Question

the specific weight of glycern is 78.6 lb/ft3. compute its density and specific gravitty. what is its specific weight in kN/m3

Answer 1

Explanation:

Specific weight is the amount of weight of substance present in a unit volume of the substance.It is given by :

$\gamma =\rho\times g$

$\rho$ = density of the substance

g = acceleration due to gravity = $9.8 m./s^2$

Specific gravity is the dimensionless ratio of the density of the substance to density of water at normal temperature.

$S_g=\frac{\rho }{\rho _{w}}$

We have : Specific weight of glycerin:

$\gamma = 78.6 lb/ft^3=78.6 lb/ft^3$

1 lb = 0.4536 kg

1 foot = 0.3048 m

$\gamma =\frac{78.6 \times 0.4536 kg}{ (0.3048 m)^3}=1,259.07 kg/m^3$

Density of glycerine = $\rho$

$\rho =\frac{\gamma }{g}=\frac{1,259.07 kg/m^3}{9.8 m/s^2}$

$=128.47 kg/m^3$

The density of the glycerin is $128.47 kg/m^3$.

The specific gravity of glycerin = $S_g$

Density of water =$\rho _w=  997 kg/m^3$

Density of glycerine = $\rho = 128.47 kg/m^3$

$S_g=\frac{128.47 kg/m^3}{997 kg/m^3}=0.128$

The specific gravity of glycerin is 0.128.

$\gamma = 78.6 lb/ft^3$

1 lb = 4.45 N

1 foot = 0.3048 m

$\gamma = \frac{78.6 \times 4.45 N}{1\times (0.3048)^3m^3}$

$=12,352.011 N/m^3=0.001 kN\times 12,352.011 N/m^3=12.352 kN/m^3$

Specific weight of glycerin is $12.352 kN/m^3$.