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In a rolling operation of mild steel plate, in a single pass the final thickness is reduced to two third of initial thickness. The entrance speed is 14 m/min and roll diameter is 450mm. If the plate width is increased by 2.5% during the operation. The exit velocity is ___ m/min.
(Important - Enter only the numerical value in the answer)
Correct answer is between '20.4,20.5'. Can you explain this answer?
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Understanding the Problem
To determine the exit velocity of the steel plate during the rolling operation, we need to consider the relationship between the initial and final thickness, as well as the entrance and exit velocities.
Given Data
- Initial thickness (h0): Let’s denote it as h.
- Final thickness (h1): h1 = (2/3)h
- Entrance speed (v0): 14 m/min
- Roll diameter (D): 450 mm
- Width increase: 2.5%
Calculating the Width Change
1. Initial Width (w0): Let w0 be the initial width.
2. Final Width (w1): The final width after a 2.5% increase can be calculated as:
- w1 = w0 × (1 + 0.025) = 1.025w0
Volume Conservation Principle
In rolling operations, the volume of the material remains constant. Thus, we can set up the equation based on initial and final dimensions:
- Initial Volume (V0) = Initial thickness × Initial width × Entrance length
- Final Volume (V1) = Final thickness × Final width × Exit length
Since V0 = V1, we have:
h × w0 × v0 = (2/3)h × (1.025w0) × v1
Canceling Out Terms
We can cancel h and w0 from both sides:
v0 = (2/3) × 1.025 × v1
Solving for Exit Velocity (v1)
Rearranging gives:
v1 = (3/2) × (v0 / 1.025)
Now, substituting the given entrance speed:
v1 = (3/2) × (14 / 1.025) ≈ 20.4 m/min
This calculation confirms that the exit velocity falls within the specified range of 20.4 to 20.5 m/min.
To determine the exit velocity of the steel plate during the rolling operation, we need to consider the relationship between the initial and final thickness, as well as the entrance and exit velocities.
Given Data
- Initial thickness (h0): Let’s denote it as h.
- Final thickness (h1): h1 = (2/3)h
- Entrance speed (v0): 14 m/min
- Roll diameter (D): 450 mm
- Width increase: 2.5%
Calculating the Width Change
1. Initial Width (w0): Let w0 be the initial width.
2. Final Width (w1): The final width after a 2.5% increase can be calculated as:
- w1 = w0 × (1 + 0.025) = 1.025w0
Volume Conservation Principle
In rolling operations, the volume of the material remains constant. Thus, we can set up the equation based on initial and final dimensions:
- Initial Volume (V0) = Initial thickness × Initial width × Entrance length
- Final Volume (V1) = Final thickness × Final width × Exit length
Since V0 = V1, we have:
h × w0 × v0 = (2/3)h × (1.025w0) × v1
Canceling Out Terms
We can cancel h and w0 from both sides:
v0 = (2/3) × 1.025 × v1
Solving for Exit Velocity (v1)
Rearranging gives:
v1 = (3/2) × (v0 / 1.025)
Now, substituting the given entrance speed:
v1 = (3/2) × (14 / 1.025) ≈ 20.4 m/min
This calculation confirms that the exit velocity falls within the specified range of 20.4 to 20.5 m/min.
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In a rolling operation of mild steelplate, in a single pass the finalthickness is reduced to two third ofinitial thickness. The entrance speedis 14 m/min and roll diameter is450mm. If the plate width isincreased by 2.5% during theoperation. The exit velocity is ___m/min.(Important - Enter only the numerical value in the answer)Correct answer is between '20.4,20.5'. Can you explain this answer?
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