To calculate the static gauge pressure at the pipe inlet, we can use Bernoulli’s equation and the power relationship. Bernoulli's equation helps us connect the energy at different points along the flow, while the power extracted by the turbine helps us account for energy loss.

Assumptions:

1. Steady flow, incompressible fluid (water).
2. The water enters the turbine from the reservoir (which is large) and exits the pipe with a velocity of 25 m/s.
3. The height of the reservoir (relative to the turbine) is 100 m.
4. All the losses are concentrated in the pipe and turbine.

Step-by-Step Approach:

We will apply Bernoulli's equation between two points:

1. Point 1: Just at the surface of the reservoir (where the velocity is negligible, and height is $h = 100$ m).
2. Point 2: At the inlet of the turbine (where the height is $h = 0$ m, and velocity is known to be $25 \, \text{m/s}$).

1. Bernoulli's equation:

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

• $P_1$ is the pressure at the surface of the reservoir (atmospheric pressure).
• $P_2$ is the static gauge pressure at the pipe inlet (what we need to find).
• $v_1$ is the velocity at the reservoir surface, which is $0 \, \text{m/s}$ (because it's large).
• $v_2 = 25 \, \text{m/s}$ is the velocity at the turbine inlet.
• $h_1 = 100 \, \text{m}$, height of the reservoir.
• $h_2 = 0 \, \text{m}$, height at the turbine inlet.
• $\rho = 1000 \, \text{kg/m}^3$, density of water.
• $g = 9.81 \, \text{m/s}^2$, acceleration due to gravity.

Simplifying Bernoulli’s equation between the reservoir surface and the pipe inlet:

$P_1 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2$ We are interested in the gauge pressure, so $P_1$ is atmospheric pressure and we take it as zero (for gauge pressure). The equation simplifies to:

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