Department of Mechanical Engineering

ME 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 p

p

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 p

p

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

0

fh 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

0

CH ,CH

CH

f

Q n h n h n h

nnQh h h

n n n

Qh h

n

Example – Methane

C

2H 2

CH 4

25 ° C

1 atm

25 ° C

1 atm

Q

5

Enthalpy of Formation Values

Using 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 0

2. The SRS is 25°C, 0.1 MPa

3. 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?

Enthalpy Values at Other States

11

0

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

0

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

Three possibilities ...

Enthalpy Values at Other States

12

0

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

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!

Enthalpy Values at Other States

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

Air

Products (P)

Q

Consider an aergonic combustion process as shown below

i i i i

P 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 h

n 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 h

n n n

The molar flow rate ratios are the molar coefficients from the

balanced combustion reaction! Therefore,

i i i i

P Rfuel

Qq h h

n

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 drawn

around the combustion chamber. Applying the First Law

results in,

Heating Values

17

Q

8 18C H

PTA 100%

ProductsCombustionChamberR SRST T

P SRST T

i i i i

P 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 nQh h

n 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 the

previous 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%

ProductsCombustionChamberR SRST T

P SRST T

i ii i

P Rfuel fuel fuel

n nQh h

n n n

i i i i

P Rfuel

Qq h h

n

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 of

normal octane in 100% theoretical air, we previously found,

Heating Values

19

Q

8 18C H

PTA 100%

ProductsCombustionChamberR SRST T

P SRST T

i i i i

P Rfuel

Qq h h

n

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 N3.67

fuel

Qq h h h h h h h

n

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

fuel

Qq h h h

n

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%

ProductsCombustionChamberR SRST T

P SRST T

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

fuel

Qq h h h

n

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%

ProductsCombustionChamberR SRST T

P SRST T

All vapor water ...

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

fuel

Qq h h h

n

MJ MJ MJ MJ

8 393.522 9 241.827 208.447 5,116.172kmol kmol kmol kmolfuel

Qq

n

All liquid water ...

MJ MJ MJ MJ

8 393.522 9 285.838 208.447 5,512.271kmol kmol kmol kmolfuel

Qq

n

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%

ProductsCombustionChamberR SRST T

P SRST T

All vapor water ...

MJ5,116.172

kmolfuel

Qq

n

All liquid water ...

MJ5,512.271

kmolfuel

Qq

n

Lower Heating Value (LHV)

Higher Heating Value (HHV)

Observations ...

1. The reactants and products are at the SRS

2. 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 fuel

HHV or LHVcomb

Example

25

Q

8 18C H

PTA 100%

ProductsCombustionChamberR SRST T

P SRST T

Back 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

90.1406

8 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%

ProductsCombustionChamberR SRST T

P SRST T

How much water will condense?

At TSRS = 25°C, the mole fraction of water vapor in the

products is,

8 47

vv

v

ny

n

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 bar0.03101

1.01325 bar

ww

Py

P