Problem 8.99 [Difficulty: 3]
Given: Flow through sudden contraction
Find: Volume flow rate
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
Hence the energy equation becomes
p1
ρ
V1
2
2
+
⎛
⎜
⎜
⎝
⎞
⎠
p2
ρ
V2
2
2
+
⎛
⎜
⎜
⎝
⎞
⎠
−K
V2
2
2
⋅=
From co ntinu ity V1V2
A2
A1
⋅= V2AR⋅=
Hence p1
ρ
V2
2AR2
⋅
2
+
⎛
⎜
⎜
⎝
⎞
⎠
p2
ρ
V2
2
2
+
⎛
⎜
⎜
⎝
⎞
⎠
−K
V2
2
2
⋅=
Solving for V 2V2
2p
1p2
−
()
⋅
ρ1AR
2
−K+
()
⋅
=AR
D2
D1
⎛
⎜
⎝
⎞
⎠
2
=1
2
⎛
⎜
⎝
⎞
⎠
2
=0.25=so from Fig. 8.14 K 0.4=
Hence V22 0.5×lbf
in2
⋅12 in⋅
1ft⋅
⎛
⎜
⎝
⎞
⎠
2
×ft3
1.94 slug⋅
×1
1 0.252
−0.4+
()
×slug ft⋅
lbf s2
⋅
×= V27.45 ft
s
⋅=
QV
2A2
⋅= πD2
2
⋅
4V2
⋅=
Qπ
4
1
12 ft⋅
⎛
⎜
⎝
⎞
⎠
2
×7.45×ft
s
⋅= Q 0.0406 ft3
s
⋅= Q 2.44 ft3
min
⋅= Q 18.2 gpm⋅=
