Chapter 9: Mechanical Properties of Fluids

Ans: Bernoulli's equation cannot describe the turbulent flow of fluids like that in a rapid river. 

Water flow in a river is chaotic. Bernoulli's equation suggests that an increase in a fluid's velocity occurs concurrently with a decrease in static pressure or simultaneously with a reduction in potential energy throughout the water flow.
This theory is only applicable to isentropic flows, wherein turbulence and other disturbances are ignored.
There are various versions of Bernoulli's equation with respect to various flow kinds. For instance, incompressible flows obey the simple form of Bernoulli's equation whereas compressible flows obey an advanced form of the equation.
Bernoulli's theory is in accordance with the principle of energy conservation. Clearly, at each point on the flowing water in the river, the sum of any and all sources of energy is equal to that as in a steady flow. The sum for kinetic energy, internal energy and potential energy should also maintain stability.
Thus, Bernoulli's equation cannot describe the flow of water through a rapid in a river.

Ans: Gauge pressure gives us the pressure difference in the fluid under consideration and some standard pressure. Using gauge pressure will only tell us about the pressure difference. 

Bernoulli's principle assumes that the pressures under consideration are linearly proportional to the square of the velocity and the height of the fluid. 

However, the difference in pressure cannot be directly shown to have this proportionality. Hence, we need to convert the gauge pressure to absolute pressure before plugging it into Bernoulli's equation. 

So yes, it matters if we use gauge instead of absolute pressures in applying Bernoulli's equation.

Ans: Provided that,

Length of the horizontal tube,  \[l=1.5\text{ }m\]

The radius of the tube, \[r=1\,cm=\,0.01m\] 

Thus, the diameter of the tube, \[\therefore d=0.02m\] 

As given in the question, Glycerine is flowing at a rate of \[M=4.0\times {{10}^{-3}}kg\,{{s}^{-1}}\] 

The density of Glycerine, \[\rho =1.3\times {{10}^{3}}kg\,{{m}^{-3}}\]

Viscosity of Glycerine\[\eta =0.83\text{ }Pa\text{ }s\]

The volume of Glycerine flowing per sec: 

\[V=M/\rho =\frac{4.0\times {{10}^{-3}}}{1.3\times {{10}^{3}}}=3.08\times {{10}^{-6}}{{m}^{3}}{{s}^{-1}}\]

According to Poiseuille's Law, we have the expression for the rate of flow:

\[V=\frac{\pi P{{r}^{4}}}{8\eta l}\]

here, P is the pressure difference between the two ends of the tube

\[P=\frac{V8\eta l}{\pi {{r}^{4}}}=\text{ }9.8\text{ }\times \text{ }{{10}^{2}}Pa\] 

Reynolds' number is given by the following expression: 

\[\frac{4\rho V}{\pi d\eta }=\frac{4(1.3\times {{10}^{3}})(3.08\times {{10}^{-6}})}{\pi (0.02)0.83}=0.3\]

Reynolds' number is about \[0.3\]; therefore, the flow is laminar.

Ans: The given speed of wind on the wing's upper surface, and the speed of wind relative to the wing's lower surface \[{{v}_{2}}=63m\text{ }{{s}^{-1}}\].

The area of the wing, \[A=2.5{{m}^{2}}\]

The density of air, \[\rho =1.3kg/{{m}^{3}}\]

Now, according to Bernoulli's theorem, we have the relation: \[{{P}_{1}}+\frac{1}{2}\rho {{v}_{1}}^{2}={{P}_{2}}+\frac{1}{2}\rho {{v}_{2}}^{2}\]

In the current case, let \[{{P}_{2}}\]  be the pressure on the lower side of the wing and \[{{P}_{1}}\] be the pressure on the upper side, and the v-s are the velocities mentioned initially. 

The required lift is the force that comes from the pressure difference in the lower and the upper surface. 

This can be obtained as:

\[Lift\,=({{P}_{1}}-{{P}_{2}})A=\left( \frac{1}{2}\rho {{v}_{2}}^{2}-\frac{1}{2}\rho {{v}_{1}}^{2} \right)A\]

\[Lift\,\,=\frac{1}{2}\rho ({{v}_{2}}^{2}-{{v}_{1}}^{2})A=\frac{1}{2}1.3({{70}^{2}}-{{63}^{2}})2.5=1.51\times {{10}^{3}}N\]

Hence, \[1.51\times {{10}^{3}}N\]is the lift on the wing of the airplane.

Ans: Considering the second figure:

Let us say A1 and A2 are the cross-section areas of the pipe and the narrow region, respectively and V1 and V2 are the speeds of the fluid in the normal region of the pipe and at the narrow region, respectively

According to the law of continuity: \[A1\times V1=A2\times V2\] 

Hence it can be noted that the velocity of the liquid at the narrow region is higher than that in the normal region. 

Now using Bernoulli's theorem, it can be said that the pressure is high in a region where the velocity is low. Hence the pressure at the normally thick (at A1) will be higher than that at A2. 

Hence the column connected at this place will also have higher pressure at the point of contact. This is why the liquid column will also rise higher than the column connected at the narrow junction. Therefore, figure (b) is possible.

Ans: Provided that,

Area of the cross-section of the spray pump \[{{A}_{1}}=8\times {{10}^{-4}}{{m}^{2}}\]

Number of holes, \[n\text{ }=\text{ }40\]

Diameter of each hole, \[d={{10}^{-3}}m\]

The radius of each hole, \[r=0.5\times {{10}^{-3}}m\]

Area of the cross-section of each hole,  \[a=\pi {{r}^{2}}=\pi {{(0.5\times {{10}^{-3}})}^{2}}\]

The total area of 40 holes, \[A=n\,\,a=n\pi {{r}^{2}}=31.41\times {{10}^{-6}}{{m}^{2}}\]

The speed of flow of liquid inside the tube, \[{{v}_{1}}=1.5m/\min =0.025m/s\]

The liquid is ejected through the holes at the speed of \[{{v}_{2}}\]

According to the continuity equation, we have \[\,{{v}_{2}}=\frac{{{A}_{1}}{{v}_{1}}}{{{A}_{2}}}=0.633\text{  }m\text{/}s\] 

Therefore, it has been calculated that the speed of ejection of the liquid through the holes is \[0.633\text{ }m/s.\]

Ans: The weight that the soap film supports, \[W=1.5\times \,{{10}^{-2}}N\]

Length of the slider, \[l=0.3m\]

A soap film has two free surfaces.

Therefore, the total length \[=2l=0.6m\]

Hence the surface tension is obtained as

\[S=\frac{Force}{2l}=\frac{1.5\times \,{{10}^{-2}}N}{0.6m}=2.5\times \,{{10}^{-2}}N/m\]

Therefore, the surface tension of the film is obtained as\[2.5\times \,{{10}^{-2}}N/m\].  

Ans: For case (a): 

The length of the liquid film is supported by the weight \[l\text{ }=\text{ }40\text{ }cm\text{ }=\text{ }0.4\text{ }m\].
The weight supported by the film, \[W=4.5\times \,{{10}^{-2}}N\] 

Every liquid film has two free surfaces. Hence the surface tension can be calculated as 

\[S=\frac{Force}{2l}=\frac{4.5\times \,{{10}^{-2}}N}{2\times \,0.4m}=5.625\times \,{{10}^{-2}}N/m\]

Since the properties of the liquid are the same in all the cases., the surface tension in all three cases is equal. And the weight supported is also the same in all three cases, i.e., \[=5.625\times \,{{10}^{-2}}N/m\].

Ans: As provided, the radius of the mercury drop, \[r=3\times {{10}^{-3}}m\] 

The surface tension of mercury, \[S=4.65\times \,{{10}^{-1}}N/m\] 

The atmospheric pressure, \[{{P}_{0}}=1.01\times \,{{10}^{5}}Pa\] 

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure 

Total pressure \[=\frac{2S}{r}+{{P}_{0}}=\frac{2\times 4.65\times \,{{10}^{-1}}}{3\times {{10}^{-3}}}+1.01\times \,{{10}^{5}}=1.013\times \,{{10}^{5}}Pa\]

Hence the excess pressure is obtained as \[=\frac{2S}{r}=310Pa\]. 

Ans: As provided in the question, the excess pressure inside the soap bubble is \[20\text{ }Pa\]; 

The pressure inside the air bubble is \[=1.06\times \,{{10}^{5}}Pa\]

A soap bubble is of radius, \[r=5\times {{10}^{-3}}m\]

The surface tension of the soap solution, \[S=2.5\times \,{{10}^{-2}}N/m\]

Also, the relative density of the soap solution \[=\text{ }1.20~\]

Thus, the density of the soap solution, \[\rho =1.2\times {{\rho }_{air}}=1.2\times {{10}^{3}}kg/{{m}^{2}}\]

It is given that the air bubble is formed at a depth, \[h\text{ }=\text{ }40\text{ }cm\text{ }=\text{ }0.4\text{ }m\]

The radius of the air bubble, \[r=5\times {{10}^{-3}}m\] 

Acceleration due to gravity, \[g=9.8m/{{s}^{2}}\] 

Therefore, it can be noted that the excess pressure inside the soap bubble is given by:

\[P=\frac{4S}{r}=20Pa\]

The excess pressure inside the air bubble can be written as: 

\[P=\frac{2S}{r}=10Pa\].

The total pressure inside the air bubble under consideration is the sum of the atmospheric pressure, the pressure due to the column of the liquid and the excess pressure:

\[{{P}_{total}}={{P}_{0}}+hg\rho +P=1.057\times \,{{10}^{5}}Pa\]

 Hence, the pressure inside the air bubble is \[1.057\times \,{{10}^{5}}Pa\].