What are the exit temperatures of bypass flow

Question

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?

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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.

Answer 1

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.

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% 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;

% Fan efficiency.  Adiabatic fan.
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)