Fluid Mechanics for MAP/Fluid Statics

Definitions

File:Fluid statics first picture new 01.svg

Fluid at rest(left) and in rigid body motion (right)

Fluid statics is the study of fluids which are either at rest or in rigid body motion with respect to a fixed frame of reference. Rigid body motion means that there is no relative velocity between the fluid particles.

In a fluid at rest, there is no shear stress, i. e. fluid does not deform, but fluid sustains normal stresses.

We can apply Newton's second law of motion to evaluate the reaction of the particle to the applied forces.

File:Fluid Statics 4.png

Forces created by pressure on the surfaces of a differential fluid volume in a static fluid

Force balance in $i^{t h}$ direction:

Template:Center top $F_{i}^{n e t} = m \cdot a_{i}$ Template:Center bottom

We can also say, $F_{i}^{b o d y} + F_{i}^{s u r f} = m \cdot a_{i}$

Force created by pressure is :

$F^{s u r f} = F^{p r e s s u r e} = - p \cdot \vec{A}$

$\vec{A}$ is the vector having the surface area as magnitude and surface normal as direction.

Thus,

$F^{p r e s s u r e} = - p \cdot \vec{A} = - p A \vec{n}$

Force caused by the pressures opposite to the surface normal.

For a differential fluid element: $d F_{i}^{b o d y} + d F_{i}^{s u r f} = d m \cdot a_{i}$

Remember Taylor Series expansion: $F (x + Δ x) = F + \frac{\partial F}{\partial x} + \frac{1}{2} \frac{\partial^{2} F}{\partial x^{2}} Δ x^{2} + \dots$

P is the pressure in the center of the fluid element, therefore the pressure on the surface in direction of $x_{i}$ is $P + \frac{\partial P}{\partial x_{i}} \frac{d x_{i}}{2}$ .

$d F_{2}^{p r e s s u r e} = [P - \frac{\partial P}{\partial x_{2}} \frac{d x_{2}}{2}] d x_{1} d x_{3} - [P + \frac{\partial P}{\partial x_{2}} \frac{d x_{2}}{2}] d x_{1} d x_{3}$: $= - \frac{\partial P}{\partial x_{2}} d x_{1} d x_{2} d x_{3} = - \frac{\partial P}{\partial x_{2}} d V$

Thus,

$d F_{i}^{p r e s s u r e} = - \frac{\partial P}{\partial x_{i}} d V$

$d F_{i}^{b o d y} = d m g_{i}$

Thus,

$- \frac{\partial P}{\partial x_{i}} d V + d m g_{i} = d m a_{i}$ or, $- \frac{\partial P}{\partial x_{i}} d V + ρ d V g_{i} = ρ a_{i}$

or, $- \frac{\partial P}{\partial x_{i}} + ρ g_{i} = ρ a_{i}$

or, $- \frac{\partial P}{\partial x_{1}} = ρ a_{1}; - \frac{\partial P}{\partial x_{2}} = ρ a_{2}; - \frac{\partial P}{\partial x_{3}} - ρ g = ρ a_{3}$

for $a_{i} = 0$

$\frac{\partial P}{\partial x_{1}} = 0; \frac{\partial P}{\partial x_{2}} = 0; \frac{\partial P}{\partial x_{3}} = - ρ g$

Pressure changes only in $x_{3}$ direction.

Pressure variation in an incompressible and static fluid

File:Hydrost press vari with hght schmtcs renew 01.svg

(a) pressure at location A and B (b)Depth is measured from the free surface (c) Pressure variation with depth of water column

$\frac{\partial P}{\partial x_{3}} = - ρ g; x_{3} = z \to \frac{\partial P}{\partial z} = - ρ g$ is constant since $ρ$ and $g$ are constants.

Template:Center top $\int_{p_{1}}^{p_{2}} d P = - \int_{z_{1}}^{z_{2}} ρ g d z$ Template:Center bottom

Template:Center top $p_{2} - p_{1} = - ρ g (z_{2} - z_{1})$ Template:Center bottom

If we take $p_{2}$ at the surface, then:

Template:Center top $p_{a t m} - p_{1} = - ρ g (z_{2} - z_{1})$ Template:Center bottom

Template:Center top $p_{1} = p_{a t m} + ρ g h; h = z_{2} - z_{1}$ Template:Center bottom

h is measured from the surface.

Any two points at the same elevation in a continuous length of the same liquid are at the same pressure. Pressure increases as one goes down in a liquid column. Remember: Template:Center top $\frac{\partial P}{\partial x_{i}} + ρ g_{i} = ρ a_{i}$ Template:Center bottom

File:Hydrostatic pressure similarity same depth rename 01.svg

similar depth in same fluid experience similar hydrostatic pressure

For incompressible flow,

Template:Center top $p_{2} - p_{1} = - ρ g (z_{2} - z_{1}) \to p_{1} = p_{a t m} + ρ g h$ Template:Center bottom

Consider 3 immiscible fluids in a container and find out a relation for the pressure at the bottom of the fluid shown in the schematics besides.

$P_{C} = ?$

$P_{A} = P_{0} + ρ_{A} g h_{A}$

$P_{B} = P_{A} + ρ_{B} g (h_{B} - h_{A})$

$P_{C} = P_{B} + ρ_{C} g (h_{C} - h_{B})$

$\Rightarrow P_{C} = ρ_{C} g (h_{C} - h_{B}) + ρ_{B} g (h_{B} - h_{A}) + ρ_{A} g h_{A} + P_{0}$

$P_{C} = ρ_{C} g H_{C} + ρ_{B} g H_{B} + ρ_{A} g H_{A} + P_{0}$

File:3 liquids in single tank new 01.svg

Hydrostatic pressure profile for fluids with different densities and height

Transmission of Pressure

Concept of transmission of pressure is very important for hydraulic and pneumatics system. Neglecting elevation changes the following relation can be written:

Template:Center top $P_{1} = P_{2} \Rightarrow \frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}} \Rightarrow F_{2} = \frac{A_{2}}{A_{1}} F_{1}$ Template:Center bottom which could be stated in the famous Pascal's law like below: Template:Quotation

File:Pascal law renew 02.svg

Pascal's law

Communicating containers

File:Communicating container renew 01.svg

variation of height of fluid column due to difference in pressure from free surface

Lets consider two closed containers(which means the free surface pressure could be different than atmospheric pressure) both contain same fluid are connected via a connector valve. When the valve is open, the heights of the fluid columns can give an indication about the pressure in both chamber.

For closed container

Template:Center top $P_{01} + ρ_{1} g (H_{1} - h_{1}) = P_{02} + ρ_{2} g (H_{2} - h_{2})$ Template:Center bottom

(of course,when we calculate the small 'h', it should be measured at the height of connecting valve for both column distinctively.)

for $ρ_{1} = ρ_{2} = ρ$

Template:Center top $P_{01} + ρ g (H_{1} - h_{1}) = P_{02} + ρ g (H_{2} - h_{2})$ Template:Center bottom

So from the picture above, we can understand that the pressure in the right column is higher than the left column.

For open Containers,

$\Rightarrow P_{01} = P_{02} = P_{a t m}$

If both fluid columns are at the same level

$\Rightarrow H_{1} - h_{1} = H_{2} - h_{2}$

so, the depth of the fluid from free surface in both column will be the same. This nice principle was used for Water-based Barometer ^[1] a.k.a 'Storm Barometer' or 'Goethe Barometer'. Try to see if you understand the device.

Pressure Measurement Equipments

Barometer

From the equations which we derived before , it is also possible to measure the pressure exerted by almost 100 km^[2] thick earth atmosphere which is above us. Since the constituents and the density varies over the height of the atmosphere , we will consider a fluid column which have free surface with no atmospheric pressure but connected with a fluid which experience atmospheric pressure like communicating container. Let consider first (from previous section), Template:Center top $P_{0} = ρ_{f} g h$ Template:Center bottom

for $P_{0} = 1$ atm = $101, 3$ kPa

water height will be:

Template:Center top $101, 3 \cdot 1 0^{3} = 1000 \cdot 9.79 \cdot h_{w a t e r}$ Template:Center bottom

$h_{w a t e r} \approx 10 m$ where as height of mercury will be around $\frac{10 m}{13, 6} \approx 0, 7 m$ So now taking this barometer to desired location and observing the mercury column height, the atmospheric pressure could be measured.

File:Barometer new 02.svg

Barometer

U-tube manometer

The volume rate of flow, Q, through a pipe can be determined by means of a flow nozzle located in the pipe as illustrated in the figure. The nozzle creates a pressure drop, $p_{A} - p_{B}$ , along the pipe which is related to the flow through the equation $Q = K \sqrt{p_{A} - p_{B}}$ where K is a constant depending on the pipe and nozzle size. The pressure drop is frequently measured with a differential U-tube manometer of the type illustrated. (a) Determine an equation for $p_{A} - p_{B}$ in terms of the specific weight of the flowing fluid, $ρ_{1}$ , the specific weight of the gage fluid, $ρ_{2}$ , and the various heights indicated. (b) For $ρ_{1} = 9.80 \frac{k N}{m^{3}}$ , $ρ_{2} = 15.6 \frac{k N}{m^{3}}$ , $h_{1} = 1.0 m$ and $h_{2} = 0.5 m$ , what is the value of the pressure drop, $p_{A} - p_{B}$ ? Solution

(a) Although the fluid in the pipe is moving, the fluids in the columns of the manometer are at rest so that the pressure variation in the manometer tubes is hydrostatic. If we start at point A and move vertically upward to level (1), the pressure will decrease by $ρ_{1} h_{1}$ and will be equal to the pressure at (2) and (3). We can move from (3) to (4) where the pressure has been further reduced by $ρ_{2} h_{2}$ . The pressures at levels (4) and (5) are equal, and as we move from (5) to B the pressure will increase by $ρ_{1} (h_{1} + h_{2})$ .

Thus, in equation form

Template:Center top $p_{A} - ρ_{1} h_{1} - ρ_{2} h_{2} + ρ_{1} (h_{1} + h_{2}) = p_{B}$ Template:Center bottom

or

Template:Center top $p_{A} - p_{B} = h_{2} (ρ_{2} - ρ_{1})$ Template:Center bottom

It is to be noted that the only column height of importance is the differential reading, $h_{2}$ . The diferential manometer could be placed 0.5 m or 5.0 m above the pipe $(h_{1} = 0.5 m$ or $h_{1} = 5.0 m)$ and the value of $h_{2}$ would remain the same. Relatively large values for the differential reading $h_{2}$ can be obtained for small pressure differences, $p A - p_{B}$ ,if the difference between $ρ_{1}$ and $ρ_{2}$ is small.

(b) The specific value of the pressure drop for the data given is

Template:Center top $p_{A} - p_{B} = (0.5 m) (15.6 \frac{k N}{m^{3}} - 9.8 \frac{k N}{m^{3}}) = 2.90 k P a$ Template:Center bottom

File:U tube manometer renew 01.svg

Application of U tube manometer to measure pressure difference

Inclined-Tube Manometer

To measure small pressure changes, a monometer of the type shown in the figure, is frequently used. One leg of the manometer is inclined at an angle $θ$ , and the differential reading $l_{2}$ is measured along the inclined tube. The difference in pressure $p_{A} - p_{B}$ can be expressed as

Template:Center top $p_{A} + ρ_{1} h_{1} - ρ_{2} l_{2} \sin θ - ρ_{3} h_{3} = p_{B}$ Template:Center bottom

or

Template:Center top $p_{A} - p_{B} = ρ_{2} l_{2} \sin θ + ρ_{3} h_{3} - ρ_{1} h_{1}$ Template:Center bottom

where the pressure difference between points (1) and (2) is due to the vertical distance between the points, which can be expressed as $l_{2} \sin θ$ . Thus, for relatively small angles the differential reading along the inclined tube can be made large even for small pressure differences. The inclined-tube manometer is often used to measure small differences in gas pressures so that if pipes A and B contain a gas then

Template:Center top $p_{A} - p_{B} = ρ_{2} l_{2} \sin θ$ Template:Center bottom

or

Template:Center top $l_{2} = \frac{p_{A} - p_{B}}{ρ_{2} \sin θ}$ Template:Center bottom

where the contributions of the gas columns $h_{1}$ and $h_{3}$ have been neglected. The equation above shows that the differential reading $l_{2}$ (for a given pressure difference) of the inclined-tube manometer can be increased over that obtained with a conventional U-tube manometer by the factor $\frac{1}{\sin θ}$ . Recall that $\sin θ \to 0$ as $θ \to 0$ .

File:Inclined manometer tube renew 02.svg

inlclined-Tube manometer to measure small difference

Fluid Mechanics for MAP/Fluid Statics

Contents

Definitions

Pressure variation in an incompressible and static fluid

Transmission of Pressure

Communicating containers

Pressure Measurement Equipments

Barometer

U-tube manometer

Inclined-Tube Manometer

Other related topics

Reference

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Fluid Mechanics for MAP/Fluid Statics

Definitions

Pressure variation in an incompressible and static fluid

Transmission of Pressure

Communicating containers

Pressure Measurement Equipments

Barometer

U-tube manometer

Inclined-Tube Manometer

Other related topics

Reference

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