If a certain liquid has a mass density of 129 msl/ m3, what are the values of its specific weight, specific gravity and specific volume in metric gravitational and metric absolute systems of units.
If 5.27 m3 of a certain oil weighs 44 kN, calculate the specific weight, mass density and specific gravity of the oil.
The Correct Answer and Explanation is:
Let’s solve each part of the question step by step.
Part 1: Given a mass density of 129 kg/m³ (129 msl/m³), find:
- Specific Weight (γ)
- Specific Gravity (SG)
- Specific Volume (v)
System 1: Metric Gravitational System
- Mass Density (ρ) = 129 kg/m³
- Acceleration due to gravity (g) = 9.81 m/s²
1. Specific Weight (γ):
This is the weight per unit volume.
$$
\gamma = \rho \cdot g = 129 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 = 1,265.49 \, \text{N/m}^3
$$
2. Specific Volume (v):
This is the volume per unit mass.
$$
v = \frac{1}{\rho} = \frac{1}{129} = 0.00775 \, \text{m}^3/\text{kg}
$$
3. Specific Gravity (SG):
This is the ratio of the density of the fluid to the density of water (1000 kg/m³).
$$
SG = \frac{\rho}{\rho_{water}} = \frac{129}{1000} = 0.129
$$
Part 2: Given 5.27 m³ of oil weighs 44 kN, find:
- Specific Weight (γ)
- Mass Density (ρ)
- Specific Gravity (SG)
1. Specific Weight (γ):
$$
\gamma = \frac{\text{Weight}}{\text{Volume}} = \frac{44\,000 \, \text{N}}{5.27 \, \text{m}^3} = 8,347.45 \, \text{N/m}^3
$$
2. Mass Density (ρ):
Using the relation $\gamma = \rho \cdot g$:
$$
\rho = \frac{\gamma}{g} = \frac{8347.45}{9.81} = 850.89 \, \text{kg/m}^3
$$
3. Specific Gravity (SG):
$$
SG = \frac{\rho}{\rho_{water}} = \frac{850.89}{1000} = 0.851
$$
Explanation (300+ words):
In fluid mechanics, several important properties describe how a fluid behaves and how it compares with water—these include specific weight, mass density, specific gravity, and specific volume. Each of these has a distinct physical meaning and mathematical relation.
For the first scenario, we’re told the mass density is 129 kg/m³. Mass density (ρ) is defined as mass per unit volume and is a fundamental property. Using this, we compute specific weight (γ) by multiplying the density by gravitational acceleration (g = 9.81 m/s²). The result, 1,265.49 N/m³, tells us how much force the liquid exerts due to gravity per unit volume.
Next, we calculate the specific volume, which is simply the reciprocal of density. This gives an idea of how much space 1 kilogram of the liquid occupies—in this case, about 0.00775 m³ per kg. Finally, specific gravity (SG) is a unitless ratio comparing the fluid’s density to water’s (typically 1000 kg/m³). A value less than 1 (here, 0.129) means the liquid is much lighter than water.
In the second part, we reverse-engineer the problem. Given the weight and volume of oil, we compute specific weight first. Then we divide by gravitational acceleration to find the mass density, followed by calculating the specific gravity as a ratio to water’s density.
This method allows engineers and scientists to understand fluid properties in both the metric gravitational system (using force units like N/m³) and metric absolute system (mass-based, like kg/m³). These conversions are crucial for accurate fluid analysis in fields such as civil engineering, hydraulics, and petroleum engineering
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