Question

Calculate the density of a hydraulic oil in units of kg/m^3 knowing that the density is 1.74 slugs/ft^3. Then, calculate the specific gravity of the oil.

Answer 1

Answer:

Density of oil will be 897.292 kg$m^3$

And specific gravity of oil will be 0.897

Explanation:

We have given density of oil is 1.74 slugs/$ft^3$

We have to convert this slugs/$ft^3$ into kg/$m^3$

We know that 1 slugs = 14.5939 kg

So 1.74 slug = 1.74×14.5939 = 25.3933 kg

And 1 cubic feet = 0.0283 cubic meter

So $1.74slug/ft^3=\frac{1.74\times 14.5939kg}{0.0283m^3}=897.292kg/m^3$

Now we have to calculate specific gravity it is the ratio of density of oil and density of water

We know that density of water = 1000 kg/$m^3$

So specific gravity of water $=\frac{897.292}{1000}=0.897$