Problem 8.97 [Difficulty: 3]
Given: Flow through gradual contraction
Find: Pressure after contraction; compare to sudden contraction
Solution:
Basic equations
p1
ρα
V1
2
2
⋅+ gz
1
⋅+
⎛
⎜
⎜
⎝
⎞
⎠
p2
ρα
V2
2
2
⋅+ gz
2
⋅+
⎛
⎜
⎜
⎝
⎞
⎠
−hlm
=hlm K
V2
2
2
⋅= QVA⋅=
Ass umption s: 1) Steady flow 2) In comp ressib le flow 3 ) α at 1 and 2 is approximately 1 4) Horizontal
Available data Q25
L
s
⋅= Q 0.025 m3
s
=D175 mm⋅= D237.5 mm⋅= p1500 kPa⋅= ρ999 kg
m3
⋅=
For an included angle of 150
o
and an area ratio
A2
A1
D2
D1
⎛
⎜
⎝
⎞
⎠
2
=37.5
75
⎛
⎜
⎝
⎞
⎠
2
=0.25= we find from Tab le 8.3 K 0.35=
Hence the energy equation becomes
p1
ρ
V1
2
2
+
⎛
⎜
⎜
⎝
⎞
⎠
p2
ρ
V2
2
2
+
⎛
⎜
⎜
⎝
⎞
⎠
−K
V2
2
2
⋅= with V1
4Q⋅
πD1
2
⋅
=V2
4Q⋅
πD2
2
⋅
=
p2p1
ρ
21K+()V
2
2
⋅V1
2
−
⎡
⎣
⎤
⎦
⋅−= p2
8ρ⋅Q2
⋅
π2
1K+()
D2
4
1
D1
4
−
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅−=
p2500 103
×N
m2
⋅8
π2999×kg
m3
⋅0.025 m3
s
⋅
⎛
⎜
⎝
⎞
⎠
2
×1 0.35+()1
0.0375 m⋅()
4
×1
0.075 m⋅()
4
−
⎡
⎢
⎣
⎤
⎥
⎦
×Ns
2
⋅
kg m⋅
×−= p2170 kPa⋅=
Repeating the above analysis for an included angle of 180o (sudden contraction) K 0.41=
p2500 103
×N
m2
⋅8
π2999×kg
m3
⋅0.025 m3
s
⋅
⎛
⎜
⎝
⎞
⎠
2
×1 0.41+()1
0.0375 m⋅()
4
×1
0.075 m⋅()
4
−
⎡
⎢
⎣
⎤
⎥
⎦
×Ns
2
⋅
kg m⋅
×−= p2155 kPa⋅=