strongly than they do on pressure. y → v, u, or h • A more accurate relation for h

• A compressed liquid may be approximated as a saturated liquid at the given temperature.

Example 3.8 Determine the internal energy of compressed liquid water at 80°C and 5 Mpa using a) Data from the compressed liquid table b) Data from saturated-liquid data What is the error involved in the second case?

2.6 The ideal gas equation of state • Equation of state: Any equation that relates the pressure, temperature, and specific volume of a substance. Ideal gas equation of state

R: gas constant M: molar mass (kg/kmol) Ru: universal gas constant

Different substances have different gas constants. 43

Mass = Molar mass × Mole number

Ideal gas equation at two states for a fixed mass

Various expressions of ideal gas equation

Properties per unit mole are denoted with a bar on the top.

44

Determine the mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25°C.

Heat and Work

Ability to acquire and explain the basic concepts in thermodynamics

The student should be able to:
Define the concept of heat and the terminology associated with energy transfer by heat.
Discuss the three mechanisms of heat transfer: conduction, convection, and radiation.
Define the concept of work and different forms of mechanical work.

3.1 Heat and energy transfer by heat 3.2 Work and energy transfer by work




Heat = A form of energy that is transferred across the boundary of a system by virtue of a temperature difference.

• Heat is energy in transition. It is recognized only as it crosses the boundary of a system.

Adiabatic process heat
How ?

A process during which there is no transfer

(1) System is well insulated (2) Both system & surroundings are at same T

6


Mechanisms of heat transfer: (a) Conduction Transfer of energy from more energetic particles of a substance to the adjacent less energetic ones as a result of interaction between particles. Fourier's law of heat conduction is

dT & Qcond = − A k t dx Q& cond

kt A dT dx

= = = =

heat flow per unit time (W) thermal conductivity (W/m⋅K) area normal to heat flow (m2) temperature gradient in the direction of heat flow (°C/m)

7

A flat wall is composed of 20 cm of brick having a thermal conductivity kt = 0.72 W/m⋅K. The right face temperature of the brick is 900°C, and the left face temperature of the brick is 20°C. Determine the rate of heat conduction through the wall per unit area of wall.

8

Tright = 900°C Tleft = 20°C

20 cm

∆T Q& cond = − kt A ∆x Q& cond ∆T W (20 − 900) K = − kt = 0.72 ∆x m⋅K 0.2m A W = 3168 2 m

9

(b)

Convection

Transfer of energy between a solid surface & adjacent fluid that is in motion & involves the combined effects of conduction & fluid motion.

10

The rate of heat transfer by convection

expressed as

Q&convis determined from Newton's law of cooling,

Q& conv = h A (Ts − Tf ) Q& conv A h Ts Tf

= heat transfer rate (W) = heat transfer area (m2) = convective heat transfer coefficient (W/m2⋅K) = surface temperature (K) = bulk fluid temperature away from the surface (K)

The convective heat transfer coefficient is an experimentally determined parameter that depends upon the surface geometry, the nature of the fluid motion, the properties of the fluid, and the bulk fluid velocity. Ranges of the convective heat transfer coefficient are given below. h W/m2⋅K free convection of gases 2-25 free convection of liquids 50-100 forced convection of gases 25-250 forced convection of liquids 50-20,000 convection in boiling and condensation

2500-100,000 11

(c)

Radiation

Transfer of energy due to the emission of electromagnetic waves(or photons)

(

4 Q& rad = εσA Ts4 − Tsurr

Q& rad

A σ ε Ts Tsurr

)

= heat transfer per unit time (W) = surface area for heat transfer (m2) = Stefan-Boltzmann constant, 5.67x10-8 W/m2K4 and 0.1713x10-8 BTU/h ft2 R4 = emissivity = absolute temperature of surface (K) = absolute temperature of surroundings (K)

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Consider a person standing in a breezy room at 20oC. Determine the rate of heat transfer from this person by radiation if the exposed surface area and the average outer surface temperature of the person are 1.6 m2 and 29oC, respectively, and the emissivity of human skin is 0.95




If energy crossing the boundary of a closed system is not heat, it must be work. work




Work is the energy transfer associated with a force acting through a distance. e.g. rotating shaft, rising piston, electric wire




Work done per unit mass of a system is denoted by w & expressed as •




Work done per unit time, W/t = Power ( W ) Unit: kJ/s or kW


Heat & work requires the specification of both magnitude & direction.
Formal sign convention: Heat transfer to a system, Qin Work done by a system, Wout (+ve)

Heat transfer from a system, Qout Work done on a system, Win (-ve)


Use subscripts in & out to indicate direction
e.g a work input = Win a heat loss = Qout
Can assume a direction for interaction
+ve assumed direction is right
-ve direction of interaction is the opposite 15

Heat vs. Work •

• • • •

Both are recognized at the boundaries of a system as they cross the boundaries. That is, both heat and work are boundary phenomena. Systems possess energy, but not heat or work. Both are associated with a process, not a state. Unlike properties, heat or work has no meaning at a state. Both are path functions (i.e., their magnitudes depend on the path followed during a process as well as the end states).

Properties are point functions have exact differentials (d ).

Properties are point functions; but heat and work are path functions (their magnitudes depend on the path followed).

Path functions have inexact differentials (δ ) 16

Electrical Work Electrical work

Electrical power

When potential difference and current change with time

Electrical power in terms of resistance R, current I, and potential difference V. When potential difference and current remain constant

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MECHANICAL FORMS OF WORK • There are two requirements for a work interaction between a system and its surroundings to exist: – there must be a force acting on the boundary. – the boundary must move. When force is not constant Work = Force × Distance

The work done is proportional to the force applied (F) and the distance traveled (s).

If there is no movement, no work is done. 18

Shaft Work

A force F acting through a moment arm r generates a torque T This force acts through a distance s

The power transmitted through the shaft is the shaft work done per unit time

Energy transmission through rotating shafts is commonly encountered in practice.

Shaft work is proportional to the torque applied and the number of revolutions of the shaft.

19

Determine the power transmitted through the shaft of a car when the torque applied is 200 N m and the shaft rotates at a rate of 4000 revolutions per minute (rpm)

20

When the length of the spring changes by a differential amount dx under the influence of a force F, the work done is

For linear elastic springs, the displacement x is proportional to the force applied

Spring Work Substituting and integrating yield

x1 and x2: the initial and the final displacements

k: spring constant (kN/m) Elongation of a spring under the influence of a force.

The displacement of a linear spring doubles when the force is doubled.

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Moving boundary work (boundary work/ P dV work)
Associated with the expansion & compression of a gas in a piston-cylinder devise. Piston is allowed to move a distance ds in a quasi-equilib. manner, the differential work done during process,

δWb = F ds = PAds = P dV (valid for quasi-equilibrium process) P = absolute pressure dV = volume change (+ve expansion) (-ve compression)
To calculate the boundary work, the process by which the system changed states must be known, i.e. functional relationship between P & V during process.
P = f(V) should be available: equation of the process path on a P-V diagram.

The boundary work = Area under the process curve plotted on diagram

a PV

Total boundary work from initial state to final state: 2

Wb = ∫ PdV 1

(kJ)

Note: P is the absolute pressure and is always positive. When dV is positive, Wb is positive. When dV is negative, Wb is negative

23

Each process gives a different value for the boundary work

The boundary work done during a process depends on the path followed as well as the end states.

24

The net work done during a cycle is the difference between the work done by the system and the work done on the system 25

Some typical processes: (a) Boundary work during a constant-volume process If the volume is held constant, dV = 0, boundary work equation becomes P

1

1

Wb = ∫ PdV = 0

2

2

V P-V diagram for V = Constant

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(b) Boundary work for a constant-pressure process

P 2

2

Wb = ∫ PdV = P ∫ dV = Po (V2 − V1 ) 1

2

1

1

V

P-V diagram for P = constant

For the constant pressure process shown above, is the boundary work positive or negative and why?

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(c) Isothermal compression of an ideal gas If the temperature of an ideal gas system is held constant, then the equation of state provides the temperature-volume relation:

mRT P= V 2

Wb = ∫ PdV = ∫

Then, the boundary work is

1

2

1

mRT dV V

 V2  = mRT ln   V1  Note: The above equation is the result of applying the ideal gas assumption for the equation of state. For real gas undergoing an isothermal process, the integral in the boundary work equation would be done numerically.

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(d) Polytropic process
During expansion & compression processes of gases, P & V are often related by: PVn = C where C are constants
The value of n -∞ to + ∞ depending on process
Some of the common values are given below: Process Exponent n Constant pressure 0 Constant volume ∞ Isothermal & ideal gas 1 Adiabatic & ideal gas k = CP/CV Cp = specific heat at constant pressure

Cv = specific heat at constant volume

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Schematic & P-V diagram for a polytropic process:

How to determine the boundary work for polytropic process? 2

Wb = ∫ PdV 1

=∫

2

1

P = CV

−n

C dV n V 30

Example 4.1 A rigid tank contains air at 500 kPa and 150oC. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65oC and 400 kPa, respectively. Determine the boundary work done during this process.

31

Example 4.2 A frictionless piston-cylinder device contains 5 kg of steam at 400 kPa and 200oC. Heat is now transferred to the steam until the temperature reaches 250oC. If the piston is not attached to a shaft and its mass is constant, determine the work done by the steam during this process.

32

Example 4.3 A piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and 80oC. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process.

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Example 4.4 A piston-cylinder device contains 0.05 m3 of a gas initially at 200 kPa. At this state, a linear spring that has a spring constant of 150 kN/m is touching the piston but exerting no force on it. Now heat is transferred to the gas, causing the piston to rise and to compress the spring until the volume inside the cylinder doubles. If the cross-sectional area of the piston is 0.25 m2, determine: (a) The final pressure inside the cylinder (b) The total work done by the gas (c) The fraction of this work done against the spring to compress it 34

Energy Analysis of Closed Systems

Course Outcome • Ability to acquire and explain the basic concepts in thermodynamics. • Ability to comprehend and apply the concept to the actual conditions and problems; i.e. closed and open systems. • Ability to apply and correlate the concept with the appropriate equations and principles to analyze and solve engineering problems.

Course Learning Outcomes The student should be able to: • State the first law of thermodynamics and identify it simply as statement of the conservation of energy. • Explain the mechanisms of energy transfer for a closed system. • Write down the energy balances for a closed system. • Calculate the boundary work for isochoric, isobaric, isothermal and polytropic processes for closed systems. • Define the specific heat at constant volume and the specific heat at constant pressure. • Relate the specific heats to the calculation of the changes in internal energy and enthalpy of ideal gases. • Describe incompressible substances and determine the changes in their internal energy and enthalpy. • Solve energy balance problems for closed systems that involve heat and work interactions for general pure substances, ideal gases, and incompressible substances.

4.1 First Law of Thermodynamics 4.2 General Energy Balance 4.3 Energy Balance for Closed Systems 4.4 Specific Heat

4.1 THE FIRST LAW OF THERMODYNAMICS •

• •

The first law of thermodynamics (the conservation of energy principle) provides a sound basis for studying the relationships among the various forms of energy and energy interactions. The first law states that energy can be neither created nor destroyed during a process; it can only change forms. The First Law: For all adiabatic processes between two specified states of a closed system, the net work done is the same regardless of the nature of the closed system and the details of the process.

Energy cannot be created or destroyed; it can only change forms.

The increase in the energy of a potato in an oven is equal to the amount of heat transferred to it. 5

The work (electrical) done on an adiabatic system is equal to the increase in the energy of the system.

In the absence of any work interactions, the energy change of a system is equal to the net heat transfer.

The work (shaft) done on an adiabatic system is equal to the increase in the energy of the system. 6

4.2 General Energy Balance The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process.

7

Energy Change of a System, ∆Esystem:

Internal, kinetic, and potential energy changes

8

4.3 Energy Balance of Closed System

The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings. The work (boundary) done on an adiabatic system is equal to the increase in the energy of the system.

9

Energy balance when sign convention is used (i.e., heat input and work output are positive; heat output and work input are negative).

For a cycle ∆E = 0, thus Q = W.

Various forms of the first-law relation for closed systems when sign convention is used.

The first law cannot be proven mathematically, but no process in nature is known to have violated the first law, and this should be taken as sufficient proof.

10

Example 2.10 A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.

11

Energy balance for a constant-pressure expansion or compression process: General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process. Q is to the system and W is from the system.

For a constant-pressure expansion or compression process:

∆U + Wb = ∆H An example of constant-pressure process

12

Problem 4.31 A substance is contained in a well-insulated rigid container that is equipped with a stirring device. Determine the change in the internal energy of this substance when 15 kJ of work is applied to the stirring device.

13

Solution

14

Problem 4.28 Saturated water vapor at 200oC is isobarically condensed to a saturated liquid in a piston-cylinder device. Calculate the heat transfer and the work done during this process in kJ/kg.

Solution

4.4 Specific Heat Specific heat at constant volume, cv: The energy required to raise the temperature of the unit mass of a substance by one degree as the volume is maintained constant. Specific heat at constant pressure, cp: The energy required to raise the temperature of the unit mass of a substance by one degree as the pressure is maintained constant.

Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way.

Constant-volume and constantpressure specific heats cv and cp (values are for helium gas). 17

• • • •

The equations in the figure are valid for any substance undergoing any process. cv and cp are properties. cv is related to the changes in internal energy and cp to the changes in enthalpy. A common unit for specific heats is kJ/kg · °C or kJ/kg · K. Are these units identical?

Formal definitions of cv and cp. The specific heat of a substance changes with temperature.

True or False? cp is always greater than cv. 18

INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF IDEAL GASES

Joule showed using this experimental apparatus that u=u(T)

For ideal gases, u, h, cv, and cp vary with temperature only.

Internal energy and enthalpy change of an ideal gas 19

•

•

At low pressures, all real gases approach ideal- • gas behavior, and therefore their specific heats depend on temperature only. • The specific heats of real gases at low pressures are called ideal-gas specific heats, or zeropressure specific heats, and are often denoted cp0 and cv0.

Ideal-gas constantpressure specific heats for some gases (see Table A– 2c for cp equations).

u and h data for a number of gases have been tabulated. These tables are obtained by choosing an arbitrary reference point and performing the integrations by treating state 1 as the reference state.

In the preparation of ideal-gas tables, 0 K is chosen as the reference temperature. 20

Internal energy and enthalpy change when specific heat is taken constant at an average value

(kJ/kg)

For small temperature intervals, the specific heats may be assumed to vary linearly with temperature. The relation ∆ u = cv ∆T is valid for any kind of process, constant-volume or not. 21

Three ways of calculating ∆u and ∆h 1. By using the tabulated u and h data. This is the easiest and most accurate way when tables are readily available. 2. By using the cv or cp relations (Table A2c) as a function of temperature and performing the integrations. This is very inconvenient for hand calculations but quite desirable for computerized calculations. The results obtained are very accurate. 3. By using average specific heats. This is very simple and certainly very convenient when property tables are not available. The results obtained are reasonably accurate if the temperature interval is not very large.

Three ways of calculating ∆u.

22

Specific Heat Relations of Ideal Gases The relationship between cp, cv and R

dh = cpdT and du = cvdT

On a molar basis

Specific heat ratio •

• The cp of an ideal gas can be determined from a knowledge of cv and R.

•

The specific ratio varies with temperature, but this variation is very mild. For monatomic gases (helium, argon, etc.), its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature. 23

Example 4.7 Air at 300 K and 200 kPa is heated at constant pressure to 600 K. Determine the change in internal energy of air per unit mass, using (a) Data from the air table (Table A-17) (b) The functional form of the specific heat (Table A-2c) (c) The average specific heat value (Table A-2b)

24

Example 4.8 An insulated rigid tank initially contains 0.7 kg of helium at 27oC and 350 kPa. A paddle wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a) The final temperature (b) The final pressure of the helium gas 25

Example 4.10 A piston-cylinder device initially contains air at 150 kPa and 27oC. At this state, the piston is resting on a pair of stops, and the enclosed volume is 400 L. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine (a) The final temperature (b) The work done by the air (c) The total heat transferred to the air 26

INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS Incompressible substance: A substance whose specific volume (or density) is constant. Solids and liquids are incompressible substances.

The specific volumes of incompressible substances remain constant during a process.

The cv and cp values of incompressible substances are identical and are denoted by c. 27

Internal Energy Changes

Enthalpy Changes

The enthalpy of a compressed liquid A more accurate relation than

28

Example 4.12 A 50-kg iron block at 80oC is dropped into an insulated tank that contains 0.5 m3 of liquid water at 25oC. Determine the temperature when thermal equilibrium is reached.

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Mass and Energy Analysis of Open Systems

Course Outcomes • Ability to acquire and explain the basic concepts in thermodynamics. • Ability to comprehend and apply the concept to the actual conditions and problems; i.e. closed and open systems. • Ability to apply and correlate the concept with the appropriate equations and principles to analyze and solve engineering problems.

Course Learning Outcomes The student should be able to : • Explain the mechanisms of energy transfer for an open system. • Write down the general energy and mass balances for an open system and simplify the energy and mass balances for steady flow systems. • Solve energy balance problems for common steady-flow devices namely, nozzle, compressors, turbine and throttling valve.

Contents 5.1 Energy Balance for Open Systems 5.2 Conservation of Mass 5.3 Flow Work and Energy of Flowing Fluid 5.4 Energy Transport by Mass 5.5 Energy Analysis of Steady-Flow Systems

5.1 Energy Balance for Open Systems

5 5

5.2 CONSERVATION OF MASS Conservation of mass: Mass, like energy, is a conserved property, and it cannot be created or destroyed during a process. Closed systems: The mass of the system remain constant during a process. Control volumes: Mass can cross the boundaries, and so we must keep track of the amount of mass entering and leaving the control volume.

6

Mass Flow Rates V& m& = ρV = v

∙

(kg/s)

The average velocity Vavg is defined as the average speed through a cross section.

Volume Flow Rates V& = Vavg Ac

(m3 /s)

The volume flow rate is the volume of fluid flowing through a cross section per unit time.

7

Conservation of Mass Principle The conservation of mass principle for a control volume: The net mass transfer to or from a control volume during a time interval ∆t is equal to the net change (increase or decrease) in the total mass within the control volume during ∆t.

Conservation of Mass for General Control Volume The conservation of mass principle for the open system or control volume is expressed as

(

)(

) (

Total mass entering − Total mass leaving = Net change in mass the CV during ∆t the CV during ∆t within the CV during ∆t

)

m in − m out = ∆ m CV (kg) m& in − m& out = dm CV /dt (kg/s) 8

Mass Balance for Steady-Flow Processes Steady-flow process, mCV = constant , Conservation of mass principle
the total amount of mass entering a control volume equal the total amount of mass leaving it.

∑ m& = ∑ m& in

m& 1 = m& 2

→

out

(kg/s)

Multiple inlets and exits

ρ 1V1 A1 = ρ 2V 2 A2 ( k g /s ) Single stream

Nozzles, diffusers, turbines, compressors and pumps involve a single stream (only one inlet and one outlet). Conservation of mass principle for a twoinlet–one-outlet steady-flow system.

9

Special Case: Incompressible Flow The conservation of mass relations can be simplified even further when the fluid is incompressible, which is usually the case for liquids.

∑

V&in =

V&1 = V&2

∑

V&out

(m3 /s)

→ V1 A1 = V2 A2

Steady, incompressible Steady, incompressible flow (single stream)

For steady flow of liquids, the volume flow rates, as well as the mass flow rates, remain constant since liquids are essentially incompressible substances.

During a steady-flow process, volume flow rates are not necessarily conserved although mass flow rates are.

10

5.3 FLOW WORK AND THE ENERGY OF A FLOWING FLUID Flow work, or flow energy: The work (or energy) required to push the mass into or out of the control volume. This work is necessary for maintaining a continuous flow through a control volume.

In the absence of acceleration, the force applied on a fluid by a piston is equal to the force applied on the piston by the fluid. Schematic for flow work.

11

Open System:

The energy content of a control volume can be changed by mass flow as well as heat and work interactions.

12 12

Total Energy of a Flowing Fluid The flow energy is automatically taken care of by enthalpy. In fact, this is the main reason for defining the property enthalpy.

h = u + Pv

Nonflowing fluid

Flowing fluid

V2 θ = Pv + u + + gz 2

V2 e=u+ + gz 2 Internal energy

Kinetic energy

Potential energy

Flow energy

Kinetic Internal energy energy The total energy consists of 3 parts for a nonflowing fluid and 4 parts for a flowing fluid.

Potential energy

13

5.4 Energy Transport by Mass

When the kinetic and potential energies of a fluid stream are negligible

When the properties of the mass at each inlet or exit change with time as well as over the cross section

& iθi is the energy The product m transported into control volume by mass per unit time. 14

Example 5.3 Steam is leaving 4-L pressure cooker whose operating pressure is 150 kPa. It is observed that the amount of liquid in the cooer has decreased by 0.6 L in 40 min after the steady operating conditions are established, and the cross-sectional area of the exit operating is 8 mm2. Determine (a) The mass flow rate of the steam and the exit velocity (b) The total and flow energies of the steam per unit mass (c) The rate at which energy leaves the cooker by steam 15

5.5 Energy Analysis of Steady-Flow Systems Many engineering systems such as power plants operate under steady conditions.

Under steady-flow conditions, the mass and energy contents of a control volume remain constant.

Under steady-flow conditions, the fluid properties at an inlet or exit remain constant (do not change with time).

16

Mass and Energy balances for a steady-flow process Mass balance

Energy balance

A water heater in steady operation.

17

Rate of all work done, excluding Wflow and Wb

Rate of heat transfer

∆h (kJ/kg) = ∆u + P∆v

∆pe (m2/s2 or J/kg)

∆ke (m2/s2 or J/kg)

18

Energy balance relations with sign conventions (i.e., heat input and work output are positive)

when kinetic and potential energy changes are negligible

Under steady operation, shaft work and electrical work are the only forms of work a simple compressible system may involve.

Some energy unit equivalents

19

STEADY-FLOW ENGINEERING DEVICES Example:

Turbines Compressors Heat exchangers Pumps

NOZZLES & DIFFUSERS

Conveniently analyzed as steady-flow devices

TURBINES & COMPRESSORS

THROTTLING VALVES 20

NOZZLES & DIFFUSERS

• Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses. • A nozzle is a device that increases the velocity of a fluid at the expense of pressure. • A diffuser is a device that increases the pressure of a fluid by slowing it down. • The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers. Energy balance for a nozzle or diffuser

Nozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies.

21

Example 5.4 Air at 10°C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m2. The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine: (a) The mass flowrate of the air (b) The temperature of the air leaving the diffuser

22

Example 5.5 Steam at 1.8 MPa and 400°C steadily enters a nozzle whose inlet area is 0.02 m2. The mass flowrate of steam through the nozzle is 5 kg/s. Steam leaves the nozzle at 1.4 MPa with a velocity of 275 m/s. Heat losses from the nozzle per unit mass of the steam are estimated to be 2.8 kJ/kg. Determine: (a) The inlet velocity (b) The exit temperature of the steam 23

TURBINES & COMPRESSORS

• Turbine drives the electric generator In steam, gas, or hydroelectric power plants. • As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. • Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. • A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. • A compressor is capable of compressing the gas to very high pressures.

Energy balance for the compressor

• Pumps work very much like compressors except that they handle liquids instead of gases.

24

Example 5.6 Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flowrate of the air is 0.02 kg/s and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor.

25

Example 5.7 The power output of an adiabatic steam turbine is 5 MW. The inlet and the exit conditions of the steam are as indicated the figure. (a) Compare the magnitudes of Δh, Δke and Δpe (b) Determine the work done per unit mass of the steam flowing through the turbine (c) Calculate the mass flowrate of the steam

P1 = 2 MPa T1 = 400°C V1 = 50 m/s z1 = 10 m

P2 = 15 MPa x2 = 90% V2 = 180 m/s z2 = 6 m 26

THROTTLING VALVES

Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid. What is the difference between a turbine and a throttling valve? The pressure drop in the fluid is often accompanied by a large drop in temperature, and for that reason throttling devices are commonly used in refrigeration and airconditioning applications.

Energy balance

The temperature of an ideal gas does not change during a throttling (h = constant) process since h = h(T).

During a throttling process, the enthalpy of a fluid remains constant. But internal and27flow energies may be converted to each other.

Example 5.8 R-134a enters the capillary tube of a refrigerator as saturated liquid at 0.8 MPa and is throttled to a pressure of 0.12 MPa. Determine the quality of the refrigerant at the final state and the temperature drop during this process.

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The Second Law of Thermodynamics







Ability to acquire and explain the basic concepts in thermodynamics. Ability to apply and correlate the concept with the appropriate equations and principles to analyze and solve engineering problems.

Course Learning Outcomes The student should be able to: • Identify valid processes as those that satisfy both the first and second laws of thermodynamics. • Explain thermal energy reservoirs, reversible and irreversible processes, heat engines, refrigerators, and heat pumps. • Describe the Kelvin–Planck and Clausius statements of the second law of thermodynamics. • Apply the second law of thermodynamics to cycles and cyclic devices. • Apply the second law to cycles, cyclic devices and to develop the absolute thermodynamic temperature scale. • Describe the Carnot cycle and examine the Carnot principles, idealized Carnot heat engines, refrigerators, and heat pumps. • Determine the expressions for the thermal efficiencies and coefficients of performance for reversible heat engines, heat pumps, and refrigerators. • Solve problems related to heat engines, heat pumps, refrigerators (both reversible and irreversible).

6.1 Introduction to the 2nd Law of Thermodynamics 6.2 Thermal Energy Reservoir 6.3 Heat Engines 6.4 Refrigerators and Heat Pumps 6.5 Reversible and Irreversible Processes 6.6 The Carnot Cycle 6.7 Thermodynamic Temperature Scale 6.8 The Carnot Heat Engines, Refrigerators and Heat Pumps

6.1 Introduction to the 2nd Law of Thermodynamics A cup of hot coffee does not get hotter in a cooler room.

Transferring heat to a paddle wheel will not cause it to rotate.

Transferring heat to a wire will not generate electricity.

These processes cannot occur even though they are not in violation of the 1st law. 5

Processes occur in a certain direction, and not in the reverse direction. A process must satisfy both the first and second laws of thermodynamics to proceed.

MAJOR USES OF THE SECOND LAW 1. To identify the direction of processes. 2. To assert that energy has quality as well as quantity. The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality. 3. To determine the degree of degradation of energy during a process. 4. To determine the theoretical limits performance of engineering systems, such as heat engines and refrigerators. 5. To predict the degree of completion of chemical reactions. 6

6.2 Thermal Energy Reservoirs SUPPLY ENERGY

ABSORB ENERGY

• •

Bodies with relatively large thermal masses can be modeled as thermal energy reservoirs. It can supply or absorb finite amounts of heat without undergoing any change in temperature.

7

6.3 Heat Engines The devices that convert heat to work. Characteristics:-

√

Χ

1. Receive heat from a high-temp. source (eg: solar energy, oil furnace, nuclear reactor, etc.) 2. Convert part of this heat to work (eg: rotating shaft.) 3. Reject the remaining waste heat to a low-temperature sink ( eg: atmosphere, rivers, etc.) 4. Cyclic operations. Remarks: Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid.

8

Steam Power Plant

A portion of the work output of a heat engine is consumed internally to maintain continuous operation.

9

Thermal efficiency (ηth)

where;

Schematic diagram of a heat engine (HE).

10

Can we save Qout?

NO. Why? Every heat engine must waste/ sink some energy by transferring it to a low-temperature reservoir in order to complete the cycle, even under idealized conditions.

Kelvin–Planck Statement 11

Example 6.1 Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine.

12

Example 6.2 An automobile engine consumes fuel at a rate of 20 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and density of 0.8 g/cm3, determine the efficiency of this engine.

13

The 2nd Law of Thermodynamics: Kelvin–Planck Statement It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.

• ηth ≠100%; • Working fluid must exchange heat with the environment; • It is a limitation that applies to both

Χ

the idealized and the actual heat engines.

A heat engine that violates the Kelvin–Planck statement of the 2nd law. 14

6.4 Refrigerators and Heat Pumps •

The transfer of heat from a lowtemperature medium to a hightemperature one requires special devices called refrigerators.

•

Refrigerators, like heat engines, are cyclic devices.

•

The working fluid: refrigerant.

•

The most frequently used refrigeration cycle is the vaporcompression refrigeration cycle. e.g.

Schematic diagram of a refrigeration system and typical operating conditions.

In a household refrigerator, the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator, and the coils usually behind the refrigerator where heat is dissipated to the kitchen air serve as the condenser. 15

Coefficient of Performance (COP) • The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP). • Objective: To remove heat (QL) from the refrigerated space.

where;

Can the value of COPR be greater than unity? 16

Heat Pumps • Objective: To supply heat (QH) into the warmer space.

Remarks:

e.g

The work supplied to a heat pump is used to extract energy from the cold outdoors and carry it into the warm indoors.

• Can the value of COPHP be lower than unity? • What does @

COPHP=1 for fixed values of QL and QH

represent?

17

Example 6.3 The food compartment of a refrigerator is maintained at 4oC by removing heat from it at a rate of 360 kJ/min. The required power input to the refrigerator is 2 kW. Determine: a) The coefficient of performance of the refrigerator. b) The rate of heat rejection to the room that houses the refrigerator.

Example 6.4 A heat pump is used to meet the heating requirements of house and maintain it at 20oC. On a day when the outdoor air temperature drops to -2oC, the house is estimated to lose heat at a rate of 80,000 kJ/h. The heat pump under these conditions has a COP of 2.5. Determine: a) The power consumed by the heat pump. b) The rate at which heat is absorbed from the cold outdoor air.

18

The 2nd Law of Thermodynamics: Clasius Statement It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a highertemperature body.

Χ

• Refrigerator cannot operate unless its compressor is driven by an external power source, such as an electric motor. • Transfer of heat from a colder body to a warmer one.

A refrigerator that violates the Clausius statement of the 2nd law. 19

Equivalence of the Two Statements

@

Violation proof of the Kelvin–Planck statement leads to the violation of the Clausius statement.

• The Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the 2nd law of thermodynamics. • Any device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa.

20

6.5 Reversible and Irreversible Processes • Reversible process: A process that can be reversed without leaving any trace on the surroundings. • Irreversible process: A process that is not reversible. • • •

Reversible processes.

All the processes occurring in nature are irreversible. Why are we interested in reversible processes? (1) they are easy to analyze and (2) they serve as idealized models (theoretical limits) to which actual processes can be compared.

Reversible processes deliver the most and consume the least work.

21

• •

•

The factors that cause a process to be irreversible are called irreversibilities. They include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions. The presence of any of these effects renders a process irreversible. E.g. 1: Friction

E.g. 2: Heat transfer (∆T)

Case 3: Unstrained expansion

22

Internally and Externally Reversible Processes • • • •

Internally reversible process: If no irreversibilities occur within the boundaries of the system during the process. Externally reversible: If no irreversibilities occur outside the system boundaries. Totally reversible process: It involves no irreversibilities within the system or its surroundings and can be restored to their original conditions. A totally reversible process involves no heat transfer through a finite temperature difference, no nonquasi-equilibrium changes, and no friction or other dissipative effects.

23

6.6 The Carnot Cycle • Carnot cycle: Reversible cycle that composed of 4 reversible processes; 2 isothermal & 2 adiabatic processes • Acts as performance indicator of a real cycles device • Execution of the Carnot cycle in a closed system is shown below:-

Reversible Isothermal Expansion TH = constant

Reversible Adiabatic Expansion Temperature drops from TH to TL

Reversible Isothermal Compression TL = constant

Reversible Adiabatic Compression Temperature rises from TL to TH

24

Carnot Cycle: P-V Diagram

P-V diagram of the Carnot cycle.

Carnot Heat Engine Cycle.

P-V diagram of the reversed Carnot cycle.

Carnot Refrigeration Cycle.

25

The Carnot Principles

The Carnot principles:1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. Forms the basis of establishing a thermodynamic temperature scale

26

6.7 Thermodynamic Temperature Scale • Thermodynamic temperature scale (Kelvin Scale): A temperature scale that is independent of the properties of the substances that are used to measure temperature.

Remarks: Temperatures on this scale are absolute temperatures.

The arrangement of heat engines used to develop the thermodynamic temperature scale.

For reversible cycles, the heat transfer ratio QH /QL can be replaced by the absolute temperature ratio TH /TL.

27

6.8 The Carnot Heat Engine, Refrigerator and Heat Pump Carnot Heat Engine • Carnot Heat Engine: Heat engine that operates on the reversible Carnot cycle • The most efficient of all heat engines operating between the same high and low-temperature reservoirs. Carnot Heat Engine

Any Heat Engine

>

Example 6.5 A Carnot heat engine receives 500 kJ of heat per cycle from a high temperature source at 652oC and rejects heat to a low temperature sink at 30oC. Determine: a) The thermal efficiency of this Carnot engine b) The amount of heat rejected to the sink per cycle

29

The Quality of Energy

Can we use °C unit for temperature here?

The fraction of heat that can be converted to work as a function of source temperature.

The higher the temperature of the thermal energy, the higher its quality.

How do you increase the thermal efficiency of a Carnot heat engine? How about for actual heat engines? 30

Carnot Refrigerator and Heat Pump • Carnot Refrigerator/ Heat Pump: Refrigerator/ Heat Pump that operates on the reversible Carnot cycle Carnot Refrigerator/ Heat Pump

Any Refrigerator/ Heat Pump

> > Remarks: • A similar relation can be obtained for heat pumps by replacing all values of COPR by COPHP in the above relation.

31

Example 6.6 An inventor claims to have developed a heat engine that receives 800 kJ of heat from a source at 400 K and produces 250 kJ of net work while rejecting the waste heat to a sink at 300 K. Determine whether this claim is reasonable or not. Verify it

32

Example 6.7 A heat pump is to be used to heat a house during winter by maintaining the temperature at 21oC at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5oC. Determine the minimum power required to drive this heat pump

Example 6.8 A Carnot refrigerator operates in a room in which the temperature is 25oC. The refrigerator consumes 500 W of power when operating and has a COP of 4.5. Determine: a) The rate of heat removal from the refrigerated space b) The temperature of the refrigerated space

33

Entropy

Course Outcomes • Ability to acquire and explain the basic concepts in thermodynamics. • Ability to apply and correlate the concept with the appropriate equations and principles to analyze and solve engineering problems.

Course Learning Outcomes The student should be able to: • Define a new property called entropy to quantify the second-law effects. • Explain the increase of entropy principle. • Calculate the entropy changes that take place during processes for pure substances, incompressible substances, and ideal gases.

Contents 7.1 Concept of Entropy 7.2 The Increase of Entropy Principle 7.3 Entropy Change of Pure Substances 7.4 Property Diagrams Involving Entropy 7.5 The T-ds Relations (Gibbs Equation) 7.6 Entropy Change of Solids and Liquids 7.7 Entropy Change of Ideal Gases

7.1 Concept of Entropy • Originated from Clausius inequality concept : • The equality holds for the reversible process and the inequality holds for the irreversible process.

• Leads to the new definition property called entropy. • Entropy: - A quantitative measure of microscopic disorder for a system/ - Measure of energy that is no longer available to perform useful work within the current environment.

• Extensive property; unit in kJ/ K @ Intensive property; unit in kJ/kg.K

5

A pure crystalline substance at absolute zero temperature is in perfect order, and its entropy is zero (the 3rd law of thermodynamics).

The level of molecular disorder (entropy) of a substance increases as it melts or evaporates.

Disorganized energy does not create much useful effect, no matter how large it is. 6

The paddle-wheel work done on a gas increases the level of disorder (entropy) of the gas, and thus energy is degraded during this process. In the absence of friction, raising a weight by a rotating shaft does not create any disorder (entropy), and thus energy is not degraded during this process.

During a heat transfer process, the net entropy increases. (The increase in the entropy of the cold body more than offsets the decrease in the entropy of the hot body.) 7

Changes in Entropy • Factors: heat transfer, mass flow and irreversibilities. • Entropy change of a system during a process:

• The entropy change between two specified states is the same whether the process is reversible or irreversible.

Special Case: Internally Reversible Isothermal Heat Transfer Processes

Where; = The constant temp of the system (K) = Heat transfer for the internally reversile process (kJ) Remarks: This equation is particularly useful for determining the entropy changes of thermal energy reservoirs. 8

7.2 The Increase of Entropy Principle

Some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities.

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A system and its surroundings form an isolated system.

where;

∆S sys = ( S2 − S1 ) sys Qnet , surr ∆S surr = Tsurr

The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.

∴ The increase in entropy principle can be summarized as follows:

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Example 7.1 A piston-cylinder device contains a liquidvapor mixture of water at 300 K. During a constant-pressure process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of water during this process.

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Some Remarks about Entropy 1. Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, Sgen ≥ 0. 2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes. 3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities during that process. It is also used to establish criteria for the performance of engineering devices.

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Problem 7.23 A rigid tank contains an ideal gas at 40oC that is being stirred by a paddle wheel. The paddle wheel does 200 kJ of work on the ideal gas. It is observed that the temperature of the ideal gas remains constant during this process as a result of heat transfer between the system and the surroundings at 30oC. Determine the entropy change of the ideal gas.

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Problem 7.27 Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which the entropy of the two reservoirs change and determine if the second law is satisfied.

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Example 7.29 A completely reversible heat pump produces heat at a rate of 100 kW to warm a house maintained at 21oC. The exterior air, which is at 10oC serves as the source. i) Calculate the rate of entropy change of the two reservoirs ii) Determine whether this heap pump satisfies the second law according to the increase of entropy principle.

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7.3 Entropy Change of Pure Substances • Entropy is a property, thus the value of entropy of a system is fixed once the state of the system is fixed. • The entropy of a pure substance is determined from the tables (like other properties).

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Example 7.3 A rigid tank contains 5 kg of refrigerant-134a initially at 20oC and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

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Example 7.4 A piston-cylinder device initially contains 1.5 kg of liquid water at 150 kPa and 20oC. The water is now heated at constant pressure by the addition of 4000 kJ of heat. Determine the entropy change of the water during this process.

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Isentropic Processes of Pure Subs. • A process during which the entropy remains constant is called an isentropic process.

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Example 7.5 Steam enters an adiabatic turbine at 5 MPa and 450oC and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

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Problem 7.37 A well insulated rigid tank contains 2 kg of a saturated liquid-vapor mixture of water at 100 kPa. Initially, threequarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporised. Determine:• The schematic diagram of the process. • The T-s diagram of the substance. • The entropy change of the substance during this process.

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Problem 7.41 An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 150 kPa. 2200 kJ of heat is being transferred to the water by electric resistance heater. Determine: • The T-s diagram of the substance • The entropy change of the substance

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Problem 7.44 Refrigerant-134a enters an adiabatic compressor as saturated vapor at 160 kPa at a rate of 2 m3/min and is compressed to a pressure of 900 kPa. Determine the minimum power that must be supplied to the compressor.

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7.4 Property Diagrams Involving Entropy The T- s diagram

The area under the process curve represents the heat transfer for internally reversible processes.

The h-s diagram (Mollier diagram)

For adiabatic steady-flow devices, ∆h is a measure of work, and the ∆s is a measure of irreversibilities. 24

T-s and h-s diagrams for water

Remarks: Mollier diagram is a useful aid in solving steam power plant problems.

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Example 7.6 Show the Carnot cycle on a T-S diagram and indicate the areas that represent the heat supplied QH, heat rejected QL, and the net work output Wnet,out on this diagram.

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7.5 The T ds Relations (GIBBS EQ.) • The T ds relations are valid for both reversible and irreversible processes and for both closed and open systems.

where ; forms:-

• The T ds eq (Gibbs eq.) can be manipulated by introducing enthalpy relationship as shown below:-

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7.6 Entropy Change of Solids and Liquids • Liquids and solids can be approximated as incompressible substances since their specific volumes remain nearly constant during a process.

; since

;

Isentropic Process of Incompressible Subs. 28

Example 7.7 Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K, and thus methane must be maintained below 191 K to keep it in liquid phase. The properties of liquid methane at various temperatures and pressures are in Table 7.1. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa (a) Using tabulated properties (b) Approximating liquid methane as an incompressible substance. What is the error involved?

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Problem 7.66 A 20 kg aluminium block initially at 200oC is brought into contact with a 20 kg block of iron at 100oC in an insulated enclosure. Determine: • The schematic diagram of the system • The final equilibrium temperature • The total entropy change for this system.

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7.7 Entropy Change of Ideal Gases

substitute by,

substitute by,

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Constant Specific Heats (Approximate Analysis)

Entropy change of an ideal gas on a unit– mole basis

Remarks Under the constant-specific heat assumption, the specific heat is assumed to be constant at some average value. 32

Isentropic Processes of Ideal Gases

where;

Leads to:-

Remarks The isentropic relations of ideal gases are valid for the isentropic processes of ideal gases only.

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Example 7.10 Air is compressed in a car engine from 22oC and 95 kPa in a reversible and adiabatic manner. If the compression ration V1/V2 of this engine is 8, determine the final temperature of the air.

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Example 7.11 Helium gas is compressed by an adiabatic compressor from an initial state of 100 kPa and 10oC to a final temperature of 160oC in a reversible manner. Determine the exit pressure of helium.

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Problem 7.85 A piston-cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27oC. The gas is now compressed slowly in a polytropic process (PV1.3 = C). The process ends when the volume is reduced by onehalf. Determine the entropy change of nitrogen during this process.

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Problem 7.89 Air is compressed in a piston-cylinder device from 100 kPa and 17oC to 800 kPa in a reversible, adiabatic process. Determine:• The final temperature of the air • The work done during this process

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Problem 7.146 Oxygen enters an insulated 12 cm diameter pipe with a velocity of 70 m/s. At the pipe entrance, the oxygen is at 240 kPa and 20oC. At the exit, it is at 200 kPa and 18oC. Calculate the rate at which entropy is generated in the pipe.

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Example 7.19 A 50 kg block of iron casting at 500 K is thrown into a large lake that is at a temperature of 285 K. The iron block eventually reaches thermal equilibrium with the lake water. The average specific heat of iron is 0.45 kJ/ kg.K. Determine:i) ii) iii) iv) v)

The schematic diagram of this process The entropy change of the iron block. The entropy change of the lake water. The entropy generated during this process. The type of this process whether it is reversible, irreversible or impossible. Verify it.

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Vapor Cycles

Course Outcomes • Ability to acquire and explain the basic concepts in thermodynamics. • Ability to apply and correlate the concept with the appropriate equations and principles to analyze and solve engineering problems.

Course Learning Outcomes The student should be able to: • Describe the principles of carnot vapor cycles and its impracticalities • Explain the principles of an ideal rankine power cycle • Explain how pressure and temperature affect thermal efficiency of an ideal rankine power cycle • Explain the principle of the reheat Rankine power cycles. • Sketch the T-s diagram for carnot,ideal rankine and ideal reheat rankine cycles. • Solve problems related to ideal rankine and reheat rankine cycles.

8.1 Carnot Vapor Cycle 8.2 Rankine Vapor Cycle 8.3 Reheat Rankine Cycle

8.1 Carnot Vapor Cycle It is most efficient cycle operating between two specified temperature limits BUT it IS NOT a suitable model for power cycles, BECAUSE: Process 1-2 Limiting the heat transfer processes to two-phase systems severely limits the maximum temperature that can be used in the cycle (374°C for water) Process 2-3 The turbine cannot handle steam with a high moisture content because of the impingement of liquid droplets on the turbine blades causing erosion and wear. Process 4-1 It is not easy to control the condensation process so precisely to achieve quality at point 1. And it is not practical to design a compressor that handles two phases. 1-2 isothermal heat addition in a boiler 2-3 isentropic expansion in a turbine 3-4 isothermal heat rejection in a condenser 4-1 isentropic compression in a compressor T-s diagram of two Carnot vapor cycles.

Those problems can be eliminated by executing Carnot cycle in figure (b), HOWEVER this cycle presents other problem.
isentropic compression to extremely high pressures
isothermal heat transfer at variable pressures. Conclusion
Carnot cycle cannot be approximated in actual devices
It is not realistic model for vapor power cycles

IDEAL CYCLE for vapor power cycle is RANKINE CYCLE

8.2 Rankine Vapor Cycle Impracticalities associated with the Carnot cycle can be eliminated by
superheating the steam in the boiler and
condensing it completely in the condenser. The ideal Rankine cycle does not involve any internal irreversibilities.

The simple ideal Rankine cycle.

Energy Analysis of the Ideal Rankine Cycle Steady-flow energy equation

( q in -q o u t ) - ( w o u t -w in ) = h e

- hi

(k J /k g )

The thermal efficiency can be interpreted as the ratio of the area enclosed by the cycle on a T-s diagram to the area under the heat-addition process.

Example 10.1 Consider a steam power plant operating on the simple ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 75 kPa. Determine the thermal efficiency of this cycle.

Problem 10.22 Consider a steam power plant that operates on a simple ideal A simple Rankine cycle and has a net power output of 45 MW. Steam enters the turbine at 7 MPa and 500°C and is cooled in the condenser at pressure of 10 kPa by running cooling water from the lake through the tubes of condenser at rate of 2000 kg/s. i. ii. iii. iv.

Show the T-s diagram with respect to saturation line Determine the thermal efficiency of the cycle Determine the mass flowrate of the steam Determine the temperature rise of the cooling water.

HOW CAN WE INCREASE THE EFFICIENCY OF THE RANKINE CYCLE? To increase the thermal efficiency…
Increase the average temperature at which heat is transferred to the working fluid in the boiler, or decrease the average temperature at which heat is rejected from the working fluid in the condenser.

Lowering the Condenser Pressure (Lowers Tlow,avg) Superheating the Steam to High Temperatures (Increases Thigh,avg) Increasing the Boiler Pressure (Increases Thigh,avg)

Lowering the Condenser Pressure (Lowers Tlow,avg)

To take advantage of the increased efficiencies at low pressures, the condensers of steam power plants usually operate well below the atmospheric pressure. There is a lower limit to this pressure depending on the temperature of the cooling medium
cannot be lower than Psat corresponding to the temperature of the cooling medium. Side effect: Lowering the condenser pressure increases the moisture content of the steam at the final stages of the turbine.

Superheating the Steam to High Temperatures (Increases Thigh,avg) Both the net work and heat input increase as a result of superheating the steam to a higher temperature. The overall effect is an increase in thermal efficiency since the average temperature at which heat is added increases. Side effect: Superheating to higher temperatures decreases the moisture content of the steam at the turbine exit, which is desirable. The effect of superheating the steam to higher temperatures on the ideal Rankine cycle.

The temperature is limited by metallurgical considerations. Presently the highest steam temperature allowed at the turbine inlet is about 620°C.

Increasing the Boiler Pressure (Increases Thigh,avg) For a fixed turbine inlet temperature, the cycle shifts to the left and the moisture content of steam at the turbine exit increases. This side effect can be corrected by reheating the steam.

The effect of increasing the boiler pressure on the ideal Rankine cycle.

Today many modern steam power plants operate at supercritical pressures (P > 22.06 MPa) and have thermal efficiencies of about 40% for fossil-fuel plants and 34% for nuclear plants.

A supercritical Rankine cycle.

Example 10.3 Consider a steam power plant operating on the ideal Rankine cycle. Steam enters the turbine at 3 MPa and 350°C and is condensed in the condenser at a pressure of 10 kPa. Determine: (a) The thermal efficiency of this power plant (b) The thermal efficiency if steam is superheated of 600°C instead of 350°C (c) The thermal efficiency if the boiler pressure is raised to 15 MPa while the turbine inlet temperature is maintained at 600°C.

8.3 Reheat Rankine Cycle How can we take advantage of the increased efficiencies at higher boiler pressures without facing the problem of excessive moisture at the final stages of the turbine? Answer: 1. Superheat the steam to very high temperatures.
It is limited metallurgically. 2. Expand the steam in the turbine in two stages, and reheat it in between (reheat)

The ideal reheat Rankine cycle.

Additional info.: • The single reheat in a modern power plant improves the cycle efficiency by 4 to 5% by increasing the average temperature at which heat is transferred to the steam. • The average temperature during the reheat process can be increased by increasing the number of expansion and reheat stages
approach an isothermal process at the maximum temperature. The use of more than two reheat stages is not practical. The theoretical improvement in efficiency from the second reheat is about half of that which results from a single reheat. • The reheat temperatures are very close or equal to the turbine inlet temperature. • The optimum reheat pressure is about onefourth of the maximum cycle pressure.

The average temperature at which heat is transferred during reheating increases as the number of reheat stages is increased.

What is the purpose of reheat cycle?

Example 10.4 Consider a steam power plant operates on the ideal reheat Rankine cycle. Steam enters the high-pressure turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure is not to exceed 10.4 percent, determine: (a) The pressure at which the steam should be reheated (b) The thermal efficiency of the cycle Assume the steam is reheated to the inlet temperature of the high-pressure turbine.