Finite Volume Analysis: Application
of Conservation of Mass (continued)
Example 2 (continued) Strategy
for solution:
Continue
with steps 2 and 3. In
general, conservation of mass is:
Note: The flow across surfaces is uniform. d/dt
∫ ρ dV
+ ∫ ρ V● n dA
= 0 cv cs Since
the river flows steadily, the first term in the above equation is zero. (steady flow) The
second term expands as follows: ∫ ρ V1● n1 dA + ∫
ρ V2● n2 dA +
∫ ρ V3●
n3 dA +
∫ ρ V4●
n3 dA = 0 ρ ( -3)(50)(6) + ρ (-4)(80)(6) + ρ
(V)(70)(6) + ρ(0.8V)(30)(6) = 0 Cancel ρ and 6 from
the above equation and solve for V. 94 V =
470 or V = 5 ft/sec (result) |
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