To calculate the local pressure at a depth of 2.5 meters in a swimming pool located at an altitude of 2400 meters, we will use the hydrostatic pressure equation:
\[ P = P_0 + \rho g h \]
Where:
- \( P \) is the total pressure at depth.
- \( P_0 \) is the atmospheric pressure at altitude (in pascals).
- \( \rho \) is the density of the fluid (for water, \(\rho = 1000 , \text{kg/m}^3\)).
- \( g \) is the acceleration due to gravity (approximately \( 9.81 , \text{m/s}^2\)).
- \( h \) is the depth of the fluid (in meters).
Step 1: Calculate the atmospheric pressure at 2400 meters
The standard atmospheric pressure at sea level is approximately \( 101325 , \text{Pa} \). However, the pressure decreases with altitude. The atmospheric pressure can be approximated at a certain altitude using the barometric formula. At 2400 meters, the approximate atmospheric pressure can be estimated as follows:
\[ P_0 = P_{\text{sea level}} - ( \text{rate of pressure decrease} ) \times \text{altitude} \]
The rate of pressure decrease can be approximated roughly as \( 12 , \text{Pa/m} \) for average conditions. Therefore:
\[ P_0 \approx 101325 , \text{Pa} - (12 , \text{Pa/m} \times 2400 , \text{m}) = 101325 , \text{Pa} - 28800 , \text{Pa} \approx 72525 , \text{Pa} \]
Step 2: Calculate the pressure at 2.5 meters depth in water
Now we can substitute the values into the hydrostatic pressure equation:
- \( \rho = 1000 , \text{kg/m}^3 \)
- \( g = 9.81 , \text{m/s}^2 \)
- \( h = 2.5 , \text{m} \)
Now, calculate the pressure due to the water:
\[ P_{\text{water}} = \rho g h = 1000 , \text{kg/m}^3 \times 9.81 , \text{m/s}^2 \times 2.5 , \text{m} = 24525 , \text{Pa} \]
Step 3: Compute the total pressure
Now, we can calculate the total pressure at this depth:
\[ P = P_0 + P_{\text{water}} = 72525 , \text{Pa} + 24525 , \text{Pa} = 97050 , \text{Pa} \]
Therefore, the local pressure at 2.5 meters depth in the swimming pool is:
\[ \boxed{97050 , \text{Pa}} \]