Problem 8.173
[2]
Given:
Flow through a venturi meter (NOTE: Throat is obviously 3 in not 30 in!)
Find:
Flow rate
Solution:
Basic equation
m
actual
C A
t
⋅
1
β
4
−
2
ρ
⋅
p
1
p
2
−
(
)
⋅
⋅
=
C A
t
⋅
1
β
4
−
2
ρ
⋅
Δ
p
⋅
⋅
=
Note that m
actual
is mass flow rate (the
software cannot render a dot!)
For Re
D1
> 2 x 10
5
, 0.980 < C < 0.995.
Assume C = 0.99, then check Re
β
D
t
D
1
=
β
3
6
=
β
0.5
=
Also
Δ
p
ρ
Hg
g
⋅
Δ
h
⋅
=
SG
Hg
ρ
⋅
g
⋅
Δ
h
⋅
=
Then
Q
m
actual
ρ
=
C A
t
⋅
ρ
1
β
4
−
⋅
2
ρ
⋅
Δ
p
⋅
⋅
=
π
C
⋅
D
t
2
⋅
4
ρ
⋅
1
β
4
−
⋅
2
ρ
⋅
SG
Hg
⋅
ρ
⋅
g
⋅
Δ
h
⋅
⋅
=
π
C
⋅
D
t
2
⋅
4
1
β
4
−
⋅
2 SG
Hg
⋅
g
⋅
Δ
h
⋅
⋅
=
Q
π
4
1
0.5
4
−
×
0.99
×
1
4
ft
⋅
⎛
⎜
⎝
⎞
⎟
⎠
2
×
2
13.6
×
32.2
×
ft
s
2
⋅
1
×
ft
⋅
×
=
Q
1.49
ft
3
s
⋅
=
Hence
V
Q
A
=
4 Q
⋅
π
D
1
2
⋅
=
V
4
π
1
1
2
ft
⋅
⎛
⎜
⎝
⎞
⎟
⎠
2
×
1.49
×
ft
3
s
⋅
=
V
7.59
ft
s
⋅
=
At 75
o
F,(Table A.7)
ν
9.96
10
6
−
×
ft
2
s
⋅
=
Re
D1
V D
1
⋅
ν
=
Re
D1
7.59
ft
s
⋅
1
2
×
ft
⋅
s
9.96
10
6
−
×
ft
2
⋅
×
=
Re
D1
3.81
10
5
×
=
Thus Re
D1
> 2 x 10
5
. The volume flow rate is
Q
1.49
ft
3
s
⋅
=

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