6: IDEAL RAMJET CYCLE

          This is essentially the Brayton cycle. Here we have a flowing fluid, so that it is not useful to draw a pressure-volume diagram. We will instead represent the engine processes on a Temperature - Entropy (T- s) diagram. This is the same as an Enthalpy-Entropy diagram (a Mollier Chart) if cp is constant.

 

Region

Process

Ideal

Actual

a to 1

No change: supersonic flow: no external compression or suction ahead of inlet.

 

1 to 2

Adiabatic compression, with no work done except volume change

Isentropic:

p, T increase; To, po constant.

s constant.

po drops due to shocks and friction.

s increases.

2 to 3

Heat addition

Constant pressure.

To, T increase;

po constant.

po drops.

3 to 4

Adiabatic expansion. No work extracted except that of volume change.

To, po constant. T, p drop.

po drops slightly.

s increases slightly.

 

T-s Diagram

 

 

Notes:

1. Lines of constant pressure slant upwards.

2. Lines of constant pressure also diverge.

3. Even in the ideal engine, entropy must increase when heat is added. We thus see that as the fluid is brought back to the ambient pressure pa, the temperature reached is higher than the ambient temperature. This is another way of expressing the Brayton cycle efficiency.

 

 

 

Ideal Ramjet Analysis

Throughout this course, we will analyze engine performance by considering one station at a time.

 

Example Problem

An ideal ramjet engine has an isentropic inlet, which decelerates incoming air at a flight Mach number of 3.0 to a low Mach number without any losses. Heat is then added by chemical reaction with a fuel whose heating value is 19,000 Btu / lbm. The altitude is 11000m.  The maximum temperature in the engine is limited to 2000K. The heat addition occurs at constant pressure, and at a low Mach number. The nozzle is fully expanded, and has an exit diameter of 1 meter. Find the thrust and thrust-specific fuel consumption of the engine, assuming constant specific heats.

 

Station 1

          Here we have undisturbed air, just before entering the inlet of the engine. The stagnation temperature can be determined from the static temperature and the Mach number using the isentropic relation.

 

           , and     

 

         

 

 

          , and

         

 

Isentropic diffuser, from Station 1 to Station 2:

Adiabatic, and no work done except that of volume change:

         

Isentropic (reversible, adiabatic, therefore no losses):

         

 

Constant-Pressure Heat Addition, from Station 2 to Station 3:

         

           is specified, for a given thrust setting. For maximum thrust, this value is limited by the limiting temperature of the combustor material.

To find the fuel/air ratio needed to achieve this temperature, we consider the enthalpy balance across the combustor.

 

Stagnation enthalpy of the air + burned products leaving the combustor = Stagnation enthalpy of the air entering the combustor + sensible enthalpy of the fuel + heat released by reaction. 

We neglect the sensible enthalpy of the fuel, since it is very small compared to the heat released by reaction.

 

         

 

Solving for the fuel/air ratio,

         

 

Isentropic Nozzle, from Station 3 to Station 4:

         

         

 

Fully expanded flow at the nozzle exit: static pressure is the same as the ambient atmospheric pressure.

         

Relating temperature ratio and pressure ratio,

         

 

         

Also, the exit Mach number can be found from:

         

Let us pause for a moment and consider what the exit Mach number will be. Since stagnation pressure is constant throughout this ideal engine, and the specific heats are assumed constant,

 

         

Obviously, the exit Mach number will be equal to the flight Mach number!

 

Exit velocity is

         

and the flight velocity is

         

 

Thrust per unit air mass flow rate:

         

and Thrust Specific Fuel Consumption

          ,  or

         

Notes:

1. As Mach number increases, the ramjet pressure ratio increases. The combustor walls must be stronger. For the same reason, high-Mach number flight at sea-level becomes difficult.

2. As Mach number increases, the temperature at the inlet to  the combustor increases. When this reaches the highest permissible combustor temperature, no heat can be added, so no thrust can be produced. Thus this provides an absolute upper limit on the Mach number at which thrust can be produced. Of course, the vehicle may not be able to fly at this Mach number, because we have not considered how much thrust is needed to overcome drag and accelerate to this Mach number.