An open tank contains 5m of water covered with 2m of oil (specific gravity = 0.8). Find the pressure at the interface and at the bottom of the tank.
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PK Bot’s answer:
We assume that the top surface of the oil is exposed to atmospheric pressure, which we take as
p_\mathrm{atm}=101300~Pa\,.
For oil with specific gravity 0.8, the density is
\rho_\mathrm{oil}=0.8\times 1000~kg/m^3=800~kg/m^3\,.
The pressure at the oil–water interface is due to the oil head alone:
p_\mathrm{interface}=p_\mathrm{atm} + \rho_\mathrm{oil}\, g\, h_\mathrm{oil}\,,
with
h_\mathrm{oil}=2~m,\quad g=9.8~m/s^2\,.
Thus,
p_\mathrm{interface}=101300~Pa+800\times9.8\times2
=101300~Pa+15680~Pa
=116980~Pa\,.
The water column has a height h_\mathrm{water}=5~m and density \rho_\mathrm{water}=1000~kg/m^3. The pressure at the bottom is
p_\mathrm{bottom}=p_\mathrm{interface}+ \rho_\mathrm{water}\,g\,h_\mathrm{water}\,,
so
p_\mathrm{bottom}=116980~Pa+1000\times9.8\times5\,.
The water contribution is
1000\times9.8\times5=49000~Pa\,,
therefore
p_\mathrm{bottom}=116980~Pa+49000~Pa=165980~Pa\,.
The calculated pressures are:
- Pressure at the interface: approximately 117~kPa
- Pressure at the bottom: approximately 166~kPa
To formulate this response, I reviewed the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked their significance below. If they are marked as relevant, I encourage you to take a look at them:
Textbook sections:
- Chapter 4 Engineering Models: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 3 Engineering Models: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 3.2: Hydrostatic Equation (HIGHLY RELEVANT)
- Chapter 3.3: Hydrostatic Distributions in Pipes and Columns (HIGHLY RELEVANT)
- Chapter 4.2: Review of Hydrostatics (HIGHLY RELEVANT)
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