Thermodynamics is a branch of physics
which deals with the energy and work of a system.
Thermodynamics deals only with the
large scale response
of a system which we can
observe and measure in experiments. In rocket science, we are most
interested in thermodynamics for the role it plays in
high speed flows.
On this slide we derive two important equations which relate the
which a gas occupies during reversible compression or expansion.
Such a process occurs in the launch of a
compressed air rocket or of a
A pump, shown in green on the figure, is used to
increase the pressure of the rocket.
As the red piston moves in the cylinder, the volume of the gas
inside the cylinder is changed. This change in volume results in a change in
pressure and temperature of the gas which determines how much
the piston can deliver.
During the motion of the piston, we are going to assume that no
is transferred into the cylinder. We are further going to neglect any
friction between the piston and cylinder and assume that there are no
energy losses of any kind. (In reality there are small losses and we
account for the losses by an "efficiency factor" applied to the result
we obtain assuming no losses.)
The resulting compression and expansion are reversible processes
in which the
of the system remains constant. We can use
equations for the entropy
to relate the flow variables of the system.
We begin our derivation by determining the value of a factor which we will
need later. From the definitions of the
specific heat coefficients,
the specific heat at constant pressure cp minus the
specific heat at constant volume cv is equal to the
gas constant R:
cp - cv = R
and we define the ratio of specific heats to be a number which we
will call "gamma"
gamma = cp / cv
If we divide the first equation by cp, and use the definition of "gamma"
R / cp = 1 - (1 / gamma) = (gamma - 1) / gamma
Now we use the equation we have derived for the
entropy of a gas:
s2 - s1 = cp ln(T2 / T1) - R ln(p2 / p1)
where the numbers 1 and 2 denote the states at the beginning and end of the
compression process, s is the entropy, T is the temperature,
p is the pressure,
and "ln" denotes the natural logarithm
Since there is no heat transferred into the cylinder and no other losses,
the change in entropy is zero. Then the equation becomes:
cp ln(T2 / T1) = R ln(p2 / p1)
We divide both sides by "cp" and take the exponential function of
both sides (this "un-does" the logarithms).
T2 / T1 = (p2 / p1) ^ (R / cp)
where the symbol "^" denotes an exponent. Now we substitute the expression
for "R / cp" to obtain:
T2 / T1 = (p2 / p1) ^ [(gamma - 1)/gamma]
During the compression process, as the pressure is increased from p1 to
p2, the temperature increases from T1 to T2 according to this exponential
equation. "Gamma" is just a number that depends on the gas. For air, at
standard conditions, it is 1.4. The value of (1 - 1/gamma) is about .286.
So if the pressure doubled, the temperature ratio is 1.219.
The key point here is that we have a function that relates the temperature
change to the pressure change during a compression process.
We can use the equation of state to derive the relation between the
volume change and the pressure change. The equation of state is:
p * v = R * T
where v is the specific volume occupied by the gas. If we substitute
this expression for T into the temperature equation, we obtain:
(p2 * v2) / (p1 * v1) = (p2 / p1) ^ [(gamma - 1)/gamma]
Multiply both sides by (p1 / p2) to get:
v2 / v1 = (p2 / p1) ^ ( - 1/gamma)
p2 / p1 = (v1 / v2) ^ (gamma)
The quantity (v1 / v2) is the
of the volume at state 1 and state 2 and
is called the compression ratio.
For v2 less than v1, the pressure p2 is greater than
With this equation we can determine
the change in pressure for a given compression ratio. And using the previous
equation we know the change in temperature as well.
The value of the compression ratio
is a function of the design of the pump used to launch our rockets.
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