Physics HYDRAULIC MACHINES, STREAMLINE FLOW, BERNOULLI’S PRINCIPLE AND SPEED OF EFFLUX

### Topic Covered

HYDRAULIC MACHINES
STREAMLINE FLOW
BERNOULLI’S PRINCIPLE
SPEED OF EFFLUX : TORRICELLI’S LAW

### Hydraulic Machines

Let's Consider a horizontal cylinder with a piston and three vertical tubes at different points. The pressure in the horizontal cylinder is indicated by the height of liquid column in the vertical tubes.It is necessarily the same in all.

If we push the piston, the fluid level rises in all the tubes, again reaching the same level in each one of them.

This indicates that when the pressure on the cylinder was increased, it was distributed uniformly throughout. We can say whenever external pressure is applied on any part of a fluid contained in a vessel, it is transmitted undiminished and equally in all directions. This is the Pascal’s law for transmission of fluid pressure and has many applications in daily life.

A number of devices such as hydraulic lift and hydraulic brakes are based on the Pascal’s law. In these devices fluids are used for transmitting pressure.

In a hydraulic lift as shown in Fig. 10.6 two pistons are separated by the space filled with a liquid. A piston of small cross section A_1 is used to exert a force F_1 directly on the liquid. The pressure P= (F_1)/(A_1) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A_2, which results in an upward force of P × A_2.

Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, directly on the liquid.

The pressure P= (F_1)/(A_1) is transmitted throughout the liquid to the larger cylinder attached with a larger piston of area A_2, which results in an upward force of P × A_2. Therefore, the piston is capable of supporting a large force (large weight of, say a car, or a truck, placed on the platform) color{red}(F_2 = PA_2 = (F_1 A_2)/(A_1)) By changing the force at A_1, the platform can be moved up or down.

Thus, the applied force has been increased by a factor of (A_2)/(A_1)and this factor is the mechanical advantage of the device. The example below clarifies it.

Hydraulic brakes in automobiles also work on the same principle. When we apply a little force on the pedal with our foot the master piston moves inside the master cylinder, and the pressure caused is transmitted through the brake oil to act on a piston of larger area.

A large force acts on the piston and is pushed down expanding the brake shoes against brake lining. In this way a small force on the pedal produces a large retarding force on the wheel.

An important advantage of the system is that the pressure set up by pressing pedal is transmitted equally to all cylinders attached to the four wheels so that the braking effort is equal on all wheels.
Q 3270601516

Two syringes of different cross sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm and 3.0 cm respectively. (a) Find the force exerted on the larger piston when a force of 10 N is applied to the smaller piston. (b) If the smaller piston is pushed in through 6.0 cm, how much does the larger piston move out?
Class 11 Chapter 11 Example 5
Solution:

Since pressure is transmitted undiminished throughout the fluid,

color{green} {F_2 = (A_2)/(A_1) F_1}

 = (pi(3//2xx10^(-2)m)^2)/(pi(1//2xx10^(-2)m)^2 xx10N

(b) Water is considered to be perfectly incompressible. Volume covered by the movement of smaller piston inwards is equal to volume moved outwards due to the larger piston
color{purple}{L_1A_1 = L_2A_2}

color{orange} {L_2 = (A_1)/(A_2) L_2}

 = (pi(1//2xx10^(-2)m)^2)/(pi(3//2xx10^(-2)m)^2) xx6xx10^(-2)m

Note, atmospheric pressure is common to both pistons and has been ignored.
Q 3200601518

In a car lift compressed air exerts a force F_1 on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm (Fig 10.7). If the mass of the car to be lifted is 1350 kg, calculate F_1 What is the pressure necessary to accomplish this task? (g = 9.8 ms^(-2)).
Class 11 Chapter 10 Example 6
Solution:

Since pressure is transmitted undiminished throughout the fluid,

color{green} {F_1 = (A_1)/(A_2)F_2}

 = (pi(5xx10^(-2)m)^2)/(pi(15xx10^(-2)m)^2) (1350Nxx9.8ms^(-2)

= 1470N
color{orange} {approx 1.5 × 10^3 N}
The air pressure that will produce this force is

color{purple} {P = (F_1)/(A_1)}

 = (1.5xx10^3N)/(pi(5xx10^(-2))^2) m = 1.9xx10^5Pa

This is almost double the atmospheric pressure.

### STREAMLINE FLOW

"So far we have studied fluids at rest." The study of the fluids in motion is known as "fluid dynamics."

When a water-tap is turned on slowly, the water flow is smooth initially, but loses its smoothness when the speed of the outflow is increased.

In studying the motion of fluids we focus our attention on what is happening to various fluid particles at a particular point in space at a particular time. The flow of the fluid is said to be steady if at any given point, the velocity of each passing fluid particle remains constant in time.

This does not mean that the velocity at different points in space is same. The velocity of a particular particle may change as it moves from one point to another.

That is, at some other point the particle may have a different velocity, but every other particle which passes the second point behaves exactly as the previous particle that has just passed that point. Each particle follows a smooth path, and the paths of the particles do not cross each other.

The path taken by a fluid particle under a steady flow is a streamline. It is defined as a curve whose tangent at any point is in the direction of the fluid velocity at that point. Consider the path of a particle as shown in Fig.10.7 (a), the curve describes how a fluid particle moves with time.

The curve PQ is like a permanent map of fluid flow, indicating how the fluid streams. No two streamlines can cross, for if they do, an oncoming fluid particle can go either one way or the other and the flow would not be steady.

Hence, In steady flow, the map of flow is stationary in time. Consider planes perpendicular to the direction of fluid flow e.g., at three points P, R and Q in Fig.10.7 (b). The plane pieces are so chosen that their boundaries be determined by the same set of streamlines.

This means that number of fluid particles crossing the surfaces as indicated at P, R and Q is the same. If area of cross-sections at these points are A_P,A_R and A_Q and speeds of fluid particles are v_Pv_R and v_Q then mass of fluid Äm_P crossing at A_P in a small interval of time Ät is ρ_PA_Pv_P Ät.

Similarly mass of fluid Äm_R flowing or crossing at A_R in a small interval of time Ät is ρ_RA_Rv_RÄt and mass of fluid Ä_mQ is
ρ_QA_Qv_Q Ät crossing at A_Q.

The mass of liquid flowing out equals the mass flowing in, holds in all cases. Therefore

color{orange}(rho_PA_Pv_P Ät = ρ_RA_Rv_R Ät = ρ_QA_Qv_Q Ät) ..........(10.9)

For flow of incompressible fluids color{red}(rho_P =rho_R =rho_Q) Equation (10.9) reduces to

color{blue}(A_Pv_P = A_Rv_R = A_Qv_Q)........(10.10)

which is called the "equation of continuity" and it is a statement of conservation of mass in flow of incompressible fluids.

In general Av = constant ..........(10.11)

Av gives the volume flux or flow rate and remains constant throughout the pipe of flow.

Thus, at narrower portions where the streamlines are closely spaced, velocity increases and its vice versa. From (Fig 10.7b) it is clear that color{navy}(A_R> A_Q \ \ "or" \ \ v_R< v_Q) , the fluid is accelerated while passing from R to Q.

This is associated with a change in pressure in fluid flow in horizontal pipes. Steady flow is achieved at low flow speeds. Beyond a limiting value, called critical speed, this flow loses steadiness and becomes turbulent.

One sees this when a fast flowing stream encounters rocks, small foamy whirlpool-like regions called ‘white water rapids are formed.
Figure 10.8 displays streamlines for some typical flows. For example, Fig. 10.8(a) describes a laminar flow where the velocities at different points in the fluid may have different magnitudes but their directions are parallel. Figure 10.8 (b) gives a sketch of turbulent flow.

### BERNOULLI’S PRINCIPLE

Fluid flow is a complex phenomenon. But we can obtain some useful properties for steady or streamline flows using the conservation of energy.

Consider a fluid moving in a pipe of varying cross-sectional area. Let the pipe be at varying heights as shown in Fig. 10.9. We now suppose that an incompressible fluid is flowing through the pipe in a steady flow.

Its velocity must change as a consequence of equation of continuity. A force is required to produce this acceleration, which is caused by the fluid surrounding it, the pressure must be different in different regions.

Bernoulli’s equation is a general expression that relates the pressure difference between two points in a pipe to both velocity changes (kinetic energy change) and elevation (height) changes (potential energy change).

The Swiss Physicist Daniel Bernoulli developed this relationship in 1738. Consider the flow at two regions 1 (i.e. BC) and 2 (i.e. DE). Consider the fluid initially lying between B and D.

In an infinitesimal time interval Delta_t, this fluid would have moved. Suppose v_1 is the speed at B and v_2 at D, then fluid initially at B has moved a distance color{navy}(v_1Delta_t to C (v1Deltat) is small enough to assume constant cross-section along BC).

In the same interval Delta_t the fluid initially at D moves to E, a distance equal to v_2Delta_t. Pressures P_1 and P_2 act as shown on the plane faces of areas A_1 and A_2 binding the two regions. The work done on the fluid at left end (BC) is color{orange}(W_1 = P_1A_1(v_1Deltat) = P_1DeltaV).

Since same volume Delta_V passes through both the regions (from the equation of continuity) the work done by the fluid at the other end (DE) is color{green}(W_2 = P_2A_2(v_2Deltat) = P_2DeltaV) or, the work done on the fluid is P_2DeltaV.

The total work done on the fluid is color{navy}(W_1 – W_2 = (P_1− P_2) DeltaV)

Part of this work goes into changing the kinetic energy of the fluid, and part goes into changing the gravitational potential energy. If the density of the fluid is r and Deltam = PA_1v_1_Deltat = rhoDeltaV is the mass passing through the pipe in time Deltat, then change in gravitational potential energy is DeltaU = PgDeltaV (h_2 − h_1) The change in its kinetic energy is DeltaU= pgDeltaV(h_2-h_1)

The change in its kinetic energy is

color{green}(DeltaK=(1/2)rho DeltaV( (v_(2)^(2) - v_(1)^(2))

We can employ the work – energy theorem (Chapter 6) to this volume of the fluid and this yields

color{navy}((P_1-P_2) DeltaV= (1/2)rho DeltaV(v_(2)^(2) - (v_(1))^(2)) + rho gDeltaV(h_2-h_1))

We now divide each term by Delta_V to obtain

color{red}((P_1-P_2)= (1/2)rho (v_(2)^(2) -v_(1)^(2) )+rho g(h_2-h_1))

We can rearrange the above terms to obtain

color{grey}(P_1+(1/2)rho v_(1)^(2) +rho gh_1=P_2+(1/2) rho v_(2)^(2) + rhogh_2) .....(10.12)

This is Bernoulli’s equation. Since 1 and 2 refer to any two locations along the pipeline, we may write the expression in general as

color{navy}(P+(1/2)rho v^2+ρgh="constant.")........(10.13)

In words, the Bernoulli’s relation may be stated as follows: As we move along a streamline the sum of the pressure (P), the kinetic energy per unit volume (ρV^2)/2 and the potential energy per unit volume (rgh) remains a constant.

"Note" that in applying the energy conservation principle, there is an assumption that no energy is lost due to friction. But in fact, when fluids flow, some energy does get lost due to internal friction.

This arises due to the fact that in a fluid flow, the different layers of the fluid flow with different velocities. These layers exert frictional forces on each other resulting in a loss of energy. This property of the fluid is called viscosity and is discussed in more detail in a later section. The lost kinetic energy of the fluid gets converted into heat energy.

Thus, Bernoulli’s equation ideally applies to fluids with zero viscosity or non-viscous fluids. Another restriction on application of Bernoulli theorem is that the fluids must be incompressible, as the elastic energy of the fluid is also not taken into consideration.

In practice, it has a large number of useful applications and can help explain a wide variety of phenomena for low viscosity incompressible fluids.

Bernoulli’s equation also does not hold for non-steady or turbulent flows, because in that situation velocity and pressure are constantly fluctuating in time. When a fluid is at rest i.e. its velocity is zero everywhere, Bernoulli’s equation becomes

color{gray}(P_1+ρgh_1=P_2+ρgh_2)

color{orange}((P_1-p_2)=ρg(h_2-h_1))

which is same as Eq. (10.6).

### Speed of Efflux : Torricelli’s Law

The word efflux means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body.

Consider a tank containing a liquid of density r with a small hole in its side at a height y_1 from the bottom (see Fig. 10.10). The air above the liquid, whose surface is at height y_2, is at pressure P. From the equation of continuity [Eq. (10.10)] we have

color{red}(v_1A_1 =v_2A_2)

color{green}(v_2=(A_1)/(A_2)v_1)

If the cross sectional area of the tank A_2 is much larger than that of the hole (A_2 > >A_1), then we may take the fluid to be approximately at rest at the top, i.e. v_2 = 0.

Now applying the Bernoulli equation at points 1 and 2 and noting that at the hole P_1 = P_a, the atmospheric pressure, we have from Eq. (10.12)

color{red}(P_a+1/2rho v_(1)^(2)+rho gy_1=P+rho gy_2)

Taking y_2 – y_1 = h we have

color{blue}(v_1= sqrt(2gh+(2(P-P_a))/rho) ........(10.14)

When P > >P_a and 2 g h may be ignored, the speed of efflux is determined by the container pressure. Such a situation occurs in rocket propulsion. On the other hand if the tank is open to the atmosphere, then P = P_a and

color{navy}(v_1= sqrt(2gh))......(10.15)

This is the speed of a freely falling body. Equation (10.15) is known as "Torricelli’s law."