In a thick film, thermal barrier coating process, spherical zirconia (ZrO_{2}) particles are heated in a thermal plasma, melted, and deposited on a substrate. This is shown in Figure. The average particle diameter is D. The particle arrives in the plasma with an initial uniform temperature T (t = 0). Assume that the surface-convection resistance is negligible such that the particle surface temperature suddenly changes to T_{s} = T_{f,∞}.
(a) Determine the temperature at the center of the particle after an elapsed time t equal to the time of particle flight L/u_{p} (elapsed time for impaction).
(b) Neglect the effect of melting on the transient conduction within the particle. Has the center of the particle melted?
(c) If the distance to the substrate is reduced by 1/2, what would be the temperature at the center of the particle at the time of impaction?
(d) What is the average temperature \left\langle T\right\rangle _{V} for (c) above? What fraction of the maximum stored energy has the particle reached in (c)?
Question 3.16: In a thick film, thermal barrier coating process, spherical ...
(a) To determine the temperature at the center of the sphere at time t, we use Figure (b)(ii). The center is marked by r = 0 and the temperature is given in the dimensionless form T − T (t = 0)/[T_{s} − T (t = 0)]. From Table, we have for zirconia
\rho =5,680 Kg/m^{3} Table
c_{p} =610 J/Kg-k Table
k = 1.675 W/m-K Table
\alpha =\frac{k}{\rho c_{p}}=4.834\times 10^{-7} m^{2}/sThe Fourier number is
Fo_{R}=\frac{\alpha t}{R^{2}}=\frac{4\alpha t}{D^{2}}=\frac{4\alpha L}{u_{p}D^{2}}=\frac{4\times (4.834\times 10^{-7})(m^{2}/s)\times 2.5\times 10^{-2}(m)}{100(m/s)\times (3\times 10^{-5})^{2}(m^{2})}
=0.5371
By examining Figure (b)(ii), we extrapolate to higher Fo_{R} and find that
\frac{T(r=0,Fo_{R}=0.5371)-\tau (t=0)}{T_{s}-T(t=0)}=1i.e., T(r=0,Fo_{R}=0.5371)=T_{s}=2,800^{\circ }C
(b) the melting of solid reduces the penetration speed. Here we neglect this effect and from the result of (a), we have
T(r=0,Fo_{R}=0.5371)=2,800^{\circ }C \gt T_{sl}=2,764^{\circ }C center of particle melts.
(c) For t = L/2u_{p} , i.e., half of the time of flight used in (a), we have
Fo_{R}=\frac{4\alpha L}{2u_{p}D^{2}}=0.2686From Figure (b)(ii), we interpolate that the temperature at the center is
\frac{T(r=0,Fo_{r}=0.2686)-T(t=0)}{T_{s}-T(t=0)}\simeq 0.81or
T(r=0,Fo_{r}=0.2686)=T(t=0)+0.81[T_{s}-T(t=0)]=100(^{\circ }C)+0.81(2,800-100)(^{\circ }C)
=2,287^{\circ }C
Here we note that this is less than the melting temperature T_{sl} = 2,760^{\circ} C. (d) From Figure (c), for spheres, we find that for Fo_{R} = 0.2686 we have
\frac{\left\langle T\right\rangle _{V}-T(t=0)}{T_{s}-T(t=0)}\simeq 0.95Note that although the dimensionless temperature was 0.81 for the center of the sphere, most of the sphere is at a much higher temperature and closer to T_{s} .
The maximum energy that is stored in the sphere occurs where the entire
sphere is at T_{s}, i.e., \left\langle T\right\rangle _{r} = T_{s} . Then
