An ideal gas flows with velocity \(V\), pressure \(p\), temperature \(T,\) and density \(\rho .\) Determine a set of equations for stagnation properties, including entropy, if the stagnation process is defined to be isothermal \((T=\text { constant ) rather than isentropic }(s=\text { constant })\).

Short Answer

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The set of equations for stagnation properties under an isothermal process include: the pressure equation which is \(p_2 = p_1 \times e^{(-\Delta s/R)}\), the temperature equation which is \(T = T_0 - V^2/2C_p\), and the density equation is \(\rho=\rho_0\). The entropy remains unchanged in an isentropic process, and thus \(\Delta s = 0\).

Step by step solution

01

State the conservation laws and definitions

Stagnation properties are defined as the conditions obtained when a flowing fluid is brought to rest isentropically. For an isothermal process, the equation for entropy change is given by \(\Delta s = c_p \ln(T_2/T_1) - R \ln(p_2/p_1)\) whereas for an isentropic process, the change in entropy is zero. Here, \(c_p\) is the specific heat at constant pressure, \(R\) is the gas constant, \(T_1\) is the initial temperature, \(T_2\) is the final temperature, \(p_1\) is the initial pressure, and \(p_2\) is the final pressure.
02

Apply the appropriate laws and definitions

In case of an isothermal process, \(T_2 = T_1\), and thus the equation for entropy change simplifies to \(\Delta s = -R \ln(p_2/p_1)\). For this process, it's also true that \(p_2 = p_1 \times e^{(-\Delta s/R)}\). This is the equation for pressure. The Bernoulli equation (in the form \(H=T+V^2/2C_p\) where \(H\) is the total enthalpy) can be used to write the equation for temperature as \(T = T_0 - V^2/2C_p\). These are the main equations for the stagnation properties pressure and temperature.
03

Derive an equation for stagnation density

By definition, the density \(\rho\) of an ideal gas is given by \(p/R/T\). Substituting our derived isothermal pressure equation into this, we obtain \(\rho = \rho_0 \times (T_0/T)\). Since \(T\) is constant, we have \(\rho=\rho_0\). This represents the equation for stagnation density under an isothermal process. Since by definition \(\Delta s = 0\) for an isentropic process, the expression for entropy remains unchanged for an isothermal process.

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Most popular questions from this chapter

Air flows isentropically through a duct to a section where \(p_{1}=25 \mathrm{kPa}, T_{1}=300 \mathrm{K},\) and \(V_{1}=900 \mathrm{m} / \mathrm{s} .\) For these conditions: (a) Determine the stagnation conditions for the flow. (b) What is the Mach number at station \(1 ?\) Show a \(T-s\) diagram displaying stagnation and static conditions. (c) Is the flow choked? Is the throat behind or ahead of section \(1 ?\) Label this state on the \(T-s\) diagram.

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