# Momentum Equation

Let us now derive the momentum equation resulting from the Reynolds Transport theorem, Eqn. 3.27. Now we have = where is the momentum. Note that momentum is a vector quantity and that it has a component in every coordinate direction. Thus,

 (3.39)

Consider the left hand side of Eqn. 3.27. We have which is proportional to the applied force as per Newton's Second Law of motion. Thus,

 (3.40)

where is again a vector. It is necessary to include both body forces, and surface forces, . Thus,

 (3.41)

Now we substitute for in the right hand side of Eqn. 3.27 giving,

 (3.42)

Writing this as three equations, one for each coordinate direction we have,

 (3.43)

The term represents the u momentum that is convected in/out by the surface in a direction normal to it. In fact momentum in other direction can also be convected out from the same area. These are given by and .

As stated before the term is replaced by .

The equation thus derived finds immense application in fluid dynamic calculations such as force at the bending of a pipe, thrust developed at the foundation of a rocket nozzle, drag about an immersed body etc. We consider some of these later.

(c) Aerospace, Mechanical & Mechatronic Engg. 2005
University of Sydney