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Heat EffectThermodynamics Course Chapter 4
Agung Ari Wibowo S.T., M.Sc
State Polytechnic of Malang
Malang
2017
Learning ObjectiveChapter 4
Sensible Heat
Heat Capacity &
Integral Evaluation of Sensible
Heat
Laten Heat of Pure
Subtance
Standard Heat of
Reaction and
Formation
Standard Heat of
Combustion
Standard Enthalpy as function of
Temperature
Heat Effect in Industrial Process
Literature UsedJ M Smith, H C Van Ness, M M Abbott, βIntroduction to Chemical Engineering Thermodynamics 7thβ
What happen when you add heat to the water ?
Sensible HeatChapter 4
Temperature
1 β’ Temperature rise
2 β’Water Start to boil
3β’ Continous Evaporation
(phase change to vapor)
Boil Water
Sensible HeatChapter 4
What is Sensible Heat ??
Heat which causing the rise in temperature to the certain degree with no change in phase and composition
ππ» =ππ»
πππ
ππ +ππ»
πππ
ππ
At constan Pressure Process
π = βπ» = ΰΆ±π1
π2
πΆπ ππ
Heat CapacityChapter 4
βπ» = ΰΆ±π1
π2
πΆπ ππ
πΆππ = π΄ + π΅π + πΆπ2 + π·πβ2
Each substance have their own A, B, C and D
constant. These constant are listed in Appendiks C
Heat capacity of ideal gas can be written as πΆπ£ππ
and πΆπππ
πΆπππ
π = π΄ + π΅π + πΆπ2 + π·πβ2
πΆπ£ππ
π =πΆπππ
π β 1
Heat CapacityChapter 4
Mixture of gasesWhat is the heat capacity of mixture?
A B C
Mixed ABC
πͺπ ππππππ
= ππ¨πͺππ¨ππ
+ ππ©πͺππ©ππ
+ ππͺπͺππͺππ
Where βyβ is mol fraction of the gas in mixture
Do you get it??
Integral Evaluation of Sensible HeatChapter 4
βπ» = ΰΆ±π1
π2
πΆπ ππ
What we know??
How we get the exact number??
ΰΆ±π0
π πΆπ
π ππ = ΰΆ±
π0
π
π΄ + π΅π + πΆπ2 + π·πβ2 ππ
It is still in Integral Equation
Can we get simpler equation??
ΰΆ±π0
π πΆπ
π ππ = ΰΆ±
π0
π
π΄ + π΅π + πΆπ2 + π·πβ2 ππ
= π΄ +π΅
2π0 π + 1 +
πΆ
3π02 π2 + π + 1 +
π·
ππ02 π β π0
πΆπ π»
π = π΄ +
π΅
2π0 π + 1 +
πΆ
3π02 π2 + π + 1 +
π·
ππ02
α»βπ― = πͺπ· π― π (π» β π»0
OK,ok.. Here much simpler form for youβ¦
π β 1 =πβπ0
π0
Lest back to our first case
Latent Heat of Pure SubtanceChapter 4
Can you see the bubles??
Lest back to our first case
Latent Heat of Pure SubtanceChapter 4
Amount of energi needed to change the phase
Latent Heat of Pure SubtanceChapter 4
Here are the equationsHow can we
know the
number??1. Clayperon Eq: βπ = πβπ
ππ¬ππ
ππ
βπ» = πΏππ‘πππ‘ π»πππ‘;βπ = Volume change occupied phase change;ππ ππ‘ = saturated vapor pressure (evaluated by Antoine Eq)
2. Troutons Eq :βπ
πππ§~ 10
Tn is absolut boiling point.
3. Riedel Eq :
βπ―
πΉπ»π=
π,πππ (ππ π·πβπ,πππα»
π,πππβπ»ππ
Pc is critical pressure and Trn is reduced temperature.(Tr = T/Tc)
4. Watson Eq:
βH2
βH1=
1 β Tr21 β Tr1
0.38
5. Using Steam tables (only for water)
βπ»ππ£ππ = π»π£ β π»π
Heat of ReactionChapter 4
1
2π2 +
3
2π»2 βΆππ»3 βπ» = β46,110 π½
Consider this reaction
βπ» = Ξπ»πππππ’ππ‘ β Ξπ»πππππ‘πππ‘
βπ»πππππ‘πππ‘ ππ βπ»πππππ’ππ‘ = ΰΆ±π1
π2
πΆπ ππ
T reference, usually 25 C
Actual T of materials
Standard Heat of ReactionChapter 4
1
2π2 +
3
2π»2 βΆππ»3 βπ»298
0 = β46,110 π½
Consider this reactionExothermic reaction
What is the meaning of this
symbol?
Standard heat of reaction is written as βH2980 , 298
means this heat were calculated at 298,15 K
And What is standard
condition?
Standard Condition :
- Gas : Pure subtance in ideal gas condition at Pressure 1 bar
- Liquid dan solid : Pure subtance in real condition at
Pressure 1 bar.
Standard Heat of FormationChapter 4
Consider this reaction
Definition :
Formation reaction was defined as reaction to produce
1 compound from its molecular element, in which the
product is 1 mole in amount as basis
πΆ +1
2π2 + 2π»2 β πΆπ»3ππ»
π»2π + ππ3 β π»2ππ4
Which one is formation reaction?
πΆ π + π2 π β πΆπ2 π βπ»π2980 = β339,509 π½
Generally formation reaction is written in form of :
β’ π stand for βformationβ
β’ The chosen condition is T 298.15 K or
25β
Standard Heat of FormationChapter 4
πΆπ2 π + π»2 π β πΆπ π + π»2π π
Calculate the heat of reaction from reaction below
πΆπ2 π β πΆ π + π2 π βπ»π2980 = 339,509 π½
πΆ π +1
2π2 π β πΆπ π βπ»π298
0 = β110,525 π½
π»2 π +1
2π2 π β π»2π π βπ»π298
0 = β285,830 π½
π»2π π β π»2π π βπ»π2980 = 44,012π½
Break down the reaction to get heat of formation of each compound
Where is H2??
Answer :
H2 is only constructed from 1 kind of
element, so its βHf2980 = 0 πΆπ2 π + π»2 π β πΆπ π + π»2π π βπ»298
0 = 41,66 π½
Wait, waittt
Standard Heat of CombustionChapter 4
Definiton : Heat of Combustion (enthalpy of combustion) of a substance is the heat liberated
when 1 mole of the substance undergoes complete combustion with oxygen at constant pressure.
Have you ever heard methanol gel?
Wedding party food heater
Standard Heat of CombustionChapter 4
The use of heat of combustion
4πΆ π + 5π»2 π β πΆ4π»10 π
Wait sir, C and H just only consist of 1 element⦠How can I know its
βπ»π2980 ?
β’ 4πΆ π + 4π2 π β 4πΆπ2 π βπ»π2980 = 4 π₯(β339,509α»π½
β’ 5π»2 π + 2,5 π2 π β 5π»2π π βπ»π2980 = 5 π₯(β285,830α» π½
β’ 4πΆπ2 π + 5π»2π π β πΆ4π»10 π + 6,5 π2 π βπ»2980 = 2877,396 π½
4πΆ π + 5π»2 π β πΆ4π»10 π βπ»π2980 = β125,790 π½
Temperature Dependent of βπ»0
Chapter 4
What about standard reaction at different temperature or not at
298,15 K?
Heat of reaction at certain temperature can be calculated if the heats of reaction at
standard condition were determined first
Reference : T = 298,15 K
What about at T = 800 β
πΆπ π + 2π»2 β πΆπ»3ππ» π
Chemical Reaction :π£1π΄1 + π£2π΄2 +β― β π£3π΄3 + π£4π΄4 +β―
βπ»0 =
π
π£ππ»ππ0
βπ»0Β° = βπ»298
Β° = β200,660 β β110,525= β90,135 π½
βHΒ° = βH0Β° + RT0
T βCPΒ°
RdT
Calculating Cp
βπͺπ·Β°
π―
πΉ= βπ¨ +
βπ©
ππ»π π + π +
βπͺ
ππ»ππ ππ + π + π +
βπ«
ππ»ππ
ΰ΅―βπ―Β° = βπ―0Β° + βπͺπ·
Β°π―(π» β π»0
i vi A 103B 106C 10-5D
CH3OH 1 2,211 12,216 -3,450 0
CO -1 3,376 0,557 0 -0,031
H2 -2 3,249 0,249 0 0,083
βπ΄=(1)(2,211)+(-1)(3,376)+(-2)(3,249) = -7,663
βπ΅=10,815 x 10-3 ; βπΆ=-3,450x10-6 ; βπ·=-0,135x105
βHΒ° = βH0Β° + RT0
T βCPΒ°
RdT
= -90,135 + 8,314 (-1615,5)
= -103,556 JT = 800 β
Heat Effect of Industrial ReactionsChapter 4
Reactor Temperature 377 K
Process Flow Diagram of Cumene Production
Hmmm,,, I smell problem hereβ¦
Heat Effect of Industrial ReactionsChapter 4
4π»πΆπ π + π2 π β 2π»2π π + 2πΆπ2 π
Material entering reactor ( R ) : HCl, O2
Material Out ( P ) : HCl, O2 Cl2, H2O
1 2
βπ» = Οππ ππ πΆππ
Β° (T β 298,15)
πΆππΒ° = π π₯
βπͺπ·Β°
π―
πΉ
βπ» = βπ»π + βπ»298Β° + βπ»π
Total enthalpy change :
βπ»π = perubahan entalpi reaktan pada suhu T1 ke suhu referens
298,15 K
βπ»π
βπ»298Β°
βπ»π
βπ»
T reference = 298,15 K
βπ»298Β° = entalpi reaksi standar pada suhu 298,15 K
βπ»π= perubahan entalpi produk dari suhu referens 298,15 K ke
suhu T