Problem 4.24

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Problem 4.24 [Difficulty: 1]
Given: Data on flow through box
Find: Velocity at station 3
Solution:
Basic equation
CS
V
→
A
→⋅( )∑ 0=
Assumptions: 1) Steady flow 2) Incompressible flow 3) Uniform flow
Then for the box
CS
V
→
A
→⋅( )∑ V1− A1⋅ V2 A2⋅+ V3 A3⋅+= 0=
Note that the vectors indicate that flow is in at location 1 and out at location 2; we assume outflow at location 3
Hence V3 V1
A1
A3
⋅ V2
A2
A3
⋅−= V3 10
ft
s
⋅ 0.5
0.6
× 20 ft
s
⋅ 0.1
0.6
×−= V3 5
ft
s
⋅=
Based on geometry Vx V3 sin 60 deg⋅( )⋅= Vx 4.33
ft
s
⋅=
Vy V3− cos 60 deg⋅( )⋅= Vy 2.5−
ft
s
⋅=
V3
→⎯
4.33
ft
s
⋅ 2.5− ft
s
⋅, ⎛⎜⎝
⎞
⎠=