The conservation of mass is a fundamental
concept of physics. Within some problem domain, the amount of mass
remains constant; mass is neither created or destroyed. The
mass of any object is simply the volume that the object
occupies times the density of the object.
For a fluid (a liquid or a gas) the
density, volume, and shape of the object can all change within the
domain with time and mass can move through the domain.
The conservation of mass (continuity) tells us that the
mass flow rate mdot
through a tube is a constant
and equal to the product of the density r,
velocity V, and flow area A:
Eq #1:
mdot = r * V * A
Considering the mass flow rate equation, it appears that for a given
area and a fixed density, we could increase the mass flow rate indefinitely by
simply increasing the velocity.
In real fluids, however, the density does not remain
fixed as the velocity increases because of
compressibility effects.
We have to account for the change in density to determine the mass flow
rate at higher velocities.
If we start with the mass flow rate equation given above and use the
isentropic flow
relations and the
equation of state,
we can derive
a compressible form of the mass flow rate equation.
We begin with the definition of the
Mach number
M,
and the
speed of sound
a:
Eq #2:
V = M * a = M * sqrt (gam * R * T)
where gam is the
specific heat ratio,
R is the
gas constant,
and T is the
temperature.
Now substitute Eq #2 into Eq # 1:
Eq #3:
mdot = r * A * M * sqrt (gam * R * T)
The equation of state is:
Eq #4:
r = p / (R * T)
where p is the
pressure.
Substitute Eq #4 into Eq # 3:
Eq #5:
mdot = A * M * sqrt (gam * R * T) * p / (R * T)
Collect terms:
Eq #6:
mdot = A * sqrt (gam / R) * M * p / sqrt(T)
From the isentropic flow equations:
Eq #7:
p = pt * (T / Tt)^(gam/(gam1))
where pt is the total pressure and Tt is the total temperature.
Substitute Eq #7 into Eq # 6:
Eq #8:
mdot = (A * pt) / sqrt(Tt) * sqrt (gam / R) * M * (T / Tt)^((gam + 1) / (2 * (gam 1 )))
Another isentropic relation gives:
Eq #9:
T/Tt = (1 + .5 * (gam 1) * M^2) ^1
Substitute Eq #9 into Eq # 8:
Eq #10:
mdot = (A * pt/sqrt[Tt]) * sqrt(gam/R) * M * [1 + .5 * (gam1) * M^2 ]^[(gam + 1)/(gam  1)/2]
This equation is shown in the red box on this slide and relates the mass
flow rate to the flow area A,
total pressure pt and temperature Tt of the flow,
the Mach number M, the
ratio of specific heats
of the gas gam, and the
gas constant R:
This equation can be further simplified to derive a
weight flow function
that depends only on the Mach number.
The compressibility effects on mass flow rate have some
unexpected results.
We can increase the mass flow through a tube by
increasing the area, increasing the total
pressure, or decreasing the total temperature.
But the effect of increasing velocity (Mach number) is a little harder to figure out.
If we were to fix the area, total pressure and temperature, and graph
the variation of mass flow rate with Mach number, we would find that a
limiting maximum value occurs at Mach number equal to one.
There is a technique in calculus to
find the maximum (or minimum) value of a function by taking the
derivative of the function and setting the resulting equation to
zero. Let us apply this technique to the mass flow rate equation. To simplify the
exercise, let us define:
Eq #11:
B = (A * pt) / sqrt(Tt) * sqrt(gam / R)
Eq #12:
C = (gam + 1)/(2 * (gam  1))
Eq #13:
D = (gam  1) / 2
Then Eq #10 can be written:
Eq #14:
mdot = B * M / ((1 + D * M^2) ^ C)
We take the derivative of this equation with respect to M
and set the result to zero to find the maximum:
Eq #15:
d mdot/dM = M * ( d [ 1 / ((1 + D * M^2) ^ C)] /dM) + 1 / ((1 + D * M^2) ^ C) = 0
(2 * C * D * M^2) / ((1 + D * M^2) ^ (C + 1)) + 1 / ((1 + D * M^2) ^ C) = 0
Using some algebra to simplify this equation::
Eq #16:
2 * C * D * M^2 = 1 + D * M^2
Now collect terms which multiply M^2:
Eq #17:
M^2 = 1 / (D * ( 2 * C  1))
We can evaluate the right side of this equation by using Eq #12 and Eq #13:
Eq #18:
D * ( 2 * C  1) = .5 * (gam 1) * (((gam + 1) / (gam  1))  1) = 1
Therefore the condition for the maximum airflow occurs at:
Eq #19:
M^2 = 1 or M = 1
There is a maximum airflow limit that occurs when the Mach
number is equal to one.
The limiting of the mass flow rate is called choking of the
flow.
If we substitute M = 1 into Eq #10 we can determine the value of the
choked mass flow rate:
mdot = (A * pt/sqrt[Tt]) * sqrt(gam/R) * [(gam + 1)/2]^[(gam + 1)/(gam  1)/2]
Mach number equal to one is called a sonic condition
because the velocity is equal to the speed of sound and
we denote the area for the sonic condition by a special symbol "A*",
pronounced
"A star".
If we have a tube with changing area, like the
nozzle
shown on the slide, the maximum mass flow rate through the system
occurs when the flow is choked at the smallest area. This
location is called the throat of the nozzle.
The conservation of mass specifies that the mass flow rate through a nozzle is a constant.
If no heat is added, and there are no pressure losses in the nozzle,
the total pressure and temperature are also constant. By substituting the sonic
conditions into the mass flow equation in the box, and doing some algebra,
we can relate
the Mach number M at any location in the nozzle to the ratio between
the area A at that location and the area of the throat A*.
Since the Mach number is related to the velocity, we can determine the exit
velocity of a nozzle, if we know the area ratio from the throat to the exit.
Knowing the exit velocity and the mass flow rate, we can determine the
thrust of the nozzle.
You can explore the operation of
a nozzle with our interactive thrust
simulator and design your own rockets!
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