Question

The absolute pressure in water at a depth of \(9 \mathrm{m}\) is read to be 185 kPa. Determine \((a)\) the local atmospheric pressure, and \((b)\) the absolute pressure at a depth of \(5 \mathrm{m}\) in a liquid whose specific gravity is 0.85 at the same location.

Short answer

Question: (a) Determine the local atmospheric pressure. (b) Calculate the absolute pressure at a depth of 5m in a liquid with a specific gravity of 0.85. Answer: (a) The local atmospheric pressure is \(96.71 \mathrm{\, kPa}\). (b) The absolute pressure at a depth of 5 m in the other liquid is \(138.395 \mathrm{\, kPa}\).

Step-by-step solution

Calculate the gauge pressure at 9 m depth in water.

To find the gauge pressure, we need to use the hydrostatic pressure equation: \(P = \rho g h\), where \(P\) is the pressure, \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth. For water, the density \(\rho_{water} = 1000 \mathrm{\, kg/m^3}\), \(g = 9.81 \mathrm{\, m/s^2}\), and \(h = 9 \mathrm{\, m}\). So we can find the gauge pressure, \(P_{gauge, water} = \rho_{water} \cdot g \cdot h = 1000 \mathrm{\, kg/m^3} \times 9.81 \mathrm{\, m/s^2} \times 9 \mathrm{\, m} = 88.29 \mathrm{\, kPa}\).

Determine the local atmospheric pressure.

Since the absolute pressure in water is given as \(P_{abs, water} = 185 \mathrm{\, kPa}\), the atmospheric pressure can be found by adding the gauge pressure to it, \(P_{atm} = P_{abs, water} - P_{gauge, water} = 185 \mathrm{\, kPa} - 88.29 \mathrm{\, kPa} = 96.71 \mathrm{\, kPa}\).

Find the pressure increase due to the depth of 5 m in the other liquid.

We are given the specific gravity of the other liquid as 0.85. Specific gravity is the ratio of the liquid's density to the density of water. Therefore, the density of the other liquid is given by: \(\rho_{liquid} = 0.85 \times \rho_{water} = 0.85 \times 1000 \mathrm{\, kg/m^3} = 850 \mathrm{\, kg/m^3}\). Now we will find the gauge pressure at the 5 m depth in the other liquid using the hydrostatic pressure equation: \(P_{gauge, liquid} = \rho_{liquid} \cdot g \cdot h_{liquid} = 850 \mathrm{\, kg/m^3} \times 9.81 \mathrm{\, m/s^2} \times 5 \mathrm{\, m} = 41.685 \mathrm{\, kPa}\)

Calculate the absolute pressure at a depth of 5 m in the other liquid.

To find the absolute pressure at the given depth, we will add the atmospheric pressure to the gauge pressure in the other liquid: \(P_{abs, liquid} = P_{atm} + P_{gauge, liquid} = 96.71 \mathrm{\, kPa} + 41.685 \mathrm{\, kPa} = 138.395 \mathrm{\, kPa}\). So, the results are: \((a)\) The local atmospheric pressure is \(96.71 \mathrm{\, kPa}\). \((b)\) The absolute pressure at a depth of 5 m in the other liquid is \(138.395 \mathrm{\, kPa}\).