# Simple Straight-Through Turbojet Engine: Formulation

In Figure 10.11, consider a CV (dashed line; note the waist-like shape of the simple turbojet) representing a straight-through, axi-symmetric turbojet engine. The CV and the component station numbers are as shown in the drawing and conventions in Figure 10.5; the gas turbine intake starts with the subscript 0, or to, and ends at the nozzle exit plane with subscript 5, or e. The free-stream airmass flow rate, ma, is inhaled into the CV at the front face perpendicular to the flow, the fuel-mass flow rate mf (from the onboard tank) is added at the combustion chamber, and the product flow rate (ma + m f) is exhaust from the nozzle plane perpendicular to flow. It is assumed that the inlet-face static pressure is pTO, which is fairly accurate. Precompression exists but, for the ideal consideration, it has no loss.

Flow does not cross the other two lateral boundaries of the CV because it is aligned with the walls of the engine. Force experienced by this CV is the thrust produced by the engine. Consider a cruise condition with an aircraft velocity of Vto. At cruise, the demand for air inhalation is considerably lower than at takeoff.

 Figure 10.11. Control volume representation of a straight-through turbojet

The intake area is sized between the two demands. At cruise, the intake-stream – tube cross-sectional area is smaller than the intake-face area – it is closer to that of the exit-plane area, A (the gas exits at a very high velocity). Because in an ideal condition there is no precompression loss, Station 0 may be considered to have free – stream properties with the subscript to.

From Newton’s second law, applied force F = rate of change of momentum + net pressure force (the momentum rate is given by the mass flow rate), where the inlet momentum rate = ma Vto and the exit momentum rate = (ma + m f )Ve.

Therefore:

the rate of change of momentum = (ma + mf )Ve – ma Vto (10.6)

The net pressure force between the intake and exit planes = peAe – pTO Ato (i. e., the axi-symmetric side pressure at the CV walls cancels out). Typically, at cruise, a sufficiently upstream Ae & Ato. Therefore:

F = (ma + mf )Ve – ma V» + Ae(pe – Pc») = net thrust (10.7)

Then:

(ma + mf )Ve + Ae(Pe – Pto) = gross thrust

and maVTO = ram drag (with – ve sign, it must be drag). It is the loss of energy seen as drag due to the slowing down of the incoming air as the ram effect. This gives:

net thrust = gross thrust – ram drag; Ae (pe – pTO) = pressure thrust

In general, subsonic commercial transport turbofans have a convergent nozzle, and the exit area is sized such that during cruise, pe & pTO (known as a perfectly expanded nozzle). This is different for military aircraft engines, especially with AB, when pe > pTO requires a convergent-divergent nozzle.

For a perfectly expanded nozzle, net thrust:

F = (ma + mf)Ve – ma V«, (10.8)

Further simplification is possible by ignoring the effect of fuel flow, m f, because m a » m f.

Then, the thrust for a perfectly expanded nozzle is:

At sea-level, static-takeoff thrust (TSLs) ratings V» = 0, which gives:

F = tha Ve+Ae(pe – p») (10.10)

Equation 10.10 indicates that the thrust increase can be achieved by increasing the intake airmass flow rate and/or increasing the exit velocity.

Equation 10.4 gives the propulsive efficiency:

Clearly, jet-propelled aircraft with low flight speeds have poor propulsive effi­ciency, np. Jet propulsion is favored for aircraft flight speeds above Mach 0.6.

The next question is: Where does the thrust act? Figure 10.12 shows a typical gas turbine engine in which the thrust is acting over the entire engine; the aircraft senses the net thrust transmitted through the engine-mounted bolts.

Figure 10.12 shows a typical straight-through turbojet pressure, velocity, and temperature variation along the length as airmass flows through. Readers may note the scale; within each component, the velocity change is negligible.