| Author | Message |
Camel
390 posts |
#177249 2008-05-20 09:47 GMT |
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A spacecraft in space has an ion thruster that ejects atoms at a velocity of 20km s-1 relative to the craft.
The thrust generated is 0.1N. (i) What is the rate of consumption (in kg s-1 ) of the xenon? (ii) The total mass of the craft is 500kg the thruster operates continuously for 6 weeks what is the change in velocity this causes? (ignore loss of mass over period). (iii) Calculate the electrical power required to run the thruster, assuming it has an efficiency of conversion of electrical to kinetic energy of 65%. |
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HocusPocusCrocus
341 posts |
#177250 2008-05-20 10:31 GMT |
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1.Mass/second, kt = F/v = 0.1/20000 = 5*10-6 kg/s
2.Since craft mass is assume to 500kg, then Mf = 500 – kt*T = 500 - 18.144 =481.85kg 3.Where T = 24*6*7*3600 seconds = 3628800sec 4.Since no losses in mass v = u + (F*T/500) = 20000 + (0.1*3628800/500) = 20725.76m/sec. 5.The change Vc = 20725.76 – 20000= 725.76 m/sec 6.The Total Power develop is Pt = FVc = 0.1*725.76 = 72.57W 7.So at 65% Po = Pt*0.65 = 72.57*0.65 = 47.17W |
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fruitsalad
355 posts |
#177251 2008-05-20 10:46 GMT |
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The equation you need is
F = R * V where F is the thust in N, R is the flow rate in kg/s and V is the exhaust velocity in m/s. 0.1 N = R * 20000 m/s R = 0.000005 kg/s For the second quesion, we use F = m a 0.1 N = 500 kg * a a = 0.0002 m/s^2 Have this acceleration for 6 week (6 * 7 * 24 * 3600 seconds = 3628800 s) means a change of velocity of V =a * t V = 0.0002 m/s^2 * 3628800 s = 725.76 m/s Every second, a 0.000005 kg mass of xenon is accelerated to 20000 m/s, this means it gains E = 1/2 m v^2 E = 1/2 0.000005 kg (20000 m/s)^2 = 1000 joules, or 1000 watt per second. With the efficiency at 65%, then one need 1538.46 watt of available electrical power. In space, at the distance similar to the Sun than the Earth is, and assuming a 30% solar cell conversion (the expensive, space worthy ones) a solar panel 3.8 m^2 would give you this much. |
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