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PROCESSING TECHNOLOGY 1

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PROCESSING TECHNOLOGY 1

Fluid Statics

Pressure, P = Force/area

Unit of pressure is the N/m2 also called the

Pascal (Pa),

1 Pa = 1 N/m2

1000 N/m2 1 kN/m2

1000000 1 MN/m2 etc

100000 N/m2 105 N/m2 1 bar 1 atm

Atmospheric pressure at sea level is

101.3 kPa or 1.013 bar.

Many instruments show gauge pressure,

i.e., the amount by which the pressure

exceeds the atmospheric pressure, the

sum of the two gives the absolute pressure.

Absolute pressure, Pabsolute = Pgauge + Patmospheric

A

A

P1

P2

Horizontal change

Consider a tank of stagnant fluid within which there exists

a horizontal element of fluid of area A at either end with

pressures P1 and P2 exerted on the ends of the element.

- End 1: F1 = P1A End 2: F2 = P2A
- Net force (F1 – F2) = 0
- = P1A - P2A = A(P1 - P2)
- A(P1 - P2) = 0
- P1 = P2
- i.e. There is nochange of pressure at
- constantheight in a static fluid.

P2

A

Consider a vertical cylindrical

element of fluid, area A on

each end with pressure P1

and P2 at the ends, which

are at heights z1 and z2.

Z2

Z1

A

P1

Mass m = density x volume = rV(1)

Volume V = A(z1-z2)(2)

Mass m = rA(z1-z2)

Weight W = mass x g

W = rA(z1-z2)g(3)

The weight of the element will act downwards,

due to the pull of gravity. The downward force

on the upper end of the element, F2 is given by,

F2 = P2A

The upward acting force F1 at the lower end of

the element is given by,

F1 = P1A

The vertical forces on the element must balance,

Fdownward-acting = Fupward -acting

F1 + W = F2

P1A + rA(z1-z2)g = P2A

(P2 - P1) = rg(z1 - z2)

Assuming a constant fluid density r,

the change in pressure DP due to a

vertical difference in height is given by,

DP = rg(z1 - z2)

Pressure measurement is an essential part

of any factory. It is used to ensure

that a tank doesn’t exceed design pressure;

or as an indicator of another factor, e.g.

pressure difference across an orifice plate

can be used to indicate fluid flowrate.

P

at

z

P

A piezometer is the simplest

example of a pressure gauge

or manometer. This is simply

an open-ended tube connected

to a pipe carrying liquid under

pressure.

P

at

z

P

With the pipe under pressure,

the fluid will rise up the tube

until the column of liquid

balances the pipe-line

pressure.

The pressure difference due to the column

of liquid within the piezometer is,

DP = rgz

The pressure within the pipe-line pushing

upwards on the column of liquid within the

piezometer is P and the pressure pushing

downwards on the column is atmospheric

pressure Pat. Hence the pressure

difference DP is,

DP = P - Pat

P

at

z

P

Equating the two expressions for DP gives,

P - Pat = rgz

Hence the pipe-line pressure is given by,

P = rgz + Pat

Reservoir 2 @ P2

Reservoir 1

@ P1

z

Manometer fluid

Reservoir 1

@ P1

Reservoir 2 @ P2

z

Manometer fluid

Used to measure the pressure

differential between two fluid reservoirs.

The unknown pressure is indicated by

the difference between the levels of

fluid in the two arms of the U-tube.

y

x

Reservoir 1

@ P1

Reservoir 2

@ P2

z

B

A

y

x

Reservoir 1

@ P1

Reservoir 2

@ P2

z

A

B

Draw a datum (A-B) across the manometer

level with the lowest liquid interface. As we

consider the manometer to be at steady state,

the downward pressure at the left hand of the

datum A is balanced by the

downward pressure

at the right hand

datum B

y

x

Reservoir 1

@ P1

Reservoir 2

@ P2

z

A

B

The pressure at A consists of the summation

of the reservoir pressure P1 and the pressure

due to the height of reservoir fluid within the

manometer leg x, =rRgx

where rR is the

density of the

reservoir fluid

y

x

Reservoir 1

@ P1

Reservoir 2

@ P2

z

A

B

The pressure at B is a summation of the

reservoir pressure P2 the pressure due to the

height of reservoir fluid within the manometer

leg y, = rRgy, and the pressure due to the

height of manometer fluid within the

manometer leg z,

= rMgz, where rM

is the density of

the manometer fluid.

PA = PB

P1 + rRgx = P2 + rRgy + rMgz

y

x

Reservoir 1

@ P1

Reservoir 2

@ P2

z

B

A

x = y + z

P1 + rRg(y + z) = P2 + rRgy + rMgz

P1 + rRgy + rRgz = P2 + rRgy + rMgz

P1 + rRgz = P2 + rMgz

P1 - P2 = rMgz - rRgz

P1 - P2 = (rM - rR)gz

Pivots

Movement

C-shaped oval cross

Rigid mount

section tube

Pressure

The terms pressure and head are often used

inter-changeably. However, the units used are

completely different i.e metres of fluid for head

and N/m2 for pressure.

It is common to describe a pressure as “being

equivalent to a head of 10m of water.” This

means that the actual pressure in N/m2 (or Pa)

is given as,

P = rgh = 1000 x 9.81 x 10 = 98.1 kPa

Where h is the head of liquid, 10m in this case.

Similarly a pressure P of 3 bar is equivalent

to a head h of mercury (r=13600 kg/m3),

P = rgh

h = P/rg = 3x105/(13600 x 9.81)

h = 2.25m