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

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

And specific gravity of oil will be 0.897

Step-by-step 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=(1.74* 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
`=(897.292)/(1000)=0.897`