Examples B

Section 9.6 Examples B

Viscosity.

Example 9.6.1.

A fluid moves along length 0.75 m with velocity 2m/s and has shearing stress of \(2 \,N/m^{2}\text{.}\) Calculate its viscosity.

Solution.

Given: Velocity, v = 2m/s, length, L = 0.75 m, shearing stress, \(p = 2 N/m^{2}.\)

The shearing stress is given by,

\begin{equation*} F = \eta A \frac{v}{L} \end{equation*}

\begin{equation*} \text{or,} p =\frac{F}{A} =\eta \frac{v}{L} \end{equation*}

\begin{equation*} \therefore \quad \eta = \frac{pL}{v} = \frac{2 \,N/m^{2}\times 0.75 \,m}{2 \,m/s} \end{equation*}

\begin{equation*} = 0.75 \,Ns/m^{2} = 0.75 \,Pa.s = 7.5 \,Poise \end{equation*}

Example 9.6.2.

A plate 0.025 mm apart from a fixed plate moves at 60 cm/s and requires force of \(2 \,N/m^{2}\) to maintain the speed. Determine the fluid viscosity between the plates.

Solution.

Given: Velocity, v = 60 cm/s = 0.6m/s, length, L = 0.025 mm = \(0.025\times 10^{-3}m\text{,}\) shearing stress, \(p = 2 N/m^{2}.\)

The shearing stress is given by,

\begin{equation*} F = \eta A \frac{v}{L} \end{equation*}

\begin{equation*} \text{or,} \quad p =\frac{F}{A} =\eta \frac{v}{L} \end{equation*}

\begin{equation*} \therefore \quad \eta = \frac{pL}{v} = \frac{2 N/m^{2}\times 0.025\times 10^{-3}m}{0.6 \,m/s} \end{equation*}

\begin{equation*} = 8.33\times 10^{-5} \,Ns/m^{2} = 8.33\times 10^{-5} \,Pa.s \end{equation*}

Example 9.6.3.

The space between two flat parallel plates is filled with an oil. Each side of the plate is 60 cm. The thickness of the oil plate is 12.5 mm. The upper plate which moves at 2.55 m/s requires a force of 98.1 N to maintain the speed. Determine, the dynamic viscosity of oil in cP.

Solution.

Given: b=60 cm=0.6 m, L=12.5 mm = \(12.5\times 10^{-3} m\text{,}\) v=2.55 m/s, F = 98.1 N, and \(A=b^{2} = 0.6 = 0.36 m^{2}.\)

The shearing stress is given by,

\begin{equation*} F = \eta A \frac{v}{L} \end{equation*}

\begin{equation*} \therefore \quad \eta = \frac{FL}{Av} =\frac{98.1 \,N\times 12.5\times 10^{-3} \,m}{0.36\, m^{2}\times 2.55 \,m/s} \end{equation*}

\begin{equation*} =1.3358 \,Ns/m^{2} =1.3358 \,Pa.s =13.36 \,P= 1336 \,cP \end{equation*}

Example 9.6.4.

The flow of blood through a large artery extending from heart to lungs with a radius of 2.5 mm and is found to be 15 cm long. The pressure across the artery ends is found to be 380 Pa then what is the blood’s average speed?

Solution.

The blood viscosity \(\eta = 0.0027 \,N.s/m^{2},\) the arterial radius, \(r= 2.5 \,mm=2.5\times 10^{-3} \,m\text{,}\) the artery length, \(l= 10 \,cm=0.10 \,m\text{,}\) and the difference of pressure, \(\Delta p= 380 Pa\text{.}\)

The volume flow rate is given by

\begin{equation*} \frac{\,dV}{\,dt} = \left(\frac{\pi \Delta p}{8\eta l}\right)r^{4} \end{equation*}

The volume flow rate is also given by

\begin{equation*} \frac{\,dV}{\,dt} = Av = \pi r^{2}v \end{equation*}

From these equantions, we get the velocity of blood flow as

\begin{equation*} \pi r^{2}v = \left(\frac{\pi \Delta p}{8\eta l}\right)r^{4} \end{equation*}

\begin{equation*} \therefore\quad v = \left(\frac{r^{2} }{8\eta l}\right)\Delta p \end{equation*}

\begin{equation*} = \left(\frac{0.0025^{2} \,m^{2}}{8(0.0027 \,Ns/m^{2}) (0.10 \,m)}\right)(380 \,N/m^{2})= 1.099 \,m/s \end{equation*}

Example 9.6.5.

A total volume of 3 ml of a viscous fluid is delivered into a patient in 10 s. When the needle is changed, 3 ml of the same fluid takes 8 s to be delivered into the patient under the identical conditions. If the radius of the original needle was 0.3 mm, what is the radius of the new needle?

Solution.

The volume flow rate is given by

\begin{equation*} \frac{\,dV}{\,dt} = \left(\frac{\pi \Delta p}{8\eta l}\right)r^{4} \end{equation*}

\begin{equation*} \text{or,}\quad \left(\frac{\pi \Delta p}{8\eta l}\right) = \frac{\frac{\,dV}{\,dt}}{r^{4}}=\frac{3 ml}{10 s\times 0.3^{4} mm^{4}} \end{equation*}

\begin{equation*} \therefore\quad \left(\frac{\pi \Delta p}{8\eta l}\right) = \frac{3 ml}{10 s\times 0.3^{4} mm^{4}} = \frac{3 ml}{8 s\times R^{4}} \end{equation*}

\begin{equation*} \text{or,}\quad R^{4} = \frac{10\times 0.3^{4}}{8} \end{equation*}

\begin{equation*} \therefore\quad R = 0.32 \,mm \end{equation*}

Example 9.6.6.

A 3 mm diameter glass ball (\(\rho=2500 \,kg/m^{3}\)) is dropped into a fluid whose density is \(875 \,kg/m^{3},\) and the terminal velocity of the ball is measured to be 0.16 m/s. Determine the viscosity of the fluid.

Solution.

Given: \(r = 1.5\,mm = 0.0015\,m, \quad \rho=2500 \,kg/m^{3}, \quad \sigma=875 kg/m^{3},\) and \(v = 0.16 m/s.\) The viscous force in a fluid is given by

\begin{equation*} F=6\pi \eta r v \end{equation*}

But ball experiences buoyant force and viscous force while falling into a liquid. After achieving the terminal velocity, we have -

\begin{equation*} F+B=mg \end{equation*}

\begin{equation*} \text{or,}\quad 6\pi \eta r v + V\sigma g = V\rho g \end{equation*}

where \(\sigma\) is a density of the fluid. Hence on solving the above equation, we get -

\begin{equation*} \eta = \frac{2 r^{2}(\rho-\sigma) g}{9 v} \end{equation*}

[Subsubsection 9.4.3.1]

\begin{equation*} \therefore \quad \eta = \frac{2 (0.0015)^{2}(2500-875) 9.8}{9 \times 0.16} = 0.049 \,Ns/m^{2} \end{equation*}

Surface Tension.

Example 9.6.7.

A 5 cm long needle is floating on the surface water without wetting. Find its weight, if surface tension \(\gamma =0.07 \,N/m.\)

Solution.

If \(W\) is the weight of the needle, then

\begin{equation*} W = 2F\cos\theta \end{equation*}

\begin{equation*} \text{or,}\quad W = 2\gamma l \cos\theta \end{equation*}

\([\because F = \gamma l]\)

\begin{equation*} \text{or,}\quad W = 2\times 0.07 \,(N/m)\times (0.05 \,m)\times 1 \end{equation*}

for a needle floating without wetting, \(\theta=0.\)

\begin{equation*} \therefore \quad W = 0.007 \,N \end{equation*}

Example 9.6.8.

A thin ring of radius 3 cm is rested flat on the liquid surface. When the ring is raised 0.03 N of force difference is noted before and after the film breaks. What is the surface tension of the liquid?

Solution.

When the ring is lifted gently form the surface of a liquid it forms a thin film around the ring. The film shrinks in it due to cohesive force as the ring pulled up and eventually the contact angle becomes zero just before it breaks. Hence in that case,

\begin{equation*} F_{1} = F_{t}+W \end{equation*}

with film adheres to ring, and

\begin{equation*} F_{2} = W \end{equation*}

when film breaks.

\begin{equation*} \therefore\quad F_{1} - F_{2} = F = F_{t}+W -W \end{equation*}

where \(F_{t}\) is a force due to surface tension of the film.

\begin{equation*} \text{or,}\quad 0.03 N = 2\gamma l \end{equation*}

The factor 2 is taken because of two surfaces of the film both inside and outside of the ring.

\begin{equation*} \text{or,} \quad 0.03 N = 2\gamma (2\pi r) \end{equation*}

\begin{equation*} \therefore\quad \gamma = \frac{0.03 \,N}{4\pi\times 0.03 \,m} = 0.08 \,N/m \end{equation*}

Example 9.6.9.

A rectangle wire frame has a length of 4 cm and breadth of 3 cm. What force does the side experience due to surface tension of the soap bubble? If \(\gamma = 0.030 \,N/m.\)

Solution.

\begin{equation*} \gamma = \frac{F\Delta x}{\Delta A} = \frac{F\Delta x}{l_{eff}\Delta x}= \frac{F}{l_{eff}} \end{equation*}

where, \(l_{eff} = 2 \times 2(l+b)\) as liquid film has two surfaces around perimeter of the frame.

\begin{equation*} \therefore\quad F = \gamma l_{eff} \end{equation*}

\begin{equation*} = (0.03 \,N/m)\times 4 \times (0.04 +0.03) \,m = 0.0084 \,N = 8.4\times10^{-3} \,N \end{equation*}

Example 9.6.10.

A U-shaped wire having a soap film on it can supports an extra mass m on its movable side. If the movable side of frame is 20 cm in length. Find the mass m, if \(\gamma =0.03 \,N/m\text{.}\)

Solution.

\begin{equation*} \gamma = \frac{F}{2l} \end{equation*}

\begin{equation*} \text{or,}\quad F = mg = 2l\gamma \end{equation*}

\begin{equation*} \text{or,}\quad m = \frac{2l\gamma }{g} \end{equation*}

\begin{equation*} = \frac{2\times 0.2 \,m \times 0.03 \,N/m }{9.8 \,m/s^{2}} = 1.2\times 10^{-3} \,kg \end{equation*}

Example 9.6.11.

A soap bubble 50 mm in diameter contains an excess pressure of 2 bar balancing the atmospheric pressure. Find the surface tension in the soap film.

Solution.

Given: \(r = \frac{50}{2}=25 \,mm= 25\times 10^{-3} \,m,\) \(\Delta p = 2 \,bar = 2\times 10^{5} \,N/m^{2}.\)

\begin{equation*} \Delta p = \frac{4\gamma}{r} \end{equation*}

\begin{equation*} \therefore \quad \gamma = \frac{r\Delta p}{4} \end{equation*}

\begin{equation*} = \frac{25\times 10^{-3}\,m\times 2\times 10^{5}\,N/m^{2}}{4} = 1250 \,N/m \end{equation*}

Example 9.6.12.

Water has a surface tension of 0.4 N/m. If the liquid rises 6 mm above the liquid surface in a 3 mm diameter vertical tube, calculate the contact angle.

Solution.

Given: \(\gamma = 0.4 \,N/m, \) \(r = \frac{d}{2} = \frac{3}{2} \,mm = 1.5\times 10^{-3}\,m,\) \(h=6mm = 6\times 10^{-3} \,m, \quad \rho = 1000 \,kg/m^{3}.\)

Capillary rise due to surface tension is given by

\begin{equation*} h = \frac{2\gamma\cos\theta}{\rho r g} \end{equation*}

\begin{equation*} \text{or,}\quad 6\times 10^{-3} = \frac{(2\times 0.4 \,N/m)\cos\theta}{(1000 \,kg/m^{3})(1.5\times 10^{-3} \,m) 9.8 \,m/s^{2}} \end{equation*}

\begin{equation*} \text{or,}\quad \cos\theta = 0.11025 \end{equation*}

\begin{equation*} \therefore \theta= \cos^{-1}(0.11025) = 83.7^{0} \end{equation*}

Example 9.6.13.

Compare the capillary rise of water and mercury in a glass tube of 2 mm diameter at \(20^{o}C\text{.}\) Given that the surface tension of water and mercury at \(20^{o}C\) are 0.0736 N/m and 0.051 N/m respectively. Contact angles of water and mercury are \(0^{o}\) and \(130^{o},\) respectively.

Solution.

Given: \(\gamma_{w} = 0.0736 \,N/m,\) \(\gamma_{m} = 0.051 \,N/m,\) specific weight of mercury \(\rho_{m} g = 9810\times 13.6 \,N/m^{3}\) and \(\theta_{m}= 130^{o}\text{,}\) specific weight of water \(\rho_{w} g = 9810 \,N/m^{3}\text{,}\) \(\theta_{w}= 0^{o},\) and \(r = 1 \,mm =0.001 \,m. \)

Capillary rise due to surface tension is given by

\begin{equation*} h_w = \frac{2\gamma\cos\theta}{\rho r g} \end{equation*}

\begin{equation*} \therefore \quad h_{w} = \frac{2\gamma\cos\theta}{\rho r g} \end{equation*}

\begin{equation*} = \frac{2\times 0.0736 \cos(0)}{9810 \times 0.001} = 15\times 10^{-3}\,m = 15 \,mm \end{equation*}

and

\begin{equation*} h_{m} = \frac{2\gamma\cos\theta}{\rho r g} \end{equation*}

\begin{equation*} = \frac{2\times 0.051 \cos(130)}{9810\times 13.6 \times 0.001} =-4.9\times10^{-4}\,m = -0.5\,mm \end{equation*}

Example 9.6.14.

Find the work done in blowing a soap bubble of radius 10 cm. If \(\gamma =0.035 N/m.\)

Solution.

Given: \(r=0.1 \,m,\quad \gamma =0.035 \,N/m.\)

\begin{equation*} \gamma = \frac{W}{A} \end{equation*}

\begin{equation*} \therefore \quad W = \gamma A \end{equation*}

\begin{equation*} = 0.035\times 2\times (4\pi r^{2}) =8.8 \times 10^{-3} \,J \end{equation*}

The factor 2 is due to two surfaces of the soap bubble.

Example 9.6.15.

A drop of mercury of radius 1 mm is broken into 8 droolets of the same size. Find the work done if the surface tension of mercury is 0.55 N/m.

Solution.

The volume of large drop = n (volume of small drops).

\begin{equation*} \frac{4}{3}\pi R^{3} = n\frac{4}{3}\pi r^{3} \qquad \Rightarrow\quad n = \left(\frac{R}{r}\right)^{3} \end{equation*}

Now, initial surface area of drop \(= 4\pi R^{2}\text{,}\)

final surface area of drop \(= n(4\pi r^{2})\)

\begin{equation*} \therefore\quad \Delta A = n(4\pi r^{2})- 4\pi R^{2} = 4\pi (nr^{2}-R^{2}) \end{equation*}

A spherical drop of liquid has only one surface, rather than two surfaces, for there is no air within it.

\begin{equation*} \therefore\quad \Delta A =4\pi (nr^{2}-R^{2}) \end{equation*}

\begin{equation*} = 4\pi R^{3} \left(\frac{1}{r} - \frac{1}{R}\right)= 4\times \pi\times R^{2} \left(n^{1/3}-1\right) = =1.24\times 10^{-5} \,m^{2} \end{equation*}

\begin{equation*} \therefore \quad W = \gamma \Delta A = 0.55 \,N/m \times1.24\times 10^{-5} \,m^{2} = 6.8\times 10^{-6} \,J \end{equation*}

Example 9.6.16.

Eight droplets of water each of radius 0.2 mm coalesce together to form a single drop. Find the change in total surface energy. \(\gamma=0.072 \,N/m. \)

Solution.

Given: \(n = 8, \quad \gamma=0.072 \,N/m, \quad r =0.2 \,mm =0.2 \times 10^{-3} \,m. \)

The volume of large drop = n (volume of small drops).

\begin{equation*} \frac{4}{3}\pi R^{3} = n\frac{4}{3}\pi r^{3} \end{equation*}

\begin{equation*} \Rightarrow\quad n = \left(\frac{R}{r}\right)^{3} \end{equation*}

\begin{equation*} \text{or,}\quad R = n^{1/3}r = 2(0.0002)= 0.0004 \,m \end{equation*}

Now, initial surface area of drop \(= 4\pi R^{2}\text{,}\)

final surface area of drop \(= n(4\pi r^{2}).\)

Since a spherical drop of liquid has only one surface, rather than two surfaces, for there is no air within it.

\begin{equation*} \therefore\quad \Delta A = n(4\pi r^{2})- 4\pi R^{2} = 4\pi (nr^{2}-R^{2}) \end{equation*}

\begin{equation*} W=E = \gamma \Delta A = 4\pi \gamma (nr^{2}-R^{2}) \end{equation*}

\begin{equation*} = 4\pi \times 0.072 [8\times (0.0002)^{2}-(0.0004)^{2}] \end{equation*}

\begin{equation*} \therefore \quad E = 1.45\times 10^{-7} \,J \end{equation*}

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