BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4

CONTROL VOLUME APPROACH

t

t + 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

t

t +t

In the limit as t goes to zero V2 approaches that of the control volume.

V2

Rate of change of X within control volume

Difference between rate at which X leaves control volume to that at which it enters.

XV3(t+t)

XV1(t)

t

t +tIn the limit as t goes to zero, V2

approaches that of the control volume.

V2

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 flowthrough surface

x = X per unit mass

V(x,y,z)

Rate of mass flowthrough surface

x = X per unit mass

t 0

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 xdV

• dV = differential mass of fluid, dm, so 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.

x = dX/dm

x = X per unit mass

• The total amount of X in the control volume = CV xVol

• dVol = differential mass of fluid, dm, so x = dX/dm

Msystem = CV dm = CV dVol, so x = 1Psystem = CV Vdm = CV VdVol, so x = VEsystem = CV edm = CV edVol, so x = e

Conservation of MassX = total mass M of system of fluid particles

dM/dt = 0 since mass can neither be created or destroyed x = 1

steady

VA cos

VA if 1-D

+–

100 m3

300oK

Air

dp/dt = ?

Example

100 m3

300oK

Air

dp/dt = ?

Example

= ? = ?

100 m3

300oK

Air

dp/dt = ?

100 /t (kg/s)– 8 (kg/s) = 0p = RTp/t = RT/t

= 0.287 (kJ/kg-K)300(K) (8 (kg/s)/100 (m3)) = 6.89 kJ/(s-m3) = 6.89 kPa/s

Example

Conservation of MomentumX = 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

Forces may be pressure, viscous, gravity,magnetic, electric, surface tension, ….

Rate of increase of linear momentum within control volume

Net rate of efflux of linear momentum through the control volume

Note – control volumecan not be accelerating

steady

steady

Air stream, 2 cm in diameter and 100 m/s

and density of 1.2 kg/m3

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

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

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.

Conservation of EnergyE = total energy of fluid particles

X = e e refers to the total energy E of the fixed mass

system per unit mass

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

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 Vor since the density = m / V, (so m = V)the flow work (p V) per unit mass (m) = p/

+ flow work

- flow work

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

30 m/s0.1 kg/s

100 m/s

h =3kJ/kg

h =2.6kJ/kg

= 0.6 kJ/s

Determine Power Output of Steam Turbine

= ?

0

{(2600 – 3000)kJ/kg + (1002- 302)/2 m2/s2} 0.1 kg/s= -0.6kJ/s – dW/dt

0

{(2600 – 3000)kJ/kg + (1002- 302)/2 m2/s2} 0.1 kg/s= -0.6kJ/s – dW/dt

4550m2/s2 = 4550kgm2/[kgs2] = 4550Nm/kg = 4.550kJ/kg

-395.45 kJ/kg x 0.1 kg/s = -39.545 kJ/s-39.545 kJ/s = -0.6 kJ/s – dW/dt

dW/dt = 38.945 kJ/s = 38.945 Watts