Question

Kerosene is mixed with 10 ft3 of ethyl alcohol so that the volume of the mixture in the tank becomes 14 ft3 . Determine the specific weight and the specific gravity of the mixture.

Answer 1

Answer:

specific weight = 49.7 lb/ft^3

specific gravity = 0.797

Explanation:

From Appendix A,

pk = 1.58 slug/ft^3

pea = 1.53 slug/ft^3

volume of kerosene is

Vk = 14 ft^3 - 10 ft^3 = 4 ft^3

The total weight of the mixture is therefore

W = pk g Vk + pea g Vea

= (1.58 slug/ft3) (32.2 ft/s^2) (4 ft^3) + (1.53 slug/ft^3) (32.2 ft/s^2) (10 ft^3)

= 696.16 lb

The specific weight and specific gravity of the mixture are

γm = W /V

γm = 696.16 lb/ 14 ft^3 = 49.73 lb/ ft^3 = 49.7 lb/ft^3

Sm = γm / γv

Sm = 49.73 lb/ft^3/ 62.4 lb/ft^3 = 0.797