solhw2 - MAE135 SOLUTIONS TO HW 2 Spring 2010 PROBLEM 1 (10...

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Unformatted text preview: MAE135 SOLUTIONS TO HW 2 Spring 2010 PROBLEM 1 (10 points) Since flow is isentropic, both T and p are conserved. In other words, T 2 = T 1 p 2 = p 1 Since we know the conditions in the freestream (1), we evaluate the total temperature and total pressure from their definitions: T 1 = T 1 (1 + - 1 2 M 2 1 ) = 560 K p 1 = p 1 parenleftbigg 1 + - 1 2 M 2 1 parenrightbigg - 1 = 367327 Pa Using the conservation of T and p , we evaluate the static temperature and static pressure at station 2: T 2 = T 2 (1 + - 1 2 M 2 2 )- 1 = 249 K p 2 = p 2 parenleftbigg 1 + - 1 2 M 2 2 parenrightbigg- - 1 = 21492 Pa PROBLEM 2 (10 points) The actual flight Mach number is M =0.9. From the exact isentropic relation, the correct pressure ratio is p p = parenleftbigg 1 + - 1 2 M 2 parenrightbigg / ( - 1) = parenleftBig 1 + 0 . 2(0 . 9) 2 parenrightBig 3 . 5 = 1 . 6913 This is the actual total-to-static pressure ratio at the outputs of the pitot-static probe. The on- board computer uses an algorithm to convert this pressure ratio to Mach number. If the algorithmboard computer uses an algorithm to convert this pressure ratio to Mach number....
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solhw2 - MAE135 SOLUTIONS TO HW 2 Spring 2010 PROBLEM 1 (10...

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