Engineering, 05.10.2019 10:30 babytony1210

Acylindrical 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

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Answer from: Lnjayy

The determined values for the fluid

a.) The volume of the fluid is 3 × π ft.³ (approximately 9.425 ft.³)b.) The mass of the fluid in pounds is 120·π lb (approximately 376.99 m) The mass of the fluid in slugs is approximately 11.72 slugsc) The specific volume, is approximately 40.02 lbf/ft.³

The reasons why the above values are correct are given as follows;

The known parameter are;

Dimensions of the cylindrical drum, are;

Diameter, D = 2 ft.

Height, h = 3 ft.

Density of the fluid, ρ = 40 lbf/ft.³

(a) The total volume of fluid is given by the formula for the volume of a cylinder which is presented as follows;

$V = \pi \times \dfrac{D^2}{4} \times h$

Therefore;

$Volume \ of \ a \ cylinder, \ V = \pi \times \dfrac{2^2}{4} \times 3 = 3 \cdot \pi$

The volume of the cylinder, V = 3·π ft.³

The volume of the cylinder = The volume of the fluid = 3 × π ft.³

(b) Mass = Density × Volume

The total mass of the fluid = The total volume of the fluid × Fluid density

∴ The total mass of the fluid = 3·π ft.³ × 40 lb/ft.³ = 120·π lb

The mass in slugs;

1 slug = 32.1740 lb

$1 \, lb = \dfrac{1 \, slug}{32.1740 }$

$120 \cdot \pi \, lb = 120 \cdot \pi \times \dfrac{1 \, slug}{32.1740 } \approx 11.72 \ slugs$

$Specific \ volume = \dfrac{V}{m}$

Therefore;

$Specific \ volume = \dfrac{3 \cdot \pi \ ft.^3}{120 \cdot \pi \ lb} = \dfrac{1}{40} \, \dfrac{ft.^3}{lb} = 0.025 \ \dfrac{ft.^3}{lb}$

(d) The specific weight, γ, is the weight per unit volume

The specific weight is given as follows;

γ = ρ·g

Where;

g = Acceleration due to gravity = 31.90 fps²

Which gives;

γ = 40 lb/ft.³ × 31.90 fps² = 1,276 lb/(ft.²·s²) ≈ 40.02 lbf/ft.³

Learn more about volume, mass, and specific volume of a fluid here:

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Answer from: bbrid8590

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

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

We know that volume V given as

$V=\pi r^2 h$

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

$V=9.42\ ft^3$

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

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

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

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

$v=0.025\ ft^3/lb$

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

w= ρ X g

w = 40 x 31.9

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

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