ce 382, hydraulic systems design (pipes, pumps and open channels)

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CE 382, Hydraulic Systems Design (pipes, pumps and open channels) Principles of hydraulics 1. Conservation of energy 2. Continuity (conservation of mass) 3. Momentum (balance of forces)

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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 Presentation

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Page 1: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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)

Page 2: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

What is conservation of energy

Energy

P/ +v2/2g +Z

E1 = E2+ hL (Bernullie equation)

hL = hf + hm

Page 3: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 4: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

What is conservation of mass continuity?

A1. V1 = A2. V2

Q1 = Q2

Page 5: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

How to calculate hf?

g

V

D

Lfhf

2.

. 2

Page 6: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

hL= hf+ hm

hL = head loss

hf = friction loss

hm = minor loss

Page 7: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

Other equations to calculate head loss

1.Darcy-Weisbach, D.W2.Manning3.Hazen-Williams, H-W

Page 8: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

Minor loss equation

hm = k. v2/2g

Page 9: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

Where does minor loss occur?

1. Valves2. Transition points3. Changes in velocity, direction or shape4. Change in flow line

Page 10: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 11: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 12: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)
Page 13: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 14: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

NR =V.D/

NR = Reynolds numberV = velocity, L/T

D= Inside Diameter, L= kinematic viscosity, L2/T

Page 15: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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.

Page 16: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 17: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)
Page 18: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 19: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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

Page 20: CE 382,  Hydraulic  Systems Design (pipes, pumps and open channels)

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