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Lecture 6. Fluid Mechanics

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

Topics Covered

- Fluid statics
- Pressure measurement
- Fluids in motion
- Pump performance parameters

Note most of the lecture comes from Lawson,

T.B., 1995.

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)

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

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

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

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

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?

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.

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)

Minor Loss Coefficients

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)

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

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

Moodys Diagram (Reynolds Number vs. Relative

Roughness)

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

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. ?

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)

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)

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

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

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)

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

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)

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.

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

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

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

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