Transcript Workshop 1: Signal Conditioning Circuit Design

Pumps and Lift
Stations
Background
Fluid Moving Equipment
Fluids are moved through flow systems using
pumps, fans, blowers, and compressors. Such
devices increase the mechanical energy of the
fluid. The additional energy can be used to
increase
• Velocity (flow rate)
• Pressure
• Elevation
Why are most
water towers
130 feet high?
Because a
column of water
2.31 ft high
exerts a
pressure of 1
psi, and most
city water
systems operate
at 50 to 60 psi
Static Head
Difference in
height between
source and
destination
destination
Static
head
source
Static
head
Independent of
Flow
Friction Head
Resistance to flow in
pipe and fittings
Depends on size, pipes,
pipe fittings, flow rate,
nature of liquid
Proportional to square
of flow rate
Friction
head
Flow
Net Positive Suction Head
The allowable limit to the
suction head on a pump
33 - friction losses through the pipe,
elbows, foot valves and other fittings
on the suction side of the pump
Power Requirement
Example
Determine the power
required to pump 2000 gpm
against a head of 14 ft, if
the efficiency of the pump
is
75% or 50%.
Power Requirement
Example
Lower efficiency means bigger motor and
electrical controls, higher operating
costs
Sump Sizing Example
DC = 0.375” per day
Area drained = 28 acres
Lift (from tile outlet in sump to
discharge pipe outlet) = 6 ft
Length of discharge pipe = 25 ft
Pump cycles per hour = 5 and 20.
Determine suitable sump size
Sump Sizing Example
Storage Volume (ft3) = 2 x Q (gpm)
n (cycles/hr)
Area of a Circle = 3.14 x (Diameter)2
4
.
Design Flowrate
Q = 18.9 x DC x A
Q = 18.9 x 0.375 x 28 = 198.5 gpm
Say use Q = 200
gpm
Sump Volume Calculation
(5 cycles/hr)
Vol = 2 x Q / n
Vol = 2 x 200 / 5 = 80
ft3
Min Required Volume = 80 ft3
-Try 4 ft diameter well
Area = 3.14 x 42 / 4 = 12.6 ft2
Storage Depth = 80 /12.6 = 6.35 ft
Try 6 ft diameter well
Area = p x 62 / 4 = 28.3 ft2
Storage Depth = 80 /28.3 = 2.83 ft
AFFINITY LAWS
Flowrate varies with rotational speed:
Q1/Q2 = N1/N2
Head varies with rotational speed
squared:
H1/H2 = (N1/N2)2
Power varies with rotational speed
cubed:
P1/P2 = (N1/N2)3
Affinity Laws Example
A pump with an efficiency of 80%,
connected to a diesel engine, pumps
200 gpm against a head of 12 ft. What
is the power output of the engine?
What will be the flow rate, head and
power output if the motor speed is
increased from 500 rpm to 600 rpm?
Affinity Laws Example
P =(200 x 12)/(3960 x 0.8)
= 0.76 hp
Q2 = 200 x (600/500) = 240 gpm
H2 = 12 x
2
(600/500)
= 17.3 ft
P2 = 0.76 x (600/500)3 = 1.3 hp