hydraulic calculations safe rain
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Hydraulic Calculations Safe RainTRANSCRIPT
SPREADSHEET
SOLVED EXAMPLE No. 2
SAFE RAIN´S DATA EXAMPLE
Basic Hydraulic calculations
AUTHOR: Juan E. González Fariñas ([email protected])
In association with Safe-Rain S.L www.saferain.com
SPREADSHEET
SOLVED EXAMPLE No. 2
SAFE RAIN´S DATA EXAMPLE
Basic Hydraulic calculations
AUTHOR: Juan E. González Fariñas ([email protected])
In association with Safe-Rain S.L www.saferain.com
El uso de este Libro
El uso de este
input data
output data
a) Introducir datos PIPE FITTINGS
Q = 12.00 l / s Staright elbow (large radius)
L straight length = 127.60 m Staright elbow (standard radius)
D interior = 94.00 mm Staright elbow (short radius)
e = 0.007 mm Elbows 450
Temperature = 25.0 (0 C) Gate valves
Globe valves
b) Results obtained Angle valves
Butterfly valves
velocity 1.7292 m/ s Check valves
Tank inlet
viscosity (n) 0.0092 cm2 / s Tank outlet
v 2 / 2g = 0.15 m Sudden enlargement
Reynolds´s Number 176,978 dimensionless Abrupt narrowing
f = 0.0183 Others
OBJECTIVE FUNCTION: -0.417 Others
TOTAL EQUIVALENT LENGTH 10.216 m
L Total = 137.816 m
Sh f = 4.091 m
SPREADSHEET: HEADLOSSES IN PIPES
STRAIGHT PIPE EQUIVALENT LENGTH FOR FITTINGS
Number K fitting l equivalent (m)
0.60 0.00
0.80 0.00
2 0.90 9.24
0.36 0.00
1 0.19 0.98
10.00 0.00
5.00 0.00
0.00
2.50 0.00
0.50 0.00
1.00 0.00
0.00
0.00
0.15 0.00
0.00
l equivalent (m) 10.22
SPREADSHEET: HEADLOSSES IN PIPES
EQUIVALENT LENGTH FOR FITTINGS
input data
output data
Jet heigth Q P
(m) (lpm) (m)
0.25 300 0.8
0.50 468 1.5
0.75 550 2.1
1.00 620 2.7
1.50 745 3.6
2.00 850 4.5
2.50 950 5.3
3.00 1032 6.1
3.50 1120 6.7
4.00 1181 7.3
4.50 1255 7.9
5.00 1304 8.3
Jet heigth Q P
(m) (lpm) (m)0.44 415 1.36
CURVILINEAR INTERPOLATION
EXAMPLE OF CURVILINEAR INTERPOLATION BY AN EXCEL SHEET
Table below shows flow and pressure head requirements in the base of the Safe Rain
F2431315 for different heights of the foam column.
Find pressure head and discharge requirements for a 0. 44 m jet heigth.
Write dessired jet heigth at DARK BLUE cell A27. Q and P are displayed at YELLOW cells
EXAMPLE OF CURVILINEAR INTERPOLATION BY AN EXCEL SHEET
Table below shows flow and pressure head requirements in the base of the Safe Rain´s nozzle
Find pressure head and discharge requirements for a 0. 44 m jet heigth.
. Q and P are displayed at YELLOW cells B27 and C27.
2. 5 m
Continue
Example 1. Determine the pressure head at point 2 if pump provides a flow rate of 415 lpm with 4.5 m head pressure at
point 1. Check for if height of the jet reaches about 1. 4 m.
pressure at
2. 5 m
p1 (m.c.a.) = 4.5
(Z1 - Z2) (m) = -2.5
Q (lpm) = 415
Continue
Solution:
The head pressure at the base of the nozzle (point 2) can be obtained as a
result of applying the Bernoulli´´ equation between points 1 and 2:
p2 (m.c.a.) = (Z1 - Z2) + p1 (m.c.a.) + V1 2/ 2g - V2
2/ 2g - hf totales
Being constant diameter pipe, the above equation becomes:
p2 (m.c.a.) = (Z1 - Z2) + p1 (m.c.a.) - Shf totales
Initial example data:
Note: Shf totales will be obtained by DATA ENTRY AND RESULTS SHEET.
The head pressure at the base of the nozzle (point 2) can be obtained as a
2. 5 m
input data
output data
a) Introducir datos
Q = 6.92
L straight length = 50.00
D interior = 94.00
e = 0.007
Temperature = 25.0
b) Results obtained
velocity 0.9967
viscosity (n) 0.0092
v 2 / 2g = 0.05
Reynolds´s Number 102,008
f = 0.0183
OBJECTIVE FUNCTION: 0.000
TOTAL EQUIVALENT LENGTH 10.216
L Total = 60.216
Sh f = 0.594
STRAIGHT PIPE
From above table,
PIPE FITTINGS Number K fitting
l / s Staright elbow (large radius) 0.60
m Staright elbow (standard radius) 0.80
mm Staright elbow (short radius) 2 0.90
mm Elbows 450 0.36
(0 C) Gate valves 1 0.19
Globe valves 10.00
Angle valves 5.00
Butterfly valves
m/ s Check valves 2.50
Tank inlet 0.50
cm2 / s Tank outlet 1.00
m Sudden enlargement
dimensionless Abrupt narrowing
D21 result is correct Others 0.15
only if D22 cell = zero. Others
m l equivalent (m)
m
m
Continue
STRAIGHT PIPE EQUIVALENT LENGTH FOR FITTINGS
SPREADSHEET OF HEADLOSSES IN PIPES
From above table, Shf totales = 0. 594 m ≈ 0. 60 m.
l equivalent (m)
0.00
0.00
9.24
0.00
0.98
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
0.00
10.22
EQUIVALENT LENGTH FOR FITTINGS
p1 (m.c.a.) = 4.5
(Z1 - Z2) (m) = -2.5
h f totales (m) = 0.59
Q (lpm) = 415
Finally: p2 (m.c.a.) = (Z1 - Z2) + p1 (m.c.a.) - hf totales
p2 (m.c.a.) = 1.4
From above table, Shf totales = 0. 594 m ≈ 0. 60 m.
Finally: p2 (m.c.a.) = (Z1 - Z2) + p1 (m.c.a.) - hf totales
RETURN
= 0. 594 m ≈ 0. 60 m.
2. 5 m