What are the exit temperatures of bypass flow

Consider a typical high bypass turbofan designed for cruise at approximately mach 0.8, with a separate nozzle for the bypass air. Ideally, the bypass air would be compressed adiabatically by the intake and the fan, and then expanded adiabatically in the nozzle to ambient pressure and temperature. But there are losses in the intake, fan and nozzle, all of which generate a little bit of heat.1 I also expect some heat is absorbed from the engine core.

Moreover, according to this answer, the bypass air exiting the nozzle in cruise is typically under-expanded, which might raise the temperature at the nozzle exit some more. The aforementioned answer also states that the bypass air is exiting near the speed of sound.

As this answer explains, the core airflow can get significantly faster than the speed of sound in ambient air and stay subsonic; the speed of sound is higher in the exiting air because of its higher temperature.

This makes me wonder, what is the temperature difference between the bypass air exiting the nozzle and the ambient air? In particular, is it enough to raise the speed of sound in the exiting air a meaningful amount?

1: The kinetic energy at 300 m/s is 3002/2=45000 j/kg. The specific heat of atmosphere is around 1 kj/(kg*K), so even 10% losses would only raise the temperature 4-5K. Hence little bit.

• The main reason temperature might change is heat exchange with the core flow. Otherwise, bypass flow just isn't being compressed enough to notice. Apr 18, 2023 at 19:45
• @Therac : Sorry for the late response. Do you have any estimates on heat transfer from the core? May 14, 2023 at 16:43
• No concrete estimates. I would guess the temperature rise to be pretty small, due to large volume of the bypass flow. It can be calculated though. You can ask it as a separate question: how much heat passes from the core to the bypass flow. May 15, 2023 at 2:53

I've attached a simple Matlab program that does a cycle analysis for a bypass flow.

With M0=0.8, fan efficiency of 0.9, inlet and nozzle pressure ratios of 0.99, and a fan pressure ratio of 1.15, the ratios of speed of sound at exit to inlet is 1.0051.

So, there is no significant change.

clear all
format compact

% Simple non-ideal 'cycle' analysis for a shaft-driven ducted fan.
% Cycle is in quotes because no work balance is required for this
% calculation.  The shaft power is assumed delivered on demand.  Component
% losses are included as appropriate for this model.

% Since this is a point-calculation, this engine does not need to be sized.
% Thrust and shaft work are calculated per mass flow.

% The nozzle is assumed to expand the flow to ambient pressure - or choked
% flow, whichever happens first.  The areas required to make this happen
% are not actually calculated.

%              ----------------------------
%          ---/    | \                     \----
%        -/        | \                          \
%                  |  \----------
%                  |   \==MMM    \
%  -- . -- . -- . -- . -- . -- . -- . -- . -- . -- . --
%   0             2     3                7       8/9

% Station numbering: (provisions for variable area are just that)
% 0   -- Freestream
% 2   -- Diffuser exit/fan front face
% 3   -- Fan exit
% 7   -- Nozzle start (fixed area)
% 8/9 -- Nozzle throat/exit (variable area) converging-only nozzle

gamma = 1.4;               % --
R = 1716.59;               % ft lbf/(R slug)
cp = R*gamma/(gamma-1.0);  % ft lbf/(R slug)

T0 = 59+459.67;            % Rankine
P0 = 2116.22;              % psf
M0 = 0.8;                  % --

% Inlet pressure ratio.  Adiabatic inlet.
IPR = 0.99;

Feta = 0.9;

% Nozzle pressure ratio.  Adiabatic nozzle.
NPR = 0.99;

a0 = sqrt(gamma*R*T0);
u0 = M0*a0;

Pt0 = P0*(1.0 + ((gamma-1.0)/2.0)*M0^2.0)^(gamma/(gamma-1.0));
Tt0 = T0*(1.0 + ((gamma-1.0)/2.0)*M0^2.0);

FPR = 1.15;
FTRi = FPR^((gamma-1.0)/gamma); % Ideal FTR
FTR = 1.0+(FTRi-1.0)/Feta;

Pt2 = Pt0 * IPR;
Tt2 = Tt0;

Pt3 = Pt2 * FPR;
Tt3 = Tt2 * FTR;

Pshaft = cp*(Tt3-Tt2);       % Shaft power per mass flow

Pt9 = Pt3 * NPR;
Tt9 = Tt3;

% Check for choked flow at nozzle
P9choked = Pt9 /(((gamma+1.0)/2.0)^(gamma/(gamma-1.0)));

if(P9choked > P0)
disp('Choked');
P9 = P9choked;
M9 = 1.0;
T9 = Tt9 /((gamma+1.0)/2.0);
else
disp('Fully expanded');
P9 = P0;
M9 = sqrt((((Pt9/P9)^((gamma-1.0)/gamma))-1.0)*(2.0/(gamma-1.0)));
T9 = Tt9 / (1.0 + ((gamma-1.0)/2.0)*M9^2.0);
end

a9 = sqrt(gamma*R*T9);

u9 = a9*M9;

rho9 = P9/(R*T9);

f = u9-u0+(P9-P0)/(u9*rho9);  % Thrust per mass flow

pprop = f*u0;              % Propulsive power per mass flow

etao = pprop/Pshaft;  % Overall efficiency

etap = 2*u0/(u9+u0);  % Classical propulsive efficiency

TSEC = f/Pshaft;             % Thrust specific energy consumption  (s/ft)

• I recently accepted this answer after long delay, sorry about that. Seems to me you show that inlet-, nozzle- and fan-losses have negligible effect on exit temperature. I just want to comment that this program doesn't take into account heat transfer from the core. According to a comment to the question this has greater potential to increase temperature than losses in compression and expansion, but perhaps estimating that is not a reasonable question. May 14, 2023 at 16:42
• The program above was designed to model an electric motor driven fan, so no core at all. I do not see any evidence that the heat transfer from the engine core and core flow is a substantial contribution to the bypass flow. If I was modeling a traditional turbofan engine, I would still most likely ignore core-to-bypass heat transfer. Modern bypass ratios of 10 really emphasize how much flow goes through the duct. May 14, 2023 at 17:42