1. BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4
WHY CONTROL VOLUME VS SYSTEM? DIFICULT TO FOLLOW SAME MASS ALL THE TIME AND OFTEN INTERESTED IN THE EFFECT OF FLUID MOTION ON AN OBJECT NOT THE MOTION OF A GIVEN MASS PARTICLE

2. CONTROL VOLUME APPROACH
Two approaches for writing the equations of motion of a fluid. The first follows a fixed mass of fluid particles as it moves through the flow field. The other considers a fixed control volume and in the flow field and relates the movement of mass, momentum and energy across the the control volume boundaries to changes taken place inside the control volume.
For steady flow, where at any particular point the flow is not changing in time, it is more convenient to use the control volume approach so time is not an independent variable. (if you were to follow an individual fluid element its properties are likely to change in time.
t + t

3. Divide by t Xt = total amount of mass, momentum
or energy of fluid particles in control
volume (V1 and V2) at time t.
At t + t same
fluid particles
now in V2 and V3.
Divide by t

4. In the limit as t goes to zero V2 approaches
t +t
XV1(t)
XV3(t+t) V2 t
In the limit as t goes to zero V2 approaches
that of the control volume.
Rate of change of X
within control volume
Difference between rate at which X leaves
control volume to that at which it enters.

5. t +t V2 t In the limit as t goes to zero, V2
approaches that of the control volume.
t +t V2 t
Rate of change of X
within control volume
Difference between rate at which X leaves
control volume to that at which it enters.
Rate of mass flow
through surface
x = X per unit mass

6. t
V(x,y,z)
Rate of mass flow
through surface
x = X per unit mass

7. x = X per unit mass X is any property of the fluid
(mass, momentum, energy)
x is the amount of X per unit mass
(in any small portion of the fluid)
The total amount of X in the control volume
=    CV xdV
dV = differential mass of fluid, dm, so x = dX/dm

8. Relates properties of a fixed mass system
x = dX/dm
Relates properties of a fixed mass system
of fluid particles to the properties of the
fluid inside of and crossing through the
boundaries of a control volume.

9. x = X per unit mass The total amount of X in the control volume
=    CV xVol
dVol = differential mass of fluid, dm, so x = dX/dm
Msystem =    CV dm =    CV dVol, so x = 1
Psystem =    CV Vdm =    CV VdVol, so x = V
Esystem =    CV edm =    CV edVol, so x = e

10. Conservation of Mass X = total mass M of system of fluid particles
dM/dt = 0 since mass can neither be created or destroyed
x = 1
steady

11. VA cos 
VA if 1-D + –

12. Example
100 m3
300oK
Air
dp/dt = ?

13. Example
100 m3
300oK
Air
dp/dt = ?
= ?
= ?

14. Example 100 /t (kg/s)– 8 (kg/s) = 0 p = RT p/t = RT/t
300oK
Air
dp/dt = ?
100 /t (kg/s)– 8 (kg/s) = 0
p = RT
p/t = RT/t
= (kJ/kg-K)300(K) (8 (kg/s)/100 (m3))
= 6.89 kJ/(s-m3) = 6.89 kPa/s

15. Conservation of Momentum
X = total lin. mom. p of system of fluid particles
dp/dt is the rate of change of lin. mom.
which equals the sum of the forces
x = V

16. Rate of increase of linear momentum within control volume
Forces may be pressure, viscous, gravity,
magnetic, electric, surface tension, ….
Note – control volume
can not be accelerating
Net rate of efflux of linear momentum
through the control volume

17. steady
steady

18. What is force on plate to keep in place? Air stream, 2 cm in
diameter and 100 m/s
and density of 1.2 kg/m3

19. Fplate= (100 m/s) [(1.2 kg/m3)(100 m/s)(/4)(0.022m2)]
Air stream, 2 cm in
diameter and 100 m/s
and density of 1.2 kg/m3
What is force
on plate ?
Fplate= (100 m/s) [(1.2 kg/m3)(100 m/s)(/4)(0.022m2)]
Fplate = 3.77 kg-m/s2 = 3.77 N

20. Conservation of Energy
Q - W’ = dE
 indicates interactions across system boundary
d indicates change of property within system
Q = heat
W’ = all work including mechanical, electric
and magnetic
E = energy such as internal, U, kinetic energy, KE,
and various forms of potential energy, PE.
Can store energy in capacitor or as chemical energy

21. Conservation of Energy
E = total energy of fluid particles
X = e
e refers to the total energy E of the fixed mass
system per unit mass

22. If E can be assumed to be U + KE + mgh
then e = u + V2/2 + gz

23. W’ = flow work (pressure) + W (viscous, shaft, electric, magnetic)
Work that acts against the external pressure at boundaries
~ if volume of mass is V then work required is p V
or since the density  = m / V, (so  m =  V)
the flow work (p V) per unit mass (m) = p/
+ flow work
Note = did divide by M to get unit mass because concerned with delta V and mass of delta V = rho x delta V
For mass flowing nto the control volume , the expression for flow work is negative, since this represents work done by the surroundings on the system.
- flow work

24. It is convenient to combine flow work per unit
mass, p/, with the internal energy per unit mass, u,
into the thermodynamic property enthalpy, h:
Energy equation
Energy equation

25. Determine Power Output of Steam Turbine
= ?
h =
3kJ/kg
h =
2.6kJ/kg
30 m/s
0.1 kg/s
100 m/s
= 0.6 kJ/s

26. {(2600 – 3000)kJ/kg + ( )/2 m2/s2} 0.1 kg/s
= -0.6kJ/s – dW/dt

27. {(2600 – 3000)kJ/kg + (1002- 302)/2 m2/s2} 0.1 kg/s = -0.6kJ/s – dW/dt
{(2600 – 3000)kJ/kg + ( )/2 m2/s2} 0.1 kg/s
= -0.6kJ/s – dW/dt
4550m2/s2 = 4550kgm2/[kgs2] = 4550Nm/kg = 4.550kJ/kg
kJ/kg x 0.1 kg/s = kJ/s
kJ/s = -0.6 kJ/s – dW/dt
dW/dt = kJ/s = Watts