Conservation of Mass

Conservation of mass equation can be developed by making the substitu-tion, m= Ω in Equation 3.44. Mass is a conserved quantity. Therefore, the generation term is zero. This results in the following expression:

d

in out d

m m msys

∑

∑

System boundary

∑Ω˙ _out

∑Ω˙ _in Ω˙ _gen Ω˙ _sys

FIGURe 3.5 Generalized system.

The mass flow rates in the above equations can be expressed in terms of the fluid density, ρ; the cross-sectional area of the flow passage, A; and the mean velocity of the fluid flow, V. In addition, the mass stored in the system is a function of the density of the fluid and the volume enclosed by the system boundary. Incorporating these ideas into Equation 3.45 gives

∑

∑

If the mass flow rate entering the system is the same as the mass flow rate leaving, then there can be no mass stored inside the system. A system operat-ing under this condition is said to be operatoperat-ing at a steady-flow condition. For a steady-flow condition, Equation 3.45 can be written as

∑

∑

To this point, the equations developed for the system are rate equations.

They are describing the condition of the system at an instant in time. If the behavior of the system over a specified time is required, then any of these equations can be integrated over time. For example, to investigate how a sys-tem’s mass might change during a process, Equation 3.45 can be integrated over time, resulting in

∑

∑

(

)

EXAMPLE 3.4

Consider the tank and water supply system as shown in Figure E3.4A. The diameter of the supply pipe is D₁= 20 mm, and the average velocity leaving the supply pipe is V1 = 0.595 m/s. A shut-off valve is located at z = 0.1 m in the exit pipe, which has a diameter of D₂ = 10 mm. The tank diameter is Dt = 0.3 m.

The density of the water is uniform at 998 kg/m³.

a. Find the time to fill the tank to H0 = 1 m, assuming the tank is initially empty and the shut-off valve is closed. Neglect the volume associated with the short pipe connecting the tank to the shut-off valve.

b. At the instant the water level reaches H₀ = 1 m, the shut-off valve is opened. The instantaneous average velocity of the outflow depends on the water depth above z and can be expressed by

[

]

=0.85 −

V2 g H t z

where g is the acceleration due to gravity. Determine whether the tank continues to fill or begins to empty immediately after the valve is opened.

c. Determine the steady-flow value of the water depth.

FIGURe e3.4a

SOLUTION: Part (a) of this problem asks how long it takes to fill the empty tank to a height of H0 = 1 m. An appropriate system boundary for this analysis is one that encompasses the inside of the tank up to a height of 1 m as shown in Figure E3.4B. During the time required to fill this system, a total mass, m_in enters and the mass inside the system changes by an amount, m2−m1. From this analysis, it is clear that we are investigating what is happening inside the system over a finite period. Therefore, the appropriate form of the conservation of mass is given by Equation 3.48.

∑

∑

(

)

FIGURe e3.4b

For the filling process, this equation reduces to

( )

= −

in 2 1 sys

m m m

Shut-off valve

z=0.1 m D₂=10 mm

D_t=0.3 m H0=1m

D₁=20 mm

V₁=0.595 m/s

V₂

Shut-off valve (closed) H₀=1m

m_in

(m2–m1)sys

Each of the terms in this equation can be expanded as follows:

The only unknown in this equation is the time required to fill, t. Solving for t gives

( )

Part (b) asks whether the water level rises or falls when the shut-off valve is opened. To determine this, the system must be analyzed at the instant in time when the valve is opened. When the valve is opened, water will flow across the system boundary at the outlet as indicated in Figure E3.4C. At that instant, we are interested in what happens to H(t), the instantaneous height of the water in the tank.

This analysis is not conducted over time, but rather at an instant in time.

Therefore, Equation 3.45 is appropriate. Since there is one mass flow entering and one mass flow leaving the system, Equation 3.45 is written as

 −  =d

in out d m m msys

t

FIGURe e3.4C

If the sign of the storage term dm_sys dt is positive, then the tank continues to fill. On the other hand, if the storage term is negative, the tank begins to drain.

The mass flow rates on the left-hand side of this equation can be rewritten in terms of the density of the fluid, cross-sectional areas, and velocities.

ρ − ρ =d

The velocity, V₂, is given in the problem statement in terms of the height of the water in the tank. Substituting this into the above equation and simplifying

( ^{[ ]} )

( )

Since the storage term is negative, the tank will begin to empty when the shut-off valve is opened.

Part (c) of this problem asks for the water depth resulting in a steady-flow sce-nario. It is important to understand that the value of the derivative computed in part (b) is instantaneous. Since the velocity of the water leaving the tank through the bottom pipe is a function of the depth of the water, this derivative will change with time. For a steady-flow condition, dm_sys dt=0. This allows for the solution of the height, H. From the analysis in part (b)

( )

When the valve is opened, the water depth will decrease from 1 to 0.9 m, at which point the incoming flow balances the outgoing flow and a steady-flow scenario is achieved.

3.7 Conservation of Energy (The First