Finite Volume Analysis: Application
of Conservation of Linear Momentum
Example 3 (continued) Use
the dot product et = cos θ i + sin θ j with (2B).
˗ ρ Vj2 A1 i +
ρ Vj2
A2 ( cos θ i
+ sin θ j ) ˗ ρ Vj2
A3 ( cos θ i
+ sin θ j ) = R n (2B) The result is: ˗ ρ
Vj2 A1 cos
θ + ρ Vj2
A2 ˗ ρ
Vj2 A3 = 0 (3) Note: dm/dt =
ρ Vj
A So ˗
dm1/dt cos
θ + dm2/dt ˗ dm3/dt = 0 And
from conservation of mass: dm1/dt = dm2/dt + dm3/dt or dm2/dt = dm1/dt ˗
dm3/dt dm1/dt
cos θ =
dm1/dt ˗
dm3/dt ˗ dm3/dt = dm1/dt ˗ dm1/dt cos θ = 2
dm3/dt
dm3/dt = (1/2) dm1/dt
[ 1 ˗ cos θ ] (result) Now dm2/dt = dm1/dt ˗ dm3/dt = dm1/dt ˗ (1/2) dm1/dt
[ 1 ˗ cos θ ] Or dm2/dt = (1/2)
dm1/dt [ 1 + cos
θ ] (result) Note: As expected dm2/dt >
dm3/dt Also note: If
θ = 90o , dm2/dt = dm3/dt = (1/2) dm1/dt
as expected. And also If
θ = 0o , dm2/dt = dm1/dt and dm3/dt =
0 as expected. |
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