Lecture 6. Fluid Mechanics

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Lecture 6. Fluid Mechanics

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

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Topics Covered

• 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

• 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
• 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)

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Fluid Statics

• 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
P ?h h depth of water (m or
ft)
P1
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Fluid Statics

• Pressure is the same at all points at equal
  height from the bottom of the tank
• Point temp doesnt 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?
• 80oF ? ? 62.22 lb/ft3 thus, P (62.22)(12)
  746.4 lb/ft 2
• 40oF ? ? 62.43 lb/ft3 thus, P (62.43)(12)
  749.2 lb/ft2

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Fluids in Motion

• Fundamental equation
• Qin Qout ? storage
• Qin quantity flowing into the system Qout
  that flowing out the difference is whats stored
• 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 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

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Closed System Fluids in Motion

• Lets say were not dealing with a system open to
  the atmosphere (e.g., a pipe vs. a pond)
• Theres 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

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Bernoullis Equation

• 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 were 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
  its 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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Reality

• 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
• If no pump (gravity flow), EP 0. EP is energy
  from the pump, hm and hf minor and frictional
  head losses, resp.

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Minor Losses

• These are losses in pressure associated with the
  fluid encountering
• 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
• Screens, foot valves also create minor losses
• A loss coefficient, K, is associated with each
  component
• total minor losses, hm, ?K(V2/2g)

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Minor Loss Coefficients
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Your Inevitable Example

• 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 (0.91.150.4)(2)2

(2)(32.2)
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Pipe Friction Losses

• Caused by friction generated by the movement of
  the fluid against the walls of pipes, fittings,
  etc.
• Magnitude of the loss depends upon
• 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
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Pipe Friction Losses

• Simplified, f 64/RN

Is known as the Reynolds number, RN, also
written as VD/v
?VD
,
?/D
?
?/D Is called the relative roughness and is
the ratio of the absolute roughness to inside
pipe diameter
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Moodys Diagram (Reynolds Number vs. Relative
Roughness)
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Absolute Roughness Coefficients
Pipe Material Absolute Roughness (in.)
Riveted steel .036-.358
Concrete .012-.122
Wood stave .007-.035
Cast iron .010
Galvanized iron .0059
Commercial steel .0018
Drawn tubing .000059
PVC .00000197
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Darcy-Weisbach Equation

• 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. ?

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Answer to Previous

• 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 Moodys Diagram ?/D aligns
  with a Reynolds Number of 150,000
• hf (.0027)(500)(1.5)2 15.5 m

(0.1)(2)(9.81)
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Reality

• This value, hf is added to hm to arrive at your
  total losses
• Alternative method for frictional losses
  Hazen-Williams 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)

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

• 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
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OK, what about PUMPING?

• Pumps performance is described by the following
  parameters
• 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)

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Pump Performance

• 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

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Pump Performance power

• 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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Pump Performance efficiency

• 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.

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Pump Performance suction head

• 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
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Pump Performance Curves

• 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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Characteristic Pump Curves

• Head ? as capacity ?
• Efficiency ? as capacity ?, up to a point
• BHP ? as capacity ?, also up to a point
• REM
• BHP 100QHS/Ep3,960