Hydrogen Burning and the Suns Nuclear Fusion

In summary, the sun is not expected to last much longer and the amount of hydrogen burnt over 10 billion years is only 14% of the mass of hydrogen in the sun now.
  • #1
Thoth
6
0
hey guys, I am new here and thought you might like to help me out. I am a first year astronomy student at Curtin University in Western Australia.

my current assignment has asked me to calculate the mass of hydrogen converted in the sun to helium and energy over 10 billion years (10 billion years being its lifespan on the HR diagram). (PS this is Question 5 of 8 haha)

now, i understand peoples reluctance to help others who don't prove they have done any work themselves. which is lovely, considering youll now be reading all my own calculations and the result I am getting!

my problem is that my amount of hydrogen burnt in 10 billion years seems to be only 14% of the mass of hydrogen in the sun now. shouldn't it be MUCH more than 14%? stars are supposed to burn of all of their hydrogen by the time they hit red giant phase.

without further ado, and recognising my use of the letter e in figures is scientific notation (10 to the power of) and nothing to do with natural logs.

FIRSTLY, the amount of seconds in 10 billion years.

60 x 60 = 3600 seconds per hour

3600 x 24 = 86400 seconds per day

86400 x 365 = 31536000 seconds per non leap year
86400 x 366 = 31622400 seconds per leap year.

since 10 billion years will consist of 75% non leap years and 25% leap years:

0.75 x 10 billion years = 7.5 e 9 years
0.25 x 10 billion years = 2.5 e 9 years

therefore:

(2.5e9 x 31622400) + (7.5e9 x 31536000) = 3.16224 e 17 seconds in 10 billion years

SECONDLY since the sun emits 3.84e26 joules per second of energy, the total power emitted over 10 billion years would be:

3.84 e 26 x 3.16224 e 17 = 1.214 e 44 watts (or joules per second) of power in 10 billion years we will need this figure later.

THIRDLY now that we have the total power output of the sun over 10 billion years we can use e=mc^2 to find out how much mass has been converted into energy in that time.

1.214 e 44 = m (3 e 8)^2

rearranging for m, gives m = 1.349 e 27 kilograms of mass converted to energy in the sun over 10 billion years

FOURTH so now we get onto the heavy stuff. knowing the masses of the hydrogen atom and helium atoms, and the reaction in the sun which converts the former to the latter + energy, i should be able to get a figure for the total hydrogen burnt. let's give it ago!

The fusion reaction summed up is 4 Hydrogen + 2 electrons ----> 1 Helium + 2 neutrinos + 6 photons. Now, remember this is a summary of the 3 steps, but they arent important. only need to know the input and final output.

the mass of a hydrogen atom is 1.673 e -27
the mass of a helium atom is 6.647 e -27

now there are 4 hydrogens going into the reaction so:

4 x 1.673 e -27 = 6.692 e -27 the mass of hydrogen going into a single full reaction

therefore:

6.692 e -27 = (6.647 e -27 + X) with X being the amount of mass converted to energy per reaction.

rearranging for X gives X = 4.5 e -29 kg mass converted into energy per reaction.

FIFTH so if the total mass converted to energy over 10 billion years is 1.349 e 27, and we know now that the mass being converted to energy per full reaction is 4.5 e -29 then:

1.349 e 27 / 4.5 e -27 = 2.998 e 55 total reactions in 10 billion years

so if 6.692 e -27 kg of hydrogen goes into a single reaction, and there is 2.998 e 55 total reactions, then:

6.692 e -27 x 2.998 e 55 = 2.006 e 29 kg of hydrogen converted into helium and energy in 10 billion years

so now you see my problem. if sun right now has 70% hydrogen, making that total hydrogen to be 1.393 e 30 kg of hydrogen in the sun right now, then how can the hydrogen burnt over 10 billion years be only one seventh (14%) of the mass of hydrogen right now?
 
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  • #2
It's been a while since I did this stuff, but I seem to remember calculations like this indicating that hydrogen fusion is not the sole energy source. Seems like I remember something about a Carbon-Nitrogen cycle as well? In other words, perhaps the inconsistency you mention was evidence that there was something else going on; i.e. that the power output of the sun was not due solely to H fusion?

If that's not the answer then I'm as confused as you, because I read somewhere the sun is only expected to last another 5 B years.
 
  • #3
merryjman: youve got me thinking now :D (or :(( I am not sure)

i reread my question, and it told me to assume hydrogen->helium is the only nuclear reaction occurring in the sun. there is another reaction which involved helium, beryllium, boron and back to helium, and there is also the CNO cycle which you mentioned. the problem is, even if i was to consider these, which I am told not to, i would still expect my figure for the burning of hydrogen over 10 billion years from the main reaction to equate to something much more than 14% of the suns current mass of hydrogen.

consider that the sun is approximately 5 billion years old, and currently has a total mass of 1.99 e 30 kg. hydrogen is 70% of this, so that's 1.393 e 30 kg of hydrogen. double this, and youve got the amount of hydrogen the sun started with at time=0 years.

thats 2.786 e 30 kg of hydrogen. now, if stars are supposed to burn all of that hydrogen in their lifetime (in this case 10 billion years for the sun), my sun is apparently burning only 7.2% of that amount in 10 billion years.

now even considering the CNO and beryllium/boron/helium reactions included, youd still expect the % of hydrogen burnt from the primary reaction to be exceedingly larger than 7.2%
 
  • #5
so...after thinking that stars burn most if not all of their hydrogen over that 10 billion year period, I am actually right in calculating that the sun will only burn 7% of its hydrogen?

many MANY thanks to both merryjman and mgb_phys for putting me back on the right track, and possibly saving me the enbarressment of my lecturer telling me I am actually correct
 
  • #6
It's a few years since I last slept through stellar evolution but I seem to remember about 10% H used

Your PM Question.
Because of the radiation pressure the material in the outer layers doesn't fall into the core to be burnt. The burn spreads out into outer layers leving behind the end-products of earlier fusion. As the fuel in the inner layers is used up, the radiation pressure decreases, the star begins to collapse, the pressure and temperature in the core rises and a new reaction kicks off. Ultimately you end up with an onion structure with different layers burning different material.

(burning of course means nuclear fusion in stellar terms)
 
  • #7
mgb to the rescue! So now, I've got a side question - if sun-like stars only burn ~10% of their H fuel, then why do we think our sun will become a red giant? I was always under the impression that stars became giants because their primary H->He reaction was done. Does this have to do with the onion thing you mentioned; i.e. the H in the star's center is used up and a new reaction begins from that center? What would that reaction be?
 
  • #8
When a star's core is depleted of hydrogen, the core contracts (due to gravity), and the star begins fusing hydrogen in a thin shell outside the core. As it turns out, there's virtually no mixing between the outer layers and the core -- hydrogen that begins near the surface of the star stays there forever and never gets fused.

The star becomes a red giant because its enormous luminosity creates enormous radiation pressure, which expands the outer layers of the star. The average surface temperature a red giant is low, but the total luminosity is immense.

- Warren
 
  • #9
thanks, all, useful and interesting information.
 

1. What is hydrogen burning and how does it relate to the sun's nuclear fusion?

Hydrogen burning, also known as hydrogen fusion, is a process in which hydrogen atoms combine to form helium atoms. This process releases a large amount of energy, which is what powers the sun's nuclear fusion. The sun's core is primarily composed of hydrogen, and the extreme pressure and temperature in the core causes hydrogen atoms to fuse together, releasing a tremendous amount of energy.

2. How does the sun's nuclear fusion process work?

The sun's nuclear fusion process involves the combination of four hydrogen nuclei (protons) to form one helium nucleus. This process releases a large amount of energy in the form of heat and light. The process requires extremely high temperatures and pressures, which are only found in the sun's core.

3. What are the products of hydrogen burning in the sun's nuclear fusion?

The main product of hydrogen burning in the sun's nuclear fusion is helium. However, other elements such as carbon, nitrogen, and oxygen are also produced in small amounts. These elements are essential for the formation of planets, including Earth.

4. How does the sun maintain its fusion process?

The sun maintains its fusion process through a delicate balance between the inward pull of gravity and the outward pressure created by the energy released from nuclear fusion. As long as there is enough hydrogen in the core and the temperature and pressure remain high enough, the sun will continue to fuse hydrogen and produce energy.

5. Can we replicate the sun's nuclear fusion on Earth?

Currently, we do not have the technology to replicate the sun's nuclear fusion on Earth. The extreme temperatures and pressures required are difficult to achieve and sustain. Scientists are working on developing controlled nuclear fusion reactions, but it is still a long way from being a viable energy source.

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