A true centrifugal casting operation needs to be performed horizontally to make copper tube section with outer diameter of 250 mm and inner diameter of 230 mm. The value of acceleration due to gravity, g = 10 m/s². If a G-factor (ratio of centrifugal force to weight) of 60 is used for casting the tube, the rotational speed required is______ rpm (round off to the nearest integer).

The correct answer is 662 (with a precise value of approximately 661.59 rpm).

In a true horizontal centrifugal casting operation, a mold rotates about its horizontal axis at high speed, creating a centrifugal force that distributes the molten metal against the inner walls of the mold. The quality and distribution of the casting depend on the G-factor, which is defined as the ratio of the centrifugal force to the weight of the metal.
The given parameters from the problem statement are:
• Outer diameter of the tube, d_o = 250 mm = 0.25 m
• Inner diameter of the tube, d_i = 230 mm = 0.23 m
• Acceleration due to gravity, g = 10 m/s²
• G-factor = 60

For horizontal centrifugal casting, the outer radius (or diameter) is used to determine the required rotational speed. This is because the outer radius experiences the maximum centrifugal force, ensuring the molten metal adheres properly to the mold wall without slipping or raining.

The G-factor (G) is mathematically expressed as the ratio of the centrifugal force (F_c) to the weight (W = mg) of the molten metal:

$G=\frac{F_c}{mg}=\frac{mr{\omega }^2}{mg}=\frac{r{\omega }^2}{g}$

Here, r is the outer radius of the casting, which is equal to half of the outer diameter (d_o / 2). The angular velocity ω can be written in terms of the rotational speed N (in rpm) as follows:

$\omega =\frac{2\pi N}{60}$

Substituting r = d_o / 2 and ω into the G-factor expression gives:

$g{\left(\frac{2\pi N}{60}\right)}^2$

Rearranging the equation to isolate the rotational speed N:

${\left(\frac{2\pi N}{60}\right)}^2=\frac{2gG}{d_o}$

Taking the square root on both sides:

$\frac{2\pi N}{60}=\sqrt{\frac{2gG}{d_o}}$

Solving for N:

$N=\sqrt{\frac{2gG}{d_o}}\times \frac{60}{2\pi }$

Now, substitute the given values (G = 60, g = 10 m/s², and d_o = 0.25 m):

$N=\sqrt{\frac{2\times 10\times 60}{0.25}}\times \frac{60}{2\pi }$

$N=\sqrt{4800}\times \frac{30}{\pi }$

$N\approx 69.282\times 9.549\approx 661.59\text{rpm}$

Rounding off the calculated rotational speed of 661.59 rpm to the nearest integer yields:
N ≈ 662 rpm