7: GAS TURBINE ENGINE ANALYSIS

At subsonic Mach numbers, and in fact upto about Mach 2.5, the pressure ratio achievable from ram compression (through deceleration of air to near-stagnation conditions) is too low for efficient cycle operation. Hence mechanical compressors are needed. To run these compressors, they are connected to turbines which extract work out of the expanding hot gas flow. Thus, we note that the turbine must be able to extract enough work from the hot gases to run the compressor.

To increase propulsive efficiency, bypass engines such as turbofans, turboprops, and propfans are used, where a large mass flow rate of air bypasses the combustor, and just goes through a propeller or fan, where the pressure of the air is increased. The air is then accelerated through a nozzle (real or virtual) until the static pressure reaches the ambient. In this case, the turbine work must be at least equal to the sum of the compressor work and the fan work.

Note that the turbojet is thus a limiting case where the fan work is zero, and the bypass ratio is zero, and the ramjet is a limiting case where the compressor pressure ratio is unity.

Region	Process	Ideal	Actual
a to 1	External acceleration or deceleration. To, Po constant.
1 to 2	Diffuser: 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	Compressor: Adiabatic work addition	Isentropic. Work added until stagnation pressure is increased to a specified level. Po, To rise. These are related by isentropic relation	Work required is greater than that for isentropic compression to specified pressure. Thus To rise is greater than that computed from isentropic.
2 to 8	Fan	same principle as compressor	same principle as compressor
3 to 4	Burner: Heat addition in the core flow	Constant pressure. To, T increase; po constant.	po drops.
4 to 5	Turbine:adiabatic work extraction	isentropic work extraction. To, po drop, related isentropically	po drops more than predicted by isentropic relation for same work extraction.
5 to 6	Afterburner	constant pressure heat addition	po drops due to: a) Rayleigh line losses of heat addition to flowing fluid b) friction and wake losses of duct and flameholders.
6 to 7	Hot nozzle: Adiabatic expansion. No work extracted except that of volume change.	To, po constant. T, p drop.	po drops slightly. s increases slightly.
8 to 9	Fan exit: cold nozzle	To, po constant. T, p drop.	po drops slightly. s increases slightly.

T-s Diagram

Example Problem

A turbofan engine operates at intermediate thrust (afterburner off) with the following characteristics:

Flight condition:

Ambient temperature = 298.15K

Ambient pressure = 101300N/m2.

Bypass ratio b = 0.6

Compressor pressure ratio p_c = 25

Fan pressure ratio p_f = 3.375

Fuel heating value q_r = 19000Btu/lbm

Turbine inlet temperature = 2000K

Diffuser pressure recovery factor r_d = 0.9

Compressor efficiency h_c = 0.89

Fan efficiency h_f = 0.91

Combustion efficiency h_b = 0.98

Burner pressure recovery factor r_b = 0.95

Turbine efficiency h_t = 0.98

Nozzle efficiency h_n = 1.0

Calculate the thrust per unit mass flow rate, and the thrust specific fuel consumption. Later, for a specified value of thrust, determine the area at the choked turbine exit, which limits the mass flow rate.

Such a problem can be solved by going station by station through the engine and using simple thermodynamic relations. Since the component efficiencies are specified, we don't have to go into detailed multi-dimensional flow and chemical analyses for each component.

ANALYSIS

Station 1: inlet

p01 = p0a isentropic external acceleration or deceleration

T01 = T0a no work done, except that of expansion or contraction.

Station 2: end of diffuser, beginning of compressor & fan:

T02 = T01 adiabatic, no work

p02 = rdp01 includes the given stagnation pressure loss

Station 3: end of compressor

p03 = p_cp02 specified.

Temperature reached if the compressor were isentropic is T03s.This can be found using the isentropic relation

Ideal compressor work per unit time per unit mass flow rate is:

Actual compressor work can be found using the compressor efficiency:

From this, the actual temperature can be found:

Station 8: end of fan compression:

The value of ideal fan work can be found just as the ideal compressor work was found, and the actual fan work is:

and

Station 4: end of burner

is specified.

Heat released per unit mass of fuel burned is

Fuel/air ratio is

Station 5: end of turbine

Actual turbine work extracted = fan work + compressor work

From this, the actual pressure can be found from the isentropic relation.

Station 6: end of afterburner

Afterburner is OFF, and we'll assume that there are no losses in the duct:

Station 7: exit of hot nozzle:

Assume that the nozzle exit pressure is the same as the ambient pressure:

The exit temperature can be found using the isentropic relation:

Exit Mach number is

Hot exit velocity is:

Station 9:Fan exit nozzle:

Thrust per unit hot mass flow rate:

Thrust-specific fuel consumption:

If the required thrust is given, the limiting mass flow rate can be found, and the engine sized accordingly.

Computing the Limiting Mass Flow Rate

Usually the mass flow rate is limited by the "choked turbine" condition: i.e., the Mach number reaches 1.0 in the stator passage at the end of the final stage of the turbine. This is because the air at this station has a relatively low stagnation pressure, low static temperature, and high flow velocity, and the area available to the flow is the area between the blades in the turbine stage annulus. Thus, the stator passage is choked. The choked area is easily found from the formula for the mass flow rate through a choked throat:

If the thrust at some flight condition is known (for example, static or takeoff thrust at sea-level), the thrust per unit mass flow rate is first computed, from which the mass flow is determined. From the choked turbine condition above, the choked area A is found. This area remains constant at all other flight conditions, and thus the limiting mass flow rate at any other flight condition can be determined, once the stagnation pressure and temperature at the turbine exit (Station 5) is found.

Note:

The limiting mass flow rate is directly proportional to the stagnation pressure, and inversely proportional to the square root of the stagnation temperature. Thus, if more work is taken out of the flow by the turbine per unit mass, the stagnation pressure drops sharply, and the mass flow and thus the thrust drop sharply as well.