Stagnation Conditions in the Tds Equations

David Scott · November 29, 2023, 12:53

Is there a proof for why the stagnation conditions (

and

) may be used in the Tds equations?

$\Delta S = c_P \ln \left( \frac{T_2}{T_1} \right) - R \ln \left( \frac{P_2}{P_1} \right)$

Denton 1993 states "the temperatures, pressures, and densities used in the Tds equations may be either all static values or all stagnation values because by definition the change from static to stagnation conditions is isentropic."

Further, some sources justify this with "the Tds equations apply between any two thermodynamic end states". Are stagnation temperature and stagnation pressure thermodynamic state variables?

FMDenaro · November 29, 2023, 13:27

Quote:

Originally Posted by David Scott

Is there a proof for why the stagnation conditions (

and

) may be used in the Tds equations?

$\Delta S = c_P \ln \left( \frac{T_2}{T_1} \right) - R \ln \left( \frac{P_2}{P_1} \right)$

Denton 1993 states "the temperatures, pressures, and densities used in the Tds equations may be either all static values or all stagnation values because by definition the change from static to stagnation conditions is isentropic."

Further, some sources justify this with "the Tds equations apply between any two thermodynamic end states". Are stagnation temperature and stagnation pressure thermodynamic state variables?

This is a jump relation for the entropy in the normal shock wave, being adiabatic the stagnation temperature is constant but not the stagnation pressure.

The states are determined indipendently from the path.

David Scott · November 29, 2023, 14:25

The Tds equations follow from the first and second laws of thermodynamics. They are applicable to many flows other than the case of a normal shock. My interest happens to be understanding losses in turbomachines.

Specifically, I'm seeking a derivation for,
$\Delta S = c_P \ln \left( \frac{T_{0_{2}}}{T_{0_{1}}} \right) - R \ln \left( \frac{P_{0_{2}}}{P_{0_{1}}} \right)$

FMDenaro · November 29, 2023, 14:47

Quote:

Originally Posted by David Scott

The Tds equations follow from the first and second laws of thermodynamics. They are applicable to many flows other than the case of a normal shock. My interest happens to be understanding losses in turbomachines.

Specifically, I'm seeking a derivation for,
$\Delta S = c_P \ln \left( \frac{T_{0_{2}}}{T_{0_{1}}} \right) - R \ln \left( \frac{P_{0_{2}}}{P_{0_{1}}} \right)$

Yes, in general you start from Gibbs relation to integrate between the state 1 and 2.

LuckyTran · November 29, 2023, 18:45

The key assumption is that the the process going from the static to stagnation conditions is via an isentropic process.

The entropy change from 1 to 01 is 0 because the change from static-to-stagnation conditions is (assumed to be) isentropic. The change from 2 to 02 is also 0. Hence the change from 1 to 2 is equal to the change from 10 to 20. Recall also that for a simple system, the thermodynamic state is fully characterized by two intensive properties (static temperature and static pressure). The stagnation conditions at state 1 or 2 are fully specified by knowing their static conditions and vice versa, hence the entropy change between 1 and 2 or 01 and 02 must be equal or you do not have a thermodynamic system.

It is common in engineering applications to assume that the change from static to a (hypothetical) stagnation conditions is isentropic. This condition is explicitly noted in your quote of Denton. The usage of stagnation properties for performance calculations in turbomachinery is also consistent with this assumption.

November 29, 2023, 12:53	Stagnation Conditions in the Tds Equations	#1
David Scott New Member Join Date: Nov 2023 Posts: 2 Rep Power: 0	Is there a proof for why the stagnation conditions ( and ) may be used in the Tds equations? $\Delta S = c_P \ln \left( \frac{T_2}{T_1} \right) - R \ln \left( \frac{P_2}{P_1} \right)$ Denton 1993 states "the temperatures, pressures, and densities used in the Tds equations may be either all static values or all stagnation values because by definition the change from static to stagnation conditions is isentropic." Further, some sources justify this with "the Tds equations apply between any two thermodynamic end states". Are stagnation temperature and stagnation pressure thermodynamic state variables?

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November 29, 2023, 14:25		#3
David Scott New Member Join Date: Nov 2023 Posts: 2 Rep Power: 0	The Tds equations follow from the first and second laws of thermodynamics. They are applicable to many flows other than the case of a normal shock. My interest happens to be understanding losses in turbomachines. Specifically, I'm seeking a derivation for, $\Delta S = c_P \ln \left( \frac{T_{0_{2}}}{T_{0_{1}}} \right) - R \ln \left( \frac{P_{0_{2}}}{P_{0_{1}}} \right)$

November 29, 2023, 18:45		#5
LuckyTran Senior Member Lucky Join Date: Apr 2011 Location: Orlando, FL USA Posts: 5,675 Rep Power: 66	The key assumption is that the the process going from the static to stagnation conditions is via an isentropic process. The entropy change from 1 to 01 is 0 because the change from static-to-stagnation conditions is (assumed to be) isentropic. The change from 2 to 02 is also 0. Hence the change from 1 to 2 is equal to the change from 10 to 20. Recall also that for a simple system, the thermodynamic state is fully characterized by two intensive properties (static temperature and static pressure). The stagnation conditions at state 1 or 2 are fully specified by knowing their static conditions and vice versa, hence the entropy change between 1 and 2 or 01 and 02 must be equal or you do not have a thermodynamic system. It is common in engineering applications to assume that the change from static to a (hypothetical) stagnation conditions is isentropic. This condition is explicitly noted in your quote of Denton. The usage of stagnation properties for performance calculations in turbomachinery is also consistent with this assumption.