Cooling rate of a molten weld droplet: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2019, Q7

Question 7 — Cooling rate of a molten weld droplet (Part B, equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Diameter	$D$	0.5 mm
Initial temperature / velocity	$T_i,\,V$	1700 K, 1 m/s
Air temperature	$T_\infty$	300 K
Droplet $\rho,\,c_p,\,\epsilon$	—	2100 kg/m³, 1100 J/kg·K, 0.20
Air at film $T_f=1000$ K	$k,\nu,Pr$	0.0667 W/m·K, 1.20×10⁻⁴ m²/s, 0.726

Find. Instantaneous cooling rate $dT/dt$ (and the convective and radiative heat losses that drive it).

The tiny droplet is nearly isothermal (small Biot number), so a lumped energy balance applies. Convection dominates radiation ~6:1 at this size despite the 1700 K surface.

Approach. Get the convection coefficient from the Ranz–Marshall sphere correlation at the film temperature, sum the convective and radiative losses, and divide by the droplet's thermal capacitance to obtain the instantaneous cooling rate.

Reynolds number and Nusselt (sphere). $Re=VD/\nu=(1)(5\times10^{-4})/1.20\times10^{-4}=4.16$; Ranz–Marshall: $$Nu=2+0.6\,Re^{1/2}Pr^{1/3}=2+0.6(2.04)(0.899)=3.10,\quad h=\frac{Nu\,k}{D}=\frac{3.10(0.0667)}{5\times10^{-4}}=414\ \text{W/m}^2\text{K}$$
Convective and radiative losses. With $A=\pi D^2=7.854\times10^{-7}\ \text{m}^2$: $$\dot Q_\text{conv}=hA(T_i-T_\infty)=414(7.854\times10^{-7})(1400)=0.455\ \text{W}$$ $$\dot Q_\text{rad}=\epsilon\sigma A(T_i^4-T_\infty^4)=0.20(5.67\times10^{-8})(7.854\times10^{-7})(1700^4-300^4)=0.074\ \text{W}$$ $$\dot Q=\dot Q_\text{conv}+\dot Q_\text{rad}=\boxed{0.529\ \text{W}}$$
Lumped cooling rate. With capacitance $\rho\,\forall\,c_p=2100\,(6.545\times10^{-11})\,1100=1.512\times10^{-4}\ \text{J/K}$ ($\forall=\pi D^3/6$): $$\frac{dT}{dt}=-\frac{\dot Q}{\rho\forall c_p}=-\frac{0.529}{1.512\times10^{-4}}=\boxed{-3.5\times10^{3}\ \text{K/s}}$$

Quantity	Result
Convection coefficient	414 W/m²K
Convective / radiative loss	0.455 W / 0.074 W
Total heat loss	0.529 W
Initial cooling rate	≈ −3500 K/s

Concept & Theory — Q7

Topic: Lumped-capacitance transient cooling of a small sphere by simultaneous convection and radiation.

Key relations: Ranz–Marshall $Nu=2+0.6Re^{1/2}Pr^{1/3}$; $h=Nu\,k/D$; energy balance $\rho\forall c_p\,dT/dt=-[hA(T-T_\infty)+\epsilon\sigma A(T^4-T_\infty^4)]$; Biot number $Bi=h(r/3)/k_\text{drop}\ll1$ justifies the lumped model.

Why this works: A 0.5 mm metal droplet has an internal conduction resistance far smaller than its surface resistance ($Bi\ll0.1$), so it stays essentially uniform in temperature and a single ordinary differential equation for $T(t)$ suffices. The instantaneous rate is just the total surface loss divided by the stored heat capacity. Even at 1700 K the radiative term is modest here because the surface area is minuscule and the emissivity low, so forced convection carries most of the load.

Common pitfall: Forgetting the "+2" floor in the sphere Nusselt number (conduction limit for a quiescent sphere), or evaluating air properties at the droplet temperature rather than the film temperature. Radiation must use absolute (kelvin) temperatures to the fourth power.

Source: Incropera 8th ed., §7.4 (flow over a sphere) & §5.1–5.3 (lumped capacitance); Çengel & Ghajar, Heat and Mass Transfer 6th ed., §4-1 & §7-3.

Question 7 of 8: Cooling rate of a molten weld droplet

Question 7 — Cooling rate of a molten weld droplet (Part B, equal value)