Step 1: Understanding the Concept:
This problem involves the dynamics of fluid flow and is solved using Bernoulli's Theorem.
Bernoulli's principle is a statement of the conservation of energy for flowing fluids.
It states that in a steady, streamline flow of an ideal fluid (non-viscous and incompressible), the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline.
For a horizontal pipe, the height (\(h\)) of the fluid remains constant at all points.
Therefore, the potential energy term (\( \rho gh \)) is the same on both sides of the equation and can be ignored.
This simplifies to the relationship: as the velocity of a fluid increases, its internal pressure must decrease.
Step 2: Key Formula or Approach:
The simplified Bernoulli's equation for horizontal flow at two points (1 and 2) is:
\[ P_{1} + \frac{1}{2}\rho v_{1}^{2} = P_{2} + \frac{1}{2}\rho v_{2}^{2} \]
Where:
\( P \) is the static pressure at a specific point.
\( \rho \) is the density of the fluid (for water, \( \rho = 1000 \text{ kg/m}^{3} \)).
\( v \) is the flow velocity at that point.
The term \( \frac{1}{2}\rho v^{2} \) represents the dynamic pressure of the fluid.
Step 3: Detailed Explanation:
Let's list the known parameters from the question for point 1 and point 2:
Point 1: \( v_{1} = 2 \text{ m/s} \), \( P_{1} = 2000 \text{ Pa} \).
Point 2: \( v_{2} = 4 \text{ m/s} \), \( P_{2} = ? \).
Density of water, \( \rho = 1000 \text{ kg/m}^{3} \).
We rearrange the Bernoulli equation to solve for the unknown pressure \( P_{2} \):
\[ P_{2} = P_{1} + \frac{1}{2}\rho v_{1}^{2} - \frac{1}{2}\rho v_{2}^{2} \]
Factor out the common term \( \frac{1}{2}\rho \):
\[ P_{2} = P_{1} + \frac{1}{2}\rho (v_{1}^{2} - v_{2}^{2}) \]
Substitute the values into the equation:
\[ P_{2} = 2000 + \frac{1}{2}(1000) (2^{2} - 4^{2}) \]
\[ P_{2} = 2000 + 500 (4 - 16) \]
\[ P_{2} = 2000 + 500 (-12) \]
\[ P_{2} = 2000 - 6000 \]
\[ P_{2} = -4000 \text{ Pa} \]
Interpretation of the Result:
A pressure of \( -4000 \text{ Pa} \) might seem physically impossible, but it is valid in the context of "gauge pressure".
Gauge pressure is measured relative to atmospheric pressure.
A negative gauge pressure indicates a "suction" or "partial vacuum" effect, where the internal pressure is lower than the surrounding atmosphere.
Step 4: Final Answer:
The pressure at the second point in the horizontal pipe is \( -4000 \text{ Pa} \).