A lot, but there can be no net energy gain in this scenario
As others have said, 100% energy capture is not physically possible with real physics. I'll skip the Carnot efficiency calculation as you're not asking for it and someone already provided it. No process gives 100% useful work. That's the easy part.
Next is the calculation of time dilation. That isn't particularly hard but to reach an actual number you have to know certain parameters going in. So I will give the formula then go through the work for you.
To safely extract energy from an H-bomb by time dilation, a time dilated second s$_1$ passes inside your time bubble in a number of real seconds according to the formula:
$$ s_1 = \frac{\textbf{H}_0}{\textbf{H}_U \textbf{H}_F(1-\textbf{M}_S)}s $$
Where:
- S$_1$ is the number of real-world seconds will pass every time one second passes inside the time bubble
- $\textbf{H}_0$ is the power density of the H-bomb at the surface of the time bubble
- $\textbf{H}_F$ is the power density that will cause your weakest part of the energy harvesting device to fail
- $\textbf{H}_U$ is the power density that will will be converted into useful work by your harvesting device (based on the device efficiency)
- $\textbf{M}_S$ is your design safety margin, or, the percentage of "slop" you want to allow your design so it will not fail
For reasons explained below your $\textbf{H}_0$ power density will have many components, but ultimately every form of energy coming from the detonation can be converted into Watts/m$^2$ and plugged into this formula. This is the number you need to find based on the yield of your bomb and the diameter of the sphere.
For the safety margin, assume you have some coupling in the collector that will be destroyed with exposure to 100,000W/m$^2$ over 5 seconds. If you want a 5% safety margin, the max power density you can allow to "safely" (per the question) operate this device would be 5kW/m$^2$. You would dilate time enough to make s$_1$ long enough that power doesn't exceed this number.
In the end, unless you are dilating time for free, you will not pull a net energy from this device, however you will make energy useful where it normally is not. Now the calculation:
You will be taking normally destructive energy and making it useful, but at a very low efficiency. Note also, that a thermonuclear device is an uncontrolled chain reaction with more energy yield as it grows larger. Someone calculated you will need a 200m diameter sphere, but the problem isn't really asking for a fixed yield so I will only touch on the why of this argument. If you bottle the bomb up into a small bubble, you will have an exponentially smaller yield compared to the uncontrolled bomb in open air. No matter, you didn't want max yield.
Disregard the thermonuclear bomb for now and just say you have a radiating body of any kind. In your case, it radiates electromagnetic waves, gamma rays, neutrons, protons, neutrinos, etc., etc. You have also introduced the novel physics that time dilation alters the wavelength of light when it is measured outside the bubble in normal time. This should not happen in your case of special relativity (we are talking about time dilation, after all) because it violates laws of conservation. An infrared photon has less energy than a red photon, for example. Where did the extra energy go?
Your question is very simple and the qualifier "safely" means we only need to find the most likely item to fail, and assume the machine fails as a whole soon afterward. That would be the safety margin. If I make this radiation come out of the bubble more slowly, smaller doses of the radiation will reach the parts of your machine per second of real-time. So you simply need to determine how many Watts per second will destroy your energy harvesting machinery. The destructive elements are many, but the calculation is very simple, and can only be given a number after knowing what your harvesting machine can withstand:
Electricity
If you have copper wiring anywhere, for example, you need to keep the temperature at least below the melting point of 1,984°F, but preferably much lower than that if you want your wires to be efficient at conducting electricity. You can have better conductors with gold or platinum, but you will also need to keep the temperature lower. You will heat up any copper parts from 25° to 1,984° ($\Delta T=1,959$°) by exposing it to your radiation. This reduces the efficiency of the electrical conductor. But how much can it take? That answer depends completely on a review of your engineering design.
Thermal radiation
We have to assume you want to harvest energy rather than destroy it. A particle leaving the blast will be moving through time more slowly but have the same energy. All you have done is change the definition of a second within that bubble. When the hydrogen's proton (or any particle) that has enough energy to undergo fusion reaches the perimeter of the bubble, unless you have caused the energy to leave the universe magically, it is still in the particle. The particle's momentum is conserved. So the proton will be traveling at the same speed as it did if time were never affected. What will change is the number of these protons reaching the perimeter per normal second. So this reduces to a simple calculation of what dosage of proton radiation will destroy your machine. As stated, the design of the machine completely determines that number. We will treat the weakest part of your machine as a black body which is absorbing heat energy (we won't use a gray body calculation because anything transmitted or reflected eventually ends up in your machine somewhere else). You need to have some heat removal method to offset the heat transfer going in.
The radiation energy per unit time from a black body is proportional to the fourth power of the absolute temperature and can be expressed with Stefan-Boltzmann Law as:
$$q=\sigma T^4A$$
where:
- q = heat transfer per unit time (W)
- σ = The Stefan-Boltzmann Constant = $5.6703e-10^{-8} \text{(W/m}^2 \text K^4$)
- T = absolute temperature in kelvin (K)
- A = area of the emitting body (m$^2$)
Now, conveniently, you are asking about a thermonuclear reaction. The sun is doing exactly what you are asking for here, and so is any fusion reactor. So let's look at the sun as an example to gauge how much energy is radiated in real time. From that, you can slow down time enough that the the heat transfer $q$ will not destroy your most delicate component.
Use a surface temperature of the sun at 5800 K, and assume that the sun can be regarded as a black body. The radiation energy per unit area can be expressed by modifying (1) to
q / A = σ T$^4$
= $(5.6703e-10^{-8}) W/m^2K^4) (5800 K)^4$
= $6.42e-10^7$ (W/m$^2$)
As you can see, the absolute temperature of your thermonuclear device—at the perimeter of the time bubble—must be known before any meaningful numbers can be drawn. That of course is determined by the diameter of your bubble, and the yield of the radiation source being considered. That can be the subject of another question, perhaps. No matter, since Watts are a unit measured against time, you are simply turning the second into a variable in this formula. You need to modify the value of the second so that the black body radiation finally coming out does not destroy your weakest link.
Let's say you have a little fancy active cooling. Now, let's say you have some thermoelectric device harvesting energy, and that device is destroyed (or becomes useless) above 1000K. That's pretty hot, but say you've designed this. Let's also say you have a small-ish time bubble and the radiation reaching the perimeter is "normally" at 5,800K, just like the sun. All you need to do is change the value of s in the Stefan-Boltzmann Law such that you are not putting more heat in than the device can dissipate.
Let's assume you can naturally dissipate heat by cooling fans at the rate of 10kW per second per meter$^2$. How long does a time-dilated second need to be to allow your cooling system to prevent overheating?
Solve the solution above ($6.42e-10^7$ (W/m$^2$)) for s by rewriting it:
Recall: 1 Watt - 1J/s, $\therefore$ 1 s = 1J/W
$$
6.42e-10^7 W/m^2 \\
= 6.42e-10^7 \frac{J}{s m^2} \\
\therefore s= \frac{1J}{6.42e-10^7 m^2}
$$
But you need to know what the slowed second is ($s_1$) that will just introduce enough heat to destroy your weakest component, at $10 \frac{kW}{s m^2}$. So we rewrite the above solution for $s_1$ as follows:
$$
s_1= \frac{1J}{1e-10^4 m^2}
$$
now we have an exact ratio for time dilation by dividing the unknown second by the normal one, $\frac{s_1}{s}$ as follows:
$$ \frac{s_1}{s}=\frac{\frac{1J}{1e-10^4 m^2}}{\frac{1J}{6.42e-10^7 m^2}} \\
s_1= \frac{6.42e-10^7}{1e-10^4}s \\
S_1 = 6,420s
$$
A:
Given these conditions for harvesting a thermonuclear detonation which has the same energy as the surface of the sun, you will need to slow time such that one second equals $1.78\overline3$ hours, or slow time by 642,000%
I made many assumptions but in simplest terms, design your energy collection device first, learn where its weakest link is, treat the time bubble as a black body radiator of a certain surface area, then modify the Stefan-Boltzmann equation to solve for your new value of a second.
