An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum

Q: Why is the minimum velocity associated with the maximum diameter?

According to the Equation of Continuity ($A \cdot v = \text{constant}$), area and velocity are inversely proportional. A larger diameter means a larger area, which forces the fluid to slow down to maintain a constant flow rate.

Q: Do we need to convert the diameters from cm to meters?

Not in this case! Since we are taking a ratio of two diameters ($\frac{d_{\text{min}}}{d_{\text{max}}}$), the units will cancel out as long as they are the same.

Q: What does "ideal fluid" mean in this context?

An ideal fluid is assumed to be incompressible (constant density) and non-viscous (no internal friction). This allows us to use the simple form of the continuity equation, $A_1 v_1 = A_2 v_2$.

Q: Why do we square the ratio of the diameters?

The area of a circle is proportional to the square of its diameter ($A \propto d^2$). Since velocity is inversely proportional to area, it becomes inversely proportional to the square of the diameter.

Q: Can we use radius instead of diameter?

Absolutely! The ratio of diameters is exactly equal to the ratio of radii ($\frac{d_1}{d_2} = \frac{2r_1}{2r_2} = \frac{r_1}{r_2}$). The final result will be exactly the same.

Question

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4 	ext{ cm}$ and $4.8 	ext{ cm}$, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is

Options:

(A) $\frac{9}{16}$

(B) $\frac{81}{256}$

(C) $\frac{3}{2}$

(D) $\frac{3}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are watering your garden. You have a hose, and water is flowing out of it steadily. What happens when you place your thumb over half of the opening? The water suddenly shoots out much faster! This everyday phenomenon is exactly what this problem is about.

We are given a pipe with a non-uniform diameter. At one end, it is narrow with a minimum diameter of $4.8 	ext{ cm}$, and at the other end, it is wide with a maximum diameter of $6.4 	ext{ cm}$. We need to find the ratio of the minimum velocity to the maximum velocity of the fluid flowing through it.

### The Master Equation

To solve this, we rely on a fundamental principle of fluid mechanics: the **Equation of Continuity**. For an ideal, incompressible fluid, the volume of fluid passing through any cross-section per unit time must remain constant.

Mathematically, this is expressed as:
$$A_1 v_1 = A_2 v_2$$

In our specific case, the product of the minimum area and the maximum velocity must equal the product of the maximum area and the minimum velocity:
$$A_{	ext{min}} v_{	ext{max}} = A_{	ext{max}} v_{	ext{min}}$$

Notice the beautiful inverse relationship here! Where the area is minimum, the velocity must be maximum to keep the flow rate constant, and vice versa.

### Setting Up the Math

We know that the cross-sectional area of a circular pipe is given by $A = \frac{\pi}{4} d^2$. Let's substitute this into our continuity equation:
$$\frac{\pi}{4} d_{	ext{min}}^2 \cdot v_{	ext{max}} = \frac{\pi}{4} d_{	ext{max}}^2 \cdot v_{	ext{min}}$$

The constant factor $\frac{\pi}{4}$ appears on both sides, so we can elegantly cancel it out. This leaves us with a much simpler relationship involving only diameters and velocities:
$$d_{	ext{min}}^2 \cdot v_{	ext{max}} = d_{	ext{max}}^2 \cdot v_{	ext{min}}$$

### Final Calculation

We are asked to find the ratio of the minimum velocity to the maximum velocity, which is $\frac{v_{	ext{min}}}{v_{	ext{max}}}$. Rearranging our simplified equation gives:
$$\frac{v_{	ext{min}}}{v_{	ext{max}}} = \frac{d_{	ext{min}}^2}{d_{	ext{max}}^2} = \left( \frac{d_{	ext{min}}}{d_{	ext{max}}} ight)^2$$

Now, it's time to plug in the given values. We have $d_{	ext{min}} = 4.8 	ext{ cm}$ and $d_{	ext{max}} = 6.4 	ext{ cm}$:
$$\frac{v_{	ext{min}}}{v_{	ext{max}}} = \left( \frac{4.8}{6.4} ight)^2$$

Let's simplify the fraction inside the parentheses before squaring. Both $4.8$ and $6.4$ are divisible by $1.6$:
$$\frac{4.8}{6.4} = \frac{3}{4}$$

Finally, we square this simplified fraction to get our answer:
$$\frac{v_{	ext{min}}}{v_{	ext{max}}} = \left( \frac{3}{4} ight)^2 = \frac{9}{16}$$

And there we have it! The ratio of the minimum to maximum velocity is $\frac{9}{16}$. This perfectly illustrates how a relatively small change in diameter leads to a significant change in velocity due to the squared relationship.

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Visualized Solution

$Non-Uniform Pipe Flow$

$Equation of Continuity$

$Area in terms of Diameter$

$Velocity Ratio Formula$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Flow of Fluid

Analyzing the Setup

The Master Equation

Setting Up the Math

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Non-Uniform Pipe Flow

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of Continuity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Area in terms of Diameter

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Velocity Ratio Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward