Thermo

http://web.itu.edu.tr/solaknu/

Dr. Nuri Solak, Asst. Prof.

www.ninova.itu.edu.tr

http://web.itu.edu.tr/solaknu/

111.09.2014 => Introduction, Definition of terms, Importance of thermodynamics in metallurgical and materials eng.

2 18.09.2014 => I. Law of thermodynamics, enthalpy, heat capacity, Kirchhoff equation

3 25.09.2014 => Heat of reaction, Hess Law, temperature dependency of heat of reaction

4 02.10.2014 => Combustion and fuel, adiabatic flame temperature

5 09.10.2014 => Adiabatic flame temperature

6 16.10.2014 => Combustion and fuel, adiabatic flame temperature, Energy balance

723.10.2014 => II. Law of thermodynamics, entropy, III. Law of thermodynamics, variation of entropy as a function of temperature

8 30.10.2014 => 1. Mid-term

9 06.11.2014 =>Standard free energy, equilibrium constant.

1013.11.2014 => Calculation of composition of reaction under equilibrium conditions,Phase equilibrium in a one-component system

1120.11.2014 => Standard free energy, equilibrium constant, calculation of composition of reaction under equilibrium conditions, Phase equilibrium in a one-component system

1227.11.2014 => Standard free energy, equilibrium constant, calculation of composition of reaction under equilibrium conditions, Phase equilibrium in a one-component system

13 04.12.2014 => 2. Mid-term

14 11.12.2014 => Ellingham diagrams, Reduction reactions of oxides

What is Thermodynamics• Thermodynamics is a science and, more importantly, an

engineering tool used to describe processes that involve changes in temperature, transformation of energy, and the relationships between heat and work.

• Thermodynamic is not only related to heat, it gives interrelation between different forms of energy.

• Thermodynamics is derived from the Greek words therme, meaning heat, and dynamis, meaning power.

• Thermodynamics is concerned only with the equilibriumstate of matter.

Basic Definitions

A thermodynamic system (sistem) is a quantity of matter of fixed identity, around which we can draw a boundary. The boundaries may be fixed or moveable.

Work or heat can be transferred across the system boundary. Everything outside the boundary is the surroundings (çevre).

Piston (boundary) and gas (system)

Basic Definitions

Isolated system (Soyut ): “have walls or boundaries that are rigid, do not permit transfer of mechanical energy, perfectly insulating, and impermeable. They have aconstant energy and mass content.

Adiabatic systems (Adyabatik): Perfectly insulated systems. No energy transfer but matter can be transferred.

Closed systems (kapalı): have walls that allow transfer of energy in or out of the system but are impervious to matter. They contain a fixed mass and composition, butvariable energy.

Open Systems (açık): have walls that allow transfer of both energy and matter to and from the system.

Basic Definitions

Homogenous System:

A finite region in the physical system across which the physical and chemical properties are uniformly constant. It is also described as phase.

Heterogeneous System:

The system when it contains two or more phases, e.g., coexisting of ice and water.

Extensive Materials Properties

An extensive property is a physical quantity whose value is proportional to the size of the system it describes.

• ENERGY !!!

• entropy

• enthalpy

• mass

• momentum

• volume

Instensive Materials Properties

It is a physical property of a system that does not depend on the system size or the amount of material in the system• temperature• density (or specific gravity)• viscosity• electrical resistivity• hardness• melting point and boiling point• magnetization

Macroscopic & Microscopic Energy

• Potential energy and kinetic energy are macroscopicforms of energy. They can be visualized in terms of the position and the velocity of objects.

• In addition to these macroscopic forms of energy, a substance possesses several microscopic forms of energy. Microscopic forms of energy include those due to the rotation, vibration, translation, and interactions among the molecules of a substance. It is so called internal energy due to atomic-molecularmotion.

E = U + KE + PEE = total energy

U = Internal energy

If we calculate change in energy

dE = q – PdV - wChange in Energy

Change in Thermodynamic Properties(Given Energy)

(Produced work)

(other, special conditions)

dE = q – PdV

d independent of the path taken (yoldan bağımsız) dependent of the path taken (yola bağımlı)

If we heat up a system the given energy is not stored in the form of heat energy. It cause increase in internal energy, e.g., increase atomic motions, therefore temperature increases .

1. Law Thermodynamics

‘the change of a body inside an adiabatic system, from a given initial state to a given final state, involves the same amount of work by whatever means the process is carried out’

‘Energy cannot be created, nor destroyed’

dE = q – PdV

energy can change forms

E = U + KE + PE E = U

dE = dU = q – PdV

Thermodynamic properties depend on : Temperature (T), Pressure (P), and Volume (V)

T = f(P,V) P = f(T,V) V = f(T,P)E = f(P,V) E = f(T,V) E = f(T,P)

𝑑𝐸 =δ𝐸

δ𝑃𝑉

𝑑𝑃 +δ𝐸

δ𝑉𝑃

𝑑𝑉

𝑑𝐸 =δ𝐸

δ𝑇𝑉

𝑑𝑇 +δ𝐸

δ𝑉𝑇

𝑑𝑉

𝑑𝐸 =δ𝐸

δ𝑇𝑃

𝑑𝑇 +δ𝐸

δ𝑃𝑇

𝑑𝑃

In a closed system where E1 and V1

q amount of energy is given

E2 and V2 are reached

dE = dU = q - PdV

E = E2 – E1 = qp – P (V2-V1)

(E2 + PV2) – (E1 + PV1) = qp

H2H1

H = E + PVH2 – H1 = qp

Enthalpy

H = qp

dependent of the path independent of the path

Units

• We use SI units!

• Calorie, the quantity of heat required to raise the temperature of 1 gram of water from 14.5°C to 15.5°C. Now, we use Joule as energy unit, 1 cal = 4.18 Joule.

• Temperature unit is Kelvin (K).

• Gas constant R = 8.314 J/mol∙K

Heat Capacity

Heat capacity (C) of a system is the ratio of the heat added to or withdrawn from the system to resultant change in the temperature of the system. Heat capacity at constant volume is Cv

and at constant pressure is Cp.

𝐶 =δ𝑞

𝑑𝑇𝐶𝑝 =

δ𝑞

𝑑𝑇𝑃

𝐶𝑉 =δ𝑞

𝑑𝑇𝑉

(δq)p = (dH)p 𝐶𝑝 =δ𝑞

𝑑𝑇𝑃

=𝑑𝐻

𝑑𝑇𝑃

Heat Capacity

𝐶𝑝 =δ𝑞

𝑑𝑇𝑃

=𝑑𝐻

𝑑𝑇𝑃

𝐻1𝐻2 𝑑𝐻 = 𝑇1

𝑇2 𝐶𝑝 𝑑𝑇

𝐻 = 𝐻2− 𝐻1 = 𝑇1𝑇2 𝐶𝑝 𝑑𝑇 Kirchhoff’s Equation

Cp = a + b T + cT -2

Required energy to increase temperature

Heat Capacity

• If there is a phase transition, e.g., melting of a metal, crystallographic (polymorphic) transition, then transition enthalpy should be included, like you did it at high school for melting latent energy (gizli ısı) for ice.

• To indicate standart conditions ° sign is used. It is 1 atm and 25°C.

Heat Capacity

Cp = a + b T + c T -2

Cp(A) = 10.5 + 3.4∙ 10-3 T + 5.6∙105 T -2

Material a b∙103 c∙10-5 Range (K)

A 10.5 3.4 5.6 200-2500

Heat Capacity

𝐻 = 𝐻2− 𝐻1 = 𝑇1𝑇2 𝐶𝑝 𝑑𝑇

Q = m C T

Q = m.L

Q = mC1 T + m.L + mC2 T

Aluminium Cp = 0.90 Water Cp = 4.18

Carbon Cp = 0.72 Ethanol Cp= 2.44

Copper Cp = 0.39 H2SO4 Cp = 1.42

Lead Cp = 0.13 HCl Cp = 0.85

Mercury Cp= 0.14 KOH Cp= 1.18

Heat Capacity Values of Selected Materials @25°C[Cp (J K-1 g-1 or J oC-1 g-1)]

What happened if the heat capacity of water were 0.4 instead 4 J/mol K

Heat Capacity at low temperature

Example - 1

α-Fe γ-Fe -Fe1187K 1664K

112 gram of α-Fe at 300K and γ–Fe at 1200KCalculate required energy to increase temperatere 100K for each case.

Cp α-Fe = 37.12 + 6.17∙ 10-3 T (298 – 1187K)

Cp γ-Fe = 24.48 + 8.45∙ 10-3 T (1187 – 1664K)

Fe = 56 gr/mol

a) 7855.9 J/112 gr b) 7008 J/112 gr

Example - 2

at 27°C, take 112 gr Al2O3

In order to increase temperature 100°C, calculate the required energy.

Cp Al2O3= 117.43 + 10.38∙ 10-3 T - 37.11 ∙ 105 T-2 (298 – 1800K)

Answer : 9903.74 J/112 gr

Hint:Be careful it is asked from 27°C to 127°CActually, it means from 300K to 400 K

Example - 3Take 1 kg and 1 mol of ZrO2 and calculate required energy to increase temperature from:

a) RT to 1200K

b) RT to 1600 K α 1450K

HT = 5900 J/mol

Ttransition = 1450 K

ZrO2 = 123.2 g/mol

Cp α-ZrO2= 69.62 + 7.53∙ 10-3 T - 14.06 ∙ 105 T -2 (298 – 1450K)

Cp -ZrO2= 74.48 (1450 – 2800 K) a) 521 781 J/kg

b) 819 980 J/kg

Homework

• Plot Temp vs Cp for Al2O3 for the giventemperature:

T = 298, 350, 550, 600, 800, 950, 1100, 1500, 1550, 1650, 1800 K

Use Excel !!!

Cp Al2O3= 117.43 + 10.38∙ 10-3 T - 37.11 ∙ 105 T-2 (298 – 1800K)

What are the advantages of high specific heat capacity of water?

http://www.tutorvista.com/physics/animations/advantages-of-high-specific-heat-of-water-animation

What are the advantages of high specific heat capacity of water?

http://www.tutorvista.com/physics/animations/advantages-of-high-specific-heat-of-water-

animation

Hess’s Law

• Hess' law states that

the energy change for any chemical or physical process is independent of the pathway or number of steps required to complete the process provided that the final and initial reaction conditions are the same. In other words, an energy change is path independent, only the initial and final states being of importance.

(dH)p = (H)p

Basic Definitions

• Heat of reaction (H) (Reaksiyon Isısı)

As a result of a reaction, the enthalpy difference between initial and final state. Released or absorbed energy of a chemical reaction.

• H>0 Enthermic reaction, energy required

• H<0 Exothermic reaction, energy release

Basic Definitions

• Enthalpy of Formation (Oluşum, Teşekkül) , it is a special form of heat of reaction.

• H does not have an absolute value (only change in H can be measured). Therefore it is convenient to introduce a convention which allows the comparison of enthalpy difference. The convention assigns the value of zero to the enthalpy of elements in their stable states at 298K. Thus the enthalpy of compound at 298K is simply the heat of formation of compound from its elements.

Basic Definitions

𝐻1𝐻2 𝑑𝐻 = 𝑇1

𝑇2 𝐶𝑝 𝑑𝑇

𝐻 = 𝐻2− 𝐻1 = 𝑇1𝑇2 𝐶𝑝 𝑑𝑇

Kirchhoff’s Equation

𝐻2 = 𝐻1 + 𝑇1𝑇2 𝐶𝑝 𝑑𝑇

Enthalpy Difference

Enthalpy @ T2

Enthalpy athigh temperature

Basic DefinitionsA + B = C T = 298K

𝐻298 = 𝐻𝐶 298 − (α𝐻𝐴 298+ 𝐻𝐵 298)

Hess’s law

Important: This a chemical reaction, due to the reaction there is an Enthalpy Change !

A + B = C T = T

𝐻𝑇 = 𝐻𝐶 𝑇 − (𝐻𝐴 𝑇+ 𝐻𝐵 𝑇)Important: This a chemical reaction, due to the reaction there is an Enthalpy Change !

𝐻𝐴 𝑇2 = 𝐻𝐴 𝑇1 + 𝑇1𝑇2 𝐶𝑝 𝑑𝑇

Fuels & Combustion

• Combustion (Yanma) is used to describe exothermic combustion reaction of fuels. Oxidation reaction such as oxidation of aluminum is an exothermic reaction but it is not considered as combustion.

• Conventional fuels can be in solid, liquid or gas state. The most commonly used fuels are hydrocarbons.

• Reaction products are water vapour and CO2.

Calorific Value (Kalorifik Güç)

• The heating value or energy value of a substance, usually a fuel (or food), is the amount of heat released during the combustion of a specified amount of it. For solid and liquids it is 1 kg. For gases 1 m3.

• Reaction is always exothermic but calorific value is taken positive.

Adiabatic Flame Temperature(Alev Sıcaklığı)

Fuel298 + Oxidation Agent298 = Reaction ProductsTF

Fuel298 + Oxidation Agent298 = Q + Reaction Products298

Q + Reaction Products298 = Reaction ProductsTF

Adiabatic Flame Temperature

298

𝑇𝑓

∆𝐶′𝑃(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)Q298= dT

Combustion at room temperature

Adiabatic Flame Temperature

𝑇

𝑇𝑓

∆𝐶′𝑃(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)QT = dT

Combustion at high temperature

@ High Temperature

Flue gas (baca gazı)

298

𝑇𝑓𝑙𝑢𝑒

∆𝐶′𝑃(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)Qflue = dT

Always 298K !Reference point !