2.4 lb/ft³ (9.81 kN/m³). The specific gravity of mercury is 13.55. Compute the density of water and the specific weight and density of mercury. Note: you should provide your answers in slug/ft³, kg/m³, lb/ft³, and kN/m³.
- The specific weight of water at ordinary pressure and temperature is 62.4 lb/ft³ (9.81 kN/m³). The specific gravity of mercury is 13.55. Compute the density of water and the specific weight and density of mercury. Note: you should provide your answers in slug/ft³, kg/m³, lb/ft³, and kN/m³.
The Correct Answer and Explanation is:
Given:
- Specific weight of water, γ<sub>water</sub> = 62.4 lb/ft³ = 9.81 kN/m³
- Specific gravity of mercury, SG<sub>Hg</sub> = 13.55
Part 1: Density and Specific Weight of Water
A. In slug/ft³:
Using the relation:Density=Specific weightg\text{Density} = \frac{\text{Specific weight}}{g}Density=gSpecific weightρwater=62.4 lb/ft332.174 ft/s2=1.94 slug/ft3\rho_{\text{water}} = \frac{62.4 \, \text{lb/ft}^3}{32.174 \, \text{ft/s}^2} = 1.94 \, \text{slug/ft}^3ρwater=32.174ft/s262.4lb/ft3=1.94slug/ft3
B. In kg/m³:γwater=9.81 kN/m3=9810 N/m3\gamma_{\text{water}} = 9.81 \, \text{kN/m}^3 = 9810 \, \text{N/m}^3γwater=9.81kN/m3=9810N/m3ρwater=9810 N/m39.81 m/s2=1000 kg/m3\rho_{\text{water}} = \frac{9810 \, \text{N/m}^3}{9.81 \, \text{m/s}^2} = 1000 \, \text{kg/m}^3ρwater=9.81m/s29810N/m3=1000kg/m3
Part 2: Specific Weight and Density of Mercury
A. Specific Weight of Mercury:γHg=SGHg×γwater\gamma_{\text{Hg}} = SG_{\text{Hg}} \times \gamma_{\text{water}}γHg=SGHg×γwaterγHg=13.55×62.4 lb/ft3=845.52 lb/ft3\gamma_{\text{Hg}} = 13.55 \times 62.4 \, \text{lb/ft}^3 = 845.52 \, \text{lb/ft}^3γHg=13.55×62.4lb/ft3=845.52lb/ft3γHg=13.55×9.81 kN/m3=132.97 kN/m3\gamma_{\text{Hg}} = 13.55 \times 9.81 \, \text{kN/m}^3 = 132.97 \, \text{kN/m}^3γHg=13.55×9.81kN/m3=132.97kN/m3
B. Density of Mercury:
- In slug/ft³:
ρHg=845.52 lb/ft332.174=26.28 slug/ft3\rho_{\text{Hg}} = \frac{845.52 \, \text{lb/ft}^3}{32.174} = 26.28 \, \text{slug/ft}^3ρHg=32.174845.52lb/ft3=26.28slug/ft3
- In kg/m³:
ρHg=132970 N/m39.81=13,560 kg/m3\rho_{\text{Hg}} = \frac{132970 \, \text{N/m}^3}{9.81} = 13,560 \, \text{kg/m}^3ρHg=9.81132970N/m3=13,560kg/m3
Final Answers Summary:
| Quantity | Value |
|---|---|
| ρ<sub>water</sub> | 1.94 slug/ft³ |
| ρ<sub>water</sub> | 1000 kg/m³ |
| γ<sub>Hg</sub> | 845.52 lb/ft³ |
| γ<sub>Hg</sub> | 132.97 kN/m³ |
| ρ<sub>Hg</sub> | 26.28 slug/ft³ |
| ρ<sub>Hg</sub> | 13,560 kg/m³ |
Explanation (Book-style, ~300 words):
In fluid mechanics, understanding the concepts of specific weight, density, and specific gravity is essential for analyzing fluid behavior under various conditions. The specific weight (γ) is the weight per unit volume and is often given in lb/ft³ or kN/m³. Density (ρ), on the other hand, is mass per unit volume, typically expressed in slug/ft³ (in the English system) or kg/m³ (SI units). Specific gravity (SG) is a dimensionless ratio comparing the density of a fluid to the density of water at standard conditions.
The specific weight of water at standard conditions is typically accepted as 62.4 lb/ft³ or 9.81 kN/m³. Using this value, we can derive the density of water by dividing the specific weight by gravitational acceleration (32.174 ft/s² or 9.81 m/s²), yielding 1.94 slug/ft³ and 1000 kg/m³, respectively.
Mercury, a dense liquid metal commonly used in barometers and thermometers, has a specific gravity of 13.55, meaning it is 13.55 times denser than water. Using this factor, we multiply the specific weight of water to find the specific weight of mercury, which equals 845.52 lb/ft³ or 132.97 kN/m³. Dividing these values by the respective gravitational constants provides the density of mercury: 26.28 slug/ft³ or 13,560 kg/m³.
These properties are crucial for hydrostatics, where pressure calculations, buoyancy, and fluid column heights depend directly on the fluid’s specific weight and density. Understanding and converting between units ensures accurate application across both engineering systems.
