MARI-5590
Aquatic System Design
Dr. Joe M. Fox

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Fluid statics
Pressure measurement
Fluids in motion
Pump performance parameters
Note: most of the lecture comes from Lawson, T.B.,
1995.

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Fluid statics : study of fluids at rest
Different from fluid dynamics in that it concerns pressure forces perpendicular to a plane (referred to as hydrostatic pressure )
If you pick any one point in a static fluid, that point is going to have a specific pressure intensity associated with it:
P = F/A where
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P = pressure in Pascals ( Pa, lb/ft 3 ) or Newtons ( N , kg/m 3 )
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F = normal forces acting on an area (lbs or kgs)
A = area over which the force is acting (ft 2 or m 2 )

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This equation, P = F/A, can be used to calculate pressure on the bottom of a tank filled with a liquid (or.. at any depth)
F =

V

= fluid specific wt
(N/m 3 ), V = volume (m 3 ) h
P
1
P =
 h h = depth of water
(m or ft)

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Pressure is the same at all points at equal height from the bottom of the tank
Point: temp doesn’t make that much difference in pressure for most aquaculture situations
Example: What is the pressure at a point 12 ft. from the bottom of a tank containing freshwater at
80 o F vs. 40 o F?
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80 o F
 
= 62.22 lb/ft 3 ; thus, P = (62.22)(12) = 746.4 lb/ft 2
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40 o F
 
= 62.43 lb/ft 3 ; thus, P = (62.43)(12) = 749.2 lb/ft 2

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Fundamental equation:
Q in
– Q out
=
 storage
Q in
= quantity flowing into the system difference is what’s stored
; Q out
= that flowing out; the
If we divide
 storage by a time interval (e.g., seconds), we can determine rate of filling or draining
Very applicable to tanks, ponds, etc.
Problem: A 100,000 m 3 pond (about 10 ha) is continuously filled with water from a distribution canal at 100 m 3 per minute. Assuming that the pond was initially full, but some idiot removed too many flashboards in the exit gate and it was draining at 200 m 3 per minute, how long will it take to be essentially empty?
Volume/flow rate = 100,000 m 3 /200 m 3 /min = 500 min

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Let’s say we’re not dealing with a system open to the atmosphere (e.g., a pipe vs. a pond)
There’s no storage potential, so
Q
1 equation
= Q
2
, a mass balance
For essentially incompressible fluids such as water, the equation becomes V and A = area (m 2 )
1
A
1
= V
2
A
2
,; where V = velocity (m/s)
Can be used to estimate flow velocity along a pipe, especially where constrictions are concerned
Example: If one end of a pipe has a diameter of 0.1 m and a flow rate of 0.05 m/s, what will be the flow velocity at a constriction in the other end having a diameter of 0.01 m?
Ans. V
2
= 0.5 m/s

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Z
1
+ ( P
1
/
 ) + (V
1
2 /2g) = Z
2
+ ( P
2
/
 ) + (V
2
2 /2g)
Wow! Z = pressure head, V 2 /2g = velocity head
(heard of these?), 2g = (2)(32.2) for Eng. System
If we’re trying to figure out how quickly a tank will drain, we use this equation in a simplified form: Z =
V 2 /2g
Example: If the vertical distance between the top of the water in a tank and the centerline of it’s discharge pipe is 14 ft, what is the initial discharge velocity of the water leaving the tank? Ans. = 30 ft/s
Can you think of any applications for this?

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In actuality, fluids have losses due to friction in the pipes and minor losses associated with tees, elbows, valves, etc.
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Also, there is usually an external power source (pump). The equation becomes
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Z
1
+ ( P
1
/
 ) + (V
1
2 /2g) + E
P
= Z
2
+ ( P
2
/
 ) + (V
2
2 /2g) + h m
+ h f
If no pump (gravity flow), E
P pump, h m and h f
= 0. E
P is energy from the
= minor and frictional head losses, resp.

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These are losses in pressure associated with the fluid encountering:
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• restrictions in the system (valves) changes in direction (elbows, bends, tees, etc.)
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• changes in pipe size (reducers, expanders) losses associated with fluid entering or leaving a pipe
Screens, foot valves also create minor losses
A loss coefficient, K, is associated with each component total minor losses, h m
, =

K(V 2 /2g)

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Calculate the total minor losses associated with the pipe to the right when the gate valve is ¾ open, D =
6 in., d = 3 in. and V =
2ft/s
Refer to the previous table
Ans: h m
= 0.15 ft h m
= (0.9+1.15+0.4)(2) 2
(2)(32.2)

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Caused by friction generated by the movement of the fluid against the walls of pipes, fittings, etc.
Magnitude of the loss depends upon:
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Internal pipe diameter
Fluid velocity
Roughness of internal pipe surfaces
Physical properties of the fluid (e.g., density, viscocity) f = function
( )
Where, f = friction factor; D = inside pipe diameter; V = fluid viscocity; density; and

= absolute roughness;

= absolute viscocity

= fluid


VD
,

/D
Is known as the Reynold’s number, RN, also written as VD/v
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Simplified, f = 64/RN

/D Is called the relative roughness and is the ratio of the absolute roughness to inside pipe diameter

Pipe Material
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel
Drawn tubing
PVC
Absolute Roughness (in.)
.036-.358
.012-.122
.007-.035
.010
.0059
.0018
.000059
.00000197

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• h f
= f(L/D)(V 2 /2g)
Where h f
= pipe friction head loss (m/ft); f = friction factor; L = total straight length of pipe
(m/ft); D = inside pipe diameter (m/ft); V = fluid velocity (m/s or ft/s); g = gravitational constant
(m/s 2 or ft/s 2 )
Problem: Water at 20 C is flowing through a 500 m section of 10 cm diameter old cast iron pipe at a velocity of 1.5m/s. Calculate the total friction losses , h f
, using the Darcy-Weisbach Equation
Ans.


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RN = VD/

; where
 or kinematic viscocity is 1 x 10 -6 (trust me on this)
RN = (1.5)(0.1)/.000001 = 150,000

= .026 (in cm) for cast iron pipe;

/D =
.00026 m/.1 = .0026
f = 0.027 where on Moody’s Diagram 
/D aligns with a Reynold’s Number of 150,000 h f
= (.0027)(500)(1.5) 2 = 15.5 m
(0.1)(2)(9.81)

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This value, h f losses is added to h m to arrive at your total
Alternative method for frictional losses: Hazen-
Williams equation h f h f
= (10.7LQ
1.852
)/(C 1.852
)(D 4.87
) metric systems
= (4.7LQ
1.852
)/((C 1.852
)(D 4.87
) English systems
Where h f
= pipe friction losses (m, ft); L = length of piping (m, ft); Q = flow rate (m 3 /s, ft 3 /s); C =
Hazen-Williams coefficient; and D = pipe diameter (m, ft)

Pipe Material
Asbestos cement
Concrete (average)
Copper
Fire hose
Cast iron (new)
Cast iron (old)
PVC
Steel (new)
C
140
130
130-140
135
140
40-120
150
120

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Estimate the friction losses in a 6-in. diameter piping system containing 200 ft of straight pipe, a half-closed gate valve, two close return bends and four ell90s. The water velocity in the pipe is 2.5 ft/s?
h f
= (10.7)(145m)(0.014) 1.852
(120) 1.852
(0.152) 4.87
= 2.6 ft

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Pump’s performance is described by the following parameters:
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Capacity
Head
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Power
Efficiency
Net positive suction head
Specific speed
Capacity , Q, is the volume of water delivered per unit time by the pump (usually gpm)

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Head is the net work done on a unit of water by the pump and is given by the following equation
H s
= SL + DL + DD + h m
+ h f
+ h o
+ h v
H s
= system head, SL = suction-side lift, DD = water source drawdown, h m
= minor losses (as previous), h f friction losses (as previous), h and h v
= velocity head (V 2 /2g) o
=
= operating head pressure,
Suction and discharge static lifts are measured when the system is not operating
DD, drawdown, is decline of the water surface elevation of the source water due to pumping (mainly for wells)
DD, h m
, h f
, h o and h v pumping capacity, Q all increase with increased

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Power to operate a pump is directly proportional to discharge head, specific gravity of the fluid (water), and is inversely proportional to pump efficiency
Power imparted to the water by the pump is referred to as water horsepower
WHP = QHS/K; where Q = pump capacity or discharge, H
= head, S = specific gravity, K = 3,960 for WHP in hp and
Q in gpm.
WHP can also equal Q(TDH)/3,960 where TDH = total dynamic head (sum of all losses while pump is operating)

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Usually determined by brake horsepower (BHP)
BHP = power that must be applied to the shaft of the pump by a motor to turn the impeller and impart power to the water
E p
= 100(WHP/BHP) = output/input
E p never equals 100% due to energy losses such as friction in bearings around shaft, moving water against pump housing, etc.
Centrifugal pump efficiencies range from 25-85%
If pump is incorrectly sized, E p is lower.

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Conditions on the suction side of a pump can impart limitations on pumping systems
What is the elevation of the pump relative to the water source?
Static suction lift (SL) = vertical distance from water surface to centerline of the pump
SL is positive if pump is above water surface, negative if below
Total suction head (H s
) = SL + friction losses + velocity head:
H s
= SL + (h m
+ h f
) + V 2 s
/2g

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Report data on a pump relevant to head, efficiency, power requirements, and net positive suction head to capacity
Each pump is unique dependent upon its geometry and dimensions of the impeller and casing
Reported as an average or as the poorest performance

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Head
 as capacity
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Efficiency
 as capacity
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, up to a point
BHP
 as capacity

, also up to a point
REM:
BHP =
100QHS/E p
3,960