This sounds like a fascinating project, and I can certainly help you with the solar part of the question. I will probably make some mistakes, so let me know if anything is unclear. This will be quite a simplistic, back of the envelope type calculation, but hopefully it helps to clarify some things.
For now, let's just assume you know the flux (power per unit area), q_out [W/m2], due to conduction, convection, phase change, chemical reaction, etc. (i.e. anything that is not radiation), required to maintain the flow of sand at the specified temperature of Ts ~ 1700 K. This could be calculated a priori but would require knowledge of how the apparatus is insulated, etc. Once you know this, it just comes down to doing an energy balance on the surface. For radiation heat transfer, you will need to consider both the flux absorbed by the surface, and the flux emitted by the surface. The absorbed flux is equal to: alpha*q_solar, where alpha is the solar absorptance (how much of the incident sunlight is absorbed) and q_solar is the incident solar flux. The emitted flux is equal to: epsilon*sigma*Ts^4, where epsilon is the total emissivity (how much energy the surface emits relative to a perfect emitter (blackbody) at Ts); sigma is the Stephen-Boltzmann constant = 5.67 e-8 W/m2-K; and Ts is the surface temperature of the sand. Both alpha and epsilon are always between 0 and 1.
We can now do an energy balance on the surface:
q_in = q_out
alpha*q_solar = q_out + epsilon*sigma*Ts^4
This can be easily solved for q_solar:
q_solar = (q_out + epsilon*sigma*Ts^4)/alpha
The most difficult part is to evaluate the q_out term. For a back of the envelope calculation, I would do the following: the radiative loss term, epsilon*sigma*Ts^4, is proportional to T^4, whereas the q_out term is *probably* proportional to T. Therefore at very high temperatures, and 1700 K is indeed high, the radiative term dominates. A *probably* conservative approach is to assume q_out = epsilon*sigma*Ts^4. Then we get:
q_solar = 2*epsilon*sigma*Ts^4/alpha
Let's plug in some numbers. I am just guessing here, but I can look up values for alpha and epsilon for sand tomorrow at my office. Both epsilon and alpha must be between 0 and 1, so let's take 0.5 for both. We get:
q_solar = 2*0.5*5.67e-8*(1700)^4/0.5 ~ 1 000 000 W/m2
Although this number seems high, it is achievable depending on the type of concentrator you use. Usually in the solar concentration field, we express things in terms of "suns" where 1 sun = 1000 W/m2, which is about the maximum direct normal solar irradiance that you can get on Earth. When expressed in terms of suns, we usually call this the "solar concentration ratio", or simply "concentration". Then:
Concentration = q_solar = 1000 suns
Therefore we need to concentrate the sunlight, at least 1000 times. Of course, depending on where you live, you may expect values closer to 500 - 800 W/m2 for the direct normal solar irradiance on a clear day. Then to be safe, you need a concentration of around 2000 suns.
In order to get a concentration this high, you will need a point-focus two-axis tracking system. I don't think you'll be able to get a Fresnel lens system to achieve these kinds of concentration, so you may want to consider parabolic dish shaped mirrored reflectors. If you are still interested, I can provide some more info on potential concentrators.
