HYDROSTATICScourses.ce.metu.edu.tr/ce272/wp-content/uploads/sites/69/2018/02/... · The lifting force due to hydrostatic pressure distribution on submerged volumes is called as ‘buoyant

METU Civil Engineering Department CE 272 FLUID MECHANICS

2.01

HYDROSTATICS

Body Forces: External forces distributed over the mass of the

fluid developed without physical contact.

kBjBiBB zyx Body force

In case of gravity being the sole body force acting:

kmgkWkBB z

Surface Forces: Forces acting on any surface defined in the

fluid.

Stress at a point:

m

F

A

x

z Ax=ABO

Ay=BCO

Az=ACO

O A

B

kFjFiFF zyx

kAjAiAA zyx

A

Flim)(Stress

0A

y

C

Given that z and g

Inertia force )kajaia(mF zyxi


2.02

Stress Tensor:

yy

x

yz

yx

xy

xz

xx

zz

zy

zx

zzzyzx

yzyyyx

xzxyxx

Stress tensor is symmetrical

zyyz

zxxz

yxxy

z

y


2.03

Thermodynamic Pressure

Without shear stresses (=0): Consider a wedge-shaped fluid

element moving as a rigid body.

x

y

z

nn

yy

zz

y

z

x

s

sinsz

cossy

2/zyx

W

W

Newton’s second law of motion in y-direction

ynnyy asinsxzx

ynnyy a2

y

In the limit as y0 while remains constant nnyy

Similarly Newton’s second law in x and z directions would

give

nnxx nnzz and


2.04

That is

In the absence of shear stresses the normal stress in a fluid is

independent of orientation of the plane thus can be represented

by a single scalar quantity. Taking the fact that fluids can

sustain only compression, pressure p is set equal to negative of

this magnitude:

pzzyyxx

With shear stresses (0):

The bulk stress )(3

1zzyyxx

is independent of rotation of coordinate axes and known as

pressure

p

σxx = σyy = σzz = σnn


2.05

PRESSURE FORCE per UNIT VOLUME

The Gradient of Pressure:

Consider a pressure field acting on an infinitesimal frictionless

fluid element moving as a rigid body or at rest (=0)

y

z

x

dyy

pp

p

The net pressure force in y-direction is

dy dx

dz

d

y

pdxdydz

y

pdxdzdy

y

pppdF

Py

The force vector due to

pressure variations in

x, y and z directions

dk

z

pj

y

pi

x

pFd

P

The pressure force per unit volume

gradppfkz

pj

y

pi

x

p

d

FdP

P

kz

jy

ix

The Gradient operator


2.06

BODY (GRAVITY) FORCE per UNIT VOLUME

Governing Equation

Consider a fluid element at rest or moving as a rigid body (i.e.

with no relative motion).

no shear stresses

a scalar pressure distribution

forces due to pressure variations and gravity only

aamF

Dividing by volume afffgP

The general equation of motion is:

akp

)kajaia(kg)kz

pj

y

pi

x

p( zyx

xax

p

ya

y

p

)ga(

z

pz

(Euler equation)

d

dBz

kj0i0fd

Bdg

Total differential of p is dzz

pdy

y

pdx

x

pdp


2.07

Since a linear element on any surface is

kdzjdyidxds

then dsGdp

When p = constant on a surface (eg. free surface):

and the resultant acceleration vector is

k)ga(jaiaG zyx

dp=0. Thus ds G

Any surface on which dp=0 (i.e. p=constant) is called an

equipotential (equipressure) surface.

.0x

p

.0

y

p

z

p

p = p(z)

Governing differential equation for

the hydrostatic pressure distribution

dz

dp

For a fluid at rest .0a

]dz)ga(dyadxa[dp zyx or


2.08

Pressure distribution in an incompressible fluid at rest or

moving as a rigid body with no acceleration

z

z

zs

p

patm

d

dz

dp

satmz

z

P

P

dzdp

d)zz(pp satm

dppp atmabsolute dppp atmabsolutegauge

z patm

gauge

pressure

absolute

pressure

w

Measurement of Pressure

Barometer

h patm

pv

hp

.0p

hpp

matm

v

mvatm

m

Pre

ssure

Local

atmospheric

pressure +gauge

-gauge

Absolute zero p


2.09

Manometers

hpp atmA

Piezometer

A1w2matm phhp

A

h1 h2 m

w

patm

U-tube manometer

Differential manometer

A

B

h1 h3

h2

1

2

3

332211AB hhhpp

A

patm

h

Inclined manometer

pA

L

x

h

patm

B332211A phhhp


2.10

Hydrostatic Forces on Plane Surfaces

hp

hc

dF

dA c

cp

F

y

y yc

yp

xc

xp

x

o

o

hdApdAdF

dAhFA

A

dAsiny

A

ydAsin

AhAysin cc

ApF c

dAysindAsinyyydFdMFyMA

2

AAA opo

Ay

Iy

Ay

AyI

Ay

I

ydAθsinγ

dAyθsinγ

F

ydFy

c

xcc

c

2cxc

c

xx

A

A2

Ap

Ay

Ix

Ay

AyxI

Ay

I

ydAsin

xydAsin

F

xdFx

c

xycc

c

ccxyc

c

xy

A

Ap

h

y


2.11

Pressure prism concept

The pressure prism is a geometric representation of the

hydrostatic force on a rectangular plane surface

FR

p

h1

h2 FR cp

b

h1

h2

The magnitude of the resultant hydrostatic force is equal to the

volume of the pressure prism and passes through its centroid.

FR = F1 + F2 = pcA

F1 = h1(h2-h1)b b)hh(2

)hh(F 12

122

h1 (h2-h1)

F2 F1 +


2.12

Hydrostatic Forces on Curved Surfaces

dF = p dA

dFx = p dAx

The horizontal component of the hydrostatic force on a curved

surface is the product of the pressure at the centroid with the

area projection on a vertical plane.

z

y

x

dAx

dAy

dAz

h

zs

dA dFz = p dAz

dFy = p dAy

p

dA

x

z

pdAcos

xx pdAcospdAdF

xcA xx AppdAF

ycA yy AppdAF

Horizontal components:

p

dA

y

z

pdAcos

yy pdAcospdAdF


2.13

The vertical component of the hydrostatic force on a curved

surface is the weight of the liquid volume between the free

surface (real or imaginary) and the curved surface.

p

x

z

zz pdAsinpdAdF

h

A zz pdAF

ddAh zA

zF

dA

Vertical component:


2.14

Buoyancy and Flotation

The lifting force due to hydrostatic pressure distribution on

submerged volumes is called as ‘buoyant force’.

h

p1

p2

dFB=(p2-p1)dAz

z

Az12B dA)hh(F

z

Az12 dA)hh(

FB = submerged

The buoyant force acts through the

center of gravity of the submerged

volume in a direction opposite to

that of gravitation.

The same result apply to

floating bodies which are

only partially submerged FB

W = FB

cp

W

cg

z


2.15

Stability of Floating Bodies

A body is said to be in stable equilibrium position if, when

displaced, it returns to its original equilibrium position.

Conversely, it is in unstable equilibrium if, when displaced, it

moves to a new equilibrium position.

cg cp

W

FB cp FB

W

cg

Restoring couple

cg

cp

W

FB

W cg

Overturning couple

cp FB


2.E01

EXAMPLES

Example 2.1 Water flows in two parallel pipes A and B as shown in the figure. A differential

manometer gives the pressure difference between A and B of 35 kN/m2 by a deflection h.

Calculate h for a = 1 m and b = 0.3 m. Take w = 9.81 kN/m3 and SGHg = 13.6.

A

H2O

H2O

B a

b

h


2.E02

Example 2.2 Calculate the force R necessary to hold the 4-m wide gate in the position shown.

(h1=10m, h2=4m, L1=5m, L2=3m)

h1

Water

Gat

e

Wall

L2

h2

Hinge

R

L1


2.E03

Example 2.3 A sluice gate consists of a quadrant of cylinder of radius 1.5 m pivoted at its center

O. Its center of gravity G is denoted in the figure below. When the water is level with the pivot

O, calculate:

a) the magnitude and direction of the resultant pressure force on the gate due to the water and,

b) the turning moment required to open this gate. The width of the gate is 3m and it has a

weight of 58860 N.

Take w=9810N/m3 and the moment arm for the vertical force 𝑙𝑣 =4𝑟

3𝜋, r=1.5m, 𝑙𝑔 = 0.6 𝑚

A

W

G

𝑙𝑔

B

r O


2.E04

Example 2.4 What is the horizontal and vertical forces acting on spherical door AB? The

specific gravity of oil is 0.8. (h1 =3 m, h2 =6 m, D = 6 m, P = 3000 Pa).

A

water

h1

air

oil

P

B

h2

D


2.E05

Example 2.5 Find hydrostatic force on the conical body shown in figure below. (H =2.7m,

L1=L2=0.6m, R =0.9m)

H

L1 L2

R

R0


2.H01

HOMEWORK PROBLEMS

2.1. Determine the gage pressure at point A in Pascal .Is it higher or lower than atmospheric?

( water=9810 N/m3 , air=12.0 N/m3,SGoil=0.85 , SGHg=13.8 , h1=45 cm , h2=30 cm , h3=25 cm ,

h4=15 cm )

2.2. A tank is constructed of a series of cylinders having diameters of 0.30, 0.25 and 0.15m as

shown in the figure. The tank contains oil, water and glycerin and a mercury manometer is

attached to the bottom as illustrated. Calculate the manometer reading, hm. SGoil=0.9,

SGGlycerin=1.26, SGmercury=13.6. (h = 0.1 m)

Water Hg

Oil

Air

h1

h2 h3

h4

A

hm

Mercury

h

h

h

h Glycerin

Oil

Water


2.H02

2.3. Determine the pressures ( gage ) at points A,B,C,D,E.

( water =1000 kg/m3 , oil =900 kg/m3 , g=9.81 m/s2 , h1=0.6m , h2=0.3m , h3=0.6m , h4=1.0m )

2.4. Determine the new differential reading along the inclined leg of the mercury manometer,

if the pressure in pipe A is decreased 12kPa and the pressure in pipe B remains unchanged. The

fluid in A has a specific gravity of 0.9 and the fluid B is water. SGmercury=13.6, h1=100mm,

h2=80mm, l1=50mm, =30.

Water

B

C

Oil

Air

D

Air

h2

h3

h1

h4

A

E

h1

A B

Mercury

l1

h2


2.H03

2.5. An air-filled hemispherical shell is attached to the ocean floor. A mercury barometer

located inside the shell reads 765 mmHg, and a mercury U-tube manometer designed to give

the outside water pressure indicates a differential reading of h2=735 mm Hg at a depth of

h1=10m as illustrated. Based on these data what is the atmospheric pressure at the ocean

surface? h3= 360mm.

2.6. The vertical cross section of a 7m long closed tank is shown in the figure. The tank contains

ethyl alcohol and the air pressure is 40kPa. Determine the magnitude of the force acting on one

side of the tank. h1=h2=2m, h3=h4=4m (ethyl_alcohol = 7740 N/m3).

Air

Ethyl

Alcohol

h1

h2

h4

h3

Ocean

Surface

h1

h3

h2

Mercury

Shell wall Sea water

Shell

Pin = 765 mm Hg


2.H04

2.7. The concrete dam given below rests on a solid foundation. The unit weight of concrete is

23.6 kN/m3. Determine the minimum coefficient of friction between the dam and the foundation

required keeping the dam from sliding at the water depth shown. Assume no fluid uplift pressure

along the base. Base your analysis on a unit length of the dam. h1 =4m, h2=5m, h3=6m, h4=2m.

2.8. Calculate the force F necessary to hold the 4 m-wide gate AB in position shown in the

figure below. The specific of oil is 0.8. ( H1=3m, H2=2.6m, H3=4m, P=49.05 kPa, L=5m)

Water h1

h2

h3

h4

hinge

H3

H1

H2

F

B

A

Water

Oil

Air

P

L


2.H05

2.9. A structure is attached to the ocean floor as shown in the figure. A 2m diameter hatch is

located in an inclined wall and hinged on one edge. Determine the minimum air pressure, p1,

within the container to open the hatch. Neglect the weight of the hatch and friction in the hinge.

=30, h=10m , SGseawater =1.03

2.10. The concrete seawall has a curved surface and restrains seawater at a depth of 7.5m. The

trace of the surface is a parabola as illustrated. Determine the moment of the fluid force (per

unit length) with respect to an axis through the toe (point A).

(h = 7.5 m, L = 7 m, y = 0.2x2, SGseawater = 1.03)

Hinge

Air pressure

Hatch

Seawater

Free surface

h

x

L

Seawater

A

y = f(x)

y

h


2.H06

2.11. The gate ABC , shown in the figure below , is weightless and has 1 m length in the normal

direction to the figure. Its lower part (BC) is semi-cylindrical in shape .As water rises at the

left-hand-side , the gate will open automatically. At what depth , h , above the hinge this will

occur ? ( water=9.81kN/m3 )

2.12. Find hydrostatic force on the conical body shown in figure below. (H =2.7m,

L1=L2=0.6m, R =0.9m)

H

L1

L2

R

H2O

R C B

A

h

hinge


2.H07

2.13. The cylinder of weight W is in equilibrium as seen in figure below. The length and radius

of the cylinder are L=5 m and R=2 m , respectively , w = 9810 N/m3 Compute;

a) the weight of the cylinder W and

b) the horizontal force exerted at point B due to the cylinder.

2.14. Find the vertical component of the hydrostatic force on hemispherical dome ABC if gage

pressure at D is 90 kPa, R = 1.5 m, L = 2 m and = 15 kN/m3.

R

B

L

D

A

B

C R


2.H08

2.15. A plug in the bottom of pressurized tank is conical in shape as shown in the figure. The

air pressure is 50kPa and the liquid in the tank has a specific weight of 27kN/m3. Determine the

magnitude, direction and line of action of the force exerted on the curved surface of the cone

within the tank due to 50kPa pressure and the liquid.

(h1 = 1 m, h2 = 3 m, Pair = 50 kPa, = 60)

2.16. A 3-m-long curved gate is located in the side of a reservoir containing water as shown in

the figure below. Determine the magnitude of the horizontal and vertical components of the

force of the water on the gate. Show that this force passes through point A.( h=6m , r =2m )

F

Air

Pair

h1

h2

r

A

h

gate


2.H09

2.17. A cube having 100-cm side length and 5000 N weight is placed in a large tank of water

as seen in the figure below. The bottom-half depth of the cube is filled with oil of specific

weight oil = 0.8w. The remaining upper part of the cube has air with a gage pressure of

Pair = 10 kPa. Determine;

a) the resultant horizontal force acting on the AB face of the cube,

b) the direction and magnitude of the resultant vertical force acting on the cube.

( a = 100 cm, w = 9810 N/m3).

2.18. A 1m diameter cylindrical mass, M, is connected to a 2m wide rectangular gate as shown

in the figure. The gate is to open when the water level, h, drops below 2.5m. Determine the

required value for M. Neglect friction at the gate hinge and the pulley.

(h1 = 4 m, h2 = 1 m, D = 1 m )

a

a / 2

a / 2

A

B

water

oil

air

D

h

h1

h2

water

Documents

HYDROSTATICScourses.ce.metu.edu.tr/ce272/wp-content/uploads/sites/69/2018/02/... · The lifting force due to hydrostatic pressure distribution on submerged volumes is called as ‘buoyant