Lecture 37

Department of Mechanical EngineeringME 322 – Mechanical Engineering

Thermodynamics

Lecture 37

Heat of Reaction

1st Law Analysis of Combustion Systems

Combustion System Analysis

2

Consider the complete combustion of octane in 150% theoretical air,

Combustion Chamber

C8H18

PTA = 150%Products (p)

Q

In the previous lecture, we found the balanced reaction,

8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N

The First Law applied to the system identified above is,

a a f f p pp

Q n h n h n h

Combustion System Analysis

3

Combustion Chamber

C8H18

PTA = 150%Products (p)

Q

reactants(air and fuel)

combustion products

Potential issue: There are no Dh values (except for O2 and N2). This has the potential to cause a datum state problem.

8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N

a a f f p pp

Q n h n h n h

Resolving the Datum State Problem

4

In combustion calculations, the enthalpy of all stable* elements is defined as zero at the standard reference state (SRS),

25 C 298.15 K at 0.1 MPa 1 atm

77 F 536.67 R at 14.5 psia 1 atm

*‘Stable’ means chemically stable at the SRS. For example, diatomic oxygen (O2) is stable at the SRS. Monatomic oxygen (O) is not stable at the SRS.

Enthalpy of Formation

0fh

Heat released in an exothermic reaction (or absorbed in an endothermic reaction) when a compound is formed from its elements. (Elements and compound at the SRS)

The enthalpy of a compound at the standard reference state

4 4 2 2

2

4 2

4 4 4

4 4

4

CH CH C C H H

HCCH C H

CH CH CH

0CH ,CH

CHf

Q n h n h n h

nnQ h h hn n n

Q h hn

Example – Methane

C

2H2

CH425°C1 atm

25°C1 atm

Q

5

Enthalpy of Formation ValuesUsing EES* …

*Unit setting = molar

7

Enthalpy of Formation Values

Conclusion: EES uses the SRS as the datum state for enthalpy for the ideal gases! Therefore, enthalpy of formation values can be calculated from EES using the ideal gas substances (except AIR)

8

Results ...

Enthalpy Values in Combustion

9

What do we know so far?

1. The enthalpy of a stable element at the SRS is 02. The SRS is 25°C, 0.1 MPa3. The enthalpy of a compound at the SRS is the

enthalpy of formation (Table 15.1 or from EES)

Enthalpy Values at Other States

10

0,i f i ih T h h D

The enthalpy of a component at any temperature in a combustion process can be evaluated by,

0,i f i i i SRSh T h h T h T

Accounts for the enthalpy difference relative to the SRS

How is the enthalpy difference in brackets determined??

1. If the heat capacity of the component can be assumed constant,

2. If the constant heat capacity assumption is not accurate enough, then use the ideal gas tables (Table C.16c). In this case the datum state for the table does not have to match the enthalpy of formation.

3. Use a set of property tables for all components that has all enthalpy values referenced to the SRS. Does such a thing exist?


11

0, ,i f i p i SRSh T h c T T


Three possibilities ...


12


Exploring Option 3 from the previous slide ...

If a thermodynamically consistent set of tables exists, then

0,f i i SRSh h T

Therefore the enthalpy of the component could simply be looked up in a table at the given temperature,

i ih T h T

If something like this were available ... combustion calculations would be EESy!


13

ALL of the ideal gas enthalpy reference states (except for the ideal gas ‘AIR’) in EES are referenced to the SRS! This is from the EES Help Menu for the ideal gas CO2 ...

All other ideal gases in EES (except AIR) say the same thing!

Significance: Combustion calculations just became EESy!

Heat of Reaction

14

Combustion Chamber

Fuel

AirProducts (P)

QConsider an aergonic combustion process as shown below

i i i iP R

Q n h n h

The First Law applied to this system results in,

Reactants (R)

i ii i

P Rfuel fuel fuel

n nQq h hn n n

Dividing by the molar flow rate of the fuel,

Heat of Reaction

15

i ii i

P Rfuel fuel fuel

n nQq h hn n n

The molar flow rate ratios are the molar coefficients from the balanced combustion reaction! Therefore,

i i i iP Rfuel

Qq h hn

This is known as the molar heat of reaction.

Heating Values of Fuels

16

Given: Gaseous octane (C8H18) is burned completely in 100% theoretical air. The reactants and the products are at the SRS.

Find: The heat released during this combustion process per mole of fuel for the following cases,

(a) the water in the products is all vapor(b) the water in the products is all liquid

Q

8 18C H

PTA 100%

ProductsCombustionChamberR SRST T

P SRST T

The system boundary is drawnaround the combustion chamber. Applying the First Law results in,

Heating Values

17

Q

8 18C H

PTA 100%


P SRST T

i i i iP R

Q n h n h

Dividing both sides of this equation by the molar flow rate of the fuel,

i ii i

P Rfuel fuel fuel

n nQ h hn n n

Number of moles of reactant per mole of fuel

Number of moles of product species per mole of fuel

How are these found?

The molar flow rate ratios on theprevious slide are the molar coefficients from the balanced combustion reaction (for one mole of fuel)! Therefore,

Heating Values

18

Q

8 18C H

PTA 100%


P SRST T

i ii i

P Rfuel fuel fuel

n nQ h hn n n

i i i i

P Rfuel

Qq h hn

Observations ...1. The importance of being able to balance the

combustion reaction is evident!2. As long as the combustion process is aergonic, the

First Law will be as written above, independent of the conditions in and out of the combustion chamber!

Notice: PTA = 100% means stoichiometric combustion

For the complete combustion ofnormal octane in 100% theoretical air, we previously found,

Heating Values

19

Q

8 18C H

PTA 100%


P SRST T

i i i iP Rfuel

Qq h hn

Applying the First Law to the system,

2 2 2 2 8 18 2 21 CO 2 H O 4 O 5 N C H 0 O N3.67fuel

Qq h h h h h h hn

2 2 8 181 ,CO 2 ,H O ,C Hf f f

fuel

Qq h h hn

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

Heating Values

20

Q

8 18C H

PTA 100%


P SRST T

2 2 8 181 ,CO 2 ,H O ,C Hf f ffuel

Qq h h hn

Now we have an interesting problem. Is the water liquid or gas (or both)?

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

Let’s consider both extremes(1) the H2O is all vapor and (2) the H2O is all liquid.

Heating Values

21

Q

8 18C H

PTA 100%


P SRST T

All vapor water ...

2 2 8 181 ,CO 2 ,H O ,C Hf f ffuel

Qq h h hn

MJ MJ MJ MJ8 393.522 9 241.827 208.447 5,116.172kmol kmol kmol kmolfuel

Qqn

All liquid water ...

MJ MJ MJ MJ8 393.522 9 285.838 208.447 5,512.271kmol kmol kmol kmolfuel

Qqn

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

Heating Values

22

Q

8 18C H

PTA 100%


P SRST T

All vapor water ...MJ5,116.172

kmolfuel

Qqn

All liquid water ...MJ5,512.271

kmolfuel

Qqn

Lower Heating Value (LHV)

Higher Heating Value (HHV)

Observations ...1. The reactants and products are at the SRS2. The reaction occurs with PTA = 100% ( i = i )3. The difference between the HHV and the LHV is the

enthalpy of vaporization of water!

Heating Values

23

1 mol fuel

Stoichiometric air

Products(vapor H2O)

Products(liquid H2O)

,fgh2 2H O H O

,HHV LHV fgh 2 2H O H O

TSRS, PSRS

TSRS

PSRS

q

LHV

HHV

Heating Values

24

The heating values represent the maximum possible heat transfer that can occur per mole of fuel.

• The reactants and products are at the SRS• The HHV represents fully condensed water vapor• The LHV represents all water vapor

These values provide a basis for the combustion efficiency,

Actual heat transferred per mole of fuelHHV or LHVcomb

Example

25

Q

8 18C H

PTA 100%


P SRST TBack to our problem ... Is there liquid water in the products at the SRS? If so, how much?

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

9 0.14068 9 47

0.1406 0.1 MPa 0.01406 MPa 14.06 kPa

52.6 C

w

w w

dp

y

P y P

T

Will water condense?

Since TSRS < Tdp, water will condense

Example

26

Q

8 18C H

PTA 100%


P SRST T

How much water will condense? At TSRS = 25°C, the mole fraction of water vapor in the products is,

8 47v

vv

nyn

8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N

The mole fraction of the water vapor at 25°C can be found,

0.03142 bar 0.031011.01325 bar

ww

Py

P

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Lecture 37