Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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Q.328. Water flows out of a big tank along a tube bent at right angles: the inside radius of the tube is equal to r = 0.50 cm (Fig. 1.87). The length of the horizontal section of the tube is equal to l = 22 cm. The water flow rate is Q = 0.50 litres per second. Find the moment of reaction forces of flowing water, acting on the tube's walls, relative to the point O. 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Solution. 328. Let the velocity of water flowing through the tube at a certain instant of time be u, then  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev where Q is the rate of flow of water and πr2 is the cross section area of the tube. 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

From impulse momentum theorem, for the stream of water striking the tube comer, in x-direction in the time interval dt,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Therefore, the force exerted on the water stream by the tube,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

According to third law, the reaction force on the tube’s wall by the stream equals Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Hence, the sought moment of force about 0 becomes

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.329. A side wall of a wide open tank is provided with a narrowing tube (Fig. 1.88) through which water flows out. The cross-sectional area of the tube decreases from S = 3.0 cm2  to s = 1.0 cm2. The water level in the tank is h = 4.6 m higher than that in the tube. Neglecting the viscosity of the water, find the horizontal component of the force tending to pull the tube out of the tank. 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Solution. 329. Suppose the radius at A is R and it decreases uniformaly to r at B where S = πR2 and s = πr2. Assume also that the semi vectical angle at 0 is α. Then

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

So   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

where y is the radius at the point P distant x from the vertex O. Suppose the velocity with which the liquid flows out is V at A, v at B and u at P. Then by the equation of continuity

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

The velocity v of efflux is given by

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

and Bernoulli's theorem gives

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

where pp is the pressure at P and p0 is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

We have subtracted p0 which is the force due to atmosphenic pressure the factor sinθ gives horizontal component of the force and ds is the length of the element of nozzle surface, ds = dx sec θ and

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev
Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Note : If we try to calculate F from the momentum change of the liquid flowing out will be wrong even as regards the sign of the force.

There is of course the effect of pressure at S and s but quantitative derivation of F fron Newton's law is difficult.


Q.330. A cylindrical vessel with water is rotated about its vertical axis with a constant angular velocity ω. Find:
 (a) the shape of the free surface of the water;
 (b) the water pressure distribution over the bottom of the vessel along its radius provided the pressure at the central point is equal to Po.

Solution. 330. The Euler’s equation is  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev in the space fixed frame where  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev downward. We assume incompressible fluid so p is constant.

Then Irodov Solutions: Hydrodynamics- 2 Notes | EduRev where z is the height vertically upwards from some fixed origin. We go to rotating frame where the equation becomes 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

the additional terms on the right are the well known coriolis and centrifugal forces. In the frame rotating with the liquid  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

On the free surface p = constant, thus

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

If we choose the origin at point r = 0 fi.e. the axis) of the free surface then “cosntant” = 0 and

  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

At the bottom z = constant

So     Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

If p = p0 on the axis at the bottom, then 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.331. A thin horizontal disc of radius R = 10 cm is located within a cylindrical cavity filled with oil whose viscosity η = 0.08 P (Fig. 1.89). The clearance between the disc and the horizontal planes of the cavity is equal to h = 1.0 mm. Find the power developed by the viscous forces acting on the disc when it rotates with the angular velocity ω = 60 rad/s. The end effects are to be neglected.

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Solution. 331. When the disc rotates the fuild in contact with, corotates but the fluid in contact with the walls of the cavity does not rotate. A velocity gradient is then set up leading to viscous forces. At a distance r from the axis the linear velocity is ω r so there is a velocity gradient  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev both in the upper and lower clearance. The corresponding force on the element whose radial width is dr is

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

The torque due to this force is

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

2nd the net torque considering both the upper and lower clearance is

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

So power developed is 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

(A s instructed end effects i.e. rotation of fluid in the clearance r > R has been neglected.)


Q.332. A long cylinder of radius R1 is displaced along its axis with a constant velocity v0 inside a stationary co-axial cylinder of radius R2. The space between the cylinders is filled with viscous liquid. Find the velocity of the liquid as a function of the distance r from the axis of the cylinders. The flow is laminar.

Solution. 332.  Let us consider a coaxial cylinder of radius r and thickness dr. then force of friction or viscous force on this elemental layer  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

This force must be constant from layer to layer so that steady motion may be possible.

or,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Integrating,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Putting

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

From (2) by (3) we get,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Note : The force F is supplied by the agency which tries to carry the inner cylinder with velocity v0.


Q.333. A fluid with viscosity η fills the space between two long co-axial cylinders of radii R1 and R2, with R1 < R2. The inner cylinder is stationary while the outer one is rotated with a constant angular velocity ω2. The fluid flow is laminar. Taking into account that the friction force acting on a unit area of a cylindrical surface of radius r is defined by the formula Irodov Solutions: Hydrodynamics- 2 Notes | EduRev find:
 (a) the angular velocity of the rotating fluid as a function of radius r;
 (b) the moment of the friction forces acting on a unit length of the outer cylinder.

Solution. 333. (a) Let us consider an elemental cylinder of radius r and thickness dr then from Newton’s formula

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

and moment of this force acting on the element,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

As in the previous problem N is constant when conditions are steady

Integrating,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or,   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev    (3)

Putting   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

From (3) and (4),

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

(b) From Eq. (4),

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.334. A tube of length 1 and radius R carries a steady flow of fluid whose density is p and viscosity η. The fluid flow velocity de- pends on the distance r from the axis of the tube as Irodov Solutions: Hydrodynamics- 2 Notes | EduRev Find:
 (a) the volume of the fluid flowing across the section of the tube per unit time;
 (b) the kinetic energy of the fluid within the tube's volume;
 (c) the friction force exerted on the tube by the fluid;
 (d) the pressure difference at the ends of the tube. 

Solution. 334. (a) Let dV be the volume flowing per second through the cylindrical shell of thickness dr then,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

and the total volume,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRevIrodov Solutions: Hydrodynamics- 2 Notes | EduRev

(b) Let, dE be the kinetic energy, within the above cylindrical shell. Then

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Hence, total energy of the fluid,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

(c) Here frictional force is the shearing force on the tube, exerted by the fluid, which  equals Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Given,    Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

So,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

And at   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Then, viscous force is given by,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

(d) Taking a cylindrical shell of thickness dr and radius r viscous force,

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Let Δp be the pressure difference, then net force on the element  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

But, since the flow is steady, Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or,   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.335. In the arrangement shown in Fig. 1.90 a viscous liquid whose density is p = 1.0 g/cm3  flows along a tube out of a wide tank A. Find the velocity of the liquid flow, if h1  = 10 cm, h2  = 20 cm, and h3 = 35 cm. All the distances l are equal. 

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Solution. 335. The loss of pressure head in travelling a distance l is seen from the middle section to be h2 - h1 = 10 cm. Since h2 - h1 = h1 in our problem and h3 - h2 = 15 cm = 5 + h2 h1, we see that a pressure head of 5 cm remains incompensated and must be converted into kinetic energy, the liquid flowing out. Thus

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.336. The cross-sectional radius of a pipeline decreases gradually as Irodov Solutions: Hydrodynamics- 2 Notes | EduRev is the distance from the pipeline inlet. Find the ratio of Reynolds numbers for two cross-sections separated by Δx = 3.2 m. 

Solution. 336. We know that, Reynold’s number (Re) is defined as, Irodov Solutions: Hydrodynamics- 2 Notes | EduRev where v is the velocity / is the characteristic length and r\ the coefficient of viscosity. In the case of circular cross section the chracteristic length is the diameter of cross-section d, and v is taken as average velocity of flow of liquid.

Now,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev (Reynold’s number at x1 from the pipe end)  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev is the velocity at distance x1 

and similarly,  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

From equation of continuity,   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or, Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev


Q.337. When a sphere of radius r1  = 1.2 mm moves in glycerin, the laminar flow is observed if the velocity of the sphere does not exceed v1 =  23 cm/s. At what minimum velocity v2  of a sphere of radius r2  = 5.5 cm will the flow in water become turbulent? The viscosities of glycerin and water are equal to η1 = 13.9 P and η2 = 0.011 P respectively. 

Solution. 337. We know that Reynold’s number for turbulent flow is greater than that on laminar flow.

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev  on putting the values.


Q.338. A lead sphere is steadily sinking in glycerin whose viscosity is equal to η = 13.9 P. What is the maximum diameter of the sphere at which the flow around that sphere still remains laminar? It is known that the transition to the turbulent flow corresponds to Reynolds number Re = 0.5. (Here the characteristic length is taken to be the sphere diameter.) 

Solution. 338. Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

(p = density of lead, p0 = density of glycerine.)

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

and Irodov Solutions: Hydrodynamics- 2 Notes | EduRev mm on putting the values, 


Q.339. A steel ball of diameter d = 3.0 mm starts sinking with zero initial velocity in olive oil whose viscosity is η = 0.90 P. How soon after the beginning of motion will the velocity of the ball differ from the steady-state velocity by n = 1.0%? 

Solution. 339.  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or    Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or  Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Since   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

So    Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

The steady state velocity is  g/k.

Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

or    Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

Thus   Irodov Solutions: Hydrodynamics- 2 Notes | EduRev

We have neglected buoyancy in olive oil.

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