Question

Calculate the minimum number of hot rolling passes required to reduce an ingot of 200 mm thickness to 100 mm thickness in two high reversible rolling mill with roll diameter of 500 mm. The coefficient of friction between rolls and the hot material for all the passes is assumed as 0.2

• A. \( 8 \)
• B. \( 9 \)
• C. \( 10 \)
• D. \( 11 \)

Hint:

The maximum possible reduction in rolling without slippage is given by \( \mu^2 R \). This relationship is crucial for determining the number of passes required to achieve a desired thickness reduction, especially in hot rolling where the friction plays a significant role.

Answer 1

The Correct Option is C

Step 1: Identify the given parameters.
Initial thickness of ingot \( h_0 = 200 \text{ mm} \).
Final thickness of ingot \( h_f = 100 \text{ mm} \).
Roll diameter \( D = 500 \text{ mm} \).
Roll radius \( R = D/2 = 500/2 = 250 \text{ mm} \).
Coefficient of friction \( \mu = 0.2 \).
Step 2: Calculate the maximum possible reduction per pass (\( \Delta h_{max} \)).
For hot rolling, the maximum possible reduction (or draft) \( \Delta h_{max} \) without slipping is given by the formula: \[ \Delta h_{max} = \mu^2 R \] Substitute the given values: \[ \Delta h_{max} = (0.2)^2 \times 250 \text{ mm} \] \[ \Delta h_{max} = 0.04 \times 250 \text{ mm} \] \[ \Delta h_{max} = 10 \text{ mm} \] This means in each pass, the thickness can be reduced by a maximum of 10 mm.
Step 3: Calculate the total reduction required.
Total reduction \( \Delta h_{total} = h_0 - h_f \). \[ \Delta h_{total} = 200 \text{ mm} - 100 \text{ mm} = 100 \text{ mm} \]
Step 4: Calculate the minimum number of passes.
Minimum number of passes \( N = \frac{\Delta h_{total}}{\Delta h_{max}} \). \[ N = \frac{100 \text{ mm}}{10 \text{ mm/pass}} \] \[ N = 10 \text{ passes} \] The final answer is $\boxed{\text{3}}$.