A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross

Q: Why didn't we convert the cross-sectional areas from square centimeters to square meters in the continuity equation?

Since the Equation of Continuity involves a ratio of areas ($v_2 = \frac{A_1}{A_2} v_1$), any conversion factor (like $10^{-4}$ for converting $\text{cm}^2$ to $\text{m}^2$) appears in both the numerator and the denominator and cancels out completely. Thus, as long as both areas are in the same units, no conversion is necessary.

Q: Does Bernoulli's equation apply to viscous fluids like honey or oil?

No, Bernoulli's equation is derived assuming an ideal fluid, which is non-viscous (no internal friction) and incompressible. For viscous fluids, energy is lost as heat due to viscous shear, and you would need to use modified equations that account for head loss.

Q: Why does pressure decrease when velocity increases? Shouldn't fast-moving water hit harder and create more pressure?

It is crucial to distinguish between static pressure (the pressure exerted on the walls of the pipe) and dynamic pressure (related to the kinetic energy of the flow). Bernoulli's equation deals with static pressure. When a fluid speeds up, its kinetic energy increases at the expense of its static pressure energy to conserve total energy.

Q: What would change if the pipe was vertical instead of horizontal?

If the pipe were vertical, the potential energy term $\rho g h$ would not cancel out. You would have to include the height difference ($h_1 - h_2$) in Bernoulli's equation: $p_2 = p_1 + \frac{1}{2}\rho(v_1^2 - v_2^2) + \rho g(h_1 - h_2)$.

Q: Is the density of water always constant at 10^3 kg/m^3?

For all standard JEE problems, water is assumed to be an incompressible liquid with a constant density of $10^3\text{ kg m}^{-3}$ (or $1\text{ g cm}^{-3}$), unless specified otherwise.

Question

A horizontal pipeline carries water in a streamline flow. At a point along the pipe, where the cross-sectional area is $10	ext{ cm}^2$, the water velocity is $1	ext{ ms}^{-1}$ and the pressure is $2000	ext{ Pa}$. The pressure of water at another point where the cross-sectional area is $5	ext{ cm}^2$, is ......Pa. (Density of water = $10^3	ext{ kg-m}^{-3}$)

Sigma Percentile · Accepted Answer

### The Magic of Fluid Dynamics

Imagine water flowing through a pipe. It seems simple, almost mundane. But beneath this everyday phenomenon lies a beautiful symphony of physics governed by two of the most elegant principles in classical mechanics: the conservation of mass and the conservation of energy.

In this problem, we are tasked with analyzing a horizontal pipeline carrying water in a steady, streamline flow. We are given the conditions at one wide section of the pipe and asked to find the pressure at a narrower section. Let's embark on this journey to uncover how fluid speed and pressure are intimately linked.

### Phase 1: The Continuity Equation (Conservation of Mass)

Before we can talk about pressure, we must understand how the speed of the water changes as it moves from the wider section to the narrower section. This is where the **Equation of Continuity** comes into play.

At its core, the Equation of Continuity is a statement of the conservation of mass. Since water is incompressible, its density remains constant throughout the flow. Therefore, the volume of water entering one end of the pipe per second must equal the volume of water leaving the other end per second.

Mathematically, we express this volumetric flow rate as the product of the cross-sectional area $A$ and the fluid velocity $v$:

$$A_1 v_1 = A_2 v_2$$

Let's look at our given values:
- Area at Section 1, $A_1 = 10	ext{ cm}^2$
- Velocity at Section 1, $v_1 = 1	ext{ ms}^{-1}$
- Area at Section 2, $A_2 = 5	ext{ cm}^2$

Substituting these values into our continuity equation, we get:

$$10 	imes 1 = 5 	imes v_2$$

Solving for $v_2$, we find:

$$v_2 = 2	ext{ ms}^{-1}$$

This result makes perfect intuitive sense. Since the cross-sectional area is halved, the water must flow twice as fast to ensure that the same amount of mass passes through every second. Think of a garden hose: when you place your thumb over the opening to narrow the exit, the water squirts out much faster!

### Phase 2: Bernoulli's Principle (Conservation of Energy)

Now that we know the velocity of the water at both sections, we can address the core question: what happens to the pressure?

To connect velocity and pressure, we use **Bernoulli's Equation**, which is essentially the work-energy theorem applied to flowing fluids. For a steady, streamline flow of an incompressible, non-viscous fluid, the total mechanical energy along a streamline is constant.

The general form of Bernoulli's equation is:

$$p_1 + \frac{1}{2}ho v_1^2 + ho g h_1 = p_2 + \frac{1}{2}ho v_2^2 + ho g h_2$$

Here, $p$ represents the pressure energy, $\frac{1}{2}ho v^2$ represents the kinetic energy per unit volume, and $ho g h$ represents the potential energy per unit volume.

Since our pipeline is explicitly stated to be **horizontal**, there is no change in height between the two points. This means $h_1 = h_2$, and the potential energy terms on both sides cancel out completely. Our equation simplifies beautifully to:

$$p_1 + \frac{1}{2}ho v_1^2 = p_2 + \frac{1}{2}ho v_2^2$$

This simplified relation tells us something profound: in a horizontal pipe, if the velocity of the fluid increases, its pressure must decrease to keep the total energy constant. This counterintuitive phenomenon is known as the **Venturi Effect**.

### Phase 3: Calculating the Final Pressure

Let's rearrange our simplified Bernoulli's equation to solve for the unknown pressure at the narrower section, $p_2$:

$$p_2 = p_1 + \frac{1}{2}ho (v_1^2 - v_2^2)$$

We are given:
- Pressure at Section 1, $p_1 = 2000	ext{ Pa}$
- Density of water, $ho = 10^3	ext{ kg m}^{-3}$
- Velocity at Section 1, $v_1 = 1	ext{ ms}^{-1}$
- Velocity at Section 2, $v_2 = 2	ext{ ms}^{-1}$

Let's substitute these values into our rearranged equation:

$$p_2 = 2000 + \frac{1}{2} 	imes 10^3 	imes (1^2 - 2^2)$$

Now, let's compute this step-by-step to avoid any silly mistakes:

$$1^2 - 2^2 = 1 - 4 = -3$$

Substituting this back in:

$$p_2 = 2000 + \frac{1}{2} 	imes 1000 	imes (-3)$$

$$p_2 = 2000 - 1500$$

$$p_2 = 500	ext{ Pa}$$

And there we have it! The pressure at the narrower section is exactly **$500	ext{ Pa}$**.

### The Venturi Effect: A Deeper Intuition

Why did the pressure drop so drastically from $2000	ext{ Pa}$ to $500	ext{ Pa}$? Let's think about the physics at a molecular level.

As the water molecules enter the constriction, they must accelerate from $1	ext{ ms}^{-1}$ to $2	ext{ ms}^{-1}$. According to Newton's second law, an acceleration requires a net force. This force can only come from a pressure difference.

Therefore, the pressure behind the fluid (at Section 1) must be higher than the pressure in front of it (at Section 2) to push and accelerate the water forward. This pressure gradient is what drives the acceleration, resulting in a lower pressure at the constriction.

This principle is not just a textbook exercise; it is the foundation of many modern technologies, from carburetors and paint sprayers to the lift generated by airplane wings. Understanding these fundamental laws of fluid dynamics opens up a whole new way of looking at the physical world!

Enter Numerical Value:

Visualized Solution

Visualizing the Flow in a Tapering Pipe

The Equation of Continuity

Substituting the Values into Continuity Equation

Calculating the Velocity $v_{2}$

Applying Bernoulli's Principle

Substituting Values into Bernoulli's Equation

Computing the Final Pressure

Conceptual Takeaways and Extensions

The Sigma Insight: Flow of Fluid

The Magic of Fluid Dynamics

Phase 1

Phase 2

Phase 3

The Venturi Effect

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