Question

Horizontal true centrifugal casting is used to make aluminum rings with length = 5 cm, outside diameter = 65 cm, and inside diameter = 60 cm. (a) Determine the rotational speed that will provide a G-factor = 60. (b) Suppose that the ring were made out of steel instead of aluminum. If the rotational speed computed in part (a) were used in the steel casting operation, determine the G-factor and (c) centrifugal force per square meter (Pa) on the mold wall. (d) Would this rotational speed result in a successful operation? The density of steel = 7.87 g/cm3

Answer 1

Answer:

a) 423 rpm

b) G-factor = 60

c) $P= 231634.882 Pa$

d) Yes, it would result in a successful operation

Explanation:

Given that:

length of aluminium ring = 5 cm

outside diameter $d_1$ = 65 cm

inside diameter $d_2$ = 60 cm

rotational G - factor = 60

G- factor = $\frac{centrifugal force}{gravitational force }$

G- factor = $\frac{mv^2}{r.mg}$

G- factor = $\frac{v^2}{rg}$

where v = r ω  and ω = $\frac{2 \pi N}{60}$

$\omega = \frac{2*3.14N}{60}$

G- factor = $\frac{ r (\frac{2(3.14}{60}N)^2 }{g}$

G- factor = $\frac{2r (0.005477)*N^2}{9.81}$

$N^2 = \frac{60}{2r*0.000558}$

$N = \sqrt{\frac{107526.882}{60*10^{-2}} }$

N = 423 rpm

b) However, the material is steel and N = 423 rpm

∴ 423 ⇒ $42.3\sqrt{\frac{G-factor}{60*10^{-2}} }$

$10 = \sqrt{\frac{G-factor}{6.0*10^{-2}} }$

G-factor = 60

Thus, the G-factor is independent of the material

c) Centrifugal force per square meter is expressed as:

$P = \frac{F}{A}$

where; F = $\frac{mv^2}{r}$

$P= \frac{mv^2}{r.A}$

Also; m = ρ. V

and  V = A. l

∴

$P = \frac{\rho (A.l)V^2}{r.A}$

$P = \frac{\rho lV^2}{r}$

 where;

ρ = 7.87 g/cm ³

ρ = 7870 kg/m³

$P = \frac{7870(5*10^{-2})(r.\omega)^2}{r}$

$P = {7870(5*10^{-2})r.\omega^2}$

$P = {7870(5*10^{-2})*30*10^{-2}.(\frac{2 \pi 423}{60} )^2$

$P= 231634.882 Pa$

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.