A fluid with a density of ρ = 1165 kg/m^3 flows through a horizontal pipe. In one segment of the pipe, the flow speed is v1 = 4.53 m/s, and in another segment, the flow speed is v2 = 1.77 m/s. What is the difference in pressure between the second segment (P2) and the first segment (P1)?

To determine the difference in pressure between two points in a flowing fluid within a horizontal pipe, we can use Bernoulli's equation, which is stated as follows for steady, incompressible, and non-viscous fluid flow: \[ P + \frac{1}{2} \rho v^2 + \rho g h = \text{constant} \]

Since the pipe is horizontal, the gravitational potential energy per unit volume (given by \( \rho g h \)) is the same for both segments, so we can ignore the height term \( h \) in the Bernoulli equation. Simplifying the equation for this scenario, we get: \[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]

Here, \( P_1 \) and \( P_2 \) are the pressures at segment one and segment two respectively, \( \rho \) is the fluid density, and \( v_1 \) and \( v_2 \) are the flow velocities at segment one and segment two respectively.

We can rearrange this equation to solve for the pressure difference \( P_2 - P_1 \): \[ P_2 - P_1 = \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 \]

Let's plug in the known values: \[ P_2 - P_1 = \frac{1}{2} \times 1165 \text{ kg/m}^3 \times (4.53 \text{ m/s})^2 - \frac{1}{2} \times 1165 \text{ kg/m}^3 \times (1.77 \text{ m/s})^2 \]

Now, calculate the individual terms: \[ \frac{1}{2} \times 1165 \times (4.53)^2 ≈ \frac{1}{2} \times 1165 \times 20.5 ≈ 11942.25 \text{ Pa} \] \[ \frac{1}{2} \times 1165 \times (1.77)^2 ≈ \frac{1}{2} \times 1165 \times 3.13 ≈ 1821.845 \text{ Pa} \]

Subtracting the second term from the first term gives us the pressure difference: \[ P_2 - P_1 ≈ 11942.25 \text{ Pa} - 1821.845 \text{ Pa} \] \[ P_2 - P_1 ≈ 10120.41 \text{ Pa} \]

Therefore, the difference in pressure between the second segment (P2) and the first segment (P1) is approximately 10120.41 Pascals (Pa).