Application of the Reynolds Transport Theorem to Conservation of Mass and Momentum

Application of the Reynolds Transport Theorem to Conservation of Mass and Momentum 

Conservation of mass The constancy of mass is inherent in the definition of a control mass system and therefore we can write

(10.13a)

To develop the analytical statement for the conservation of mass of a control volume, the Eq. (10.11) is used with N = m (mass) and η = 1 along with the Eq. (10.13a).

This gives

(10.13b)

The Eq. (10.13b) is identical to Eq. (10.6) which is the integral form of the continuity equation derived in earlier section. At steady state, the first term on the left hand side of Eq. (10.13b) is zero. Hence, it becomes

(10.13c)

Conservation of Momentum or Momentum Theorem The principle of conservation of momentum as applied to a control volume is usually referred to as the momentum theorem.

Linear momentum The first step in deriving the analytical statement of linear momentum theorem is to write the Eq. (10.11) for the property N as the linear - momentum  and accordingly η as the velocity . Then it becomes

(10.14)

The velocity  defining the linear momentum in Eq. (10.14) is described in an inertial frame of reference. Therefore we can substitute the left hand side of Eq. (10.14) by the external forces  on the control mass system or on the coinciding control volume by the direct application of Newton’s law of motion. This gives

(10.15)

This Equation is the analytical statement of linear momentum theorem.

In the analysis of finite control volumes pertaining to practical problems, it is convenient to describe all fluid velocities in a frame of coordinates attached to the control volume. Therefore, an equivalent form of Eq.(10.14) can be obtained, under the situation, by substituting N as and accordingly η as  , we get

(10.16)

With the help of the Eq. (10.12) the left hand side of Eq. can be written as

where  is the rectilinear acceleration of the control volume (observed in a fixed coordinate system) which may or may not be a function of time. From Newton’s law of motion

Therefore,	(10.17)

The Eq. (10.16) can be written in consideration of Eq. (10.17) as

(10.18a)

At steady state, it becomes

(10.18b)

In case of an inertial control volume (which is either fixed or moving with a constant rectilinear velocity),  and hence Eqs (10.18a) and (10.18b) becomes respectively

	(10.18c)
and	(10.18d)

The Eqs (10.18c) and (10.18d) are the useful forms of the linear momentum theorem as applied to an inertial control volume at unsteady and steady state respectively, while the Eqs (10.18a) and (10.18b) are the same for a non-inertial control volume having an arbitrary rectilinear acceleration.

In general, the external forces  in Eqs (10.14, 10.18a to 10.18c) have two components - the body force and the surface force. Therefore we can write

(10.18e)

where  is the body force per unit volume and  is the area weighted surface force.

Angular Momentum

The angular momentum or moment of momentum theorem is also derived from Eq.(10.10) in consideration of the property N as the angular momentum and accordingly η as the angular momentum per unit mass. Thus,

(10.19)

where A_{Control mass system} is the angular momentum of the control mass system. . It has to be noted that the origin for the angular momentum is the origin of the position vector

The term on the left hand side of Eq.(10.19) is the time rate of change of angular momentum of a control mass system, while the first and second terms on the right hand side of the equation are the time rate of increase of angular momentum within a control volume and rate of net efflux of angular momentum across the control surface.

The velocity  defining the angular momentum in Eq.(10.19) is described in an inertial frame of reference.Therefore, the term   can be substituted by the net moment ΣM applied to the system or to the coinciding control volume. Hence one can write Eq. (10.19) as

(10.20a)

At steady state

	(10.20b)