sxL1,()
x2
2g⋅4L1
d
⋅3.4735 1.5635 ln 2 ks
d
⋅
⎛
⎜
⎝
⎞
⎠
1.11
63.635 ν
dx⋅
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
2−
⋅1+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅H−:=
Now a program is written to calculate the exit velocity over a range of pipe
lengths (L varies from 50 meters to 1000 meters).
Q 2.1002=
Qπ
4d2
⋅V⋅:=
Q is the volume flow rate in meters cubed per second
Re 5.3482 106
×=
Re d V
ν
⋅:=
f 0.0121=
f is the friction factor
f 4 3.4735 1.5635ln 2 ks
d
⋅
⎛
⎜
⎝
⎞
⎠
1.11
63.635 ν
dV⋅
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
2−
⋅:=
V 10.6964=
V is the exit velocity in meters/second
V root s x()x,():=
x is the "guess" for the root solver
x3:=
sx() x2
2g⋅4L
d
⋅3.4735 1.5635 ln 2 ks
d
⋅
⎛
⎜
⎝
⎞
⎠
1.11
63.635 ν
dx⋅
⋅+
⎡
⎢
⎣
⎤
⎥
⎦
⋅−
⎡
⎢
⎣
⎤
⎥
⎦
2−
⋅1+
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
⋅H−:=
Extended Bernoulli Equation
with f by Haaland Correlation
ks .000046:=
(SI units)
L 100:=ν .0000010:=d 0.5:=g 9.8:=H20:=
Calculate the exit velocity for a 100 meter long pipe.
Computer Solution to Example 10.6
ENGR335
