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A perfect gas stored in a large reservoir exhausts into the atmosphere through a convergent duct. The reservoir pressure is \(P_{0}\) and temperature is \(T_{0}\). The jet emerges from the nozzle at choked conditions with average velocity \(u\), Mach number \(M\), pressure \(p\), temperature \(T\), and density \(\rho\). If the reservoir pressure is increased, then

  • A.\(u\), \(M\), \(p\), \(T\), and \(\rho\) increase
  • B.\(u\), \(p\), \(T\), and \(\rho\) increase, but \(M\) remains the same
  • C.\(u\), \(M\), and \(T\) remain the same, but \(p\) and \(\rho\) increase
  • D.\(u\), \(M\), \(T\) and \(\rho\) remain the same, but \(p\) increases

Answer: C

Raising the reservoir pressure leaves \(u\), \(M\) and \(T\) unchanged, while \(p\) and \(\rho\) increase.

A choked convergent nozzle has \(M = 1\) at its exit, and once choked it stays at \(M = 1\) however high the reservoir pressure goes. With adiabatic flow, the exit temperature is fixed by the reservoir temperature alone, and so is the exit velocity:

\[ \begin{aligned} T &= \frac{2T_{0}}{\gamma + 1} \\ u &= \sqrt{\gamma RT} \end{aligned} \]

Neither depends on \(P_{0}\). The exit pressure, however, is a fixed fraction of the reservoir pressure, \(p = P_{0}(2/(\gamma+1))^{\gamma/(\gamma-1)}\), so it rises with \(P_{0}\); then \(\rho = p/RT\) rises too, and that is how the mass flow of a choked nozzle increases.

Any option in which \(u\), \(M\) or \(T\) increases is therefore wrong, and the option keeping \(\rho\) constant contradicts the ideal gas law with \(p\) rising at fixed \(T\).

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