Transcript Lecture 6. Fluid Mechanics MARI-5590 Aquatic System Design Dr. Joe M. Fox

Lecture 6. Fluid Mechanics
MARI-5590
Aquatic System Design
Dr. Joe M. Fox
Topics Covered
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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.
Fluid Statics
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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/ft3) or Newtons (N, kg/m3)
F = normal forces acting on an area (lbs or kgs)
A = area over which the force is acting (ft2 or m2)
Fluid Statics
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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/m3), V = volume (m3)
h
P1
P = h
h = depth of water
(m or ft)
Fluid Statics
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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
80oF vs. 40oF?
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80oF   = 62.22 lb/ft3; thus, P = (62.22)(12) = 746.4 lb/ft 2
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40oF   = 62.43 lb/ft3; thus, P = (62.43)(12) = 749.2 lb/ft2
Fluids in Motion
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Fundamental equation:
Qin – Qout =  storage
Qin = quantity flowing into the system; Qout = that flowing out; the
difference is what’s stored
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If we divide  storage by a time interval (e.g., seconds), we
can determine rate of filling or draining
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Very applicable to tanks, ponds, etc.
Problem: A 100,000 m3 pond (about 10 ha) is continuously filled with
water from a distribution canal at 100 m3 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 m3 per minute,
how long will it take to be essentially empty?
Volume/flow rate = 100,000 m3/200 m3/min = 500 min
Closed System Fluids in Motion
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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 Q1 = Q2, a mass balance
equation
For essentially incompressible fluids such as water, the
equation becomes V1A1 = V2A2,; where V = velocity (m/s)
and A = area (m2)
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. V2 = 0.5 m/s
Bernoulli’s Equation
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Z1 + (P1/) + (V12/2g) = Z2 + (P2/) + (V22/2g)
Wow! Z = pressure head, V2/2g = 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 =
V2/2g
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?
Reality
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In actuality, fluids have losses due to
friction in the pipes and minor losses
associated with tees, elbows, valves, etc.
Also, there is usually an external power
source (pump). The equation becomes
Z1 + (P1/) + (V12/2g) + EP = Z2 + (P2/) + (V22/2g) + hm + hf
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If no pump (gravity flow), EP = 0. EP is energy from the
pump, hm and hf = minor and frictional head losses, resp.
Minor Losses
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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.)
changes in pipe size (reducers, expanders)
losses associated with fluid entering or leaving a pipe
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Screens, foot valves also create minor losses
A loss coefficient, K, is associated with each
component
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total minor losses, hm, = K(V2/2g)
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Minor Loss Coefficients
Your Inevitable Example
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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: hm = 0.15 ft
hm = (0.9+1.15+0.4)(2)2
(2)(32.2)
Pipe Friction Losses
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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
(
VD , 

D
)
Where, f = friction factor; D = inside pipe diameter; V =
fluid viscocity;  = absolute roughness;  = fluid
density; and  = absolute viscocity
Pipe Friction Losses
VD ,/D

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Is known as the Reynold’s number, RN, also
written as VD/v
Simplified, f = 64/RN
/D
Is called the relative roughness and is the
ratio of the absolute roughness to inside pipe diameter
Moody’s Diagram (Reynold’s
Number vs. Relative Roughness)
Absolute Roughness Coefficients
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
Darcy-Weisbach Equation
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hf = f(L/D)(V2/2g)
Where hf = 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/s2 or ft/s2)
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 , hf, using the Darcy-Weisbach Equation
Ans. 
Answer to Previous
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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
hf = (.0027)(500)(1.5)2 = 15.5 m
(0.1)(2)(9.81)
Reality
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This value, hf is added to hm to arrive at your total
losses
Alternative method for frictional losses: HazenWilliams equation
hf = (10.7LQ1.852)/(C1.852)(D4.87) metric systems
hf = (4.7LQ1.852)/((C1.852)(D4.87) English systems
Where hf = pipe friction losses (m, ft); L = length
of piping (m, ft); Q = flow rate (m3/s, ft3/s); C =
Hazen-Williams coefficient; and D = pipe
diameter (m, ft)
Hazen-Williams Values
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
Example
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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?
hf = (10.7)(145m)(0.014)1.852
(120)1.852(0.152)4.87
= 2.6 ft
OK, what about PUMPING?
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Pump’s performance is described by the following
parameters:
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Capacity
Head
Power
Efficiency
Net positive suction head
Specific speed
Capacity, Q, is the volume of water delivered per
unit time by the pump (usually gpm)
Pump Performance
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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
Hs = SL + DL + DD + hm + hf + ho + hv
Hs = system head, SL = suction-side lift, DD = water
source drawdown, hm = minor losses (as previous), hf =
friction losses (as previous), ho = operating head pressure,
and hv = velocity head (V2/2g)
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, hm, hf, ho and hv all increase with increased
pumping capacity, Q
Pump Performance: power
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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)
Pump Performance: efficiency
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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
Ep = 100(WHP/BHP) = output/input
Ep 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, Ep is lower.
Pump Performance: suction
head
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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 (Hs) = SL + friction losses + velocity
head:
Hs = SL + (hm + hf) + V2s/2g
Pump Performance Curves
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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
Characteristic Pump Curves
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Head  as capacity 
Efficiency  as
capacity , up to a
point
BHP  as capacity ,
also up to a point
REM:
BHP =
100QHS/Ep3,960