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AAE450 Spring 2009 Slide 1 of 7 Orbital Transfer Vehicle (OTV) Thermal Control Ian Meginnis February 26, 2009 Group Leader - Power Systems Phase Leader.

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Presentation on theme: "AAE450 Spring 2009 Slide 1 of 7 Orbital Transfer Vehicle (OTV) Thermal Control Ian Meginnis February 26, 2009 Group Leader - Power Systems Phase Leader."— Presentation transcript:

1 AAE450 Spring 2009 Slide 1 of 7 Orbital Transfer Vehicle (OTV) Thermal Control Ian Meginnis February 26, 2009 Group Leader - Power Systems Phase Leader - Translunar Injection OTV Power Systems OTV Thermal Control Ian Meginnis Power Systems

2 AAE450 Spring 2009 Slide 2 of 7 Ian Meginnis Power Systems Note: Not to scale Acronym Definitions: PCDU - Power Conditioning and Distribution Unit PPU - Power Processing Unit Batt - Battery DC - Direct Current PPU PCDU Batt DC/DC Converter Aluminum Heat Pipes with Ammonia Aluminum Mount 2 Radiators Hall Thruster Thruster Support Radiator (Exhaust) Electronics Board Thermal Control Hall Thruster Thermal Control

3 AAE450 Spring 2009 Slide 3 of 7 OTV Thermal Control  Approximately 250W needs to be removed from electronics board –Heat conducts from devices to aluminum mounting plate –Heat pipes with ammonia transfer heat to 2 radiators outside OTV –Thermal System Mass: 11.1kg  Xenon needs to be stored at gaseous state –5W heater maintains xenon to constant temperature of 300K –Thermal System Mass: 0.2kg  490W needs to be removed from Hall thruster –Heat conducts from thruster to aluminum mounting plate –Heat conducts from plate to 2 radiators that radiate heat to space –Thermal System Mass: 3kg  Total Thermal Control System Mass: 14.3kg Ian Meginnis Power Systems

4 AAE450 Spring 2009 Slide 4 of 7 Calculation of Electronics Board Thermal Control  Aluminum (Al) Mounting Plate: –Max. operating temperature of components: 323K –Total area of plate: 0.22m 2 –Thermal conductivity for Al: 236W/(m*K) –Density of Al: 2700kg/m 3 –From conduction: Δx < AK(T 1 -T 2 ) / q Δx = thickness of plate T 1 = max. operating temperature = 323K T 2 = temperature of plate (assume constant at 300K) q = heat dissipation from components = 250W –Δx < 4.77m (thickness of plate must be less than this)  selected to be 0.5cm to minimize mass –Mass of plate = 2700kg/m 3 * (0.005m * 0.22m 2 ) = 3kg Ian Meginnis Power Systems

5 AAE450 Spring 2009 Slide 5 of 7 Calculation of Electronics Board Thermal Control (continued)  Sizing of heat pipes –Mass of ammonia: Latent heat of vaporization of ammonia: 1371kJ/kg Mass = 0.25kW * 450sec / (1371kJ/kg) = 0.0821kg –Assumes ammonia boils in 7mins (450sec)  Mass of Al pipes: –Diameter of pipes: 2.54cm (OD); 2.44cm (ID) –Length of pipes: 3m –Mass = π[(1.27cm) 2 – (1.22cm) 2 ] * 300cm * 0.027kg/cm 3 = 3.17kg  Mass of radiators –Area of radiators: A = q / (εσT 4 ) = 1.44m 2 For aluminum with white paint (Z93): Emissivity (ε) = 0.92 σ = 5.67E-8 J/(K 4 *m 2 *s) q = 250W T = boiling point of ammonia @ 1 atm = 240K –Mass of radiators = 2700kg/m 3 * 1.44m 2 * 0.002m = 7.8kg Ian Meginnis Power Systems

6 AAE450 Spring 2009 Slide 6 of 7 Calculation of Hall Thruster Thermal Control Ian Meginnis Power Systems  Find q radiated to space from EP structure –Radiated power: q = AεσT 4 = 491W Hall thruster surface area: A = 3 * (0.25m*0.25m) = 0.188m 2 –3 sides available to radiate power on Hall thruster For aluminum with white (Z93) paint: Emissivity (ε) = 0.92; Absorptivity (α) = 0.20 T = max. operating temperature of Hall thruster = 473K

7 AAE450 Spring 2009 Slide 7 of 7 Calculation of Hall Thruster Thermal Control (continued)  Need to dissipate 509W (1000W-491W) –Conduction transports heat from thruster to 2 radiators –Radiation dissipates heat from radiators Area of radiators: A = q / (εσT 4 – α*1300W/m 2 ) = 0.216m 2 –q = 509W –T = 473K –α*1300W/m 2 = power input from sun Mass of radiators = 2700kg/m 3 * 0.005m * 0.216m 2 = 2.92kg Ian Meginnis Power Systems


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