Question

The gage pressure in a liquid at a depth of \(3 \mathrm{m}\) is read to be \(42 \mathrm{kPa}\). Determine the gage pressure in the same liquid at a depth of \(9 \mathrm{m}\)

Short answer

Answer: The gage pressure at a depth of 9 meters in the same liquid is approximately 123.538 kPa.

Step-by-step solution

Recall the formula for gage pressure in a liquid

The gage pressure (P) in a liquid at a given depth (h) can be calculated using the following formula: P = ρgh Here, ρ represents the density of the liquid, g is the acceleration due to gravity, and h is the depth.

Determine the density of the liquid

We are given the gage pressure (P1) and depth (h1) at one point in the liquid: P1 = 42 kPa h1 = 3 m Using the above formula, we can write: P1 = ρgh1 Now, we can solve for the density (ρ): ρ = P1 / (gh1)

Convert the given pressure units

To find the density in kg/m³, we need to convert the given gage pressure from kilopascals (kPa) to pascals (Pa). 1 kPa is equal to 1000 Pa. Therefore, P1 = 42 kPa * 1000 = 42000 Pa

Calculate the density of the liquid

Assuming the acceleration due to gravity (g) is 9.81 m/s², we can now calculate the density of the liquid: ρ = (42000 Pa) / (9.81 m/s² * 3 m) ρ = 1400 kg/m³

Calculate the gage pressure at a depth of 9 meters

Now that we have the density of the liquid, we can calculate the gage pressure (P2) at a depth of 9 meters (h2) using the formula: P2 = ρgh2 Substitute the values: P2 = (1400 kg/m³) * (9.81 m/s²) * (9 m)

Calculate the final gage pressure

Now, let's calculate the gage pressure (P2) at the depth of 9 meters: P2 = 123,538 Pa

Convert the gage pressure back to kPa

As our final step, we need to convert the gage pressure back to kilopascals (kPa). To do this, divide the gage pressure in pascals (Pa) by 1000: P2 = 123.538 kPa Thus, the gage pressure in the same liquid at a depth of 9 meters is approximately 123.538 kPa.