VALENCIA RAMIREZ LUIS MARCELINO.
Adiabatic process
The term adiabatic is used describe any process Turing which heat is prevented from
crossing the Boundary of the system. That is, an adiabatic process is one undergone by a
system which is thermally insulated from its surroundings. For adiabatic non-flow
process, the energy equation reduces to
W = (u2 – u1)
In an adiabatic expansion, work is done by the system at the expense of its internal
energy; in an adiabatic compression, the internal energy is increased by an amount equal
to the work done on the system. If the inicial and final states are known, the values u1
and u2 are determined and the work done can be calculated. We shall proceed to show
that in the case of a reversible adiabatic process there is only one process which can join
the end states, i.e. that there is a definite relation between p and v. In this case, therefore,
the work done can be predicted when the initial state and only one property of the
system in its final state are known.
We shall suppose that p1 and v1 are known, and that it is required to find the work done
by the system during a reversible adiabatic expansion to p2. For this process the energy
equation may be written in differential form as
p dv = -du
If a know algebraic equation u = Φ(p,v) relates p, v and u may be rewritten as
p dv = -dΦ(p,v)
and integrated. The result will be a definite relation between p and v, e.g. v = f(p).v2 can
be found by putting p = p2 in f(p). Consequently u2 and u1 can be found from Φ(p,v),
and the work can be calculated using equation W = ( u2 – u1 ).
If no simple algebraic equation relates p, v and u, the following graphical method could,
in principie, be used:
 First, a series of constant pressure lines and drawn on a diagram using u and v as
coordinates, from the information contained in the tableo f properties. For each
pressure in the table a series of values of u and v are given, and these pairs of
values can be represented by points on the diagram. All points relating to one
pressure pa, say, are joined up to form a constant pressure line.
pa
pb
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py
pz
va vb
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vy vz
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Values of
temperature.
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 Secondly, the path of the reversible process is traced out on the diagram starting
from the inicial state 1. Recasting equation p dv = - du in the form du/dv = -p, it
is seen that at any point in the process the slope of the curve is negative and
equal in magnitude to the pressure at the point. Through each constant pressure
line a series of short parallel lines can be drawn with a slope equal –p. The path
of the process will be represented by a curve drawn from point 1 to that its slope
is always the same as that of the short parallel line where it intersects each
constant pressure line.
 Lastly, since the final pressure p2 is known, the final state can be located on th
curve, and the value of u2 can be read off the diagram.
The foregoing graphical procedure is very lenghty. Looking ahead, it is easy to show
that the calculation can be considerably simplified if we can write dQ = T ds for a
reversible process, where T is the absolute thermodynamic temperature and s is the
property entropy. Then, since dQ = 0 for an adiabatic process, ds = 0 when it is
reversible and the entropy must remain constant. The foregoing problem may be solved
by finding s1 corresponding to p1 and v1 from the tables of properties, and then, since s2
= s1, finding v2 corresponding to p1 and p2 and s2.
For a fluid undergoing an adiabatic process the energy equation per unit mass reduces to
W = (u2 – u1)
When the fluid is a perfect gas,
W = cv (T2 – T1)
For a perfect gas undergoing a reversible adiabatic process, we can deduce the
following relations between properties:
0 = cv dT + p dv
R dT = p dv + v dp
Eliminating dT from these two equations, we obtain
0 = (1 + cv/R)p dv + (cv/R)v dp
=cpp dv + cvv dp
Writing cp/cv = γ, this reduces to
γ(dv / v) + (dp / p) = 0
and on integration,
γ ln v + ln p = constant
pvγ = constant
Since we also have pv / T = constant, we can eliminate v and p in turn to give
Tp(1-γ)/γ = constant or (T2/T1) = (p2/p1)(γ-1)/γ
Tvγ-1 = constant or (T2/T1) = (v2/v1)1-γ
It is evident from (9.4)that, for aperfect gas, the reversible adiabatic process is a special
case of the reversible polytropic process, and (9.5) and (9.6) can be obtained by putting
n = γ in
(T2 / T1) = (p2 / p1)(n-1)/n
(T2 / T1) = (v2 / v1)1-n
γ is called the index of isentropic expansion or compression. As in the case of the
specific heat capacities, for real gases γ varies with temperature and pressure (mainly
with the former ),but a mean value can be used for most purposes. If we put n = γ in the
expression for the work transfer in a polytropic process, we have
W = (R/γ-1)(T2 – T1)
This can easily be shown to be identical with the expression already derived, as follows:
(R/γ-1)(T2 – T1) = (cp – cv/(cp/cv-1))(T2 – T1)=cv(T2 – T1)
In a reversible adiabatic process s2 = s1, and equation (9.4) could have been obtained by
putting this result in the general expression
s2 – s1 = cv ln (p2/p1) + cp ln (v2/v1)
If this is done we have
cp ln (v2/v1) + cv ln (p2/p1) = 0
and hence
(p2/p1) = (v1/v2)γ
Since the process is reversible, the foregoing relation applies to any pair of states
between the end states, and consequently throughout the process
pvγ = constant