ce 382, hydraulic systems design (pipes, pumps and open channels)
DESCRIPTION
CE 382, Hydraulic Systems Design (pipes, pumps and open channels). Principles of hydraulics Conservation of energy Continuity (conservation of mass) Momentum (balance of forces). What is conservation of energy. Energy P/ +v2/2g +Z E1 = E2+ hL (Bernullie equation) hL = hf + hm. - PowerPoint PPT PresentationTRANSCRIPT
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CE 382, Hydraulic Systems Design
(pipes, pumps and open channels)
Principles of hydraulics
1. Conservation of energy2. Continuity (conservation of mass)3. Momentum (balance of forces)
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What is conservation of energy
Energy
P/ +v2/2g +Z
E1 = E2+ hL (Bernullie equation)
hL = hf + hm
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The complete form of Bernullies equation
E1 = E2 + hL- hp +ht
hL = head loss = sum of friction loss +minor losses
hp = head produced by a pump
ht =Head taken out by turbine
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What is conservation of mass continuity?
A1. V1 = A2. V2
Q1 = Q2
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How to calculate hf?
g
V
D
Lfhf
2.
. 2
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hL= hf+ hm
hL = head loss
hf = friction loss
hm = minor loss
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Other equations to calculate head loss
1.Darcy-Weisbach, D.W2.Manning3.Hazen-Williams, H-W
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Minor loss equation
hm = k. v2/2g
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Where does minor loss occur?
1. Valves2. Transition points3. Changes in velocity, direction or shape4. Change in flow line
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A
B
C
Elev. A= 120 ftElev. B= 115 ftElev. C = 108 ftPipe B-C: 6 inch PVCL= 1000 ft
How much water will flow to point C?
If you want to reduce the flow, what would you do?
Draw the EGL
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A
B
C
Elev. A= 120 ftElev. B= 115 ftElev. C = 108 ftPipe B-C: 6 inch PNCL=1000 ft
How much water will flow to point C?
If you want to reduce the flow, what would you do?
Draw the EGLE1=120
E2= v2/2g
EGL
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Calculating Reynolds number
...
ReDV
NR
= density of water Mass per unit volume
V= Velocity of flow
D = diameter
µ = Dynamic viscosity lb.s/ft2 or N.M/m2
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NR =V.D/
NR = Reynolds numberV = velocity, L/T
D= Inside Diameter, L= kinematic viscosity, L2/T
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Values of Viscosity for Water
At 70 F, µ = 2.037 x 10-5 lb.s/ft2 or 1.002 x10-3 N.S/m2
At 70 F, = 1.05 x 10-5 ft2/sec or 1.006 x 10-6 m2/sec.
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How to Calculate f?
Example:
Pipe: Commercial steel, newID= 6 inch =0.5 ft
V= 8.6 ft/s =1.2x10^-6 ft2/se = 0.00015 fte/D = 3x10^-4= 0.0003NR= (V.D)/ = 3.67x10^6
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A
B
C
Elev. A= 120 ftElev. B= 115 ftElev. C = 108 ftPipe B-C: 6 inch steelf = 0.02
How much water will flow to point C?
If you want to reduce the flow, what would you do?
Draw the EGL
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E1 = E2 +(f.L/D).V2/2g
0+0+120=0+V2/2g +108 +(f.L/D).V2/2g
12 = V2/2g [1+f.L/D)
Function = 12-V2/2g[1+f.L/D)
Solve for V
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What is a good number for V?
Assume v = 7 ft/s
NR = 3.5 x10^5f = 0.014
Function, F = -10
Assume a lower number, V = 5 ft/s
NR = 2.5x10^5
f = 0.015
Function, F = -0.03, good enough