2. “
1. N.B. Weber, 1971, Fluid Mechanics for
Civil Engineers, Chapman & Hall
2. Albert T. Fromhold Jr., 2011, Quantum
Mechanics for Applied Physics and Engineering,
Dover Books on Physics
2
4. Contoh: Pentingnya ukuran pipa
4
Tujuan dari setiap sistem distribusi fluida adalah untuk
memasok fluida pada tekanan yang benar ke titik
penggunaan. Oleh karena itu, penurunan tekanan pada
sistem distribusi merupakan hal penting.
7. Semi empirical theory of pipe
resistance
→ Velocity-Deficiency Equation
→ Velocity distribution adjacent to pipe wall
8. pipe resistance definition
The fluid-to-pipe resistance, sometimes referred simply
as pipe resistance, is the sum of the resistance of the
pipe wall and the resistance of the fluid flowing
inside the pipe.
10. Bernoulli’s Theorem (Daniel Bernoulli 1700 - 1782)
relates changes in the total energy of a flowing fluid to
energy dissipation expressed either in terms of a head
loss hf (m) or specific energy loss g hf (J/kg).
11. the most important mechanisms of energy dissipation
within a flowing fluid is introduced, that is, the loss in
total mechanical energy due to friction at the wall of a
uniform pipe carrying a steady flow of fluid.
12. ∙ The loss in the total energy of fluid flowing through a circular
pipe must depend on:
∙ L = The length of the pipe (m)
∙ D = The pipe diameter (m)
∙ u = The mean velocity of the fluid flow (m/s)
∙ μ = The dynamic viscosity of the fluid (kg/m s = Pa s)
13. ∙ ρ= The fluid density (kg/m³)
∙ kS = The roughness of the pipe wall* (m)
∙ * Since the energy dissipation is associated with shear stress
at the pipe wall, the nature of the wall surface will be
influential, as a smooth surface will interact with the fluid in a
different way than a rough surface.
14. ∙ All these variables are brought together in the D’Arcy-
Weisbach equation (often referred to as the D’Arcy equation).
15. Contoh kasus satu
The high frequency (0.1 kHz) pressure and velocity measurements of
two-phase unsteady flow are used to determine the hydraulic-grade-
line (HGL) along the pipeline with a complex layout. The semi-empirical
method is applied to estimate the pressure drop for the outflow, and
thus determine the pipe downstream-end conditions for the modeling
purposes.
16. Laanearu, Janek & Annus, I. & Sergejeva, M. & Koppel, Tiit. (2014). Semi-empirical
Method for Estimation of Energy Losses in a Large- scale Pipeline. Procedia
Engineering. 70. 969-977. 10.1016/j.proeng.2014.02.108.
16
17. Semi empirical theory of pipe
resistance
→ Velocity-Deficiency Equation
→ Velocity distribution adjacent to pipe wall
20. Pipe Flow Summary
∙ Dimensionally correct equations fit to the empirical results can be
incorporated into computer or calculator solution techniques.
∙ Minor losses are obtained from the pressure coefficient based on the
fact that the pressure coefficient is _______ at high Reynolds
numbers.
∙ Solutions for discharge or pipe diameter often require iterative or
computer solutions
constant
21. Solve for the Pressure Head, Velocity Head, and Elevation Head at each
point, and then plot the Energy Line and the Hydraulic Grade Line
Contoh soal satu
1
2
3 4
1’
4’
γH2O= 62.4 lbs/ft3
Assumptions and Hints:
P1 and P4 = 0 --- V3 = V4 same diameter tube
R = .5’
R = .25’
22. 1
2
3 4
1’
4’
γH2O= 62.4 lbs/ft3
Point 1:
Pressure Head : Only atmospheric P1/γ = 0
Velocity Head : In a large tank, V1 = 0 V1
2/2g = 0
Elevation Head : Z1 = 4’
R = .5’
R = .25’
23. 1
2
3 4
1’
4’
γH2O= 62.4 lbs/ft3
Point 4:
Apply the Bernoulli equation between 1 and 4
0 + 0 + 4 = 0 + V4
2/2(32.2) + 1
V4 = 13.9 ft/s
Pressure Head : Only atmospheric P4/γ = 0
Velocity Head : V4
2/2g = 3’
Elevation Head : Z4 = 1’
R = .5’
R = .25’
24. 1
2
3 4
1’
4’
γH2O= 62.4 lbs/ft3
Point 3:
Apply the Bernoulli equation between 3 and 4 (V3=V4)
P3/62.4 + 3 + 1 = 0 + 3 + 1
P3 = 0
Pressure Head : P3/γ = 0
Velocity Head : V3
2/2g = 3’
Elevation Head : Z3 = 1’
R = .5’
R = .25’
25. 1
2
3 4
1’
4’
γH2O= 62.4 lbs/ft3
Point 2:
Apply the Bernoulli equation between 2 and 3
P2/62.4 + V2
2/2(32.2) + 1 = 0 + 3 + 1
Apply the Continuity Equation
(Π.52)V2 = (Π.252)x13.9 V2 = 3.475 ft/s
P2/62.4 + 3.4752/2(32.2) + 1 = 4 P2 = 175.5 lbs/ft2
R = .5’
R = .25’
Pressure Head :
P2/γ = 2.81’
Velocity Head :
V2
2/2g = .19’
Elevation Head :
Z2 = 1’
26. Plotting the EL and HGL
Energy Line = Sum of the Pressure, Velocity and Elevation heads
Hydraulic Grade Line = Sum of the Pressure and Velocity heads
EL
HGL
Z=1’
Z=1’
Z=1’
V2/2g=3’
V2/2g=3’
Z=4’
P/γ =2.81’
V2/2g=.19’
27. Contoh soal dua
1
2
Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ? γoil= 8.82 kN/m3
f = .035
If oil flows from the upper to lower reservoir at a velocity of
1.58 m/s in the 15 cm diameter smooth pipe, what is the
elevation of the oil surface in the upper reservoir?
Include major losses along the pipe, and the minor losses
associated with the entrance, the two bends, and the outlet.
Kout=1
r/D = 0
28. 1
2
Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?
γoil= 8.82 kN/m3
f = .035
Kout=1
r/D = 0
Apply Bernoulli’s equation between points 1 and 2:
Assumptions: P1 = P2 = Atmospheric = 0 V1 = V2 = 0 (large tank)
0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin
Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)
Hmaj= 5.85m
29. 1
2
Z2 = 130 m
130 m
7 m
60 m
r/D = 2
Z1 = ?
γoil= 8.82 kN/m3
f = .035
Kout=1
r/D = 0
0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin
Hmin= 2KbendV2/2g + KentV2/2g + KoutV2/2g
From Loss Coefficient table: Kbend = 0.19 Kent = 0.5 Kout = 1
Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8)
Hmin = 0.24 m
