Chapter_2

Chemical Engineering Thermodynamics

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 Textbooks 1.

Koretsky, M.D. (2004) Engineering and Chemical Thermodynamics, 1st edition, John Wiley &Sons. (ISBN: 0-471-38586-7) (New Edition) Koretsky, M.D. (2013) Engineering and Chemical Thermodynamics, 2nd edition, John Wiley &Sons.

2. Smith J.M., Van Ness H.C., Abbott M.M.(2004)

Introduction to Chemical Engineering Thermodynamics, 7th Ed. Mc-Graw Hill (ISBN: 0072402962) 2

Marks Allocation  COURSE WORKS:

1 Test: 20 %, Assignment: 20 %  FINAL EXAMINATION: 60 %  PASSING MARK:

50 %

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CONSULTATION HOURS:  Wednesday: 9 am – 11 am; 2 pm – 4 pm  *please make appointment at least 2 days prior to the

consultation

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Chapter 2 The First Law of Thermodynamics Learning Objectives  Thermodynamic property, extensive and intensive properties  Introduce the laws of thermodynamics and the forms of energy  Themochemical data for internal energy & enthalpy  Describe the energy changes associated with sensible heat, latent heat, and chemical reaction on both a macroscopic and a molecular level.  Calculate their enthalpy changes using available data such as heat capacity, enthalpies of vaporization, fusion and sublimation, and enthalpies of formation.

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Classification of Thermodynamics properties  Extensive properties (K)

Depend on the mass/size of the system Example: V, G, U, H, S  Intensive properties (k)

Does not depend on the mass/size of the system (1) Intensive molar (𝑘 =

𝐾 ) 𝑛

Example: v, g, u, h, s (2) Intensive specific (𝑘 = Example: 𝑣, 𝑔, 𝑢, ℎ, 𝑠

𝐾 ) 𝑚 6

Energy (1) Macroscopic kinetic energy (Ek)  energy associated with the bulk (macroscopic) motion of the system as a whole (2) Macroscopic potential energy (Ep)  energy associated with the bulk (macroscopic) position of the system in a potential field (3) Internal energy (U)  energy associated with the motion, position, and chemical-bonding configuration of the individual molecules of the substances within the system. 7

(4) Enthalpy (H)  associated with both internal energy and flow of work  Formula: h = u + pv intensive Unit: kJ/kg or H = U + PV extensive Unit: kJ (5) Entropy (S)  Clausius: Heat absorbed during reversible process Formula: 𝑑𝑠 =

𝛿𝑞𝑟𝑒𝑣 𝑇

intensive

Unit: kJ/kg.K

 Boltzmann’s molecular view: Degree of disorder in a

system 8

+Q

-Q Q given off by the system

SYSTEM

Q received by the system

(6) Heat (Q)  A form of energy transferred between the surrounding and the system due to the driving force of temperature difference  A process in which there is no heat (Q = 0) is called an adiabatic process (7) Work (W)  A form of energy transferred across the boundary associated with a force acting through a distance * W and Q are process functions! (i.e., their magnitudes depend on the path followed during a process as well as the end states)

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Law of Thermodynamics (1) Zeroth Law of Thermodynamics: Two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other

(2) First Law of Thermodynamics: Conservation of energy principle i.e. energy cannot be created or destroyed

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 The law of conservation of energy +Q

+W SYSTEM ∆Ek ∆Ep ∆Et ∆U

SURROUNDINGS

BOUNDARY

 For a closed system undergoing a thermodynamic

process, the energy balance equation

∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 + 𝑊 11

(3) Second Law of Thermodynamics: It is impossible to build an engine which operating in a cycle can convert all the heat it absorbs into work. Kelvin and Planck Heat cannot be caused to flow from a cooler body to a hotter body without producing some other effect. Clausius (4) Third Law of Thermodynamics: The entropy of a perfect crystal is zero at a temperature of absolute zero 12

Heat Capacity: Cv and Cp Heat capacity at constant volume, cv

𝑐𝑣 =

𝛿𝑢 ( )𝑣 𝛿𝑇

sensible heat: “sense” the result of the heat input with the thermocouple

at constant volume

(2.50) (2.24) ∆u=q+w ∆u=q (closed system, const v)

Figure 2.10 Schematics of the experimental determination of heat capacity. (a) constant-volume calorimeter to obtain cv 13

Heat capacity at constant volume, cv  By taking the slope of the curve as a function of temp,

cv = cv (T)  Fit the data to a polynomial expression cv = a + BT + CT2 + DT-2 + ET3

(2.51)(-pg. 68) (2 .52) (2.25)

 Then, we can find ∆u by integration

∆u =

𝑇2 𝑐 𝑑𝑇 𝑇1 𝑣

=

𝑇2 𝑇1

a + BT + CT2 + DT−2 + ET3 𝑑𝑇 (2.53)(2.26)

 Heat capacity should only be used for temperature

changes between the same phase. 14

Heat capacity at constant pressure, cp 𝑐𝑝 =

𝛿ℎ ( )𝑝 𝛿𝑇

at constant pressure

(2.58) (2.29) ∆u=q+w ∆u=q-P∆v or ∆u+∆(Pv)=q ∆(Pv)= P∆v+v∆P =P∆v ∆h = q (closed system, const v)

Figure 2.10 Schematics of the experimental determination of heat capacity. (a) constant-pressure calorimeter to obtain cp 15

Heat capacity at constant pressure, cp  Fit the data to a polynomial expression

cp = A + BT + CT2 + DT-2 + ET3

(2 .59) (2.30)

 The parameters A, B, C, D, and E of some ideal gasses are

reported in Appendix A.2.  Heat capacity parameters at constant pressure of some liquids and solids are reported too.

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Relationships between cpandcv  cp ≈ cv  𝑐𝑝 =

liquids and solids

(2.60) (- pg. 71)

𝛿ℎ 𝛿(𝑢+𝑃𝑣) 𝛿𝑢 𝛿𝑅𝑇 𝛿𝑢 ( )𝑝 = [ ]𝑝 = ( )𝑝 +( )𝑝 = ( )𝑝 +𝑅 𝛿𝑇 𝛿𝑇 𝛿𝑇 𝛿𝑇 𝛿𝑇 𝛿𝑢 𝑑𝑢 𝛿𝑢 ( )𝑝 = = ( )𝑣 𝛿𝑇 𝑑𝑇 𝛿𝑇

 cp= cv + R

for ideal gases

(2.63) (- pg. 71)

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Mean Heat Capacity, 𝑐𝑝  Definition: The average of cpbetween two temperatures.  It is usually reported between 298 K and a given

temperature, T.

∆h = 𝑐𝑝 (T– 298)

(2.64) (- pg. 71)

 Solving for𝑐𝑝 :

𝑐𝑝 =

𝑇 𝑐 𝑑𝑇 298 𝑝

𝑇−298

(2.65) (2.33)

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EXAMPLE 2.6 (2.7) Heat Input Calculations Using Different Data Sources Consider heating 2 moles of steam from 200 oC and 1 MPa to 500 oC and 1 MPa. Calculate the heat input required using the following sources for data:

(a) Heat capacity (b) Steam tables

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SOLUTION (a) At constant pressure, heat input required is q = ∆h or Q = n∆h Assume water is ideal gas, heat capacity values given in Appendix A.2 5 𝑐𝑝 0.121𝑥10 = A + BT + DT−2 = 3.470 + 1.450 𝑥 10−3 𝑇 + 𝑅 𝑇2 By using the definition of heat capacity, 0 0 ∆h=

𝑇2 𝑐 𝑑𝑇 𝑇1 𝑝

=𝑅

𝑇2 𝑇1

A + BT + CT2 + DT−2 + ET3 𝑑𝑇 773 𝐵 2 𝐷 ∆ℎ = 𝑅[𝐴𝑇 + 𝑇 − ]473 2 𝑇

∆h = 8.314 3.470 300 + 0.735𝑥10−3 (7732 −4732 ) − 0.121𝑥10−5

1 773

−

1 473

= 10,991 J/mol

Q = n∆h = 2(10,991) = 21,981 J

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(b) From steam tables, Appendix B.2: ℎ1 (at 1 MPa, 200oC) = 2827.9 kJ/kg ℎ2 (at 1 MPa, 500oC) = 3478.4 kJ/kg

∆ℎ = ℎ2 -ℎ1 = 650.5 kJ/kg Molecular weight for water, MWH2O = 0.018 kg/mol Q = m∆ℎ = (2 mol)(0.018 kg/mol)(650.5 kJ/kg)(103 J/kJ) = 23,418 J  The answer for part (b) is approximately 6% higher

than for part (a). At 1 MPa, water is not an ideal gas

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EXAMPLE 2.7 (2.8) Determination of Mean Heat Capacity for Air Use the data available in Appendix A.2 to calculate the mean heat capacity cp, for air between T1 = 298 K and T2 = 300 to 1000 K, in intervals of 100 K.

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SOLUTION TABLE E2.8 Calculated Values for Mean Heat Capacity of Air at Different Temperatures ∆ℎ =

𝑇2 𝑇1

𝑐𝑝 𝑑𝑇 = 𝑅

𝑇2

A + BT + DT−2 𝑑𝑇

298

𝐵 𝐷 = 𝑅[𝐴𝑇 + 𝑇 2 − ]298 2 𝑇

T

𝑇

𝑐 𝑑𝑇 ∆ℎ 298 𝑝 𝑐𝑝 = = (𝑇 − 298) 𝑇 − 298

Appendix A.2: Heat capacity for air: A = 3.355; B = 0.575x10-3; C = -0.016x105

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Latent Heats Definition: The change of enthalpy during a phase transition at constant pressure (Heat absorbed as substance A changes phase) (1) Enthalpy of vaporization, hvap = hv - hl  The enthalpy of the vapor minus the enthalpy of the liquid (2) Enthalpy of fusion, hfus = hs - hl  The change in enthalpy from the liquid phase to the solid phase (3) Enthalpy of sublimation, hsub = hv - hs  The change from the solid phase to the vapor phase 24

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Calculation of hvap, T at any T Step 1: Change in enthalpy of the liquid from T to Tb Step 2: Vaporize the liquid at the Tb Step 3: Change in enthalpy of the vapor from Tb to T

Figure 2.12 Hypothetical path to calculate hvap at temperature T from data available at normal boiling point, Tb and heat capacity data.

 where

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EXAMPLE 2.11 (2.10) Determination of Heat Required to Evaporate Hexane 10 mol/sec of liquid hexane flows into a steady-state boiler at 25 oC. It exits as vapor at 100 oC. What is the required heat input to the heater. Take the enthalpy of vaporization at 68.8 oC to be hvap, 68.8 oC = 28.88 [kJ/mol]

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SOLUTION: Hexane (25oC 100oC) hvap, 68.8 oC= 28.88 kJ/mol  Heat-transfer rate, 𝑄 = 𝑛(ℎ2 − ℎ1 )

Enthalpy change involves (1) sensible heat bring hexane (l) to boiling point; (2) enthalpy of vaporization (Latent heat) (3) sensible heat to bring it to vapor state of 100oC ℎ2 − ℎ1 = ℎ1 + ℎ2 + ℎ3 l,25oC  68.8oC

vap, 68.8oC

v,68.8oC  100oC

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h1 = h2 =

=

342 𝐽 23.695 8.314 𝑑𝑇 = 8668 298.2 𝑚𝑜𝑙 373.2 2 𝑅(𝐴 + 𝐵𝑇 + 𝐶𝑇 )𝑑𝑇 342

= 8.668 𝑘𝐽/𝑚𝑜𝑙

53.722𝑥10−3 8.314[3.025(373.2-342)+ (373.22 -3422 ) − 2 −6 16.791𝑥10 (373.23 −3423 )] 3

= 5.20 kJ/mol 𝑄 = 𝑛(ℎ2 − ℎ1 ) = (10 mol/s)(8.668 + 28.88 + 5.20)kJ/mol

= 427 kJ/s

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Enthalpy of Formation, hf

 Definition: Enthalpy difference between a given molecule and

the reference state (pure elements in their stable form at 298 K and 1 bar) hf

Elements  Species i

(2.69) (-pg. 82)

 The enthalpy of formation of a species containing only one

element, as it is found in nature, is identically zero, eg. H2, O2 H2(g) + ½ O2(g)  H2O(l)

(2.70) (-pg. 82)

The value for the enthalpy of formation for this reaction is found in Appendix A.3.2 (p.503) (pg. 646) Water ∆ hof,298 = -285.83 kJ/mol (phase : L) Water ∆ hof,298 = -241.82 kJ/mol (phase: G)

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Enthalpy of Reactions, hrxn Definition: Amounts of energy can be absorbed or liberated during chemical reactions

h1: Reactants  elements h2: Elements  products Note: Stoichiometric coefficients Reactants are negative Products are positive

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 Equating the 2 paths yields

∆horxn298 = ∆h1 + ∆h2 = ∑ νi (∆hof,298)i + ∑ νi (∆hof,298)i reactants

= ∑ νi (∆hof,298)i

products

(2.71) (-pg. 83)

 hrxn can be determined by scaling each species’ hf by

its stoichiometric coefficient ∆horxn = ∑ νih0i = ∑ νi (∆hof)i

(2.72) (2.35)

 Enthalpy of formation data can be found in Appendix

A3 32

EXAMPLE 2.12 (2.13) Determination of Enthalpy of Reaction Calculate the enthalpy of reaction at 298K for the following reaction: SOLUTION: Enthalpy of formation from App. A.3 & Eq (2.72) (2.35) 0

(hof,298)H2 = 0 The form hydrogen takes at 298 K and 1 bar.

ho298 = (-393.51) + 3(0) – (–241.82) – (– 200.66) = 49.0 kJ/mol  ∆h is positive = Reaction is endothermic

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Example 2.13 (2.15) Adiabatic Flame Temperature Calculations Propane enters an adiabatic constant-pressure combustion chamber at 25oC. It is mixed with a stoichiometic amount of air. Assume complete combustion and that the carbon distribution in the product stream contains 90% CO2 and 10% CO. What is the exit temperature?

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SOLUTION:

Figure E2.13B Hypothetical path for calculation of enthalpy in complete CO2, CO, N2, combustion of propane. HO 2

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 Using a basis of 1 mol C3H8, chemical reaction for CO2 production

The hrxn can be found based on the reaction stoichiometry given: Using the values from Appendix 3



Using a basis of 1 mol C3H8, chemical reaction for CO production

 The carbon distribution in the product stream contains 90% CO2 and 10% CO,

we multiple the 1st reaction by nIC3H8 = 0.9 and the 2nd reaction by nIC3H8 = 0.1. The total enthalpy reaction is: 36

 The balanced chemical reactions are

The carbon distribution are 90% CO2 and 10% CO, thus, the species concentration are

 An energy balance on the closed system at constant pressure

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Appendix A.2 Heat capacity data

T2 is found to be T2= 2345 K 38

Tutorial 1  Problems:  No. 2.9, -, 2.49, 2.53, 2.54, 2.55

(Koretsky, M.D. (2004) Engineering and Chemical Thermodynamics, 1st edition, John Wiley &Sons) No. 2.23, 2.55, 2.74, 2.78, 2.79, 2.80

(Koretsky, M.D. (2013) Engineering and Chemical Thermodynamics, 2nd edition, John Wiley &Sons)

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