Question

A cylindrical drum (2 ft. dia ,3 ft height) is filled with a fluid whose density is 40 lb/ft^3. Determine (a. the total volume of a fluid, (b) its total mass in pound and Slugs, (c) its specific volume, and (d) its specific weight where g= 31.90 fps^2

Answer 1

Answer:

a)$V=9.42\ ft^3$

b)Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)$v=0.025\ ft^3/lb$

d)$w=1276 \ lb/ft.s^2$

Explanation:

Given that

d= 2 ft

r= 1 ft

h= 3 ft

Density

$\rho = 40\ lb/ft^3$

a)

We know that volume V given as

$V=\pi r^2 h$

$V=\pi \times 1^2\times 3$

$V=9.42\ ft^3$

b)

Mass = Density x volume

$mass =40\times 9.42\ lb$

mass= 376.8 lb

We know that

1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

$v=\dfrac{1}{\rho}\ ft^3/lb$

$v=\dfrac{1}{40}\ ft^3/lb$

$v=0.025\ ft^3/lb$

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

w = 40 x 31.9

$w=1276 \ lb/ft.s^2$