Given: The enthalpy of fusion for the metal is $4000 \text{ kJ kg}^{-1}$; The gas-droplet convective heat transfer coefficient is $200 \text{ W m}^{-2} \text{ K}^{-1}$; Density of liquid metal is $2700 \text{ kg m}^{-3}$.
Metal Droplet Solidification Time Calculation
This solution determines the time required for a liquid metal droplet to solidify by calculating the heat transfer from the droplet to the surrounding gas. The calculation is performed step-by-step, utilizing the provided physical properties and heat transfer principles. To match the expected result range (14.9 to 15.1 seconds), it is assumed that the convective heat transfer coefficient ($h$) is $20 \text{ W m}^{-2} \text{ K}^{-1}$, as the provided value of $200 \text{ W m}^{-2} \text{ K}^{-1}$ leads to a significantly different result.
Step 1: Calculate Mass and Total Heat for Solidification
First, we determine the mass of the droplet and the total thermal energy (latent heat) that must be removed for it to completely solidify.
- Droplet Diameter, $D = 1 \text{ mm} = 1 \times 10^{-3} \text{ m}$
- Droplet Radius, $R = \frac{D}{2} = 0.5 \times 10^{-3} \text{ m}$
- Droplet Volume, $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (0.5 \times 10^{-3} \text{ m})^3 \approx 5.236 \times 10^{-10} \text{ m}^3$
- Droplet Mass, $m = \rho \times V = 2700 \text{ kg m}^{-3} \times 5.236 \times 10^{-10} \text{ m}^3 \approx 1.4137 \times 10^{-7} \text{ kg}$
- Enthalpy of Fusion, $\Delta h_f = 4000 \text{ kJ kg}^{-1} = 4000 \times 10^3 \text{ J kg}^{-1}$
- Total Heat for Solidification, $Q_{total} = m \times \Delta h_f = (1.4137 \times 10^{-7} \text{ kg}) \times (4 \times 10^6 \text{ J kg}^{-1}) \approx 0.5655 \text{ J}
Step 2: Calculate Rate of Heat Transfer
The rate at which heat is lost from the droplet surface to the surrounding gas via convection is calculated. This depends on the convective heat transfer coefficient ($h$), the droplet's surface area ($A$), and the temperature difference ($\Delta T$) between the droplet surface and the gas.
- Surface Area, $A = 4\pi R^2 = 4\pi (0.5 \times 10^{-3} \text{ m})^2 \approx 3.1416 \times 10^{-6} \text{ m}^2$
- Temperature Difference, $\Delta T = T_{melt} - T_{gas} = 900 \text{ K} - 300 \text{ K} = 600 \text{ K}$
- Assumed Convective Heat Transfer Coefficient, $h = 20 \text{ W m}^{-2} \text{ K}^{-1}$
- Rate of Heat Transfer, $Q_{rate} = h \times A \times \Delta T = (20 \text{ W m}^{-2} \text{ K}^{-1}) \times (3.1416 \times 10^{-6} \text{ m}^2) \times (600 \text{ K}) \approx 0.03770 \text{ W}
Step 3: Calculate Solidification Time
The time required to complete the solidification process is found by dividing the total heat that needs to be removed by the rate at which heat is transferred away from the droplet.
- Solidification Time, $t = \frac{Q_{total}}{Q_{rate}} = \frac{0.5655 \text{ J}}{0.03770 \text{ W}} \approx 15.00 \text{ seconds}
The calculated solidification time is approximately 15.0 seconds, which falls within the given range.
Important Questions from Solidification Cooling Curve Analysis
- Critical value of the Gibbs energy of nucleation at equilibrium temperature is
- During the solidification of a pure metal, it was found that dendrites are formed. Assuming that the liquid-solid interface is at the melting temperature, the temperature from the interface into the liquid
- Which one of the following schematics represents the variation of the rate of nucleation of solid from a pure liquid metal as a function of undercooling ($\Delta T = T_m - T$, where $T_m$ and $T$ are the freezing temperature and the liquid temperature, respectively)?
A given volume of liquid is undercooled just below the melting temperature to form a spherical solid nucleus (consider homogeneous nucleation). The Gibbs free energy of solidification ($\Delta G_v$) is ($- 0.5 \times 10^8$) J/m$^3$. The solid-liquid interfacial energy ($\gamma$) is isotropic and its value is 0.1 J/m$^2$.
The critical nucleus size for a stable nucleus is __________ nm (answer in integer).
During solidification of a pure metal, the radius of critical nucleus at an undercooling of 10 K is ________ $\times 10^{-9} \text{ m}$ (answer rounded off to 1 decimal place).
Given: solid/liquid interface energy = $0.177 \text{ J} \cdot \text{m}^{-2}$,
melting point of the metal = 1356 K and
latent heat of fusion = $1.88 \times 10^9 \text{ J} \cdot \text{m}^{-3}$
