FredGarvin said:
Are you going on the assumption that the air:fuel ratio is close to stoich? The relative ratio is actually on the order of 100:1. In a typical installation, the combustion products are mixed in the dillution zone of the burner prior to entering the HP turbine to control inlet temperatures. So you have a small amount of nitrogen from the combustion process that is still being mixed with unburnt air.
It seems to me the air:fuel ratio being so high would make there be even more nitrogen that is combined with the combustion products.
I did a calculation for how much the nitrogen would degrade the performance copied below:
Firstly, the oxygen is thorougly mixed with the nitrogen already so it
shouldn't be a problem of applying the heat of combustion to the
nitrogen as well.
I also looked at a textbook that gave a calculation for a turbine jet
engine and it calculated the thrust produced by simply applying the
heat produced to the total air.
However, we can estimate the loss of efficiency by the fact that the
nitrogen is not contributing to the heat produced, i.e., it is in
effect just being carried along.
Let's say the temperature *increase* of the oxygen+fuel mix would be
3000K above the temperature of the incoming oxygen if it were just pure
oxygen being mixed with the fuel. The space shuttle main engines for
example produce temperatures in this range. However, the space shuttle
combustion chamber operates at 200 atm's, much higher than that of
ram/scram -jets so our temperature increase probably won't be this
much.
The oxygen coming in is quite hot for (sc)ramjets because it has to be
severely slowed down from supersonic or hypersonic speeds. Let's call
it 1000K. Then the total temperature would be 4000K. I'll assume the
fuel is hydrogen for simplicity.
Now we imagine this heat is being distributed to the nitrogen as well
as the exhaust which is now water vapor. Now the heat given to a gas is
proportional to the temperature increase, the specific heat of that
gas, and the mass. The specific heat of water (vapor) is rather high,
higher than for nitrogen gas. This means it is actually easier to raise
the temperature of nitrogen for a given heat input than water.
However, taking into account water's higher molecular weight over
oxygen, the mass ratio of the nitrogen to the water is about 3 to 1. So
the total mass is about 4 times that of the water. So I'll estimate the
temperature change for the mixture as only half. (The specific heat for
the mixture can be calculated from those of the nitrogen and of the
water and therefore the exact temperature increase for the mixture can
be calculated knowing the mass is 4 times as much, but for simplicity
I'll estimate it as about half of the pure water vapor case.) Then the
temperature for the nitrogen+water vapor mix is 1000K + 1500K = 2500K.
This page gives the formula for the exhaust velocity dependent on
temperature:
ROCKET PROPULSION.
Combustion & Exhaust Velocity
http://www.braeunig.us/space/propuls.htm#combustion
The formula is often written in approximate form as Ve = sqrt(2RT/M),
where R is the universal gas constant, T is the temperature in kelvin
and M is the molecular weight of the exhaust. So if the temperature is
smaller by a factor of 2500/4000 and the molecular weight is larger by
a factor of 30/18, the exhaust velocity will be smaller by a factor of:
sqrt[(2500/4000)*(18*30) ] = sqrt(.375) = .6124
This may seem acceptable when the mass of the exhaust is 4 times as
high, making the exhaust thrust 4*.6124 = 2.45 times as high as the
oxygen+fuel only case. But the drag due to the larger incoming mass is
also 4 times as high. The thrust then is not increased by as large
amount as is the drag.
Bob Clark