Question

The gauge pressure at the surface of a liquid of density $900 \, \text{kg/m}^3$ is $0.4 \, \text{bar}$. If the atmospheric pressure is $1 \times 10^5 \, \text{Pa}$, calculate the absolute pressure at a depth of $50 \, \text{m}$.

• A. 5.8145 bar
• B. 2.5245 bar
• C. 7.3242 bar
• D. 4.3647 bar

Hint:

To calculate absolute pressure at a depth, use: \[ P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} + \rho g h. \] Convert all units consistently (e.g., Pa to bar if required).

Answer 1

The Correct Option is A

The absolute pressure is calculated as:
\[P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}} + \rho g h\]
where:
 $P_{\text{gauge}} = 0.4 \, \text{bar} = 0.4 \times 10^5 \, \text{Pa}$,
   $P_{\text{atm}} = 1 \times 10^5 \, \text{Pa}$,
   $\rho = 900 \, \text{kg/m}^3$,
   $g = 9.81 \, \text{m/s}^2$,
   $h = 50 \, \text{m}$.
Substitute the values:
\[P_{\text{abs}} = (0.4 \times 10^5) + (1 \times 10^5) + (900 \times 9.81 \times 50).\]
\[P_{\text{abs}} = 0.4 \times 10^5 + 1 \times 10^5 + 4.4145 \times 10^5 = 5.8145 \times 10^5 \, \text{Pa}.\]
Convert to bar:
\[P_{\text{abs}} = 5.8145 \, \text{bar}.\]