# If the inventory of the reactor of about 20,000kg

1. Oct 12, 2011

### matt222

1. The problem statement, all variables and given/known data
If the inventory of the reactor of about 20,000kg of Zr given the reaction:
Zr+2H2O---> ZrO2+2H2, Q=616MJ/Kg-mole of Zr

1- find the 10% of the heat generated from Zr oxidation by steam in MJ
2- Find the 10% hydrogen generated from Zr oxidation by steam in kg and the volume fraction (given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas.
3- what is the energy released from the combustion of H generated and what will be the power if it is released within 10s, assumption is the amount of Oxygen is enough to burn the total H generated, heat of combustion is 240MJ/Kg-mol
4- what will be the pressure increased by the H generation from 10% oxidation given initial pressure 1 bar and volume 90,000m^3, treat hydrogen as an ideal gas, ignorw the T rise.

2. Relevant equations

3. The attempt at a solution
1- it is simple :
molar mass for Zr is 91.22 kg/kmole

10%x 20000kgx 616/91.22 =13.5x10^3 MJ

2- it is by: 2kg/kmol of 2H, so 2x13.5x10^3/(616)=43.8kg
I didnt understand the second part of volume fraction?

3- energy release from the combustion of H:

H2+1/2O2---->H2O+ 240MJ/Kgmol

240x43.8=10.51x10^3MJ

the power is going to be:

10.51x10^3MJ/10=10.51x10^2MW

4-I have no idea but I need a hint

is my work true

2. Oct 12, 2011

### Staff: Mentor

Re: chemistry

I guess they mean volume fraction in the mixture after the reaction - it still contains remaining water AND produced hydrogen. However, it is not clear to me what is the temperature. Perhaps just 100°C, as that's enough for water to be a steam at 1 bar.

Convert between mass/moles and volume using ideal gas equation, nothing more fancy.

3. Oct 12, 2011

### matt222

Re: chemistry

so PV=nRT, so in this case 90,000m^3 going to be the total volume right?

n=2kg/kmol for hydrogen once we find the voulme of hydrogen just divide it by the total volume of 90,000m^3 is it right?

4. Oct 12, 2011

### matt222

Re: chemistry

the T=100c you are right

5. Oct 12, 2011

### Staff: Mentor

Re: chemistry

Yes.

Note that this case is specific - total number of moles of gas doesn't change (one mole of H2O is replaced with one mole of H2 - this is in the stoichiometric coefficients). It doesn't have to be true, which may add an additional step to calculations.

6. Oct 12, 2011

### matt222

Re: chemistry

so the is going to be V of hydrogen=8.32x10^3x373x2/1x10^5=31m^3

which is very small!

volume fraction=31/90,000=3.45x10^-4!!

What do you think

7. Oct 12, 2011

### Staff: Mentor

Re: chemistry

There is in general something wrong with your numbers.

Why do you mix 616 (which is enthalpy change) with masses, when calculating stoichiometry?

8. Oct 12, 2011

### matt222

Re: chemistry

oops yes you are right its mistake it should be 13.5x10^3/616=21.9kg>>what about now?

9. Oct 12, 2011

### Staff: Mentor

Re: chemistry

No, it shouldn't be. You can't calculate mass produced using enthalpy change (which would be obvious if you were paying attention to the units). You start with masses, you end with masses. This is a simple stoichiometry.

10. Oct 12, 2011

### matt222

Re: chemistry

I confused really because this is not my background field!
can you refer me to a wiki link for understanding more because I really lost

11. Oct 12, 2011

### Staff: Mentor

12. Oct 12, 2011

### matt222

Re: chemistry

Zr+2H2O---> ZrO2+2H2
91.1kg/kmol for Zr.....2kg/kmol of 2H
10%x20,000kg of Zr......Y

in this case Y is the required mass for this problem! which is equal to 43.91kg!! almost the same as what Igot before!

13. Oct 12, 2011

### Staff: Mentor

Re: chemistry

No, it is still wrong. 2 kg is a correct mass of a kilomole of hydrogen, but you are not producing 1 kilomole of hydrogen (H2) per kilomole of Zr.

14. Oct 12, 2011

### matt222

Re: chemistry

91.1kg/kmol for Zr.....4kg/kmol of 2H
10%x20,000kg of Zr......Y

87.82kg what about now? this is what i understand from the link? really almost will be crazy from chemistry lol

15. Oct 12, 2011

### matt222

Re: chemistry

can you give me example to understand this problem plz

16. Oct 12, 2011

### Staff: Mentor

Re: chemistry

Not 4kg/kmol of 2H but 2 times 2kg/kmol of H2. Think what each 2 in 2H2 means.

That's the correct mass of hydrogen which is a first part of the answer. Not you have to convert it to the volume.

17. Oct 12, 2011

### matt222

Re: chemistry

so the is going to be V of hydrogen=8.32x10^3x373x4/1x10^5=124m^3

18. Oct 12, 2011

### Staff: Mentor

Re: chemistry

Units please - I am not going to guess what you are doing.

I think you are wrong, but without knowing what each number represents I can't be sure.

19. Oct 12, 2011

### matt222

Re: chemistry

so the is going to be V of hydrogen=8.31 x10^3(pa m^3/kmolK)x373 (K) x4 (Kmol)/1x10^5(pa)=124m^3

this is after showing the units!

20. Oct 12, 2011

### Staff: Mentor

Re: chemistry

Why 4 kmols? Is that amount of hydrogen produced?