A turning operation is performed with HSS tooling on mild steel, with Taylor tool life parameters n = 0.12, C = 60 m/min. Work part length = 450 mm and diameter = 80 mm. Feed = 0.20 mm/rev. Handling time per piece = 4.0 min, and tool change time = 1.5 min. Cost of machine and operator = $27/hr, and tooling cost = $2 per cutting edge. Find the a. cutting speed for maximum production rate and b. cutting speed for minimum cost Equations used n *-=c(") * Vmax = C 1-n Tt 1 Vmin = C Co n Pad-C(₁) n CoTt + Ct )"

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**Educational Content on Cutting Speed Optimization in Turning Operations**

A turning operation is performed using High-Speed Steel (HSS) tooling on mild steel. The Taylor tool life parameters are given as \( n = 0.12 \) and \( C = 60 \) m/min. The work part has a length of 450 mm and a diameter of 80 mm. The feed rate is 0.20 mm/rev. Handling time per piece is 4.0 minutes, and the tool change time is 1.5 minutes. The cost of the machine and operator is $27 per hour, and the tooling cost is $2 per cutting edge. Your task is to find:

a. The cutting speed for maximum production rate.
b. The cutting speed for minimum cost.

**Equations Used**

1. **Cutting Speed for Maximum Production Rate (\(v_{\text{max}}\))**:

   \[
   v_{\text{max}} = C \left( \frac{n}{1 - n} \cdot \frac{1}{T_t} \right)^n
   \]

2. **Cutting Speed for Minimum Cost (\(v_{\text{min}}\))**:

   \[
   v_{\text{min}} = C \left( \frac{n}{1 - n} \cdot \frac{C_o}{C_o T_t + C_t} \right)^n
   \]

In these equations:
- \( C \) is the Taylor constant (60 m/min).
- \( n \) is the Taylor tool life exponent (0.12).
- \( T_t \) is the total operation time per part.
- \( C_o \) is the cost of operation (machine and operator cost).
- \( C_t \) is the tooling cost.

**Understanding the Parameters**:

- **Taylor tool life parameters**: Used to predict the tool life based on cutting speeds and material properties.
- **Work part dimensions**: Important for calculating the material removal rate.
- **Feed and handling times**: Crucial for determining total machining and production times.
- **Costs**: Affect the calculation of economic cutting speeds that minimize overall production costs.

This setup reflects real-world industrial applications where optimizing costs and production rates is essential.
Transcribed Image Text:**Educational Content on Cutting Speed Optimization in Turning Operations** A turning operation is performed using High-Speed Steel (HSS) tooling on mild steel. The Taylor tool life parameters are given as \( n = 0.12 \) and \( C = 60 \) m/min. The work part has a length of 450 mm and a diameter of 80 mm. The feed rate is 0.20 mm/rev. Handling time per piece is 4.0 minutes, and the tool change time is 1.5 minutes. The cost of the machine and operator is $27 per hour, and the tooling cost is $2 per cutting edge. Your task is to find: a. The cutting speed for maximum production rate. b. The cutting speed for minimum cost. **Equations Used** 1. **Cutting Speed for Maximum Production Rate (\(v_{\text{max}}\))**: \[ v_{\text{max}} = C \left( \frac{n}{1 - n} \cdot \frac{1}{T_t} \right)^n \] 2. **Cutting Speed for Minimum Cost (\(v_{\text{min}}\))**: \[ v_{\text{min}} = C \left( \frac{n}{1 - n} \cdot \frac{C_o}{C_o T_t + C_t} \right)^n \] In these equations: - \( C \) is the Taylor constant (60 m/min). - \( n \) is the Taylor tool life exponent (0.12). - \( T_t \) is the total operation time per part. - \( C_o \) is the cost of operation (machine and operator cost). - \( C_t \) is the tooling cost. **Understanding the Parameters**: - **Taylor tool life parameters**: Used to predict the tool life based on cutting speeds and material properties. - **Work part dimensions**: Important for calculating the material removal rate. - **Feed and handling times**: Crucial for determining total machining and production times. - **Costs**: Affect the calculation of economic cutting speeds that minimize overall production costs. This setup reflects real-world industrial applications where optimizing costs and production rates is essential.
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