The numbers:
Code:
Solar Output: P(S)=4E26 watts
Solar Radius (mean): R(S)=6.96E8 m
Collector Radius: R(C)= 2.5E2 m
Collector Area: A(C)= 1.9635E5 m^2
Cell Efficiency: µ(C)= 0.8
Sol-Collector Distance- 25 Solar Diameters: D(S-C)= 1.745E10 m
First, lets determine the energy cross-section at the collectors distance (which is quite close). A(S-C) is the area of the sphere with radius D(S-C). P(S-C) is the power-per-square-meter at the orbit of the collector. P(C) is the power delivered over the area of the collector.
Code:
A(S-C)= 4π*D(S-C)^2
= 4π*(1.745E10m)^2
= 3.826E21 m^2
P(S-C)= P(S)/A(S-C)
= 4E26W/3.826E21m^2
= 1.045E5 W/m^2
P(C)= P(S-C)*A(C)
= 1.045E5 W/m^2 * 1.9635E5 m^2
= 2.052E10 W
With that solved for, now find the useful power, P(E), intercepted by the collector and the waste power, P(W).
Code:
P(E)= µ(C)*P(C)
=0.8*2.052E10 W
=1.642E10 W
P(W)= (1-µ(C))*P(C)
= (1-0.8)*2.052E10 W
= 4.104E9 W
Now we have to solve for the power requirements for antimatter production. Particle accelerator production operates by increasing the thermal energy of a region of space to the pair-producing temperature (which is twice that of the rest-mass energy of the desired anti-particle). When this energy level is reached matter-antimatter pairs are created and can be separated easily due to their opposite charges.
First things first, here the reference numbers that will be used.
Code:
Speed of Light: c=2.99792458E8 m/s
Usable Power: P(E)= 1.642E10 W
What we want to determine here is the power required to produce a unit-mass of antimatter per unit time. In a best-case scenario you the particle accelerator only needs to produce half of the pair-producing energy. Inefficiency of the accelerator and antimatter collector assemblies is going to be ignored. M(A) is the rate of antimatter production from the accelerator.
Code:
M(A)= P(E)/c^2
= 1.642E10 W/(2.99792E8 ms^-1)^2
= 1.826971E-7 kg*s^-1
So per second the collector is able to produce, assuming ideal conditions, 0.183 mg/s. This is
657.7 mg/hour.
Now, we also need to consider the waste energy, which is also substantial. For ease of comparison, let us determine at what rate this energy output could melt homogenous iron that started at 0*C. The environment is assumed to be at 1 atm (actual vacuum would increase the mass vaporized).
Code:
Fe, Density: p=0.055845 kg*mol^-1
Fe, heat of fusion: h(f)=13.81 kj*mol^-1
Fe, heat of vaporization: h(v)=340 kj*mol^-1
Fe, specific heat capacity: S=25.10 J*mol^-1*C^-1
T(initial)= 0*C
T(final)= 2862*C
P(W)= 4.104E9 W
T(0*C-1538*C)
E(1)= 1538*C * 25.10 J*mol^-1*C^-1
= 3.86038E4 J*mol^-1
T(S-L)
E(2)= 1.381E4 J*mol^-1
T(1538*C-2862*C)
E(3)= (2862-1538)*C *25.10 J*mol^-1 *C^-1
= 3.32324E4 J*mol^-1
T(L-V)
E(4)= 3.4E5 J*mol^-1
E(total)= 4.2565E5 J*mol^-1
E(total-M)= 4.2565E5 J*mol^-1 / 0.055845 kg*mol^-1
= 7.621989E6 J/kg
M(W)= P(W)/ E(total-M)
= 4.104E9 J/s / 7.621989E6 J/kg
= 438 kg/s
Summary
At a orbital distance of 25 solar diameters, a collector with a diameter of 500 meters, a cell efficiency of 80%, and otherwise perfect efficiency the system will produce antimatter at the rate of 0.183 mg/s or 657.7 mg/hour. The waste heat output during production (with all efficiencies other than the collector cells being perfect) will be sufficient to vaporize 438 kg of Iron per second or 1,576.8 tons per hour.
Edit: I should note that despite the above numbers systems in this style
can produce useful amounts of antimatter, they just need to be much, much larger than the one submitted.