Fluid Mechanics Concept Page - 9

Example

Deriving Dynamic Lift using Bernoulli's Equation

In a test experiment on a model aeroplane in a wind tunnel, the flow speed on the upper and lower surfaces of the wings are v1$v_1$ and v2$v_2$ respectively. If ρ$\rho$ is the density of air, then find the dynamic lift? ( Given: A = cross-sectional area of the wing)
Solution:Applying Bernoulli's principle, we have
P1+12ρv12=P2+12ρv22$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2$
⇒P2−P1=12ρ(v21−v22)$\Rightarrow P_2-P_1=\frac{1}{2}\rho (v_1^2-v_2^2)$
The difference in pressure provides the lift to the aeroplane. So
lift on the aeroplane=pressure difference ×$\times$ area of wings
=12ρ(v21−v22)×A=12ρA(v21−v22)$$\begin{gathered} =\frac{1}{2}\rho (v_1^2-v_2^2)\times A \\ =\frac{1}{2}\rho A(v_1^2-v_2^2) \end{gathered}$$

Result

Mercury in a barometer

The advantages of use of mercury in a barometer are :
1) its density is very high. 
2) it has low vapour pressure.
3) its rate of evaporation is not very high.
4) it neither wets nor sticks to glass.
5) its surface is shining and opaque.
The disadvantages of use of mercury in a barometer are :
1) it is bulkier in nature.
2) the chances of failure of glass tube is very high.

Result

Fortin barometer

The arrangement is a modification of a simple barometer. There is a leather cup at the bottom which contains mercury and behaves like a trough. At the bottom of the enclosing case, there is a screw, then end of which supports the leather cup of the glass vessel. This can be raised or lowered using the screw. The mercury level is adjusted to the zero reading of the main scale using the screw. An ivory pointer is provided to indicate the level of mercury.

Result

Aneroid Baromter

The aneroid barometer has no liquid. It is light and portable.
A diaphragm separates a partially evacuated box with atmospheric pressure. The circular scale is calibrated to read atmospheric pressure. When the pressure increases it presses the diaphragm and the pointer moves to show the changed pressure. When the pressure decreases the diaphragm bulges, and the pointer moves again to show the changed pressure.

Result

Weather forecast with barometer

The pressure at a certain place depends on the density of air at that place. As the temperature increases, the density decreases and so does the pressure. Also density depends on the amount of moisture present in air. The following are a few indications of change in temperature with change in barometric pressure,
1) If the barometric height suddenly falls, it is an indication of a coming storm.
2) If the barometric height falls gradually, it indicates possibility of rain
3) If the barometric height increases gradually, it indicates the advent of dry weather.
4) If the barometric height suddenly increases, this indicates flow of air from that place to surrounding area which will mean coming of extremely dry weather.

Definition

Different heads in Bernoulli's equation

As we know the Bernoulli's equation is : 
P+ρgh+(12)ρv2=constant$P+\rho gh+({\displaystyle \frac{1}{2}})\rho v^2=constant$
Here the symbols have their standard meanings.
For a particular fluid ρ$\rho$ (density of fluid) is constant. Hence we can divide the whole equation with the term ρg$\rho g$. 
The equation hence obtained will be: 
Pρg+v22g+h=constant${\displaystyle \frac{P}{\rho g}}+{\displaystyle \frac{v^2}{2g}}+h=constant$
In fluid mechanics, energy per unit weight of fluid is termed as head. Now this energy can be in the form of pressure energy, kinetic energy or gravitational potential energy.
Accordingly, the first term in the above equation is pressure energy per unit weight of fluid. Hence it is termed as pressure head.
Similarly, the second term represents kinetic energy per unit weight. Hence it is termed as velocity head.And the third term represents gravitational potential energy per unit weight. Hence it is termed as gravitational head.Note that each of the above heads has unit of length. 

Formula

Conditions for applicability of Bernoullis equation

Conditions for applicability of Bernoulli's principle:

• the flow must be steady, i.e. the fluid velocity at a point cannot change with time,
• the flow must be incompressible even though pressure varies, the density must remain constant along a streamline;
• friction by viscous forces has to be negligible.

Definition

Bernoulli's Equation

Bernoulli's principle states that, in a steady flow, the sum of all forms of energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy, potential energy and internal energy remains constant.

In the given diagram,
P1+12ρv21+ρgz1=P2+12ρv22+ρgz2$P_1+{\displaystyle \frac{1}{2}}\rho v_1^2+\rho g{z}_1=P_2+{\displaystyle \frac{1}{2}}\rho v_2^2+\rho g{z}_2$

Example

Use of Bernoulli's equation to explain physical situations

Observation: The stream of water flowing at high speed from a garden hose pipe tends to spread like a fountain when held vertically up, but tends to narrow down when held vertically down.

Explanation: In a hose pipe, using the equation of continuity, the volumetric flow
rate of the fluid remains the same. However, using this with Bernoulli's equation, 
P+0.5ρv2+ρgh=constant$P+0.5\rho v^2+\rho gh=constant$
When the hose pipe is faced vertically, upwards, the height increase, therefore velocity of fluid decreases and so the area of cross section for spreading increases. The case is opposite, when the hose in faced vertically downwards.
In any steady flow of an incompressible fluid, the volume flow rate of the fluid remains constant.

Example

Application of Bernoulli's equation in open atmosphere

Example: A plane is in a level flight at a constant speed and each of its two wings has an area of 25 m2${}^2$. If the speed of air is 180 km h−1${}^{-1}$ over the lower wing and 234 km h−1${}^{-1}$ over the upper wing surface, the plane's mass is. (Take density of air =$=$ 1 kg m−3${}^{-3}$.)

Solution:Applying bernouillis theorem above and below the wing,
p1+12fv21=p2+12fv22$p_1+{\displaystyle \frac{1}{2}}f{v}_1^2=p_2+{\displaystyle \frac{1}{2}}f{v}_2^2$
⇒p1−p2=12(652−502)$\Rightarrow p_1-p_2=\frac{1}{2}({65}^2-{50}^2)$
⇒p1−p2=862.5 Pa$\Rightarrow p_1-p_2=862.5Pa$
So,
upthrust on wings = weight of plane
⇒862.5×2×2510=$\Rightarrow {\displaystyle \frac{862.5\times 2\times 25}{10}}=$ mass plane
⇒Mass=4312kgs≈4400kgs

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