ELEMENTARY FLUID MECHANICS REVJEW

The subject offluid mechanics is usually divided into fluid statics and fluid dynamics. In static

fluids, the only force exerted by the fluid is a pressure force~ nonnal to the surface. The basic equation

is F == f area pd~ and the pressure may be found by the fundamental equation of hydrostatics:

dp = .:ydz. Fluid dynamics problems are solved using con!;ervation of mass, energy, and momentum.

Elementary fluid mechanics problems usually deal only with incompressible fluids in either the

static or the steady, one-dimensional flow condition. There are five general categories of problems:

three types of forces (hydrostatic, dynamic, and viscous) and two types of flow (confined and

unconfined). Many complicated problems involve several categories.

Almost all equations found in fluid mechanics te~:tbooks and handbooks were derived using

certain simplifying assumptions. Do not use an equation unless you are sure that in making these

assumptions you are not introducing an intolerable error to the accuracy of your answer. Check the

answer to insure the assumption was valid. For example, if you use a laminar flow equation to find

the velocity in a pipe, check the Reynolds' number using that velocity to be sure that it is within the

laminar range.

HYDROST AllC FORCES AND PRESSURES

A For incompressible fluids, the fundamental equation (dp = -ydz) simplifies to:

p + yz = constant.

B.

The only force exerted by a static fluid is a pressure force normal to the surface:

F = f area pdA = volume of pressure prism. It'the fluid is incompressible, the force on a

1

2

submerged plane area is equal to (pressure at the centroid) x (area) and is located below the centroid

a distance of

I c

yAe =

The horizontal component of the force on a submerged curved surface is equal to the forcec.on the area projected on a vertical plane.

The vertical component of the force on a submerg(~d curved surface is equal to the weight of

D.

the fluid (real or imaginary) above the surface.

Draw a free body diagram and use the basic equations of static equilibrium: ~F = 0 andE.

~M=O.

The bt;i°yant force on a body is equal to the wei:ght of the volume of fluid displaced by theF.

body.

Any manometer problem can be worked by using these 2 rules:G.

At a given elevation the pressure is the same at all points (if fluid iscontinuous and static).

Pressure increases by y ~ z as you go down and decreases by y ~ z as you go

up.

Note: Sometimes the problem is more easily handled using feet of water as the unit

of pressure and rule (2) becomes: Pressure ~ in feet of water increases by

s~z as you go down and decreases by S~z as you go up. (S is the specific

gravity of the fluid through which you are moving.)

3

VISCOUS FORCES

A. Viscous shear force: F = A-r = AJ.L(dJl!dy). If the distance is small, we assume

d'fl!dy = J1. 'fl! J1.y.

Viscous drag force is usually calculated by empiJical equations such as F = Y2CdpAu2,

B.

DYNAMIC FORCES

Momentum equation: the net external force Q!1 an isolated fluid volume is equal to theA

increase in momentum flux through the volume:

Forces and velocities are vectors, but density and flow rate are scalars.B

For the dynamic force exerted Q.y ajet stream (\1'1):c.

(1) On a single vane moving at velocity U:

(2) On a series of vanes moving at velocity U:

Relative velocities are always tangent to the vane.

D.

For ideal fluids, Vl = V2 (in magnitude).

E.

5

OPEN CHANNEL FLOW

A The continuity. energy and power equations are used as above. but the cross sectional area

of the flow is not so easily obtained. We usually g(~t a cubic equation which has two positive

roots for the depth of flow.

B. Empirical equation for head loss vs. velocity:

1.49n

R213 81/2Manning: V = , whl~re

v = velocity (£l:/s)

R = hydraulic radius = Area I Wetted Pelimeter (ft)

S = slope of energy grade line (channel slope for uniform flow)

n = roughness coefficient

A closed tank contains compressed ai:t, and oil (SGoil =: 0.90) asshown in the figure. AU-tube mahomete2:' using mercury (SGH9 = 13.6)is connected to the tank. For column hE~ights h, = 36 in, hz = 6 in,and h3 = 9 in, determine the pressure re~ading (J.n psi) of the gage.

~ h~~= 0-f- -

y"~I

0-

~I'"-

~Q!,

~

7JQ~'1e.. -:: 3. O(P ?S; ~s.

If gate AS is rectangular and is two meters wide (normal to thepage) I what force F is needed to open the gate? Region C containsair at atmospheric pressure. The water surface lies 2 m above thehinge at Band t is 5 M.

s", / oS';' 1.r 0 = 707"

~a.te.-h

- 4.5" ~

Fw =

0;;:07"2-

..2""--

~,n 'Ii.

='J-G. I- ~ .3,~ n1::

If--

'c. AfG,.3 ~,.., - J: OZ rr}=

(C;.

~ ""g= 0

The curved surface AB is 6 ft long perpendicular to the plane ofthe figure. Determine (a) the resultant horizontal force on AB,its direction and line of action and (b) the resultant verticalforce on AB, its direction and line of action.

f7

4'

!_-5£A wItT6'.£.

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J:c..-=--h<;.A

h = s. 0 7 .{-f:"P A,~.ht.. 5/1-f::

f (st+J (z. F+-) (r, 1=1-)

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Fred

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t( == 0 .7& II rlJlf 0 Ji__l11io

The partially submerged wood pole is attached to the wall by ahinge as shown. The pole is in equilib:rium under the action of theweight and buoyant forces. Determine the density of the wood.

p = Il... A r--H 03 ~

L..(f L.. tD'~ 30. )

--

s-

/8 ~oto-

s-l:f ~i)

1-

- -r woo~

Y...;-~

.---/J, =- 55_I!;I 5 Jq //"11- 11- _I 3

f ko/' -.? :::> ,.)..::J .~~. /"11 ,.~ '-

A horizontal 150 rom pipe, in which 62 lis of water are flowing,contracts to a 75 rom diameter. If the pressure in the 150 mm pipeis 275 kpa, calculate the magnitude and direction of the horizontalforce exerted on the contraction.

= O.or--z. ,..,I/sq

.o,z.. m't' 3. ~-/ ",/s~ : --

.°'

z..11.0 "'/$y =.

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=

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I..

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(/82.3 XIOJ lI/m&) ( it (:~7~M)~1.tf

=-

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=. ~S-O.S' /'I

F = 3'/otf N A,,~

Water is flowing in the system shown.a flowrate of 28 lis.

Calculate the pump power for

~o~

2..

1;~r-r

~# f -Z =-I

'i"f £p

30 r71= ~ f -lz-'Z-:j

~~...~

=. GO (?'1~ f ~'l-r.: 30,S-2. ~£ = '0,.. -30,., + O. ~.\-n') -0./1. 9...,

f"

(O,Ol8 ,.,.I/s)('ir/tJ ;Al/",1 ) (10, S-l.fPJ)1000

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rt-~./~ ff,)~ ('2..cJ /J./r-t)-!

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~=~ 170 ~ S" 1'/.'iI L-

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.2..9' /,~'/2: f:'1 =-J:> :::

"18.~0( =

"F = 935 /6 ~~~ /8.$ ~s.

-( ( .z'1~),-t4n ""iU =

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Is-h =. 57'-./

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