# Homework Help: Ion Thruster Problem

1. Nov 2, 2007

### ColdFusion85

1. The problem statement, all variables and given/known data

A Xenon (mass = 131 AMU) ion propulsion system has specific impulse 4920.0 sec.

Calculate how much Xenon (in kg) required to maintain GEO orbit for 12 years for 8000 kg spacecraft. Assume thrust = .046 N.

Calculated this to be 360 kg.

Asked to find other values such as:

exit velocity of ions (48.265 km/s)
ion acceleration voltage (1581 V)
mass flow (9.5307 E-7 kg/s)
# ions exiting per sec (4.381 E18 ions/sec)
ion current (.70198 A)
ion beam power (1110 W)

I did not want to write out all the equations for this problem, so I attached the sheet that had the formulas on it. Anyway, assuming my answers are correct I am stuck on the last question:

Calculate the number of hours the thruster should work, on average, each day to maintain the spacecraft in GEO orbit.

Any ideas?

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2. Nov 3, 2007

### ColdFusion85

nobody has any input?

3. Nov 3, 2007

### D H

Staff Emeritus
Regarding the first part of the problem: Since the underlying equations are hidden in your attachments I have to wait until a moderator approves them.

Regarding the part on which you are stuck: You know the Isp, the thrust, the total fuel consumption over 12 years. Those three values alone give you the number of hours over the 12 year period, from which you can easily compute hours of operation per day.

4. Nov 3, 2007

### ColdFusion85

Well, I used the thrust and Isp to calculate the mass flow rate, and hence the total mass consumption over 12 years (I assumed constant thrusting, that is I multiplied the mass flow rate by (60 s/min *60 min/hr *24 hr/day *365 day/yr *12 yrs). I guess I am confused on what is meant by how much "should the truster work"

Regarding my attachments, might I have done something wrong by posting them? Why does a moderator need to approve?

Last edited: Nov 3, 2007
5. Nov 3, 2007

### Astronuc

Staff Emeritus
Just a precaution. All attachments must be approved.

A quick check indicates these are correct:
exit velocity of ions (48.265 km/s)
mass flow (9.5307 E-7 kg/s)
# ions exiting per sec (4.381 E18 ions/sec)

the other values are probably correct. Is one assuming singly charged ions?

6. Nov 3, 2007

### ColdFusion85

I was wondering the same at first, but plugging in my values for the Voltage, mass of Xenon ion, and assuming singly charged ion, I got back the Isp given in the problem. So I'm assuming we are assuming (poor wording there!) singly charged ions.

$$I_sp = \frac{1}{g_e}\sqrt\frac{2qV_a}{M_I}$$

Any chance you can clarify my question above, with regard to what is meant by "how many hours should the ion thruster work, on average, each day"?

7. Nov 3, 2007

### D H

Staff Emeritus
Note well: I still can't see the attachments.

I just looked over your answer once again and did the calculation myself. You are assuming a 100% duty cycle! This assumption is the answer to your second question that is consistent with 360 kg of fuel (but probably not the correct answer).

To maintain a geosynchronous orbit you need to counter the perturbations. In GEO, the Earth's J2 coefficent is the dominant perturbing factor. The Sun and Moon also "help".

ColdFusion, did you worry about perturbations and happen to get a 100% duty cycle, or did you just assume a 100% duty cycle?

8. Nov 3, 2007

### ColdFusion85

I didn't worry about perturbations, but we didn't discuss them in the lecture pertaining to this homework. I'm not sure that he would make the homework the assignment that difficult without at least hinting at it on the homework assignment. I'm not sure we even would have learned about the J2 coefficient except in problems that explicitly asked for the regression of nodes for an orbit. Anyway, assuming it is implied that we use 100% duty cycle, how would I go about determining the answer to my question?

9. Nov 3, 2007

### D H

Staff Emeritus
A 100% duty cycle means that the thrusters are operating 24 hours a day.

Your instructor "explicitly asked for the regression of nodes for an orbit". You have to counteract this. It takes a lot less than a 100% duty cycle (or 360 kg of fuel) to accomplish this.

10. Nov 3, 2007

### Astronuc

Staff Emeritus
I'm thinking about that. I would expect some orbital correction rather than continuous thrust. Is there any other information related to station keeping or perturbation along the lines of what D H mentioned?

11. Nov 4, 2007

### D H

Staff Emeritus
The attachments have been approved but only address ion thrusters.

ColdFusion, it appears you have simply assumed a 100% duty cycle. This is not a valid assumption.

GEO is far above the Earth's atmosphere, so the orbit won't decay. The orbit will however shift from the desired location. The chief culprit at GEO altitude is the non-spherical nature of Earth's gravitational field. Other culprits include solar and lunar gravity and solar radiation pressure.

As you have already been exposed to non-spherical gravity to some extent (precession induced by the Earth's oblateness) in other homework problems, you should start with this. The Earth's oblateness causes the node of the orbit to precess by a certain amount at geosynchronous altitude. What does the satellite need to do to counteract this effect?

12. Nov 5, 2007

### ColdFusion85

Alright, I talked with the TA and what we had to do was calculate the delta-V required for North-South station-keeping for the 12 year period. From this we could obtain the mass of propellant needed (it was about 100 kg). For the part I was stuck on, we find the mass of propellant needed per day by dividing the 100 kg / (12 years* 365 days/year) = kg/day. Then, we take the mass flow rate and find how much is needed per hour: kg/s * 3600 s/hr = kg/hr. Then, we divide (kg/day)/(kg/hr) to obtain hr/day needed. Incidentally, with my new value for total mass of propellant needed, I got about 6.67 hours/day needed for operation.

13. Nov 5, 2007

### D H

Staff Emeritus
That looks much better. Doing the same calculation with your initial value of 360 kg of fuel will yield 24 hours/day, but that results because you assumed 24 hours/day at the outset.

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