MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Regarding the 1st example in Lecture 33, the hT , h298 values for
CH4 could be obtained from tables or estimated from the cp and
the temperature difference from the reference state. For the case
of CH4 , the tables are not in the course textbook and the latter
approach must be used. The discussion below shows you how to
approximate the ∆h for CH4 without tables. ( ∆h = hT − h298 )

For CH4

hT = 10965
From Tables (not in text)
h298 = 10025
∆h = 940 h = h fo + ∆h

Since these tables do not exist in the text use the following
approximation:

h = h fo + ∆h
where ∆h ≈ c p ( ave ) ∆T
h = h + c p ∆T
f
o

c p (average) given in question OR calculated as per

information given in Table A-21 for Tave .

∆T = 25 K in this case
For Tave = (298 K + 323 K ) / 2 = 310.5 K

Evaluate c p by Table A-21 data then ∆h = c p ∆T and h = h fo + c p ∆T

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

For CH4 (from Table A-21)

cp
= α + βT + γ T 2 + δ T 3 + εT 4
R

c p (310.5 K )
= 3.826 + (−3.979 x10−3 )(310.5) + (24.558 x10−6 )(310.5) 2 +
R
+ (−22.733x10−9 )(310.5)3 + (6.963x10−12 )(310.5) 4

c p = 36.1 kJ / kmol ⋅ K

∆h ≈ c p ∆T
∆h ≈ (36.1)(25 K )
∆h ≈ 903 kJ / kmol

Close enough to Tabular approach (940 kJ/kmol).

Note: you could have integrated the above expression for cp from
298 to 323 K. Try this as an exercise and compare the results.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Enthalpy of Combustion and Heating Values

Often h fo data is not available. The sheer variation in composition

for fuels such as oil and coal makes this impractical. However,
what is sometimes available is experimentally–determinable.

Enthalpy of Combustion ( hRP )

which is defined as:

hRP = ne he − ni hi
P R

n ’s = coefficients from reaction equation

This is ∆h between products and reactants when combustion

occurs at a given temperature and pressure. Can also be per unit
mass of fuel basis ( hRP ), or given at standard reference state
( hRPo , hRPo ).

hRP can be measured in a calorimeter. After combustion, Qcv

required to return products to initial temperature is:

Qcv
= ne he − ni hi = hRP Eq. 33.1
n fuel P R

And this is how tabulated values of hRP can be determined.

N.B. for combustion, Q is negative and thus hRP is usually

negative.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

We can use hRP for energy analyses.

Recall h = h fo + ∆h

We can thus expand Eq. 33.1 as:

Qcv Wcv
− = ne (h fo + ∆h )e − ni (h fo + ∆h )i
nf nf P R

Qcv Wcv
− = ne (h fo )e − ni (h fo )i + ne (∆h )e − ni (∆h )i
nf nf P R P R

o
hRP

Thus

Qcv Wcv
− = hRP
o
+ ne (∆h )e − ni (∆h )i
nf nf P R

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Heating Values (given in Table A-25)

Heating value of a fuel is a positive number equal in magnitude to

hRP (the enthalpy of combustion).

Higher Heating Value (HHV) – all of the water formed by

combustion is liquid.

Lower Heating Value (LHV) – all of the water formed by

combustion is vapour.

HHV exceeds LHV by the energy required to vapourized the liquid

formed.

Values also depend upon whether the fuel is a liquid or gas.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Example Text Q. 13.50

Problem: Find enthalpy of combustion ( hRP ) of butane (C4H10) in

kJ/kmol and kJ/kg of fuel, at 25oC and1 atm.

Assuming: (a) water vapour in products

(b)liquid water in products

Assume: Complete reaction in theoretical amount of air.

Solution:

Step 1: Balance chemical reaction:

C4 H10 + 6.5 (O2 + 3.76 N 2 ) → 4 CO2 + 5 H 2 0 + 24.44 N 2

Step 2: Write enthalpy of combustion:

hRP = ne he − ni hi
P R

0 0
hRP = ne ( h fo ) e − ni (h fo )i + ne ( ∆h )e − ni ( ∆h )i
P R P R
0 0

hRP = hRP
o
= 4( h fo )CO2 + 5(h fo ) H 2O − (h fo )C4 H10 − 6.5( h fo )O2 − 24.44(h fo ) N 2

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

(a) if H2O is vapour: (use A-25)

hRP = 4( −393520) + 5( −241820) − ( −126150) = −2.7 x106 kJ / kmol

(b) if H2O is liquid:

hRP = 4( −393520) + 5( −285830) − ( −126150) = −2.9 x106 kJ / kmol

Divide each by M = 58.12 (molar mass of C4H10) to calculate on

mass basis.

Adiabatic Flame Temperature

There are two ways the energy from a combustion reactor is

transferred:

1) with the exiting combustion products

2) heat transfer to surroundings (e.g., through reactor walls)

Energy carried out

with products

Qcv to surroundings

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

The lower the Q , the greater the energy carried out with the
products (and the higher the temperature of the products). In the
limit of Q = 0 (adiabatic operation of the reactor), this energy and
temperature of the products should be maximized.

This temperature obtained under adiabatic operation is the

Adiabatic Flame Temperature (or adiabatic combustion
temperature). In this circumstance, the energy balance reduced
to:

0 0 since adiabatic
Q − W = hP − hR

OR

hP = hR
ne he = ni hi
P R

However to calculate the adiabatic flame temperature:

ne (h fo + ∆h )e = ni (h fo + ∆h )i
P R

ne (∆h )e = ni (∆h )i + ni (h fo )i − ne (h fo )e
P R R P

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Steps to solve:

1) Obtain h fo values from Table A-25. Use enthalpy of

combustion values if h fo values are not tabulated.
2) ∆h for each reactant evaluated as discussed in Lecture 33.
3) Solve right hand-side gives you a # there.
4) Expand out left side
e.g., 8(∆h )CO2 + 9(∆h ) H 2O + 42.3(∆h ) N2 = #
5) Solve iteratively: Pick a reasonable T . See if you get a #
close to the right hand-side. Repeat as per previous iterative
approaches.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Example Text Q. 13.56

Problem: Liquid octane (C8H18) enters insulated reactor and

burns with 90% theoretical air at 25oC, 1 atm.
Products are CO2, CO, H2O, N2
Find temperature of products

C8H18 Tp = ?

90% theoretical air CO2, CO,

H2O, N2

Solution: Write balanced reaction for complete combustion

C8 H18 + 12.5 (O2 + 3.76 N 2 ) → 8 CO2 + 9 H 2 0 + 47 N 2

so for 90% air,

C8 H18 + 11.25 (O2 + 3.76 N 2 ) → 5.5 CO2 + 2.5 CO + 9 H 2 0 + 42.3 N 2

Energy rate balance is:

0
ne (∆h )e = ni (∆h )i + ni (h fo )i − ne (h fo )e
P R R P

LHS: 5.5(∆h )CO 2 + 2.5(∆h )CO + 9(∆h ) H 2O + 42.3 (∆h ) N2

(h fo )C8 H18 + 11.25(h fo )O2 + 42.3(h fo ) N2 − 5.5(h fo )CO 2 − 2.5(h fo )CO

RHS:
− 9(h fo ) H 2O − 42.3(h fo ) N 2

No ∆h terms in RHS because reactants at 298 K and 1 atm.

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MECH 330: APPLIED THERMODYNAMICS II LECTURE 34

Then using ideal gas tables (A-23)

5.5[h (TP ) − 9364]CO 2 + 2.5[h (TP ) − 8669]CO + 9[h (TP ) − 9904]H 2O

+ 42.3 [h (TP ) − 8669]N2

= − 249910 − 5.5(−393520) − 2.5(−110530) − 9(241820)

= 4367155 kJ / kmol

Now we need to guess some value of TP , look up h (TP ) values in

Table A-23 and see if we get close to 4367155 kJ/kmol.

Example

TP = 2000 K LHS of eqn = 3673733 kJ/kmol (too low)

TP = 2500 K LHS of eqn = 4892224 kJ/kmol (too high)

Correct answer approximately 2300 K.

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