Question

A ring frame is scheduled to run with the following settings:
Linear speed of traveller = 30 m/s
Front roller delivery speed = 18 m/min
Bobbin diameter = 25 mm
Spindle rpm = 15430
What ring diameter (in mm) will be appropriate for the above settings?

• A. 37.7
• B. 62.8
• C. 35.5
• D. 0.628

Hint:

In ring spinning problems, remember that traveller speed is governed mainly by ring diameter and spindle speed. Always convert rpm into per-second terms using division by 60.

Answer 1

The Correct Option is A

Step 1: Understanding the concept.
In a ring frame, the linear speed of the traveller is equal to the peripheral speed of the ring. The traveller speed depends on the ring diameter and the spindle speed. The relationship between traveller speed, ring diameter, and spindle speed is given by:
\[ V = \frac{\pi \times D \times N}{60} \] where:
$V$ = linear speed of traveller (m/s)
$D$ = ring diameter (m)
$N$ = spindle speed (rpm)
Step 2: Substituting the given values.
Given:
\[ V = 30 \text{ m/s}, \quad N = 15430 \text{ rpm} \] Rearranging the formula to find ring diameter:
\[ D = \frac{V \times 60}{\pi \times N} \] Step 3: Calculation.
\[ D = \frac{30 \times 60}{\pi \times 15430} \] \[ D = \frac{1800}{48473} \] \[ D \approx 0.0371 \text{ m} \] Step 4: Converting into millimetres.
\[ D = 0.0371 \times 1000 = 37.1 \text{ mm} \] This value is closest to 37.7 mm among the given options.
Step 5: Conclusion.
Hence, the appropriate ring diameter for the given operating conditions is approximately 37.7 mm.

Answer 2

Instead of starting from a memorized formula, we can build the relationship from the physical picture of what's happening: the traveller races around the ring once per revolution of the spindle, so the distance it covers in one revolution is the ring's circumference.

First, convert the spindle speed into revolutions per second:
\[ N = 15430 \text{ rpm} = \frac{15430}{60} \approx 257.17 \text{ rev/s} \]

Since the traveller completes one full trip around the ring (a distance equal to the ring's circumference, \( \pi D \)) for every one revolution, and it does this 257.17 times every second at a linear speed of 30 m/s, the circumference must satisfy:
\[ \pi D \times 257.17 = 30 \text{ m/s} \]

Solving for the circumference:
\[ \pi D = \frac{30}{257.17} \approx 0.1167 \text{ m} \]

Solving for the diameter:
\[ D = \frac{0.1167}{\pi} \approx 0.0371 \text{ m} = 37.1 \text{ mm} \]

Among the given choices, this value sits closest to 37.7 mm, with the other options (62.8 mm, 35.5 mm, and 0.628 mm) being far less consistent with the calculated ring size.

Therefore, the correct answer is 37.7 mm.