Figure 10.9. Simple straight-through jet

10.6 Formulation and Theory: Isentropic Case

Gas turbine equations relevant to this book are provided in this section and are valid for all types of processes. For more details, see [2] through [6].

10.6.1 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 ∞, and ends at the nozzle exit plane with subscript 5, or e. The free-stream airmass flow rate, m˙ a, is inhaled into the CV at the front face perpendicular to the flow, the fuel-mass flow rate m˙ f (from the onboard tank) is added at the combustion chamber, and the product flow rate (m˙ a + m˙ f ) is exhaust from the nozzle plane perpendicular to flow. It is assumed that the inlet-face static pressure is p∞, 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 V∞. At cruise, the demand for air inhalation is considerably lower than at takeoff.

Figure 10.10. Real and ideal Joule (Brayton) cycle comparison of a straight-through jet

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 freestream properties with the subscript ∞.

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 = m˙ a V∞ and the exit momentum rate = (m˙ a + m˙ f ) Ve.

Therefore:

The net pressure force between the intake and exit planes = pe Ae − p∞ A∞ (i.e., the axi-symmetric side pressure at the CV walls cancels out). Typically, at

and m˙ a V∞ = 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 − p∞) = pressure thrust

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

For a perfectly expanded nozzle, net thrust:

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:

Figure 10.12. Where does the thrust act?

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

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, ηp. 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.

10.6.2 Bypass Turbofan Engine: Formulation

Typically, in this book, a long-duct nacelle is preferred to obtain better thrust and fuel economy and to offset the weight gain as compared to short-duct nacelles (see Figure 10.21). The pressure increase across the fan (i.e., secondary cold flow) is substantially lower than the pressure increase of the primary airflow. The secondary airflow does not have the addition of heat as in the primary flow. The cooler and lower exit pressure of the fan exit – when mixed with the primary hot flow within the long duct – reduces the final pressure to lower than the critical pressure, favoring a perfectly expanded exit nozzle ( pe = p∞). Through mixing, there is a reduction in the jet velocity, which provides a vital benefit in noise reduction (see Chapter 15) to meet airworthiness requirements. The long-duct nacelle exit plane can be sized to expand perfectly.

Primary flow has a subscript designation of p and secondary flow has a subscript designation of s. Therefore, Fp and Vep denote primary flow thrust and exit velocity, respectively, and Fs and Vsp denote bypass flow thrust and its exit velocity, respectively. Thrust (F) equations of perfectly expanded turbofans are computed separately for primary and secondary flows and then added to obtain the net thrust, F, of the engine (i.e., a perfectly expanded nozzle):

F = Fp + Fs = [(m˙ p + m˙ f ) × Vep − m˙ p × V∞] + [m˙ s × (Ves − V∞)]

Specific thrust in terms of primary flow becomes (f = fuel-to-air ratio), or:

If the fuel flow is ignored, then:

For kinetic energy (KE):

At a given design point (i.e., flight speed V∞), BPR, fuel consumption, and m˙ p are held constant. Then, the best specific thrust and KE are found by varying the fan exit velocity for a given Vep, setting the differentiation relative to Ves equal to zero. (This may be considered as trend analysis for ideal turbofan engines; real engine analysis is more complex.)

Then, by differentiating Equation 10.13:

Equation 10.14 becomes:

Combining Equations 10.15 and 10.16:

−BPR × Vep + BPR × Vsp = 0

Because BPR =0, the optimum is when:

That is, the best specific thrust is when the primary (i.e., hot core) exit-flow velocity equals the secondary (i.e., cold fan) exit-flow velocity.

Equation 10.4 gives the turbojet propulsive efficiency, ηp = , for a simple turbojet engine; however, for the turbofan, there are two exit-plane velocities – for the hot-core primary flow (Vep) and for the cold-fan secondary flow (Ves ). Therefore, an equivalent mixed turbofan exit velocity (Veq) can substitute for Ve in the previous equation. Fuel-flow rates are minor and can be ignored. The equivalent turbofan exit velocity (Veq) is obtained by equating the total thrust (i.e., a perfectly expanded nozzle) as if it were a turbojet engine with total airmass flow (m˙ p + m˙ s ).