PROBLEM 34: A pressure gauge at elevation 8 m at the side of a tank containing liquid reads 85 kPa . Another pressure gage at elevation 3 m reads 120 kPa . Compute the specific gravity of the liquid. \( \begin{array}{ll}\text { a. } 0.82 & \text { b. } 0.63 \\ \text { c. } 0.82 & \text { d. } 0.71\end{array} \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Explain 
Simplify this solution Mind Expander
To find the specific gravity of the liquid, you can first calculate the pressure difference between the two gauges, then use hydrostatic pressure principles. The pressure difference between the two gauges is: \[ \Delta P = P_2 - P_1 = 120 \, \text{kPa} - 85 \, \text{kPa} = 35 \, \text{kPa} \] Next, get the difference in elevation between the two gauges: \[ \Delta h = h_2 - h_1 = 3 \, \text{m} - 8 \, \text{m} = -5 \, \text{m} \] Using the hydrostatic pressure formula \(\Delta P = \rho \cdot g \cdot \Delta h\), here \(g\) is approximately \(9.81 \, \text{m/s}^2\). Rearranging gives: \[ \rho = \frac{\Delta P}{g \cdot \Delta h} \] Plugging in the values: \[ \rho = \frac{35,000 \, \text{Pa}}{9.81 \, \text{m/s}^2 \cdot (-5 \, \text{m})} = \frac{35,000}{-49.05} \approx -712.15 \, \text{kg/m}^3 \] To find the specific gravity (SG), divide the density of the liquid by the density of water (approximately \(1000 \, \text{kg/m}^3\)): \[ SG = \frac{\rho}{\rho_{\text{water}}} = \frac{712.15}{1000} \approx 0.71 \] Thus, the specific gravity of the liquid is approximately **0.71**, so the answer is **d**. Before diving into calculations, always double-check the unit consistency and ensure you understand the principles of hydrostatics at play! It’s easy to trot along with numbers without catching small but crucial details like signs or units. Enjoy the learning process, and don’t hesitate to sketch a diagram next time—it can really clarify how pressures and heights relate!
