Stagnation properties - Gas Dynamics Notes

Up until this point, we have exclusively referred to static properties, which are measured as though we are moving along with the fluid.

A useful concept is the stagnation state, which is a reference state associated with the fluid being brought to rest and at zero potential, isentropically, and with no work done. This means that there is no energy exchange with surroundings ( $Q = W = 0$ , and $dS_e = 0$ ), there are no losses ( $dS_i = 0$ ).

Stagnation enthalpy¶

We can apply the energy equation from any point in the flow a to the stagnation reference point b:

\begin{gather*} h_a + \frac{V_a^2}{2} + g z_a + q = h_b + \frac{V_b^2}{2} + g z_b + w_s \\ \rightarrow h_b = h_a + \frac{V_a^2}{2} + g z_a \end{gather*}

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and thus we can define the stagnation enthlapy, or total enthalpy, as

h_t = h + \frac{V^2}{2} + g z \;.

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The one-dimensional steady energy equation can then be written as

h_{t1} + q = h_{t2} + w_s \;,

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or $\delta q = \delta w_s + d h_t$ .

Stagnation pressure¶

We can also consider a stagnation pressure, which is the pressure at the stagnation state. Applying our property relationship to the stagnation state:

\begin{gather*} T \, ds = dh - v \, dp \\ \rightarrow T_t \, ds_t = d h_t - v_t \, d p_t \;, \end{gather*}

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where $d s_t = ds = ds_e + ds_i$ , since the stagnation process is isentropic. We can combine this with the (stagnation) energy equation:

\begin{gather*} T_t \left( ds_e + ds_i \right) = dh_t - v_t \, dp_t \\ \delta q = \delta w_s + d h_t \\ \delta q = \delta w_s + T_t \left( ds_e + ds_i \right) + v_t \, dp_t \\ \rightarrow \frac{dp_t}{\rho_t} + ds_e (T_t - T) + T_t \, ds_i + \delta w_s = 0 \;, \end{gather*}

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which is the stagnation pressure-energy equation.

This looks complicated, but tells us something useful. If we consider a process with no shaft work ( $\delta w_s = 0$ ), no heat transfer ( $\delta q = 0$ ), and no losses ( $ds_i = 0$ ), then

\frac{d p_t}{\rho_t} = 0 = d p_t \;,

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or stagnation pressure is constant.

Losses will be present in most real systems, and so the more realistic version of this equation is

\frac{d p_t}{\rho_t} + T_t \, ds_i = 0 \,

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which shows us that losses cause the stagnation pressure to decrease (since $ds_i > 0$ ).