# Is there a real temperature change along a streamline when its velocity changes?

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?

• 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 ? – Quentin Chester Oct 28 '16 at 7:42
• 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 – Quentin Chester Oct 30 '16 at 2:45