Solved Examples on Fluid Mechanics

Problem 1:-The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is T

_{0}when the containing vessel (see below figure) is at rest. Show that the tension T, when the vessel has an upward vertical acceleration a, is given by T_{0}(1+a/g).

Concept:-The various forces acting on the block when it is inside the water are the weight, the buoyant force and the tension in the string. When the vessel is at rest, there is no net force acting on the block. The sum of the three forces is equal to zero when the vessel is at rest.

When the vessel moves upward with an acceleration, the buoyant force and the tension in the string will be differ. There is a net force acting vessel. Thus, the sum of the three forces is equal to the net force acting on the block.

Solution:-The weight of the block having mass

mwhich is acting downward isW = mg

Here,

Acceleration due to gravity at the point of observation is g.

The magnitude of the buoyant which acts upwards is

F

_{b}= VρgHere,

Volume of the water displaced by the block is

VDensity of the water is ρ.

The various forces acting on the block which is placed inside the vessel is shown below:

As the vessel is rest, the three forces are in equilibrium. Thus, the sum of the three forces is equal to zero.

Thus,

ΣF = 0

This makes

W – F

_{b}– T_{0}= 0Here,

Tension in the string when the vessel is at rest is T

_{0}.Insert the values of the various terms involved in the above equation gives

F

_{b}– W – T_{0}= 0Vρg – mg – T

_{0 }= 0So, T

_{0}= Vρg – mgThis represents the tension in the string when the vessel is at rest.

When the vessel is moving with an upward vertical acceleration

a, there is net forces acting on the block. So, there is change in the buoyant force and the tension in the string.The figure which shows the forces acting on the block when the vessel is accelerating upward is

The buoyant forces acting on the block is

F

_{b,a}= Vρ (g+a)Thus, the net forces is given as

ΣF = ma

It is equal to the sum of the buoyant force, the weight of the block and tension (

T).Thus,

F

_{b,a}– W -T = MaSubstitute the values of F

_{b,a}andWin the above equation givesF

_{b,a}– W – T = maVρ (g+a) – mg – T = ma

So, T = Vρa +Vρg – mg – ma

To obtain the tension in the string, substitute T

_{0}= Vρg – mg in the above equationT = Vρg – mg + Vρa – ma = T

_{0}+ a (Vρ – m)Dividing both numerator and denominator in the second part in the right hand side of the equation by

ggivesT = T

_{0}+ [a/g (Vρg – mg)]= T

_{0}+[(a/g) T_{0}] = T_{0}(1+a/g)This represents the tension in the string when the vessel is accelerating upward.

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Problem 2:-A cubic object of dimensions L = 0.608 m on a side and weight W = 4450 N in a vacuum is suspended by a wire in an open tank of liquid of density ρ = 944 kg/m

^{3}, as shown in below figure.(a) Find the total downward force exerted by the liquid and the atmosphere on the top of the object. (b) Find the total upward force on the bottom of the object. (c) Find the tension in the wire. (d) Calculate the buoyant force on the object using Archimedes principle. What relation exists among all these quantities?

Concept:-The pressure acting on the top of the object is

p

_{T}= p_{0}+ρg (L/2)Here, atmospheric pressure is p

_{0}, density of the liquid is ρ, acceleration due to gravity is g and length of the side of the object is L.The total downward force exerted by the liquid and the atmosphere on the top of the object is equal to pressure times the surface area of the object.

Thus,

F

_{T}= p_{T}L^{2 }= [p_{0}+ ρg (L/2)] L^{2}The pressure acting on the bottom of the object is

p

_{B}= p_{0}+ρg (3L/2)The total upward force acting on the object is

F

_{B}= p_{B}L^{2 }= [p_{0}+ρg (3L/2)]L^{2}According to Archimedes’ principle the buoyant force acting on the object is

F

_{Buoyant}= L^{3}ρgThe tension in the wire is

T = W – F

_{T}+ F_{B}Here, weight of the object is W.

Solution:-(a)

One atmospheric pressure is equal to 1.01×10^{5}PaTo obtain the force acting on the bottom of the object, substitute 1.01×10

^{5}Pa for p_{0}, 944 kg/m^{3}for ρ , 9.81 m/s^{2}for g and 0.608 m for L in the equation F_{T}= [p_{0}+ ρg (L/2)]L^{2},F

_{T}= [p_{0}+ ρg (L/2)]L^{2}= [1.01×10

^{5}Pa + (944 kg/m^{3}) ( 9.81 m/s^{2}) (0.608 m/2)] (0.608 m)^{2}= (3.8376×10

^{4}kg.m/s^{2}) [1 N/(1 kg.m/s^{2})] = 3.8376×10^{4}NRounding off to three significant figures, the total downward force exerted by the liquid on the object is 3.8376×10

^{4 }N.(b) To obtain the total downward force exerted by the liquid, substitute 1.01×10

^{5}Pa for p_{0}, 944 kg/m^{3}for ρ, 9.81 m/s^{2}for g , and 0.608 m for L in the equation F_{B}= [p_{0}+ρg (3L/2)]L^{2},F

_{B}= [p_{0}+ρg (3L/2)]L^{2}= [1.01×10

^{5}Pa + (944 kg/m^{3}) ( 9.81 m/s^{2}) (3(0.608 m)/2)] (0.608 m)^{2}= (4.0455×10

^{4}kg.m/s^{2}) [1 N/(1 kg.m/s^{2})] = 4.0455×10^{4}NRounding off to three significant figures, the total downward force exerted by the liquid on the object is 4.0455×10

^{4}N.(c) To obtain the buoyant force acting on the object, substitute 0.608 m for L, 944 kg/m

^{3}for ρ, 9.81 m/s^{2}for g in the equation F_{Buoyant}= L^{3}ρg,F

_{Buoyant}= L^{3}ρg= (0.608 m)

^{3}(944 kg/m^{3}) (9.81 m/s^{2})= (2.0793×10

^{3}kg.m/s^{2}) [1 N/(1 kg.m/s^{2})] = 2.0793×10^{3}NRounding off to three significant figures, the buoyant force acting on the object due to the liquid is 2.08×10

^{3}N .(d) The various force acting on the object in the liquid is shown below:

To obtain the tension in the wire, substitute 4450 N for W, 3.84×10

^{4 }N for F_{T}and 4.05×10^{4}N for F_{B}in the equation T = W – F_{T}+ F_{B},T = W – F

_{T}+ F_{B}= 4450 N – 3.84×10

^{4 }N + 4.05×10^{4}N = 2350 NRounding off to three significant figures, the tension in the wire is 2350 N.

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Problem 3:-A cylindrical barrel has a narrow tube fixed to the top, as shown with dimensions in below figure. The vessel is filled with water to the top of the tube. Calculate the ratio of the hydrostatic force exerted on the bottom of the barrel to the weight of the water contained inside. Why is the ratio not equal to one? (Ignore the presence of the atmosphere.)

Concept:-The pressure at the bottom of the cylindrical barrel is

p = ρgh

Here, density of the liquid is ρ, acceleration due to gravity is g and height of the cylindrical barrel is h.

The hydrostatic force acting at the bottom of the barrel is

F = PA

Here, surface area of the barrel at the bottom is A.

The weight of the liquid is

W = ρgV

Here, total volume of the cylindrical barrel is V.

Solution:-Now, the total height of the cylindrical barrel is

h = 1.8 m + 1.8 m

From above figure, substitute 3.6 m for h in the equation p = ρgh gives,

p = ρgh = ρg (3.6 m)

This gives the pressure at the bottom of the barrel.

The volume of the thin barrel at the top is

V

_{1}= (4.6 cm^{2}) (1.8 m)= (4.6 cm

^{2}) (10^{-2}m/1 cm)^{2}(1.8 m) = 8.24×10^{-4}m^{3}The radius of the barrel at the lower part is

r = 1.2 m/2 = 0.6 m

The volume of the barrel at the lower part is

V

_{2}= πr^{2}(1.8 m)= (3.14) (0.6 m)

^{2}(1.8 m) = 2.03472 m^{3}Now, the total volume of the cylindrical barrel is

V = V

_{1}+ V_{2}= 8.24×10

^{-4}m^{3}+ 2.03472 m^{3}= 2.035548 m^{3}The surface area of the barrel at the bottom is

A = πr

^{2}= (3.14) (0.6 m)^{2 }= 1.1304 m^{2}To obtain the hydrostatic force exerted by the liquid at the bottom, substitute ρg (3.6 m) for p and 1.1304 m

^{2}for A in the equation F = pA gives,F = pA

= ρg (3.6 m) (1.1304 m

^{2}) = ρg (4.06944 m^{3})To obtain the weight of the liquid in the cylindrical barrel, substitute 2.035548 m

^{3}for V in the equation W = ρgV gives,W = ρgV = ρg ( 2.035548 m

^{3})Now, the ratio of the hydrostatic force to the weight of the liquid is

F/W = [ρg (4.06944 m

^{3})]/[ρg ( 2.035548 m^{3})] = 1.9992Rounding off to two significant figures, the ratio of the hydrostatic force to the weight of the liquid is 2.0.

The hydrostatic pressure is depending on the height of the liquid column. So, the same amount of liquid when taken in different volume of containers having different heights will not be the same. Weight is the volume times the density of the liquid. Therefore, the ratio is not equal to one.

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Problem 4:-The Goodyear blimp Columbia as shown in below figure is cruising slowly at low altitude, filled as usual with helium gas. Its maximum useful payload, including crew and cargo, is 1280 kg. How much more payload could the Columbia carry if you replaced the helium with hydrogen? Why not do it? The volume of the helium-filled interior space is 5000 m

^{3}. The density of helium gas is 0.160 kg/m^{3}and the density of hydrogen is 0.0810 kg/m^{3}.

Concept:-The mass of the object having density ρ and volume V is given by relation as;

m = ρV

The volume of the hydrogen and helium will be equal to the volume of the interior space.

The denser body will have larger mass.

Solution:-The mass of the hydrogen gas that should filled in the interior space is

m

_{H}= ρ_{H}VTo obtain the required mass, substitute 0.0810 kg/m

^{3}for the density of the hydrogen ρ_{H},V = 5000 m

^{3}for volume of the interiorV in the given equation m

_{H}= ρ_{H}V ,m

_{H}= ρ_{H}V= (0.0810 kg/m

^{3}) (5000 m^{3}) = 405 kgThe amount of the hydrogen required is 405 kg.

The mass of the helium gas that should filled in the interior space is

m

_{He}= ρ_{He}VTo obtain the required mass, substitute 0.160 kg/m

^{3}for the density of the helium ρ_{He},V = 5000 m^{3}for volume of the interior V in the given equation m_{He}= ρ_{He}V,m

_{He}= ρ_{He}V= ( 0.160 kg/m

^{3}) (5000 m^{3}) = 800 kgThe amount of the hydrogen gas required is 800 kg.

It is found that the mass of the helium gas is greater than the mass of the helium gas.

The difference in the mass of the gases will give the amount of payload that could be carrying the payload.

To obtain the mass of the hydrogen required, the equation required is

∆m = m

_{He}– m_{H}Substitute 800 kg for m

_{He}and 405 kg for m_{H}in the above equation, the mass of the hydrogen required is∆m =m

_{He}– m_{H}= 800 kg – 405 kg = 395 kg

Therefore, the amount of payload that the Columbia could carry is 395 kg.

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Problem 5:-In 1654 Otto von Guericke, Burgermeister of Magdeburg and inventor of the air pump, gave a demonstration before the Imperial Diet in which two teams of horses could not pull apart two evacuated brass hemispheres. (a) Show that the force F required to pull apart the hemispheres is F = πR

^{2}∆p, where R is the (outside) radius of the hemispheres and ∆p is the difference in pressure outside and inside the sphere as shown in the below figure. (b) Taking R equal to 0.305 m and the inside pressure as 0.100 atm, what force would the team of horses have had to exert to pull apart the hemispheres? (c) Why were two teams of horse used? Would not one team prove the point just as well?

Concept:-The two brass hemispheres with an open flat can be replaced with two hemispheres with a closed flat end.

The force required to pull apart the hemispheres is equal to pressure times the surface area.

Solution:-(a) It is given that the pressure difference between outside and inside the sphere is ∆p . The radius of the circle formed by the half hemispheres which face each other is R.

Now, the surface area of the hemisphere is

A = πR

^{2}The force required to pull the apart the brass hemisphere is

F = ∆pA

= ∆p (πR

^{2}) = πR^{2}∆pTherefore, the force required to pull apart the two hemispheres is πR

^{2}∆p.(b) It is given that the inside pressure is 0.11 atm.

Thus, the pressure difference between the inside and outside of the hemispheres is

∆p = 1.00 atm – 0.11 atm = 0.89 atm

Substitute 0.305 m for R and 0.89 atm for ∆p in the equation F = πR

^{2}∆pF = πR

^{2}∆p= (3.14) (0.305 m)

^{2}(0.89 atm) (1.01×10^{5}Pa/1 atm) [(1 kg/m.s^{2})/1 Pa]= (2.5997×10

^{4}kg.m/s^{2}) [1 N/(1 kg.m/s^{2})] = 2.5997×10^{4}NRounding off to three significant figures, the force required to pull apart the two hemispheres by the team of horses is 2.5997×10

^{4}.(c) The two teams of horses were used so as to pull apart the brass hemispheres. They could not do so as the pressure difference is too high. For such pressure difference, large amount of force is required.

Problem 6:-The below figure displays the phase diagram of carbon, showing the ranges of temperature and pressure in which carbon will crystallize either as diamond or graphite. What is the minimum depth at which diamonds can form if the local temperature is 100ºC and the subsurface rocks have density 3.1 g/cm

^{3}. Assume that, as in a fluid, the pressure is due to the weight of material lying above.

Concept:-The gauge pressure of the carbon lysing at a depth h from the surface of the Earth is

p = ρgh

Here, density of the subsurface rocks is ρ and acceleration due to gravity is g.

Solution:-The corresponding phase diagram of carbon is represented as

From the graph, the point A gives the corresponding gauge pressure at the temperature of 1000ºC. It is found to be 4 GPa.

From the equation p = ρgh, the minimum depth at which diamonds can form at the temperature of 1000ºC is

h = p/ρg

Substitute 4 GPa for p, 3.1 g/cm

^{3}for ρ and 9.8 m/s^{2}for g in the equation h = p/ρg givesh = p/ρg

= (4 GPa) (10

^{9}Pa/1 GPa) [(1 kg/m.s^{2})/1 Pa] / (3.1 g/cm^{3}) (10^{-3}kg/1 g) (10^{2}cm/1 m)^{3}(9.8 m/s^{2})= 1.3167×10

^{5}mRounding off to two significant figures, the minimum depth at which diamond can form from the carbon depth below the Earth is 1.3167×10

^{5}m.___________________________________________________________________________________

Problem 7:-In analyzing certain geological features of the Earth, it is often appropriate to assume that the pressure at some horizontal level of compensation, deep in the Earth, is the same over a large region and is equal to the exerted by the weight of the overlying material. That is, the pressure on the level of compensation is given by the hydrostatic (fluid) pressure formula. This requires, for example, that mountains have low-density roots; as shown in below figure. Consider a mountain 6.00 km high. The continental rocks have a density of 2.90 g/cm

^{3}; beneath the continent is the mantle, with a density of 3.30 g/cm^{3}. Calculate the depth D of the root. (Hint: Set the pressure at points a and b equal; the depth y of the level of compensation will cancel out.)

Concept:-The gauge pressure at a depth h from the surface of the Earth is

p = ρgh

Here, acceleration due to gravity is g and density of the material inside the Earth is ρ .

Solution:-Referring the figure 15-25 given in the problem, the pressure at the point a should be considered from the top of the mountain.

So, the corresponding pressure is

p

_{a}= ρ_{c}g_{ }(6.0 km + 32 km + D) + ρ_{M}g (y – D)Here, density of the material of the continent is ρ

_{c}, density of the material of the mantle is ρ_{M}and depth of the mantle is y.p

_{b}= ρ_{c}gh_{c}+ ρ_{M}g yReferring the figure 15-25 given in the problem, the pressure at the point b is

Here, depth of the continent is h

_{C}.The hydrostatic pressures at the points a and b is equal.

Thus,

p

_{a}= p_{b}ρ

_{c}gh_{c}+ ρ_{M}gy = ρ_{c}g (6.0 km + 32 km + D) + ρ_{M}g (y – D)ρ

_{c}h_{c}+ ρ_{M}y = ρ_{c}(6.0 km + 32 km + D) + ρ_{M}y –ρ_{M}DD = (ρ

_{c }/ ρ_{M}– ρ_{c}) [(6.0 km + 32 km) – h_{C}]Substitute 2.9 g/cm

^{3}for ρ_{c}, 3.3 g/cm^{3}for ρ_{M}and 32 km for h_{C}in the above equation givesD = (ρ

_{c }/ ρ_{M}– ρ_{c}) [(6.0 km + 32 km) – h_{C}]= {[2.9 g/cm

^{3}] / [3.3 g/cm^{3}– 2.9 g/cm^{3}]} [(6.0 km + 32 km) – 32 km]= (7.25) 6.0 km) (10

^{3}m/1 km) = 43.5×10^{3}mTherefore, the depth D of the root is 43.5×10

^{3}m.

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