Conservation laws
• Laws of conservation of mass, energy, and momentum.
• Conservation laws are first applied to a fixed quantity of
matter called a closed system or just a system, and
then extended to regions in space called control
volumes.
• The conservation relations are also called balance
equations since any conserved quantity must balance
during a process.
Continuity Equation
Conservation of mass for a system
Time rate of change of the system mass =0
DM sys
Dt
0
where Msys =mass of the system
M sys =   dV
sys
For a system and a fixed nondeforming control volume that are coincident
at an instant of time ,with B - mass and b=1 we see at
D


dV

dV   V ndA


CV
CS
Dt sys
t
Figure 5.1 (p. 194)
System and control volume at three different instances of time. (a) System
and control volume at time t – δt. (b) System and control volume at time t,
coincident condition. (c) System and control volume at time t + δt.
• Time rate of change of the mass of the system = time
rate of change in the mass of the control volume + net
rate of flow of mass through control surface.

• Time rate of change in the mass of the control volume = t CV dV
• Net rate of flow of mass through control surface = CS V ndA

• For steady flow, CV dV  0
t
• Net mass flow rate through the control surface =

CS
V ndA = mout  min
where m mass flow rate( slug/s;kg /s)
Continuity Equation
• Conservation of mass: for a fixed, non deforming control
volume

 dV   V ndA =0
t
• More commonly used:
CV
CS
m   Q   AV
where =fluid density,Q=volume flow rate,
V=component of fliud velocity perpendicular to area
• Average velocity:

V
A
V ndA
A
For steady flow, mout  min  0
For incompressible flow, Qout  Qin  0
For uniformly distributed flow, m = AV
For non-uniformly distributed flow, m = AV ,V is the average velocity
m
out
 min
For steady flow involving more than one stream of specific fluid,
m   A1V1   A2 V2
Example 1
• Seawater flows steadily through a simple conical-shaped
nozzle. If the nozzle exit velocity =20 m/s, determine the
minimum pumping capacity in m3/s.
Example 2
• Airflows steadily between two sections in a long straight
portion of 4-in diameter pipe. The uniformly distributed
temperature and pressure at each section are given. If
the average air velocity at section (2) is 100 ft/s,
calculate the average air velocity at section (1).
Example 3
Moist air enters a dehumidifier at the rate of 22 slugs/hr.
Determine the mass flow rate of the dry air and the water
vapor leaving the dehumidifier.
Example 4
A bathtub is being filled with water from a faucet. The
rate of flow from the faucet is steady at 9 gal/min. The
tub volume is approximated by a rectangular space.
Figure E5.5b (p. 199)
Moving, Non deforming Control Volume
V = W +VCV
V= absolute velocity, V
W= relative velocity of fluid seen by an observer moving
with the control volume velocity, Vcv

dV   W ndA =0

CV
CS
t
Continuity equation for a moving
non-deforming control volume
Example 5. An airplane moves forward at a speed of 971
km/hr. The frontal intake area of the jet engine is 0.80 m2,
and the entering air density is 0.736 kg/m3. A stationary
observer determines that relative to the earth, the jet
engine exhaust gases moves away from the engine with a
speed of 1050 km/hr. The engine exhaust area is 0.558
m2, and the exhaust gas density is 0.515 kg/m3. Estimate
the mass flow arte of fuel into the engine in kg/hr.