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# solhw7 - MAE135 Solutions to HW 7 Spring 2010 PROBLEM 1(10...

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MAE135 Solutions to HW 7 Spring 2010 PROBLEM 1 (10 points) First we calculate that, in region 1, p 01 /p 1 =36.7. Case (a) From 1 to 2 : M 1 = 3 . 0 and θ = 20 o Oblique shock charts: β = 37 . 5 o M 1 n = M 1 sin β = 1 . 83 Normal shock table: M 1 n = 1 . 83 p 2 /p 1 = 3 . 74 , p 02 p 01 = 0 . 80 , M 2 n = 0 . 605 M 2 = M 2 n sin( β - θ ) = 2 . 03 p 2 = p 1 p 2 p 1 = (1 atm )(3 . 74) = 3 . 74 atm p 02 = p 01 p 02 p 01 = (36 . 7 atm )(0 . 8) = 29 . 4 atm Case (b) From 1 to 2 : M 1 = 3 . 0 and θ = 10 o Oblique shock charts: β = 27 . 5 o M 1 n = M 1 sin β = 1 . 385 Normal shock table: M 1 n = 1 . 385 p 2 /p 1 = 2 . 09 , p 02 p 01 = 0 . 962 , M 2 n = 0 . 746 M 2 = M 2 n sin( β - θ ) = 0 . 746 sin 17 . 5 = 2 . 48 p 2 = p 1 p 2 p 1 = (1 atm )(2 . 09) = 2 . 09 atm p 02 = p 01 p 02 p 01 = (36 . 7 atm )(0 . 962) = 35 . 3 atm From 2 to 3 : M 2 = 2 . 48 and θ = 10 o Oblique shock charts: β = 32 o M 2 n = M 2 sin β = 1 . 31 Normal shock table: M 2 n = 1 . 31 p 3 /p 2 = 1 . 835 , p 03 /p 02 = 0 . 978 , M 3 n = 0 . 781 M 3 = 0 . 781 sin(22) = 2 . 08 p 3 = p 3 p 3 p 2 = (2 . 09 atm )(1 . 835) = 3 . 83 atm p 02 = p 02 p 03 p 02 = (35 . 3 atm )(0 . 978) = 34 . 5 atm For efficient operation of an airplane engine, we seek inlet configurations that minimize loss in total pressure (or increase in entropy). The one-step inlet reduced total pressure by 20%, while the

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