Laboratory No.
Bernoulli’s Equation
Name: Gero, Micah Jhilyn T.
Group Members:
Beloro, Idrelle
Buenafe, Kathleen Faith
Credito, Rother
Dela Torre, Kenan
Melaya, Michael
Peralta, Jhovert
Rockwell, France
Sotoza, Kerrrvinjay
Objective
To determine that the energy between two points are equal and to determine the existence of Head Loss
equation.
Equipment
Hydraulic Bench
Bernoulli’s Apparatus
Constant and Variable Head tank
Beaker or Graduated cylinder
Stop watch
Procedures
(1) Assemble the Hydraulic bench as seen from the diagram
(2) Turn on the bench and allow water to flow, wait until the water level stabilizes
(3) Measure the flow rate by allowing water to flow at a certain volume over time.
Data
Pipe Width L (distance from pt. 6) Manometer Head Area of Flow Flow Rate Velocity Head
(mm) (mm) (m) (m2) (m3/s) (m/s)
1 125 0.325 1.026 x10-4 0.374
2 100 0.305 0.901 x10-4 0.425
3 75 0.290 0.777 x10-4 0.493
4 56 0.268 0.682 x10-4 0.562
5 25 0.233 0.528 x10-4 0.726
6 6.35 0 0.161 0.403 x10-4 3.833 x10-5 0.951
7 25 0.185 0.528 x10-4 0.726
8 56 0.223 0.682 x10-4 0.562
9 75 0.238 0.777 x10-4 0.493
10 100 0.246 0.901 x10-4 0.425
11 125 0.250 1.026 x10-4 0.374
Computations
Flow Rate = volume/time = 0.004 m3/104.35s = 3.833 x10-5 m3/s
Height
h1&11 = 125 tan (4°29’) = 9.801 mm
h2&10 = 100 tan (4°29’) = 7.841 mm
h3&9 = 75 tan (4°29’) = 5.881 mm
h4&8 = 56 tan (4°29’) = 4.391 mm
h5&7 = 25 tan (4°29’) = 1.960 mm
h6 = 0 tan (4°29’) = 0 mm
Area of Flow = (width) (6.35+h)
A1&11 = (6.35) (6.35+9.801) = 102.559 mm 2 = 1.026 x10-4 m2
A2&10 = (6.35) (6.35+7.841) = 90.113 mm2 = 0.901 x10-4 m2
A3&9 = (6.35) (6.35+5.881) = 77.667 mm2 = 0.777 x10-4 m2
A4&8 = (6.35) (6.35+4.391) = 68.205 mm2 = 0.682 x10-4 m2
A5&7 = (6.35) (6.35+1.960) = 52.769 mm2 = 0.528 x10-4 m2
A6 = (6.35) (6.35+0) = 40.323 mm2 = 0.403 x10-4 m2
Velocity Head = flow rate/area
V1&11 = (3.833 x10-5) / (1.026 x10-4) = 0.374 m/s
V2&10 = (3.833 x10-5) / (0.901 x10-4) = 0.425 m/s
V3&9 = (3.833 x10-5) / (0.777 x10-4) = 0.493 m/s
V4&8 = (3.833 x10-5) / (0.682 x10-4) = 0.562 m/s
V5&7 = (3.833 x10-5) / (0.528 x10-4) = 0.726 m/s
V6 = (3.833 x10-5) / (0.403 x10-4) = 0.951 m/s
𝑃1 𝑉2
1 𝑃2 𝑉2
2
Head Loss = + 2𝑔 = + 2𝑔 + 𝐻𝐿1−2
𝛿 𝛿
0.325 (0.374)2 0.250 (0.374)2
Pt. 1 vs. Pt. 11 =
9.81
+ 2(9.81)
= 9.81
+ 2(9.81)
+ HL1−11 = 0.00765 m
0.305 (0.425)2 0.246 (0.425)2
Pt. 2 vs. Pt. 10:
9.81
+ 2(9.81)
= 9.81
+ 2(9.81)
+ HL2−10 = 0.00601 m
0.290 (0.493)2 0.238 (0.493)2
Pt. 3 vs. Pt. 9: + = + + HL3−9 = 0.00530 m
9.81 2(9.81) 9.81 2(9.81)
0.268 (0.562)2 0.223 (0.562)2
Pt. 4 vs. Pt. 8:
9.81
+ 2(9.81)
= 9.81
+ 2(9.81)
+ HL4−8 = 0.00459 m
0.233 (0.726)2 0.185 (0.726)2
Pt. 5 vs. Pt. 7:
9.81
+ 2(9.81)
= 9.81
+ 2(9.81)
+ HL5−7 = 0.00489 m
Documentation
Manometer 1 (0.325m) Manometer 2 (0.305m) Manometer 3 (0.290m) Manometer 4 (0.268m)
Manometer 5 (0.233m) Manometer 6 (0.161m) Manometer 7 (0.185m) Manometer 8 (0.223m)
Manometer 9 (0.238m) Manometer 10 (0.246m) Manometer 11 (0.250m)
Conclusion
Recommendation