Question

A 36 Ne ring spun yarn is produced from 1.2 Ne roving. The total draft and the break draft of the roving frame are 12 and 1.2, respectively. The diameters of back bottom roller, middle bottom roller and front bottom roller of the roving frame are 28 mm, 25 mm and 28 mm, respectively. If the middle bottom roller of the roving frame is eccentric, then the wavelength (m) of the periodic fault in the yarn, neglecting twist contraction, (rounded off to 2 decimal places) is \underline{\hspace{2cm}.}

Hint:

For periodic faults from a particular roller, $\lambda_{\text{final}} = C_{\text{roller}} \times (\text{draft from that roller to final delivery})$. Identify the disturbance zone, compute the relevant draft to the final stage, and multiply by the roller circumference.

Answer 1

Correct Answer: 23

Step 1: Draft distribution in the roving frame
Total draft $=12$, break draft $=1.2 \Rightarrow$ main draft (middle$\to$front) $= \dfrac{12}{1.2}=10$.
Step 2: Disturbance source and base circumference
Eccentric roller $=$ middle bottom roller; circumference $C_m=\pi D_m=\pi \times 25\ \text{mm}=78.54\ \text{mm} =0.07854\ \text{m}$.
Step 3: Wavelength at roving-frame delivery due to middle roller
At front nip of roving frame, the length delivered per one revolution of the middle roller $= C_m \times (\text{main draft}) = 0.07854 \times 10=0.7854\ \text{m}$.
Step 4: Amplification through ring frame
Roving $1.2$ Ne $\to$ yarn $36$ Ne $\Rightarrow$ spinning draft $= \dfrac{36}{1.2}=30$.
Therefore wavelength in the final yarn: $\lambda = 0.7854 \times 30 = 23.562 \approx 23.56\ \text{m}$.