FIGURE 3 Derivation of the Reynolds transport theorem.

The volume on the right with the hatching slanting upward to the left plus Volume bcde is the control mass, and the volume with the hatching is Volume II. Control mass is that part of the universe composed of masses identified for analysis; as such, if a boundary is used to enclose the masses, no other masses are allowed to pass through this boundary. In other words, the system is closed. Thus, control mass is also called a closed system or simply system. The control volume, on the other hand, is a specific volume in space where masses are allowed to pass through, implying that its boundary is permeable to the "traffic" of masses crisscrossing through it. Thus, the control volume is an open system; it is, in fact, also called open system.

Imagine the control mass and the control volume coinciding at an initial time t, that is, the boundaries of both the control volume and the control mass coincide at this initial time. From this position, let them move at different speeds but with the control mass moving faster. After some time At later, the control mass and the control volume take different positions; this is the configuration portrayed in the figure.

Let O represent the value of any property common to both the closed and open systems. The change of this property for the closed system when the system goes in time from t to t + At is

where A is a symbol for change and M refers to control mass. OM,(+a( is the value of O for the control mass at t + At and OM( is the value of O for the control mass at t. At the initial time, because control mass and control volume coincide with each other, the O of the control mass OM( is equal to the O of the masses inside the control volume OV t. V refers to the control volume.

Referring to the figure, at time At later, OM t + at is given by

where OV t+at, OIt+at, and OIIt+at are, respectively, the values of O of the masses in the control volume, Volume I, and Volume II at time t + At. Substituting Equation (77) in Equation (76), with OVt equals OM>(, produces

Rearranging and dividing all terms by At produces aom = ^yi+aj-^yj Oi^aí + Qiv+aí (79)

At At AtAt

Because it was derived in the closed system, the function O on the left-hand side of Equation (79) is a function of only one independent variable, t. There is only one independent variable, so a partial derivative for O is not possible, and there can only be a full or total derivative. Thus, in the limit, the term will produce a total derivative as follows:

at — 0 At dt dt where t is the value of the property O per unit mass of the system and p is the mass density of the system. Note that jMfp dV is OM.

Because of the presence of V (space) and t and because it was derived for the control volume, the numerator of the first term on the right-hand side of Equation (79), OV (+a( - OVt, is a function of both space and time. Therefore, in the limit, the derivative will be a partial derivative. Thus, lim = dU*PdV (81)

The property, O, of the mass inside Volume I results from the masses entering the holes of the control volume. Thus, this property must be a function of space as well as, of course, a function of time. The second term on the right-hand side of Equation (79), -O7 ,t+At/At, therefore, in the limit, results in a partial derivative, lim i-°r-l = -d (82)

at — 0^ At ) dt where -dOj/dt is simply the rate of inflow of the property O across the boundary, where this boundary is dab. The volume rate of inflow is the product of velocity and cross-sectional area of flow. If the velocity is v and the area is A, then the volume rate of inflow is -f a v • n dA, where n is the unit normal vector to dA. Note that in the integration is over Ain, because the integration pertains to the inflow. Note, also, the negative sign, which is the result of the dot product, v and n. This is explained as follows: The unit normal n always points away to the outside of surface (area) A. In this direction, any vector is considered positive, by convention. On the other hand, vector v, because it is with the mass entering the control volume through the area A, is opposite to the direction of the unit normal. Because v and n are in opposite directions, their dot product will be negative, thus, -Ja v • n dA.

The mass rate of inflow is, accordingly, the product of the mass density and the volume rate of inflow. If the mass density is p, the mass rate of inflow is -JA pv • n dA Now, if the property per unit mass is 0,

dt [ Ainin

The third term of Equation (79), t+At/At, is a result of mass passing through the boundary of the control volume, out into the control mass. This boundary is bed, the "exit boundary" of the control volume. The velocity at this boundary will be in the same direction as the unit normal vector. Thus, the corresponding dot product will be positive. In the limit, the expression for ®iit+At/At will be the same as ,t+At/At , except that the sign will be positive. Also, the integration will be over Aout, because the integration pertains to the outflow. Therefore, lim r= r p. n dA

at^Q V At ) dt JAout The last two terms of the right-hand side of Equation (79) are then, in the limit, lim (-%+a- + = r p . n dA +f <pV • n dA

where j>A now represents that the terms inside the parenthesis on the left-hand side are summed over the area A that surrounds the control volume V.

Substituting Eqs. (80), (81), and (84) into Equation (79),

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