Answer:
a) 423 rpm
b) G-factor = 60
c) [tex]P= 231634.882 Pa[/tex]
d) Yes, it would result in a successful operation
Explanation:
Given that:
length of aluminium ring = 5 cm
outside diameter [tex]d_1[/tex] = 65 cm
inside diameter [tex]d_2[/tex] = 60 cm
rotational G - factor = 60
G- factor = [tex]\frac{centrifugal force}{gravitational force }[/tex]
G- factor = [tex]\frac{mv^2}{r.mg}[/tex]
G- factor = [tex]\frac{v^2}{rg}[/tex]
where v = r ω and ω = [tex]\frac{2 \pi N}{60}[/tex]
[tex]\omega = \frac{2*3.14N}{60}[/tex]
G- factor = [tex]\frac{ r (\frac{2(3.14}{60}N)^2 }{g}[/tex]
G- factor = [tex]\frac{2r (0.005477)*N^2}{9.81}[/tex]
[tex]N^2 = \frac{60}{2r*0.000558}[/tex]
[tex]N = \sqrt{\frac{107526.882}{60*10^{-2}} }[/tex]
N = 423 rpm
b) However, the material is steel and N = 423 rpm
∴ 423 ⇒ [tex]42.3\sqrt{\frac{G-factor}{60*10^{-2}} }[/tex]
[tex]10 = \sqrt{\frac{G-factor}{6.0*10^{-2}} }[/tex]
G-factor = 60
Thus, the G-factor is independent of the material
c) Centrifugal force per square meter is expressed as:
[tex]P = \frac{F}{A}[/tex]
where; F = [tex]\frac{mv^2}{r}[/tex]
[tex]P= \frac{mv^2}{r.A}[/tex]
Also; m = ρ. V
and V = A. l
∴
[tex]P = \frac{\rho (A.l)V^2}{r.A}[/tex]
[tex]P = \frac{\rho lV^2}{r}[/tex]
where;
ρ = 7.87 g/cm ³
ρ = 7870 kg/m³
[tex]P = \frac{7870(5*10^{-2})(r.\omega)^2}{r}[/tex]
[tex]P = {7870(5*10^{-2})r.\omega^2}[/tex]
[tex]P = {7870(5*10^{-2})*30*10^{-2}.(\frac{2 \pi 423}{60} )^2[/tex]
[tex]P= 231634.882 Pa[/tex]
d)
The rotational speed will possibly result into a successful operation. This is so because the range of speed of operation of centrifugal casting is between (300- 3000) rpm and N = 423 rpm which tells us that; it is still in range and will definitely result into a successful operation.
