Fluid Mechanics Concept Page - 10

Example

Use of Bernoulli's equation in rotating frame

Example: A liquid is kept in a cylindrical vessel, which is being rotated about a vertical axis through the center of the circular base. If the radius of the vessel is r and angular velocity of rotation is ω$\omega$, then what is the difference in the heights of the liquid at the center of the vessel and the edge?

Solution:When the cylindrical vessel is rotated at angular speed ω$\omega$ about its axis, the velocity of the liquid at the sides is maximum, given by
vs=rω$v_s=r\omega$
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have P+12ρv2=$P+{\displaystyle \frac{1}{2}}\rho v^2=$ constant
Ps+12ρvs2=Pc+12ρvc2$P_s+{\displaystyle \frac{1}{2}}\rho {v_s}^2=P_c+{\displaystyle \frac{1}{2}}\rho {v_c}^2$
where
Ps=$P_s=$ pressure at the sides
vs=$v_s=$ velocity of the liquid at the sides
Pc=$P_c=$ pressure at the center
vc=$v_c=$ velocity of the liquid at the center
Pc−Ps=12ρvs2=12ρr2ω2$P_c-P_s={\displaystyle \frac{1}{2}}\rho {v_s}^2={\displaystyle \frac{1}{2}}\rho r^2{\omega }^2$   ...(I)$(I)$     ∵vc=0$\because v_c=0$
Since
Pc$P_c$ is greater than Ps$P_s$, the liquid rises at the sides of the vessel. Let h$h$ be the difference in the levels of the liquids at the sides and at the center, so we have
Pc−Ps=ρgh$P_c-P_s=\rho gh$ ...(II)$(II)$
from (I)$(I)$ and (II)$(II)$ we have
ρgh=12ρr2ω2⇒h=r2ω22g$$\begin{gathered} \rho gh={\displaystyle \frac{1}{2}}\rho r^2{\omega }^2 \\ \Rightarrow h={\displaystyle \frac{r^2{\omega }^2}{2g}} \end{gathered}$$

Definition

Magnus Effect

The Magnus effect is the commonly observed effect in which a spinning ball (or cylinder) curves away from its principle flight path. It is important in many ball sports. It affects spinning missiles, and has some engineering uses, for instance in the design of rotor ships and Flettner aeroplanes.
Example: In terms of ball games, topspin is defined as spin about an axis perpendicular to the direction of travel, where the top surface of the ball is moving forward with the spin. Under the Magnus effect, topspin produces a downward swerve of a moving ball, greater than would be produced by gravity alone, and backspin has the opposite effect. Likewise side-spin causes swerve to either side as seen during some baseball pitches, e.g. slider. The overall behaviour is similar to that around an aerofoil with a circulation which is generated by the mechanical rotation, rather than by airfoil action

Definition

Limitations of Bernoulli's principle

Limitations of Bernoulli's equation are as follows:
1. The Bernoulli equation has been derived by assuming that the velocity of every element of the liquid across any cross-section of the pipe is uniform. Practically,it is not true. The elements of the liquid in the innermost layer have the maximum velocity. The velocity of the liquid decreases towards the walls of the pipe. Therefore, we should take into account the mean velocity of the liquid.
2. While deriving Bernoulli's equation, the viscous drag of the liquid has not been taken into consideration. The viscous drag comes into play, when a liquid is in motion.
3.Bernoulli's equation has been derived on the assumption that there is no loss of energy, when a liquid is in motion. In fact, some kinetic energy is converted into heat energy and a part of it is lost due to shear force.
4. If the liquid is flowing along a curved path, the energy due to centrifugal force should also be taken into consideration.

Definition

Principle and Working of aspirator pump

An aspirator is a type of ejector-jet pump, that produces vacuum by means of the Venturi effect. In an aspirator, fluid (liquid or gaseous) flows through a tube which narrows then expands in cross-sectional area, and thus volume. When the tube narrows the fluid's pressure decreases.

The cheap and simple water aspirator is the most common type of aspirator. It is used in chemistry and biology laboratories and consists of a tee fitting which is attached to a tap and has a hose barb at one side.
The flow of water passes through the straight portion of the tee, which has a restriction at the intersection, where the hose barb is attached. The vacuum hose should be connected to this barb. While previously common for low-strength vacuums needed in chemistry bench work, they use a lot of water, and depending on what the vacuum is being used for, i.e. solvent removal, they can violate environmental protection laws such as RCRA by mixing these potentially-hazardous solvents into the water stream, then flushing them down a drain that often leads directly to the municipal sewer. 
 If a liquid is used as the working fluid, the strength of the vacuum produced is limited by the vapor pressure of the liquid (for water, 3.2$3.2$ kPa or 0.46$0.46$ psi or 32$32$ mbar at 25$25$ 0C${}^{0}C$ or 77 F$77F$). If a gas is used, however, this restriction does not exist. The industrial steam ejector (also called the 'steam jet ejector', 'steam aspirator', or 'steam jet aspirator') uses steam as a working fluid.

Result

Altimeter

An altimeter is an aneroid barometer, but it is used in aircraft to measure its altitude. Since atmospheric pressure decreases with increase in height, therefore a barometer which measures the atmospheric pressure can be used to determine the altitude above sea level.

Example

Working of pumps

An ink dropper has a bulb at one end of a tube. When the bulb is pressed, the pressure inside the tube drops. Now the tube is dipped inside the ink. Atmospheric pressure acts at the surface of the ink level in the ink pot. When the bulb is released atmospheric pressure pushes the ink up the tube towards the low pressure area.

Definition

Principle and Working of Pitot tube

A pitot tube is an open-ended right-angled tube pointing in opposition to the flow of a fluid and used to measure pressure.
Working:
The liquid flows up the tube and when equilibrium is attained, the liquid reaches a height above the free surface of the water stream. Since the static pressure, under this situation, is equal to the hydrostatic pressure due to its depth below the free surface, the difference in level between the liquid in the glass tube and the free surface becomes the measure of dynamic pressure. Therefore, we can write, neglecting friction,
po−p=ρv22=hρg$p_o-p={\displaystyle \frac{\rho v^2}{2}}=h\rho g$
where p0,p$p_0,p$ and V$V$ are the stagnation pressure, static pressure and velocity respectively at point A(fig. a);
V=2gh−−−√$V=\sqrt{2gh}$
Such a tube is known as a Pitot tube and provides one of the most accurate means of measuring the fluid velocity. For an open stream of liquid with a free surface, this single tube is sufficient to determine the velocity. But for a fluid flowing through a closed duct, the Pitot tube measures only the stagnation pressure and so the static pressure must be measured separately. Measurement of static pressure in this case is made at the boundary of the wall (Fig. b). The axis of the tube measuring the static pressure must be perpendicular to the boundary and free from burrs, so that the boundary is smooth and hence the streamlines adjacent to it are not curved. This is done to sense the static pressure only without any part of the dynamic pressure. A Pitot tube is also inserted as shown (Fig. b) to sense the stagnation pressure. The ends of the Pitot tube, measuring the stagnation pressure, and the piezometric tube, measuring the static pressure, may be connected to a suitable differential manometer for the determination of flow velocity and hence the flow rate.

Example: A Pitot tube is inserted in an air flow (at STP) to measure the flow speed. The tube is inserted so that it points upstream into the flow and the pressure sensed by the tube is the stagnation pressure. The static pressure is measured at the same location in the flow, using a wall pressure tap. If the pressure difference is 30$30$ mm of mercury, determine the flow speed.

 A Pitot tube inserted in a flow as shown. The flowing fluid is air and the manometer liquid is mercury. Determine the flow speed.

Solution:
Governing equation: pρ+V22+gz=${\displaystyle \frac{p}{\rho }}+{\displaystyle \frac{V^2}{2}}+gz=$ constant 
Assumptions: (1) Steady flow.
                       (2) Incompressible flow.
                       (3) Flow along a streamline. 
                       (4) Frictionless deceleration along stagnation streamline. 
Writing Bernoulli's equation along the stagnation streamline (with Δz=0$\mathrm{\Delta }z=0$) yields                      
p0ρ=pρ+V22${\displaystyle \frac{p_0}{\rho }}={\displaystyle \frac{p}{\rho }}+{\displaystyle \frac{V^2}{2}}$
p0$p_0$is the stagnation pressure at the tube opening where the speed has been
reduced, without friction, to zero. Solving for V$V$ gives
                     V=2(p0−p)ρair−−−−−−−−√$V=\sqrt{{\displaystyle \frac{2(p_0-p)}{{\rho }_{air}}}}$ From the diagram,              
                     p0−p=ρHggh=ρH2OghSGHg$p_0-p={\rho }_{Hg}gh={\rho }_{H_{2}O}ghS{G}_{Hg}$and              
                    V=2ρH2OghSGHgρair−−−−−−−−−−−−√$V=\sqrt{{\displaystyle \frac{2{\rho }_{H_{2}O}ghS{G}_{Hg}}{{\rho }_{air}}}}$                 
                        =2×1000kgm3×9.81ms2×30mm×13.6×m3123kg×1m1000mm−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√$=\sqrt{2\times 1000{\displaystyle \frac{kg}{m^3}}\times 9.81{\displaystyle \frac{m}{s^2}}\times 30mm\times 13.6\times {\displaystyle \frac{m^3}{123kg}}\times {\displaystyle \frac{1m}{1000mm}}}$            
                    V=80.8m/s$V=80.8m/s$ V
At T=20oC$T={20}^{o}C$, the speed of sound in air is 343$343$ m/s. Hence, M=0.236$M=0.236$ and the assumption of incompressible flow is valid.

Definition

Principle and Working of Atomizer

Atomizer: An Atomizer is a device used for reducing a liquid to a fine spray, such as the nozzle used to feed oil into a furnace or an enclosed bottle with a fine outlet used to spray perfumes or medicines.

Principle of Operation: When a fast gas stream is injected into the atmosphere and across the top of the vertical tube, it is forced to follow a curved path up, over and downward on the other side of the tube. This curved path creates a lower pressure on the inside of the curve at the top of the tube.

 This curve-caused lower pressure near the tube and the atmospheric pressure further up is the net force causing the curved, velocity-changed path (radial acceleration) shown by Bernoulli's Principle.

The difference between the reduced pressure at the top of the tube and the higher atmospheric pressure inside the bottle pushes the liquid from the reservoir up the tube and into the moving stream of air where it is broken up into small droplets (not atoms as the name suggests) and carried away with the stream of air.

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