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I am wondering how to calculate accurately the loss of total temperature of a moving gas which has exchanging heat through walls and pressure drop.


More precisely, let's consider a gas in motion in a duct, and let's consider two small cells, 1 and 2, of the duct next to each other (cell 2 is right after cell 1). And I will assume that thermodynamic properties are homogeneous in each cell (it's like meshing the duct).

The frame of reference is the static, motionless duct, and the gas is moving inside the duct at constant mass flow rate $\dot m$.

As the gas is moving in that frame, we can define static and total temperatures ($T_s$ and $T_i$ resp.), as well as static and total pressures ($P_s$ and $P_i$ resp.).

When gas is flowing from cell 1 to cell 2, there is a pressure drop $\Delta P_i$, such that $P_{i_2} = P_{i_1} - \Delta P_i$.

Because of the walls, heat is exchanged through the walls between the moving gas and the ambient, outer air surrounding the duct. The specific exchange heat is $q_1$.

Finally, I know the dependence of the specific heat capacity ratio with static temperature : $c_p = f(T_s)$.

The problem is summarized in the image below, with the notation :
$c_p$ : specific constant-pressure heat capacity (J/kg/K)
$\gamma$ : specific heat ratio
$M$ : mach number
$q_1$ : specific heat exchanged with outer atmosphere through the wall at cell 1 (J/kg)
$T_{out}$ : temperature of outer atmosphere

enter image description here


I would like to estimate the loss in total temperature, that is to say estimate $T_{i_2}$ knowing thermodynamic state at cell 1. But I am not sure which of the two following options I have thought is the right one (maybe none of them are right ?) :

1 - Energy at cell 1 = Energy at cell 2 + energy loss through the wall
So : $c_{p_1}*T_{i_1} = c_{p_2}*T_{i_2} +q_1$
Then I resolve the following system of equations of unknows $T_{i_2}$, $T_{s_2}$, $P_{s_2}$, $M_2$, $c_{p_2}$

$\begin{Bmatrix} c_{p_2}(T_{s_2})*T_{i_2}=c_{p_1}(T_{s_1})*T_{i_1}-q_1 \\ T_{i_2} = T_{s_2}*(1+\frac{\gamma_2-1}{2}M_2^2) \\ P_{i_2}=P_{s_2}*(1+\frac{\gamma_2-1}{2}M_2^2)^{\frac{\gamma_2}{\gamma_2-1}} \\ M_2^2=\frac{\dot{m}^2\,r\,T_{s_2}}{\gamma_2A^2P_{s_2}^2} \\ c_{p_2}=f(T_{s_2}) \end{Bmatrix}$

2 - First law of thermodynamic, assuming no mechanical work : $\Delta H +\Delta E_{kinetic} = -q_1$
So $\int\limits_1^2 c_p(T_s)\,\mathrm{d}T_s + \frac{1}{2}(V_2^2-V_1^2) = -q_1$

As $V = \frac{\dot{m}}{\rho\,A} = \frac{\dot{m}\,r\,T_s}{P_s\,A}$

and given $P_{i_2}=P_{s_2}*(1+\frac{\gamma_2-1}{2}M_2^2)^{\frac{\gamma_2}{\gamma_2-1}}$

I find $T_{s_2}$, $P_{s_2}$, $M_2$, then I calculate $T_{i_2} = T_{s_2}*(1+\frac{\gamma_2-1}{2}M_2^2)$

Thank you for your help

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  • $\begingroup$ Is the anything close to a flow where kinetic energy effects are going to be important (significant Mach numbers), or can they in practice be neglected? Do you know how to write a differential balance on a section of duct? $\endgroup$ Jun 28, 2021 at 13:58
  • $\begingroup$ @ChetMiller Mach number is not significant, yet the flow is compressible, so I thought that kinetic energy should be considered. No, I don't think I know how to write a differential balance on a section duct. How should I do ? $\endgroup$
    – Jonses
    Jun 28, 2021 at 14:04
  • $\begingroup$ Are you considering viscous frictional drag as the cause of the pressure variation? $\endgroup$ Jun 28, 2021 at 16:19
  • $\begingroup$ @ChetMiller Yes I do consider viscous frictional drag, therefore I have an equation for the pressure drop depending mainly from Reynolds number and gas speed among others $\endgroup$
    – Jonses
    Jun 29, 2021 at 6:14

1 Answer 1

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If you are neglecting kinetic energy effects and do a differential energy balance on the section of the duct between x and $x + \Delta x$, you get $$\dot{m}C_p[T(x+\Delta x)-T(x)]=-PU\Delta x(T(x)-T_{atm})$$where $\dot{m}$ is the mass flow rate, P is the wetted perimeter of the duct, U is the overall heat transfer coefficient between the bulk gas and the surrounding atmosphere, and $T_{atm}$ is the temperature of the surrounding atmosphere. If we divide by $\Delta x$ take the limit as $\Delta x$ approaches zero, we obtain: $$\dot{m}C_p\frac{dT}{dx}=-PU(T-T_{atm})$$

To apply this equation, you need how to determine the overall heat transfer coefficient U, which accounts for forced convection within the duct and natural convection outside the duct. The methods of estimating this are presented in Transport Phenomena by Bird, Stewart, and Lightfoot.

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  • $\begingroup$ But what if $C_p=C_p(T)$ was temperature dependent, and kinetic energy was not negligible ? And how about pressure drop and the subsonic compressible flow, how these would interact and impact with the loss of temperature $T$ ? $\endgroup$
    – Jonses
    Jun 28, 2021 at 20:18
  • $\begingroup$ If Cp is not constant, then you treat it as a function of temperature in the final equation. If the KE is not negligible, you can check on that a postiori, and, if necessary, that term can be included. If there is a pressure drop with subsonic compressible flow, using the enthalpy in the equation automatically accounts for that. $\endgroup$ Jun 28, 2021 at 20:48
  • $\begingroup$ But in the first equation, it looks like you have assumed that $C_p$ is constant. Therefore, if $C_p$ isn't constant, the last equation shouldn't be like $\dot{m}\frac{d[C_p\,T]}{dx} = -PU(T-T_{atm})$ ? $\endgroup$
    – Jonses
    Jun 29, 2021 at 6:04
  • $\begingroup$ In the first equation, I should have written the enthalpy as the integral of CpdT $\endgroup$ Jun 29, 2021 at 14:10
  • $\begingroup$ Just to be sure, in that case it becomes the equation of the second option I have thought in my post ? $\endgroup$
    – Jonses
    Jun 29, 2021 at 14:57

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