1. Flow in Channels (수로흐름)
Pipe (closed channel) flow (관수로 흐름): No free-surface flow (fluid-full), pressurized, flow driven by pressure difference (piezometric pressure) – waterways (상수도), penstock (댐 여수로), petroleum pipeline, etc.
Open channel flow (개수로 흐름): Free surface flow (surface is open to the atmosphere), not pressurized (atmospheric pressure on the flow surface), flow driven by gravity – culvert (배수로), irrigation channel, canal, sewer (하수도), rainwater drain, etc.
Pipe (closed channel) flow
Laminar vs. turbulent flow in a circular pipe
Basic criteria for the identification of laminar/turbulent characteristics of a flow – Reynolds number (Re = )
In most engineering applications,
if Re < 2000, laminar flow
if Re > 3000, turbulent flow

2. Head losses in a pipe flow
1. Major loss: Friction – 관내 마찰에 의한 손실
2. Minor loss:
Inlet – 유체가 수조로부터 관으로 유입될 때의 turbulence로 인한 손실
Bend – 관수로의 방향 변화로 인한 손실
Contraction – 관수로 직경의 감소로 인한 손실
Expansion – 관수로 직경의 확대로 인한 손실
Fittings/Valves – 관수로 내의 장애물로 인한 손실
In general, friction head loss induces the largest energy loss (> 90% of total head loss).

3. 1. Head loss due to friction
Generalized by the Darcy-Weisbach equation
Velocity distribution in a pipe = f (shear stress distribution)
Shear stress applied on a control volume of a fluid flowing through a circular pipe (steady, uniform flow)
Flow direction → direction of decreasing piezometric pressure
Shear stress → linearly increases from the center of a pipe ( = 0) to the pipe wall

4. Local shear stress coefficient (국부 전단응력계수, dimensionless, ratio of shear stress force/inertial force),
Typically, resistance (friction) coefficient (마찰계수, f) used instead of Cf (Cf = f/4)

5. Shear stress applied on a control volume of a fluid flowing through a non-circular pipe (steady, uniform flow)
where P: perimeter, A: cross-sectional area, A/P≡ hydraulic radius (Rh)
4Rh used in a non-circular pipe as D used in a circular pipe
 Reynolds number =
Class quiz) Air (20oC, 101 kPa abs.) flows at a rate of 2.5m3/s in a steel rectangular duct (0.3m0.6m) of which length is 100m. Is this air flow laminar or turbulent? What is the friction head loss (hf)? 단, f = (given value in this problem)

6. Roughness (smoothness) of a pipe
Expressed by the equivalent sand roughness (ks, 등가모래조도)
Roughness of a pipe – evaluated by the relative roughness (ks/D)
In a laminar flow, friction is independent of pipe’s roughness (“0” velocity at the pipe’s surface and shear stress (resistance) is
 friction in a rough pipe = friction in a smooth pipe, f = f(Re))
In a turbulent flow,
If Re and relative roughness (ks/D) were small, the height of viscous sublayer > those of roughness elements.
 friction in a rough pipe = friction in a smooth pipe, f = f(Re)
if Re were large, the viscous sublayer becomes thin, resulting in projection of the roughness elements into the main flow.
 friction in a rough pipe = f (drag force around the roughness elements: i.e., ks/D)  f (Re)
Size of roughness < viscous sublayer → pipe is “smooth”
Size of roughness > viscous sublayer → pipe is “rough”
Note) Non-circular pipes (channels) - 4Rh used instead of D (pipe diameter)
 Relative roughness = ks/4Rh

7. Turbulent flow
Resistance coeff. vs. Re
– Nikuradse’s experiment

8. Resistance (friction) coefficient (f) in a smooth pipe
Resistance (friction) coefficient (f) in a rough pipe (commercial pipe – irregular roughness)
Laminar flow 
Turbulent flow  Moody diagram
(empirical approximation)

9. Moody diagram How to find f using Moody diagram Calculate Re and ks/D
Find a point where the Re line meets the ks/D line.
Read f value from the joint point
If hf is known, but V is not given, then use Ref1/2 instead of Re.
 black slanted lines

10. Moody diagram (cont’)

11. Summary for friction head loss
Darcy-Weisbach equation
Friction head loss is proportional to;
pipe length (L), 1/pipe diameter (1/D)
resistance (friction) coeff. (f)
1/Re for laminar flow
velocity (v2)
independent of pressure
Class quiz) Water (20oC) flows at a rate of 0.05 m3 in a 20-cm asphalted cast-iron pipe. What is the head loss per kilometer of pipe? (Ex 10.4)
Class quiz) The head loss per kilometer of asphalted cast-iron pipe (20-cm dia.) is 12.2 m. What is discharge of water? (Ex 10.5)

12. In a practical situation, we’d like to compute the discharge of a particular pipe.
→ We don’t know V, so that we don’t know Re.
→ 1. Assume an appropriate value of f, and then compute V from energy eqn. and compute Re
2. Find a new f1 from the Moody diagram and compute V and Re again
3. Find a f2 from the Moody diagram again
4. Repeat such calculations until f1  f2
→ select V at this step and compute Q = AV
Class quiz) How much is the discharge of water through the pipe shown below?

13. 2. Head loss due to transitions and fittings/valves
General form
Pipe inlet
Bend (elbow)
Contraction
Expansion
Pipe fittings/valves

14. 2. Head loss due to transitions and fittings/valves

15. Class quiz) Oil (=410-5 m2/s, S=0
Class quiz) Oil (=410-5 m2/s, S=0.9) flows in a smooth pipe as shown below (Q=0.028 m3/s, D=0.15m). What is elevation of the oil surface in the upper reservoir? (Ex 10.9)

16. EGL/HGL in a conduit (including all losses)
Example 1 – Head loss in entrance
Example 2 – All losses in a conduit delivering a fluid between two reservoirs
Commonly, local gradual changes are simplified as abrupt changes.

17. Class quiz) What is the elevation of the water surface in the upper reservoir if Q=0.15m3/s? and draw the EGL and HGL (Prob 10.97)

18. Open channel flow (uniform flow) – 개수로에서의 등류
Flow surface is open to the atmosphere
(piezometric head) = depth of flow (수심) + elevation of flow bottom
(하상높이 = y + z)
HGL (동수경사선) = flow surface
Uniform flow → channel shape, depth of flow, velocity along the channel (flow direction) are the same (velocity across the channel section varies)
→ slope of HGL = channel slope
Artificial and natural open channels

19. Geometric elements of cross section of open channels
Depth – distance btw. flow surface and channel bottom
Wetted perimeter (P, 윤변) – perimeter of channel cross-section with which the fluid flow in contact
Wetted area (A, 흐름단면) – cross-sectional area of an open channel flow
Hydraulic radius (Rh, 경심) – A/P
Example) Rh for circular pipe 100% full / half full, rectangular channel (B  y) and trapezoidal channel B B D y
1:1 y

20. Friction head loss in open channels
→ same as one in a closed channel (Darcy-Weisbach equation)
4Rh used in open channels as D used in a circular pipe (closed channel)
 Relative roughness = ks/4Rh, Reynolds number =
 Reynolds number criterion for open channels is similar to the case of closed channels
if Re(= ) < 2000, laminar flow
if Re > 3000, turbulent flow
Reynolds number for open channels, Reopen ≡
if Reopen < 500 , laminar flow
if Reopen Re >750, turbulent flow
Class quiz) Water (60oF) flows in a 10-ft wide rectangular channel at a depth of 6 ft and a velocity of 0.1 ft/sec. What is the Re? and What is the max. depth that keeps the flow laminar? (Ex 10.14)

21. Empirical equations for the determination of discharges (velocities) in an open channel uniform flow (constant depth)
1. Chezy equation
From the Darcy-Weisbach equation,
Chezy coeff. (C = f(ks/D, Re)) can be obtained from Moody diagram.
Class quiz) How much is the discharge of water (60oF) that flows through a rectangular concrete channel (10-width, 6-ft flow depth, channel slope, ks=0.005ft, Ex 10.15)

22. Empirical equations for the determination of discharges (velocities) in an open channel uniform flow (constant depth)
2. Manning’s equation
The usual design formula for C in the SI system,
Class quiz)
1. What is the discharge in the concrete, unfinished? – 10-ft width, 6-ft flow depth, channel slope – (Ex 10.18)
2. What is the Chezy coefficient if Manning’s n is for a 4 cm-D pipe (full flow)?
3. What is the appropriate channel slope for a 0.4 m-D concrete (troweled, half full) circular channel to have 100 l/sec discharge?

23. 3. Hagen-Williams’ equation
For a relatively large diameter pipe (D>5cm) and slow flow (V3m/s) – water supply systems,
where CHW: Hagen-Williams coefficient (150 for a smooth pipe, 80 for a rough pipe)
관재료
CHW
석면 시멘트
140 납 청동
130 – 140
플라스틱 벽돌
100 강철
140 – 150 주철
130
리벳강철
110
콘크리트, 강철
64 – 83 주석
거푸집
나무관
120
합판 거푸집 구리
아연철 토관 유리

24. Best hydraulic section (최량수리단면)
According to Manning’s equation, Q  ARh2/3 (=A5/3/P2/3). For a given cross-sectional area, A, the minimum P provides the maximum discharge. A cross-section that has a minimum P → the best hydraulic section.
Rectangular channel – depth : width = 1:2
Trapezoidal channel – half of a hexagon
Circular channel – half circle
Triangular channel – half of a square
Class quiz) Determine the best hydraulic section for a rectangular channel (Ex 10.19) B y

25. Pump in a pipe (closed channel)
Pumps (commonly used)
1. Radial flow pump (반경류/방사류 펌프) – centrifugal pump (원심 펌프), the most commonly used in water treatment systems, flow in the direction of rotational axis and flow out in radial direction
2. Axial flow pump (축류 펌프) – good for low heads and high rate flows, flow in/out in the axial direction
3. Mixed pump (혼류형 펌프) – flow out in both radial and axial direction
Radial flow pump
Axial flow pump

26. Pump in a pipe (closed channel)
Create an enough piezometric head (including head losses) to deliver a fluid from low elevation to high
From the energy equation,
For a system with one size pipe, the pump head (양정),
 The required head produced by a pump = elevation difference + all head losses
Required pump power

27. The optimized (the most efficient) operating point (operating discharge, Q)
1. System curve (system head-discharge curve)
Total head (pump head required in a given system) in a pumping system = f (discharge, Q (=AV))  Q → hp  (from energy equation)
2. Pump curve (pump head-discharge curve)
Pump head (produced by pump) = f (discharge, Q (=AV))  Q → hp 
Provided by pump manufacturers (via pumping test)
At junction point where two curves meet → “operating discharge, Q  the point where the head produced by the pump is just the amount needed to overcome the head loss in the pipe
This allows us to select or operate a pump in an efficient and economic way

28. Class quiz) Find an optimal discharge considering the pump characteristics shown in the previous slide. (Ex 10.11)

29. Net positive suction head (NPSH, 순양흡입수두)
The pressure decreases at the suction side of a pump with increasing flow velocity (Bernoulli effect), which may result in a significant pressure drop, “cavitation”.
NPSH = pressure head difference btw. the suction side of a pump and the vapor pressure head of a fluid pumped
From energy equation btw. points 1 and 2 (suction side),

30. Net positive suction head (NPSH)
The critical NPSH of a pump – determined by pump manufacturers via pumping test
If NPSH < critical NPSH → cavitation
Typically, critical NPSH values for centrifugal pumps ~ 10 ft
In the application of a given pump, if discharge (Q=AV), suction height (H), or head loss,→ the possibility of cavitation
Class quiz) The pump below delivers 2cfs flow of 80oC water and the intake pipe diameter is 8 in. The pipe length bet. pipe inlet and pump intake is 10 ft. The pump intake is located 6 ft above the water surface level in the reservoir. What is the NPSH for this system? (entrance loss coeff.=0.1, bend loss coeff.=0.2, f=0.015, Ex 14.7)