A 15.5-kg mass of copper has an apparent mass of 8.7 kg in oil whose SG is 0.77. What is the volume of the copper and its specific weight?
A dam is 283 m high when it is full of water. What is the pounds per square inch absolute at the bottom of the reservoir?
The Correct Answer and Explanation is :
Problem 1: Volume and Specific Weight of Copper
Given Data:
- Mass of copper, ( m_c = 15.5 \, \text{kg} )
- Apparent mass of copper in oil, ( m_a = 8.7 \, \text{kg} )
- Specific gravity of oil, ( SG_{\text{oil}} = 0.77 )
Solution:
- Buoyant Force:
The buoyant force ( F_b ) is the difference between the actual weight and the apparent weight of the copper:
[
F_b = (m_c – m_a) \cdot g
]
Substituting values:
[
F_b = (15.5 – 8.7) \cdot 9.81 = 66.46 \, \text{N}
] - Volume of Copper:
The buoyant force equals the weight of the displaced oil. Using the oil’s density, the volume of displaced oil (and copper) is:
[
F_b = \rho_{\text{oil}} \cdot V \cdot g
]
Since ( \rho_{\text{oil}} = SG_{\text{oil}} \cdot \rho_{\text{water}} ) and ( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 ):
[
V = \frac{F_b}{\rho_{\text{oil}} \cdot g} = \frac{66.46}{(0.77 \cdot 1000) \cdot 9.81} = 0.00876 \, \text{m}^3
] - Specific Weight of Copper:
[
\gamma_c = \frac{\text{Weight of Copper}}{\text{Volume of Copper}} = \frac{m_c \cdot g}{V} = \frac{15.5 \cdot 9.81}{0.00876} = 17333 \, \text{N/m}^3
]
Problem 2: Pressure at the Bottom of the Reservoir
Given Data:
- Height of water, ( h = 283 \, \text{m} )
- Density of water, ( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 )
Solution:
- Gauge Pressure:
[
P_g = \rho_{\text{water}} \cdot g \cdot h
]
Substituting:
[
P_g = 1000 \cdot 9.81 \cdot 283 = 2.775 \times 10^6 \, \text{Pa} = 2.775 \, \text{MPa}
] - Convert to PSI:
Using ( 1 \, \text{Pa} = 0.0001450377 \, \text{psi} ):
[
P_g = 2.775 \cdot 10^6 \cdot 0.0001450377 = 402.7 \, \text{psi}
] - Absolute Pressure:
Adding atmospheric pressure (( P_{\text{atm}} \approx 14.7 \, \text{psi} )):
[
P_{\text{abs}} = P_g + P_{\text{atm}} = 402.7 + 14.7 = 417.4 \, \text{psi}
]
Explanation:
For the first problem, the apparent mass difference in oil reveals the buoyant force, from which we calculate the displaced oil volume and hence the volume of copper. Specific weight is derived as weight per unit volume.
For the second problem, hydrostatic pressure is due to the height of the water column, calculated with the density of water and acceleration due to gravity. Converting to PSI ensures practical comprehension, and adding atmospheric pressure gives the absolute value.