Strategy in the Analysis of Fluid
Mechanics using Control Volumes
(continued)
|
The
sample calculations become: Apply: ∂/∂t ∫ ρ dV +
∫ ρ V . n dS = 0 cv cs ∂/∂t ∫ ρ dV +
∫ ρ V . n dS1 +
∫ ρ V . n dS2 + ∫ ρ V . n dS3
= 0 cv cs cs cs If
the flow is steady, then ∂/∂t ∫ ρ dV =
0 i.e. no change with time
(flow is established)
cv Assume
the fluid is incompressible. Then the
mass density, ρ, is constant. Assume
the flow across the control surface is uniform (constant across opening). Then
the three remaining terms become: ˗
ρV1 A1
+ ρV2 A2 +
ρV3 A3
= 0 Or ρV1 A1 =
ρV2 A2
+ ρV3 A3 (mass flow in equals mass flow
out) Or V1 A1 = V2
A2 + V3 A3 (volumetric flow in equals
volumetric flow out) Note: For applications involving conservation of
linear momentum, vector terms enter.
So you
will, in general, have x and y- components as part of the calculation
including surface forces
and body forces. Also
for conservation of energy, energy is a scalar quantity. There will be no vector terms. Click
here for applications involving
conservation of mass. |
All rights reserved.