**Step 1: Conceptual Introduction** **Conceptual Introduction:** Quenching is a rapid cooling process used in heat treatment to achieve desired mechanical properties in metals. For cylindrical billets, cooling occurs primarily by convection, where heat is transferred from the billet surface to the surrounding air. The effectiveness of quenching depends on properties such as thermal conductivity, specific heat, density, billet geometry, and the convection heat transfer coefficient, which varies with air velocity (still or forced). The Biot number determines whether temperature gradients inside the billet are significant and influences the choice of analysis method. **Explanation Block:** The heat transfer from the billet to the air involves both conduction within the billet and convection at the billet surface. For cylindrical objects, transient heat conduction theory is used when internal temperature gradients are significant (high Biot number). The convection heat transfer coefficient (`h`) is crucial in determining how quickly the billet cools. Depending on whether the air is still or forced, `h` changes, impacting the cooling rate and temperature distribution within the billet. --- **Step 2: Formulae** **Key Formulae Used:** 1. **Convection Heat Transfer Coefficient (for horizontal cylinder):** \[ \text{Nu}_D = C \times \text{Ra}_D^n \] \[ h = \frac{\text{Nu}_D \cdot k_{air}}{D} \] 2. **Transient Heat Conduction in a Cylinder (Heisler Chart Solution):** \[ \frac{T(r, t) - T_\infty}{T_i - T_\infty} = f(\text{Bi}, \text{Fo}, r/R) \] Where: - \( \text{Bi} = \frac{hR}{k_{solid}} \) (Biot number) - \( \text{Fo} = \frac{\alpha t}{R^2} \) (Fourier number) 3. **Heat Removed from Billet:** \[ Q = m c (T_i - T_f) \] Where: - \( m = \rho V \) (mass) - \( V = \pi (D/2)^2 L \) (volume of cylinder) 4. **Centerline Temperature (from Heisler solution at \( r = 0 \)):** 5. **Biot Number:** \[ \text{Bi} = \frac{hR}{k_{solid}} \] **Explanation Block:** The Nusselt number (`Nu_D`) is a dimensionless number used to estimate `h` for convection. For transient analysis, the Heisler solution gives temperature at any point and time, using Biot and Fourier numbers. The heat removed is calculated using the mass, specific heat, and temperature change. The Biot number tells us whether we can use lumped capacitance (Bi < 0.1) or must consider spatial temperature gradients. --- **Step 3: Still Air Quench Calculations** ### (a) Estimate the convection heat transfer coefficient (`h`) Given: - D = 0.25 m (Diameter) - T_surface ≈ T_initial = 500°C - T_air = 25°C For still air, use empirical correlations for natural convection over a horizontal cylinder: - \( h \approx 7.32 \, \text{W/m}^2\cdot\text{K} \) (given in figure for still air) **Explanation Block:** For natural convection around a horizontal cylinder, the heat transfer coefficient is typically in the range 5–10 W/m²·K for air at these sizes and temperature differences. The value provided (7.32 W/m²·K) is acceptable for further calculations. --- ### (b) Estimate the time to reach 325°C from 500°C 1. **Calculate Biot Number:** \[ R = D/2 = 0.125 \, \text{m} \] \[ \text{Bi} = \frac{hR}{k} = \frac{7.32 \times 0.125}{13.4} \approx 0.068 \] 2. **Calculate Fourier Number for Lumped Capacitance:** \[ \text{Fo} = \frac{\alpha t}{R^2} \] Since Bi < 0.1, we can use the lumped capacitance method: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho c V} t\right) \] For a cylinder: \( V = \pi R^2 L \), \( A = 2\pi RL + 2\pi R^2 \) For a long cylinder, neglect the ends: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{hA}{\rho c V} t\right) \] With \( A/V = \frac{2}{R} \) for a long cylinder: \[ \frac{T(t) - T_\infty}{T_i - T_\infty} = \exp\left(-\frac{2h}{\rho c R} t\right) \] Substitute values: \[ \ln\left(\frac{325 - 25}{500 - 25}\right) = -\frac{2 \times 7.32}{8238 \times 468 \times 0.125} t \] \[ \ln\left(\frac{300}{475}\right) = -\frac{14.64}{481986} t \] \[ \ln(0.632) = -3.037 \times 10^{-5} t \] \[ -0.459 = -3.037 \times 10^{-5} t \] \[ t = \frac{0.459}{3.037 \times 10^{-5}} \approx 15113 \, \text{s} \approx 252 \, \text{min} \] **Explanation Block:** Since Bi < 0.1, lumped capacitance is valid. The temperature ratio is set up for the desired final temperature. The time is solved directly. The answer exceeds the allowable 220 min, indicating still air is too slow. --- ### (c) Centerline Temperature after 25 min \[ t = 25 \, \text{min} = 1500 \, \text{s} \] \[ \frac{T(t) - 25}{500 - 25} = \exp\left(-3.037 \times 10^{-5} \times 1500\right) \] \[ = \exp(-0.0456) = 0.955 \] \[ T(t) = 25 + 0.955 \times 475 \approx 25 + 454 = 479^\circ C \] **Explanation Block:** The lumped capacitance model is used again, as Bi < 0.1. The temperature after 25 min remains very high, showing slow cooling in still air. --- ### (d) Heat Removed from Billet \[ m = \rho V = 8238 \times \pi (0.125)^2 \times 1 \approx 404 \, \text{kg} \] \[ Q = m c (T_i - T_f) = 404 \times 468 \times (500 - 325) = 404 \times 468 \times 175 \approx 33,040,920\, \text{J} \approx 33,041\, \text{kJ} \] **Explanation Block:** The mass is calculated using the cylinder volume and density. The energy removed is the product of mass, specific heat, and temperature change. --- ### (e) Biot Number \[ \text{Bi} = \frac{hR}{k} = \frac{7.32 \times 0.125}{13.4} \approx 0.068 \] **Explanation Block:** Bi < 0.1 confirms that temperature gradients within the billet are negligible, validating the use of the lumped capacitance method. --- **Step 4: Forced Air Quench Calculations** ### (Repeat steps with forced air) #### (a) Estimate `h` for forced air Given: \( h \approx 70 \, \text{W/m}^2\cdot\text{K} \) (from figure for forced air at 3 m/s) #### (b) Time to Reach 325°C \[ \text{Bi} = \frac{70 \times 0.125}{13.4} \approx 0.654 \] Now Bi > 0.1, so use Heisler chart (transient conduction in cylinder). But for estimation, let's try lumped capacitance for a quick comparison. \[ \frac{T(t) - 25}{500 - 25} = \exp\left(-\frac{2 \times 70}{8238 \times 468 \times 0.125} t\right) \] \[ = \exp\left(-\frac{140}{481986} t\right) \] \[ = \exp(-0.000290 t) \] Set up: \[ \ln(0.632) = -0.000290 t \] \[ -0.459 = -0.000290 t \] \[ t = \frac{0.459}{0.000290} \approx 1583 \, \text{s} \approx 26.4 \, \text{min} \] Since Bi > 0.1, this result is approximate, but it shows forced air is much faster. #### (c) Centerline Temperature after 25 min \[ t = 1500\, \text{s} \] \[ \exp(-0.000290 \times 1500) = \exp(-0.435) = 0.647 \] \[ T = 25 + 0.647 \times 475 \approx 25 + 307 = 332^\circ C \] #### (d) Heat Removed (same as before): \[ Q = 404 \times 468 \times (500 - 325) = 33,041\, \text{kJ} \] #### (e) Biot Number \[ \text{Bi} = 0.654 \] **Explanation Block:** Forced air has a much higher heat transfer coefficient, greatly reducing the cooling time. However, Bi > 0.1 means that temperature gradients inside the billet are now significant, so a more detailed (Heisler) analysis may be needed for precise results. --- **Step 5: Appropriate Quenching Medium** Forced air easily cools the billet to 325°C in less than 220 min, while still air cannot achieve this within required time. Therefore, forced air is the appropriate choice. **Explanation Block:** The forced air option provides a much higher heat transfer rate, ensuring the billet cools quickly enough to meet metallurgical requirements. Still air is too slow to meet the process specification. --- ### **Final Answers:** - **Still air quench:** - Heat transfer coefficient: **7.32 W/m²·K** - Time to 325°C: **252 min** (too slow) - Centerline temp after 25 min: **479°C** - Heat removed: **33,041 kJ** - Biot number: **0.068** - **Forced air quench:** - Heat transfer coefficient: **70 W/m²·K** - Time to 325°C: **26.4 min** - Centerline temp after 25 min: **332°C** - Heat removed: **33,041 kJ** - Biot number: **0.654** - **Recommended quenching medium:** **Forced air.**