The hydrostatic equation holds not just for air, but for other fluids such as water (air, though it is a gas, is often referred to as a 'fluid') as well. In the case of water the equation becomes even easier, as water (as opposed to air) can be treated as incompressible, meaning the density is constant. Use a value of 1000 kilogram per cubic metre as density.
Using this information, compute the local pressure (in [Pa]) at 2.5 metres depth in a swimming pool situated at 2400 metres altitude in the mountains. Assume a standard atmosphere.
ur supposed to add the nearly 25kpa tot the 75.6kpa so the answer should be around 100139.2 pa
The pressure will be 1*2.5/10 atmospsheres below the atmospheric pressure at the surface of the pool.
At 2400 meters altitude, atmospheric pressure should be p = 101325 (1 - 2.25577e-5 h)^5.25588 or
p=101325(1-2.25577e-5*2400)^5.25588=75625Pa
so, pressure 2.5m below surface should be 75.6kPa-2.5*101.3/10 kPa
=50.3 kPa
100139.2 is the correct answer.
Well, well, well, let's dive into this question, shall we? So, we have a swimming pool at 2400 meters altitude in the mountains, and we want to calculate the pressure at a depth of 2.5 meters in the pool. One thing we know is that at such heights, the air is as thin as my hairline, but fear not, my friend, we won't let that water you down!
Now, we're told to treat water as incompressible, which means its density remains constant at 1000 kilograms per cubic meter. That's one dense liquid, I tell ya!
Before we proceed, let me just share a little secret with you - the pressure in a fluid depends not only on the depth but also on the atmospheric pressure. So, let's consider the standard atmosphere in this case.
Ahem! Now, according to the hydrostatic equation, the pressure at a certain depth in a fluid is given by:
Pressure = atmospheric pressure + (density × gravity × depth)
We'll need to convert the altitude of 2400 meters to atmospheric pressure, but instead of boring you with that, I'll just tell you it's about 76 percent of the standard atmospheric pressure. Now, remember, we're adding that to the pressure due to water at 2.5 meters depth.
Since the density of water is 1000 kilograms per cubic meter, and gravity is approximately 9.8 meters per second squared (oh, gravity, you always bring me down), we can now do the calculations.
Pressure = (0.76 × standard atmospheric pressure) + (1000 × 9.8 × 2.5)
And voila! Crunch those numbers, and you'll have your local pressure in pascals, my friend. Just remember to have a good time swimming and not let the pressure get to you!
To compute the local pressure at a specific depth in a swimming pool, we can utilize the hydrostatic equation. The hydrostatic equation states that the pressure in a fluid at a given depth is directly proportional to the density of the fluid, the acceleration due to gravity, and the depth. In this case, as mentioned, we will assume that water is being used and can be treated as incompressible.
The hydrostatic equation can be written as:
P = ρ * g * h
Where:
P is the pressure at depth h,
ρ is the density of the fluid,
g is the acceleration due to gravity.
Given:
Density of water (ρ) = 1000 kg/m³
Depth (h) = 2.5 m
The first step is to determine the value of g. Since the pool is situated at an altitude of 2400 meters, we need to account for the variation in gravity with altitude. At sea level, the standard value for gravity is 9.81 m/s². However, as altitude increases, gravity decreases slightly. We can use the following equation to calculate g at the given altitude:
g = g₀ / (1 + α * h)
Where:
g₀ is the standard gravity at sea level (9.81 m/s²)
α is the gravitational acceleration constant (-0.00003 m^-1)
h is the altitude (2400 m)
Substituting the given values:
g = 9.81 / (1 - 0.00003 * 2400)
Calculating g:
g ≈ 9.79 m/s²
Now, we can substitute the values of ρ, g, and h into the hydrostatic equation to calculate the local pressure:
P = 1000 * 9.79 * 2.5
Calculating P:
P ≈ 24,475 Pa
Therefore, the local pressure at a depth of 2.5 meters in the swimming pool situated at 2400 meters altitude in the mountains would be approximately 24,475 Pa (Pascals).