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Unformatted text preview: MAE135 SOLUTIONS TO HW 4 Spring 2010 PROBLEM 1 (15 points) The bottle flow rate determines the flow rate of the leak ˙ m i under the initial cabin conditions p i and T i . Since the pressure ratio across the rupture orifice is infinite, the flow must be choked: ˙ m i = C p i radicalbig T i The constant C involves the orifice area and gas constants. It is known from the above relation. Now let’s look at the situation when the bottles are empty. The pressure drops and so does the temperature (they are linked throough the isentropic relations). The orifice is always choked, so the mass flow rate is ˙ m = C p √ T This can be rewritten as follows: ˙ m = C p i radicalbig T i p p i parenleftbigg T T i parenrightbigg- 1 / 2 = ˙ m i p p i parenleftbigg T T i parenrightbigg- 1 / 2 Using the isentropic relation T ∼ p ( γ- 1) /γ ˙ m = ˙ m i parenleftbigg p p i parenrightbigg ( γ +1) / 2 γ The final step is to establish a relation between p and the mass inside the chamber. Given that the volume is fixed, the isentropic relation gives p p i = parenleftbigg ρ ρ i parenrightbigg γ = parenleftbigg m m i parenrightbigg γ or m m i = parenleftbigg p p i parenrightbigg 1 /γ From this step will also realize that the critical condition, when the cabin pressure drops by 50%,...
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- Spring '08
- Thermodynamics, 50%, total pressure, 5.99 kg, 9.16 kg