Tech Production Fluidmechanics

7/28/2019 Tech Production Fluidmechanics

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Lecture 6. Fluid Mechanics

MARI-5590

Aquatic System DesignDr. 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 fromfluid dynamics in that it concerns

pressure forces perpendicular to a plane (referred toas hydrostatic pressure)

If you pick any one point in a static fluid, that pointis 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)

P1

h

F = V = fluid specific wt

(N/m3), V= volume (m3)

P = h h = depth of water

(m or ft)


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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 inpressure 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:

QinQout= storage

Qin = quantity flowing into the system; Qout= that flowing out; thedifference is whats stored

If we divide storageby a time interval (e.g., seconds), wecan determine rate of filling or draining

Very applicable to tanks, ponds, etc.

Problem: A 100,000 m

3

pond (about 10 ha) is continuously filledwith water from a distribution canal at 100 m3 per minute. Assumingthat the pond was initially full, but some idiot removed too manyflashboards 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 theatmosphere (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 anda flow rate of 0.05 m/s, what will be the flow velocity at aconstriction 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/) + (V2

2/2g)

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 ofthe water in a tank and the centerline of its discharge

pipe is 14 ft, what is the initial discharge velocity ofthe 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 thefluid 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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Your Inevitable Example

Calculate the total minor

losses associated with the

pipe to the right when the

gate valve is 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)


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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

VD

,Is known as the Reynolds number, RN, also

written as VD/v

/D Is called the relative roughness and is the

ratio of the absolute roughness to inside pipe diameter

/D


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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 .0059Commercial 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 = fluidvelocity (m/s or ft/s); g = gravitational constant(m/s2 or ft/s2)

Problem: Water at 20 C is flowing through a 500m section of 10 cm diameter old cast iron pipe at avelocity of 1.5m/s. Calculate the total frictionlosses , hf, using the Darcy-Weisbach Equation

Ans.


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Answer to Previous

RN = VD/; where or kinematic viscocityis 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 /Daligns 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 totallosses

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 = lengthof piping (m, ft); Q = flow rate (m3/s, ft3/s); C=Hazen-Williams coefficient; andD = pipediameter (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) 140Cast 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, twoclose 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 followingparameters:

Capacity

Head

Power

Efficiency

Net positive suction head

Specific speed

Capacity, Q, is the volume of water delivered perunit time by the pump (usually gpm)


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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 andimpart power to the water

Ep= 100(WHP/BHP) = output/input

Epnever equals 100% due to energy losses such as

friction in bearings around shaft, moving wateragainst 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) + V

2s/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 impellerand casing

Reported as an average or as the poorestperformance


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Characteristic Pump Curves

Head as capacity Efficiency as

capacity, up to apoint

BHP as capacity,also up to a point

REM: BHP =

100QHS/Ep3,960

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