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Stagnation enthalpies are given as

$H_T=H+\frac{v^2}{2}\ $

Along an adiabatic inviscid steady flow streamline Stagnation Enthalpies are said to be constant.

$H_{T1}=H_{T2}$

$H_1+\frac{{v_1}^2}{2}\ =H_2+\frac{{v_2}^2}{2}\ $

$H_1-H_2=\frac{{v_1}^2}{2}\ - \frac{{v_2}^2}{2}\ $

$H_1-H_2=\frac{{v_1}^2-{v_2}^2}{2}\ $

Stagnation temperatures are also constant

$T_0=T_1+ \frac{{v_1^2}}{2CP}$

$T_0=T_2+ \frac{{v_2}^2}{2CP}$

$T_1+ \frac{{v_1}^2}{2CP}=T_2+ \frac{{v_2}^2}{2CP}$

$T_1-T_2= \frac{{v_1}^2}{2CP}- \frac{{v_2}^2}{2CP}$

$T_1-T_2= \frac{{v_1}^2-{v_2}^2}{2CP}$

In both of these equations there is derived a notion of a static enthalpy and static temperature at a point of interest along a streamline. These can be different by the value of the flow velocity v

This is stated to be applicable to compressible and incompressible flow

Is it correct to say that there is a real temperature change $T_1-T_2$ along a streamline when its velocity changes?

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Don't know what you mean by real temperature. You will certainly experience a change of static temperature in an adiabatic flow if the velocity changes. Stagnation enthalpy describes the total energy contained in the fluid which comprises inner (thermal) energy, kinetic energy (and potential energy which is often neglected). Stagnation enthalpy is constant for an adiabatic flow with no energy exchange. Thus, thermal and kinetic energy are antagonistic in the sense that increases of the kinetic energy need to be countered by reducing thermal energy in order to keep a constant value of total i.e. stagnation enthalpy, and vice versa. The stagnation temperature of a fluid is a virtual entity, temperature entities that are observable i.e. measurable are static temperatures. However, in a stagnating point the static temperature ideally features the value of the stagnation temperature. Back to your question: If you measure temperatures in an adiabatic flow at points of different flow velocities, you will experience different temperature readings. Your measurement setup has to ensure that your readings are not falsified by slowing down the fluid at the points of measurement.

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  • $\begingroup$ Thanks for your reply Chris that’s exactly what I would expect but I also keep getting referred by some very smart people to the work/energy theorem (applied to Bernoullis incompressible adiabatic) saying that $dU=dQ+dW =0$ and therefore if $dU=0$, $dT=0$. This does not account for the conversion along a streamline of Static enthalpy $H$ to Dynamic enthalpy $v$ , which as you say for constant total enthalpy $H_T$ must mean a reduction in Static enthalpy $H$ If $PdV=0$ this must come from static $U$ which explains how the temperature changes with no $dQ$ or $P.dV$ Which one is right ? $\endgroup$
    – Quentin Chester
    Commented Oct 28, 2016 at 7:42
  • $\begingroup$ U is for closed systems, H describes open systems as you regard it. Both entities are linked via H = U + pdV. $\endgroup$
    – Chris
    Commented Oct 28, 2016 at 8:01
  • $\begingroup$ Hi Chris thanks for your reply , I don't understand the open and closed distinction, researching it seems both energy balances can have a change in the kinetic term while $dQ$ and $P.dV$ are zero $\endgroup$
    – Quentin Chester
    Commented Oct 30, 2016 at 2:45

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