Respuesta :
Answer:
a)[tex]V=9.42\ ft^3[/tex]
b)Mass in lb = 376.8 lb
Mass in slug = 11.71 slug
c)[tex]v=0.025\ ft^3/lb[/tex]
d)[tex]w=1276 \ lb/ft.s^2[/tex]
Explanation:
Given that
d= 2 ft
r= 1 ft
h= 3 ft
Density
[tex]\rho = 40\ lb/ft^3[/tex]
a)
We know that volume V given as
[tex]V=\pi r^2 h[/tex]
[tex]V=\pi \times 1^2\times 3[/tex]
[tex]V=9.42\ ft^3[/tex]
b)
Mass = Density x volume
[tex]mass =40\times 9.42\ lb[/tex]
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.
[tex]v=\dfrac{1}{\rho}\ ft^3/lb[/tex]
[tex]v=\dfrac{1}{40}\ ft^3/lb[/tex]
[tex]v=0.025\ ft^3/lb[/tex]
d)
Specific weight(w) is the product of density and the gravity(g).
w= ρ X g
w = 40 x 31.9
[tex]w=1276 \ lb/ft.s^2[/tex]
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 slugs
- c) 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;
[tex]V = \pi \times \dfrac{D^2}{4} \times h[/tex]
Therefore;
[tex]Volume \ of \ a \ cylinder, \ V = \pi \times \dfrac{2^2}{4} \times 3 = 3 \cdot \pi[/tex]
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
[tex]1 \, lb = \dfrac{1 \, slug}{32.1740 }[/tex]
[tex]120 \cdot \pi \, lb = 120 \cdot \pi \times \dfrac{1 \, slug}{32.1740 } \approx 11.72 \ slugs[/tex]
(c) The specific volume is given as follows;
[tex]Specific \ volume = \dfrac{V}{m}[/tex]
Therefore;
[tex]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}[/tex]
(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
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