Minimum energy and power, break-even; inertial fusion

In summary: Plugging in these values, we get:W = 7.02*10^-11 J/1 s = 7.02*10^-11 WIn summary, the minimum energy and power of the laser device required to heat the pellet to T = 10^4 eV and achieve the energy break-even condition is 7.02*10^-11 J and 7.02*10^-11 W, respectively. This calculation takes into account the density of the pellet, which can be calculated using the mass density formula and the volume of the pellet.
  • #1
smiler2505
2
0

Homework Statement


A small 50:50 D-T pellet of density p=3g/cm^3 is used for inertial fusion. Estimate minimum energy and power of the laser device required to heat the pellet to T=10^4 eV and to achieve the energy break-even condition


Homework Equations


Lawson criteria: nt>10^20
E=3nTV
W=E/t


The Attempt at a Solution


nt>10^20
t=R/v
T=0.5mv^2 => v=10^6 m/s
=> nR>10^26
V=(4/3)piR^3
=> E=4piR^2.T.nR=2.01.10^12.R^2

My problem is the density; from wikipedia, I've found pR>1g/cm^3; how can that be derived? I haven't seen it used in any of our lectures and google hasn't proved very helpful. Number density n=M/p, but when I use a rough mass (M=2.5*proton mass), I get an absurd number density.

I'm pretty sure my energy and power equations are correct.

Thanks in advance!
 
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  • #2


Thank you for your question. In order to calculate the density of the pellet, we need to use the formula for mass density, which is defined as the mass per unit volume. In this case, we know the density (p) and the volume (V), so we can rearrange the equation to solve for the mass (M):

p = M/V

M = pV

Now, we know that the volume of the pellet is given by V = (4/3)piR^3, where R is the radius of the pellet. We can plug this into the equation for mass to get:

M = p(4/3)piR^3

Now, we need to find the radius of the pellet. We can use the density and the mass to calculate the volume of the pellet (V = M/p). We can then rearrange the equation for volume to solve for the radius:

V = (4/3)piR^3

R = (3V/4pi)^(1/3)

Plugging in the values for V and p, we get:

R = (3(2.5*proton mass)/4pi*3g/cm^3)^(1/3)

R = 1.22*10^-10 cm

Now, we can use this value for the radius to calculate the energy required to heat the pellet to T = 10^4 eV. We can use the formula for thermal energy (E = 3nTV) to do this. Plugging in the values for n, T, and V, we get:

E = 3(10^20)(10^4)(4/3)pi(1.22*10^-10)^3 = 7.02*10^-11 J

To achieve the energy break-even condition, the energy input must be equal to the energy output. This means that the minimum energy of the laser device must be equal to the energy required to heat the pellet. Therefore, the minimum energy of the laser device is also 7.02*10^-11 J.

To calculate the power of the laser device, we can use the formula W = E/t, where W is the power, E is the energy, and t is the time. We know the energy (7.02*10^-11 J), and we can assume a time of 1 second (since it is a minimum
 

1. What is minimum energy and power in relation to inertial fusion?

Minimum energy and power refers to the minimum amount of energy and power required to initiate and sustain a fusion reaction in an inertial fusion system. This is an important factor to consider in designing and operating a fusion reactor, as it determines the efficiency and feasibility of the fusion process.

2. How is break-even achieved in inertial fusion?

Break-even in inertial fusion is achieved when the amount of energy produced by the fusion reaction is equal to or greater than the amount of energy required to initiate and sustain the reaction. This is a significant milestone in the development of fusion technology, as it indicates the potential for a self-sustaining fusion reaction.

3. What is the role of inertial confinement in achieving break-even in fusion?

In inertial fusion, the fusion fuel is compressed and heated to extremely high temperatures through the use of powerful lasers or particle beams. This process is known as inertial confinement and is crucial in achieving the high energy and density necessary for a fusion reaction to occur. Without effective inertial confinement, break-even would not be possible.

4. How does minimum energy and power affect the feasibility of fusion as an energy source?

The minimum energy and power requirements for inertial fusion impact the feasibility of fusion as an energy source in several ways. These factors influence the cost, efficiency, and size of a fusion reactor, as well as the overall sustainability of fusion as a long-term energy solution.

5. What are the current challenges in achieving break-even in inertial fusion?

Despite significant progress in inertial fusion research, there are still several challenges that need to be overcome in order to achieve break-even. These include improving the efficiency and stability of inertial confinement, finding suitable fusion fuel sources, and developing cost-effective and sustainable methods for initiating and sustaining fusion reactions.

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