A Modest Temperature Gradient

A supersonic aircraft generates a shock wave. At the wave's origin, the temperature is measured to be at \(300\text{ K}\), but in its direction of travel, the temperature gradient can be modeled by the expression \(\left( -\frac { 2 }{ 75 } d \right)\text{ K}\cdot \text{ km}^{ -2 }\), where \(d\) is the distance from the wave's origin in kilometers. (Note: For this expression, plug in the value and the units for \(d\).)

Using the given approximations, calculate the time (in seconds) that it will take an observer \(15 \text{ km}\) away from the aircraft to hear the shock wave after it is created. If this can be written in the form \(a\sqrt { b }\) seconds, where \(a\) and \(b\) are positive integers with \(b\) square-free, write your answer in the form \(a+b\).

Details and Assumptions

  • \(\arcsin { \theta \approx \theta }\) for small values of \(\theta\).
  • Use the following approximation: \(\sqrt { \dfrac { { { M }_{ a } } }{ { \gamma }_{ a }R } } \approx \frac { 1 }{ 20 }\).
  • \({ M }_{ a }\approx 2.897\times { 10 }^{ -2 }\text{ kg}\cdot \text{ mol}^{ -1 }\), molar mass of dry air.
  • \({ \gamma }_{ a }\approx 1.400\), adiabatic index/heat capacity ratio of dry air at \(293\text{ K}\).
  • \(R\approx 8.315J\cdot \text{ K}^{ -1 }\cdot \text{ mol}^{ -1 }\), ideal gas constant.
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