CONSERVATION: MASS, MOMENTUM, AND ENERGY BALANCES - Biomass as a Sustainable Energy Source for the Future - page 85

Thus,

c p dT x
dx

dT x
T 0 −

r h
φ V ρ

2 π

φ V ρ = h T 0 − T x 2 π r ) =

c p

dx

T x

Integration from the inlet to the outlet for temperature and length of the tube gives
ð T out

= ð x=L
x=0

T out
T in =

dT x
T 0 −

r h
φ V ρ

2 π r h

φ V ρ

2 π

dx ) − ln T 0 − T x ð L − 0 Þ

c p c p

T x

T in

So

=

ð T 0 − T out Þ 2 r h L

φ V ρ

π

ln −

ð T 0 − T in Þ c p

Substituting the data in this equation gives L = 5.53 m.

3.3.2 Energy Balances for Systems with Chemical Reactions

Here, we focus on the conservation of energy in reaction systems. This context has
impact on the evaluation of the thermodynamic properties. When no chemical reaction
takes place, as for the systems discussed in Section 3.3.1, the composition of a stream
does not change and differences in thermodynamic properties such as internal energy
(u), enthalpy (h), and entropy (s) are of importance, but the absolute values can be
considered with respect to arbitrary reference states. In contrast, chemical reactions
in systems cause species to be formed and destructed; therefore, changes cannot be
calculated in general for all species in such systems. The thermodynamic properties
now need to be univocally measured with respect to a unique standard state.
For this purpose, the choice has been made by international convention to define
the enthalpy of the elements in their stable state to be zero (0) at a reference temper-
ature, T ref , of 298.15 K and reference pressure, p ref , of 1 atm (absolute). For example,
the stable state of some relevant, abundant species, hydrogen, nitrogen, and oxygen, is
the molecular state: H 2 ,N 2 , and O 2 .
The enthalpy for a species i on a mole-specific basis can be expressed as

= h 0

= h 0

h i T, ðÞ h i T,ð − h i T ref , p ref

Δ h i

f , i +

ð Þ

f , i +

ð Eq : 3 : 34 Þ

3.3.2.1 Open-Flow Systems For open-flow systems, the steady-state energy bal-
ance Equation (3.28) is rewritten in terms of moles and molar properties. With the
assumption that changes in KE and PE are neglected, the following equation results:

(

X

Q cv − W cv + h f +

Δ h φ n , fuel in n in per mole fuel

in

Reactants

) =0

X

h 0

Δ h

− n out per mole fuel

f +

ð Eq : 3 : 35 Þ

out

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