A fluid is flowing through a horizontal pipe of varying cross-section with speed v ms^-1 at a point

Q: Why do we completely ignore the potential energy term ($\rho g h$) in this problem?

The problem explicitly states that the pipe is horizontal. This means the center of the pipe at both sections is at the exact same height from any reference level ($h_1 = h_2$). Because they are equal, the potential energy terms on both sides of Bernoulli's equation cancel each other out.

Q: Does the density of the fluid change as it moves from the wider to the narrower section?

No. In standard fluid dynamics problems at this level, we assume the fluid is incompressible (like water). This means its density $\rho$ remains perfectly constant throughout the entire flow.

Q: Why does the pressure decrease at the second point?

According to Bernoulli's principle for a horizontal flow, an increase in fluid speed must be accompanied by a decrease in pressure to conserve total energy. Since the pressure dropped from $p$ to $p/2$, we know the fluid must have sped up. This is often due to a narrowing of the pipe, known as the Venturi effect.

Q: Can we use the Equation of Continuity ($A_1v_1 = A_2v_2$) to solve this?

We could use the Equation of Continuity if we were given the cross-sectional areas ($A_1$ and $A_2$) of the pipe. However, since we only have pressure and speed data, Bernoulli's equation is the correct and sufficient tool to use.

Q: What would happen if the pipe was not horizontal?

If the pipe was inclined, the heights $h_1$ and $h_2$ would be different. You would then have to use the full version of Bernoulli's equation, keeping the $\rho g h$ terms, to account for the change in potential energy as the fluid moves up or down.

Question

A fluid is flowing through a horizontal pipe of varying cross-section with speed $v 	ext{ ms}^{-1}$ at a point where the pressure is $p 	ext{ Pa}$. At another point, where pressure is $\frac{p}{2} 	ext{ Pa}$ its speed is $V 	ext{ ms}^{-1}$. If the density of the fluid is $ho 	ext{ kg m}^{-3}$ and the flow is streamline, then $V$ is equal to

Options:

(A) $\sqrt{\frac{p}{ho}} + v$

(B) $\sqrt{\frac{2p}{ho} + v^2}$

(C) $\sqrt{\frac{p}{2ho} + v^2}$

(D) $\sqrt{\frac{p}{ho} + v^2}$

Sigma Percentile · Accepted Answer

### The Elegant Dance of Pressure and Speed: Unlocking Bernoulli's Principle

Imagine you are watching a river flow. Where the river is wide, the water meanders slowly and peacefully. But when the river narrows into a gorge, the water suddenly rushes forward with immense speed. This intuitive observation is the heart of fluid dynamics, and it is beautifully captured by **Bernoulli's Principle**.

In this problem, we are dealing with a similar situation: a fluid flowing through a horizontal pipe of varying cross-section. We are given the pressure and speed at two different points and asked to find the unknown speed at the second point. Let's break down the physics and the math behind this elegant phenomenon.

### Analyzing the Setup

We have a streamline flow of an incompressible fluid with a constant density $\rho$.

At the first section of the pipe, the fluid has a speed of $v$ and exerts a pressure $p$.

As the fluid travels to a second section, the pressure drops to $\frac{p}{2}$. We need to find the new speed, $V$.

Because the pipe is horizontal, the average height of the fluid from any reference level remains constant. This is a crucial detail that will simplify our mathematical journey.

### The Master Equation

To connect pressure, speed, and height in a flowing fluid, we invoke **Bernoulli's Equation**. It is essentially the principle of conservation of energy applied to a fluid. It states that the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume is constant along a streamline:

$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$$

Since our pipe is strictly horizontal, the heights are equal ($h_1 = h_2$). This means the potential energy terms ($\rho g h_1$ and $\rho g h_2$) are identical on both sides of the equation. We can gracefully cancel them out, leaving us with a much cleaner relationship between pressure and speed:

$$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$

This simplified equation tells us a profound physical truth: if the pressure drops, the kinetic energy (and thus the speed) must increase to keep the total energy constant.

### The Final Calculation

Now, we substitute the specific values given in our problem into this simplified equation.

For the first section, $P_1 = p$ and $v_1 = v$.

For the second section, $P_2 = \frac{p}{2}$ and $v_2 = V$.

Plugging these in, we get:

$$p + \frac{1}{2}\rho v^2 = \frac{p}{2} + \frac{1}{2}\rho V^2$$

Our goal is to isolate $V$. Let's group the pressure terms on one side and the velocity terms on the other:

$$\frac{1}{2}\rho V^2 - \frac{1}{2}\rho v^2 = p - \frac{p}{2}$$

Subtracting the pressures gives us $\frac{p}{2}$:

$$\frac{1}{2}\rho (V^2 - v^2) = \frac{p}{2}$$

We can cancel the $\frac{1}{2}$ from both sides and divide by the density $\rho$:

$$V^2 - v^2 = \frac{p}{\rho}$$

Finally, we move $v^2$ to the right side and take the square root to find our unknown speed $V$:

$$V = \sqrt{\frac{p}{\rho} + v^2}$$

And there we have it! The final expression beautifully shows how the new speed depends on the initial speed, the pressure difference, and the density of the fluid. It's a perfect demonstration of energy conservation in motion.

Select Answer:

Visualized Solution

Visualizing the Fluid Flow

Defining the Parameters

Bernoulli's Principle

Horizontal Pipe Constraint

Substituting the Values

Solving for $V$

Final Expression

The Sigma Insight: Flow of Fluid

The Elegant Dance of Pressure and Speed

Analyzing the Setup

The Master Equation

The Final Calculation

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