Question

Problem 2: If 6m3 of oil weighs 47kN, calculate the specific weight, density, specific volume, and specific gravity. It is given that thespecific weight of water at 4 0C is 9810N/m3.

Answer 1

Answer on Question 52993, Physics, Mechanics | Kinematics | Dynamics

Question:

If $6m^3$ of oil weights $47kN$, calculate the specific weight, density, specific volume, and specific gravity. It is given that the specific weight of water at $4{}^{\circ}\mathrm{C}$ is $9810N/m^3$.

Solution:

1) The specific weight of oil is its weight per unit volume:

$$\gamma = W/V = 47 \cdot 10^3 N / 6m^3 = 7.833 kN/m^3.$$

2) The density of oil is its mass per unit volume:

$$m = W/g = 47 \cdot 10^3 N / 9.81 \, m/s^2 = 4791 kg,$$ $$\rho = m/V = 4791 kg / 6m^3 = 798.5 \, kg/m^3.$$

3) The specific volume of oil is the ratio of its volume to its mass:

$$v = V/m = \rho^{-1} = (798.5 \, kg/m^3)^{-1} = 0.00125 \, m^3/kg.$$

4) The specific gravity is defined as the ratio of the specific weight of fluid to the specific weight of a standard fluid. For liquids, the standard fluid is taken water. So, let's obtain the specific gravity of oil:

$$s.g. = \gamma / \gamma_{water \, at \, 4{}^{\circ}\mathrm{C}} = 7.833 \cdot 10^3 N/m^3 / 9.81 \cdot 10^3 N/m^3 = 0.8.$$

Answer:

1) $\gamma = 7.833 \, kN/m^3$.

2) $\rho = 798.5 \, kg/m^3$.

3) $v = 0.00125 \, m^3/kg$.

4) $s.g. = 0.8$.

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