Answer
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Hint: The dependency between fluid flow and velocity can be deduced from the continuity equation for a dynamic (flowing) fluid, while the behaviour of fluid flow and pressure with respect to velocity can be derived by analysing Bernoulli’s principle (aka Bernoulli’s effect) for a horizontal pipe. Do not forget that the narrowest part of the pipe is characterized by its lowest cross-sectional area.
Formula Used:
Continuity equation:- $Av = constant$
Bernoulli’s principle:- $P + \dfrac{1}{2}\rho v^2 + \rho gh= contant$
Complete Solution:
Let us begin by first establishing the basis of fluidic streamline flow, following which we shall analyse the behaviour of the fluid in the context of our question.
The flow in which a fluid flows in parallel layers such that there is no intermixing of any layers or disruptions between layers is called streamline or steady flow. It means that at any particular instant in time, the velocities of all constituent particles at a given point are the same but the velocities might not necessarily be the same across all points in space.
The continuity equation in fluid dynamics is used to characterize the flow rate of a steady fluid in a pipe. It states that the product of the cross-sectional area of the pipe and the fluid speed at any point along the pipe is always contacted. It is given as,
$A.v = constant \Rightarrow A \propto \dfrac{1}{v}$
From the above relation, we can thus establish that at the narrowest part of the tube, i.e., when the cross-sectional area is minimum, the velocity of the flow of fluid is maximum.
Now, let us try and deduce a relationship between velocity and pressure of the fluid by looking at Bernoulli’s principle.
Bernoulli’s principle is essentially a principle of conservation of energy for dynamic fluids which states that at any instant, the total mechanical energy of a flowing fluid remains constant. It is given as the sum of fluid pressure, kinetic energy of fluid motion and gravitational potential energy of elevation, i.e.,
$P + \dfrac{1}{2}\rho v^2 + \rho gh= constant$, where,
P is the fluid pressure, $\rho$ is the fluid density, v is the fluid velocity, g is the acceleration due to gravity and h is the height of the fluid container.
For a horizontal pipe, the height of the fluid remains the same throughout its cross-section, therefore, the gravitational potential energy remains the same at all points. The Bernoulli equation now becomes:
$P + \dfrac{1}{2}\rho v^2= constant$
This means that the sum of fluid pressure and the kinetic energy of fluid motion has to be the same at all points in the fluid.
Now, at the widest part of the pipe, the cross-sectional area is maximum. From the continuity equation, we know that this means that the fluid velocity is minimum in this region. This decrease in velocity (and consequently, the kinetic energy) must be countered by an increase in fluid pressure, so as to keep the algebraic sum of the fluid pressure and kinetic energy constant.
This means that pressure and velocity are inversely proportional to each other. As the velocity decreases the fluid molecules are able to exert more pressure than when they are moving quickly where their pressure may partially get dissipated towards their motion/velocity.Thus, the pressure is maximum at the widest region of the pipe where the fluid velocity is minimum.
We can thus conclude that the correct choice would be A. The velocity is maximum at the narrowest part of the pipe and the pressure is maximum at the widest part of the pipe.
Note:
It is important to understand that when we talk about pressure in Bernoulli’s equation and its relation with flow velocity, we are strictly referring to ‘fluid pressure’ and not any other pressure. The difference is that fluid pressure is generated by the random motion of fluid molecules whereas other forms of pressure entail external pressure or atmospheric pressure applied at any point on the container that may be transmitted throughout the fluid. The inverse relation between pressure and velocity that we established in the problem is applicable only under the influence of fluid pressure and not other forms of pressure.
Formula Used:
Continuity equation:- $Av = constant$
Bernoulli’s principle:- $P + \dfrac{1}{2}\rho v^2 + \rho gh= contant$
Complete Solution:
Let us begin by first establishing the basis of fluidic streamline flow, following which we shall analyse the behaviour of the fluid in the context of our question.
The flow in which a fluid flows in parallel layers such that there is no intermixing of any layers or disruptions between layers is called streamline or steady flow. It means that at any particular instant in time, the velocities of all constituent particles at a given point are the same but the velocities might not necessarily be the same across all points in space.
The continuity equation in fluid dynamics is used to characterize the flow rate of a steady fluid in a pipe. It states that the product of the cross-sectional area of the pipe and the fluid speed at any point along the pipe is always contacted. It is given as,
$A.v = constant \Rightarrow A \propto \dfrac{1}{v}$
From the above relation, we can thus establish that at the narrowest part of the tube, i.e., when the cross-sectional area is minimum, the velocity of the flow of fluid is maximum.
Now, let us try and deduce a relationship between velocity and pressure of the fluid by looking at Bernoulli’s principle.
Bernoulli’s principle is essentially a principle of conservation of energy for dynamic fluids which states that at any instant, the total mechanical energy of a flowing fluid remains constant. It is given as the sum of fluid pressure, kinetic energy of fluid motion and gravitational potential energy of elevation, i.e.,
$P + \dfrac{1}{2}\rho v^2 + \rho gh= constant$, where,
P is the fluid pressure, $\rho$ is the fluid density, v is the fluid velocity, g is the acceleration due to gravity and h is the height of the fluid container.
For a horizontal pipe, the height of the fluid remains the same throughout its cross-section, therefore, the gravitational potential energy remains the same at all points. The Bernoulli equation now becomes:
$P + \dfrac{1}{2}\rho v^2= constant$
This means that the sum of fluid pressure and the kinetic energy of fluid motion has to be the same at all points in the fluid.
Now, at the widest part of the pipe, the cross-sectional area is maximum. From the continuity equation, we know that this means that the fluid velocity is minimum in this region. This decrease in velocity (and consequently, the kinetic energy) must be countered by an increase in fluid pressure, so as to keep the algebraic sum of the fluid pressure and kinetic energy constant.
This means that pressure and velocity are inversely proportional to each other. As the velocity decreases the fluid molecules are able to exert more pressure than when they are moving quickly where their pressure may partially get dissipated towards their motion/velocity.Thus, the pressure is maximum at the widest region of the pipe where the fluid velocity is minimum.
We can thus conclude that the correct choice would be A. The velocity is maximum at the narrowest part of the pipe and the pressure is maximum at the widest part of the pipe.
Note:
It is important to understand that when we talk about pressure in Bernoulli’s equation and its relation with flow velocity, we are strictly referring to ‘fluid pressure’ and not any other pressure. The difference is that fluid pressure is generated by the random motion of fluid molecules whereas other forms of pressure entail external pressure or atmospheric pressure applied at any point on the container that may be transmitted throughout the fluid. The inverse relation between pressure and velocity that we established in the problem is applicable only under the influence of fluid pressure and not other forms of pressure.
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