A graphical version of this page is also available. In this text only version, * denotes multiplication, / denotes division, and del denotes the Greek symbol delta signifying a small change. d preceding a variable denotes a differential change

The conservation of momentum is a fundamental concept of physics
along with the
conservation of energy
and the
conservation of mass.
Momentum is simply the mass of an object multiplied by the velocity
of the object.
Within some problem domain, the amount of momentum remains constant -
momentum is neither created nor destroyed, but only changed
through the action of **forces** as described by Newton's
laws of motion.
Dealing with momentum is more difficult than dealing with mass and energy because
momentum is a
vector quantity
having both a magnitude and a direction. Momentum is conserved in all three
physical directions. It is even more difficult when dealing with a
gas
because forces in one direction can affect the momentum in another direction
through the collisions of many molecules.
On this page, we will present a very simplified flow problem
where properties can only change in one direction.
The problem is further simplified by considering a steady flow which does
not change with time and by limiting the forces to only those
associated with the
pressure.
**Be aware that real flow problems are much more complex than this simple
example.**

Let us consider the flow of a gas through a domain in which flow properties
only change in one direction, which we will call "x". The gas enters the domain
at station 1 with some velocity (u) and some pressure (p) and exits at station
2 with a different value
of velocity and pressure. For simplicity, we will assume that the density (r)
remains constant within the domain and that the area (A) through which the
gas flows also remains constant. The location of stations 1 and 2 are separated
by a distance called "del x". (Delta is
the Greek letter "d". Mathematicians often use this symbol to denote
a change or variation of a quantity.) From station 1 to 2 the velocity
and pressure change. A change with distance is referred to as a **gradient**
to avoid confusion with a change with time which is called a **rate**.
The velocity gradient is called "del u / del x"; the change in velocity per
change in distance. So at station 2, the velocity is given by the velocity
at 1 plus the gradient times the distance.
A similar expression gives the pressure
at the exit. We will denote conditions at station 2 by the word "sub2" and
the conditions at station 1 by the word "sub1". At the exit of the
domain we have two equations:

Equation 1: u sub2 = u sub1 + (del u / del x) * del x

Equation 2: p sub2 = p sub1 + (del p / del x) * del x

Newton's second law of motion states that force (F) is equal to mass (m) times acceleration (a).

Equation 3: F = m * a

An acceleration is a change in velocity with a change in time (delta u / delta t).

Equation 4: F = m (del u / del t)

The force in this problem comes from the pressure gradient. Since pressure is a force per unit area, the net force on our fluid domain is the pressure times the area at the exit minus the pressure times the area at the entrance.

Equation 5: -[(p * A) sub2 - (p * A) sub1] = m * (u sub2 - u sub1) / del t

The minus sign at the beginning of this expression is used because gases move from a region of high pressure to a region of low pressure; if the pressure increases with x, the velocity will decrease. Now substitute Equation 1 and Equation 2 into Equation 5:

Equation 6: -{[p + (del p / del x) * del x] * A - p * A} = m [u + (del u / del x) * del x - u] / del t

Simplify Equation 6:

Equation 7: - (del p / del x) * del x * A = m * (del u / del x) * (del x / del t)

Noting that :

Equation 8: (del x / del t) = u

Equation 9: m = r * del x * A

Substitute Equation 8 and Equation 9 into Equation 7:

Equation 10: - (del p / del x) = r * u * (del u / del x)

The "del p / del x" and "del u / del x" are kept in parentheses because they represent the pressure and velocity gradients. If we shrink our domain down to differential sizes we obtain the differential equation.

Equation 11: - dp / dx = r * u * du / dx

This is a one dimensional, steady form of **Euler's Equation**. It is interesting
to note that the pressure drop of a fluid (the term on the left) is proportional
to both the value of the velocity and the gradient of the velocity.

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*byTom
Benson
Please send suggestions/corrections to: benson@grc.nasa.gov *