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Unformatted text preview: MAE135 SOLUTIONS TO HW 5 Spring 2010 PROBLEM 1 (10 points) In the case where heat transfer is ignored, and since flow is frictionless, both p and T are preserved. The pressure ratio across the injector, p 1 /p a =3 is more than sufficient (higher than 1.893) to create choked flow at the exit. Therefore, ˙ m = C p * √ T * = C p 1 radicalbig T 1 where C includes the area and all the gas-dynamic constants in the mass flow rate expression. Heating the injector tube has two effects: increasing T and decreasing p . Assuming the exit is choked (we will need to verify this) we use the table for frictionless 1D flow with heat transfer: T T * = 0 . 5 → p p * = 1 . 114 The pressure ratio at the injector exit is p * /p a =3.0/1.114=2.693 which is still above the critical value 1.893. So the assumption of sonic flow was correct, and we can still use the sonic mass flow rate relation ˙ m = C p * √ T * = C p 1 / 1 . 114 radicalbig 2 * T 1 = 0 . 635 C p 1 radicalbig T 1 So the actual mass flow rate is 0...
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- Spring '08
- Mach number, Order theory, 1 m, Monotonic function, Sonic boom, 0.0635 kg