# Jet engine design point performance
## Design point performance calculation

In fixed-wing aircraft driven by one or more jet engines, the performance of the jet engine is important to the operation of the aircraft. Strictly speaking, the performance of the jet engine includes among other parameters the measurement/estimation of thrust, fuel consumption, noise and engine emissions. This article addresses mainly the calculation of thrust.
Suggested reading is the topic Jet Engine Off-Design and Transient Performance .

### TS diagram

A reciprocating engine, such as the petrol or diesel unit in a vehicle, has a Constant Volume (during combustion) Cycle. Similarly a Jet Engine (and gas turbine engines in general) has a Constant Pressure (during combustion) Cycle. Also, the Jet Engine has a continuos flow process, whereas the reciprocating engine cycle is intermittent.
Temperature vs. entropy (TS) diagrams (see example RHS) are usually used to illustrate the cycle of gas turbine engines. Entropy represents the degree of disorder of the molecules in the fluid. It tends to increase as energy is converted between different forms, i.e. chemical and mechanical.
The TS diagram shown on the RHS is for a single spool turbojet, where a single drive shaft connects the turbine unit with the compressor unit.
Apart from stations 0 and 8s, stagnation pressure and stagnation temperature are used. Station 0 is ambient. Stagnation quantities are frequently used in gas turbine cycle studies, because no knowledge of the flow velocity is required.
The processes depicted are:

#### Freestream (stations 0 to 1)

    In the example, the aircraft is stationary, so stations 0 and 1 are coincident. Station 1 is not depicted on the diagram.

#### Intake (stations 1 to 2)

    In the example, a 100% intake pressure recovery is assumed, so stations 1 and 2 are coincident.

#### Compression (stations 2 to 3)

    The ideal process would appear vertical on a TS diagram. In the real process there is friction, turbulence and, possibly, shock losses, making the exit temperature, for a given pressure ratio, higher than ideal. The shallower the positive slope on the TS diagram, the less efficient the compression process.

#### Combustion (stations 3 to 4)

    Heat (usually by burning fuel) is added, raising the temperature of the fluid. There is an associated pressure loss, some of which is unavoidable

#### Turbine (stations 4 to 5)

    The temperature rise in the compressor dictates that there will be an associated temperature drop across the turbine. Ideally the process would be vertical on a TS diagram. However, in the real process, friction and turbulence cause the pressure drop to be greater than ideal. The shallower the negative slope on the TS diagram, the less efficient the expansion process.

#### Jetpipe (stations 5 to 8)

    In the example the jetpipe is very short, so there is no pressure loss. Consequently, stations 5 and 8 are coincident on the TS diagram.

#### Nozzle (stations 8 to 8s)

    These two stations are both at the throat of the (convergent) nozzle. Station 8s represents static conditions. Not shown on the example TS diagram is the expansion process, external to the nozzle, down to ambient pressure.

### Design point performance equations

In theory, any combination of flight condition/throttle setting can be nominated as the engine performance Design Point. Usually, however, the Design Point corresponds to the highest corrected flow at inlet to the compression system (e.g. Top-of-Climb, Mach 0.85, 35,000 ft, ISA).
The design point net thrust of any jet engine can be estimated by working through the engine cycle, step by step. Below are the equations for a single spool turbojet.

#### Freestream

The stagnation (or total) temperature in the freestream approaching the engine can be estimated using the following equation, derived from the Steady Flow Energy Equation:
$T_{1}=t_{0}\cdot (1+({\gamma }_{c}-1)\cdot M^{2}/2)$
The corresponding freestream stagnation (or total) pressure is:
$P_{1}=p_{0}\cdot (T_{1}/t_{0})^{{\gamma }_{c}/({\gamma }_{c}-1)}$

#### Intake

Since there is no work or heat loss in the intake under steady state conditions:
$T_{2}=T_{1}\,$
However, friction and shock losses in the intake system must be accounted for:
$P_{2}=P_{1}\cdot \mathrm {prf}$

#### Compressor

The actual discharge temperature of the compressor, assuming a polytropic efficiency is given by:
$T_{3}=T_{2}\cdot (P_{3}/P_{2})^{{(\gamma }_{c}-1)/({\gamma }_{c}\cdot {\eta }pc)}$
Normally a compressor pressure ratio is assumed, so:
$P_{3}=P_{2}\cdot (P_{3}/P_{2})$

#### Combustor

A turbine rotor inlet temperature is usually assumed:
$T_{4}=\mathrm {RIT} \,$
The pressure loss in the combustor reduces the pressure at turbine entry:
$P_{4}=P_{3}\cdot (P_{4}/P_{3})$

#### Turbine

Equating the turbine and compressor powers and ignoring any power offtake (e.g. to drive an alternator, pump, etc), we have:
$w_{4}\cdot C_{\mathrm {pt} }(T_{4}-T_{5})=w_{2}\cdot C_{\mathrm {pc} }(T_{3}-T_{2})$
A simplyfying assumption sometimes made is for the addition of fuel flow to be exactly offset by an overboard compressor bleed, so mass flow remains constant throughout the cycle.
The pressure ratio across the turbine can be calculated, assuming a turbine polytropic efficiency:
$P4/P5=(T4/T5)^{{\gamma }_{t}/(({\gamma }_{t}-1).{\eta }_{\mathrm {pt} })}$
Obviously:
$P_{5}=P_{4}/(P_{4}/P_{5})\,$

#### Jetpipe

Since, under Steady State conditions, there is no work or heat loss in the jetpipe:
$T_{8}=T_{5}\,$
However, the jetpipe pressure loss must be accounted for:
$P_{8}=P_{5}\cdot (P_{8}/P_{5})\,$

#### Nozzle

Is the nozzle choked? The nozzle is choked when the throat Mach number = 1.0. This occurs when the nozzle pressure ratio reaches or exceeds a critical level:
$(P_{8}/p_{\mathrm {8s} })crit=(({\gamma }_{t}+1)/2)^{{\gamma }_{t}/({\gamma }_{t}-1)}\,$
If $(P_{8}/p_{0})>=(P_{8}/p_{\mathrm {8s} })crit\,$ then the nozzle is CHOKED.
If $(P_{8}/p_{0})<(P_{8}/p_{\mathrm {8s} })crit\,$ then the nozzle is UNCHOKED.

#### Choked Nozzle

The following calculation method is only suitable for choked nozzles.
Assuming the nozzle is choked, the nozzle static temperature is calculated as follows:
$t_{\mathrm {8s} }=T_{8}/(({\gamma }_{t}+1)/2)\,$
Similarly for the nozzle static pressure:
$p_{\mathrm {8s} }=P_{8}/(T_{8}/t_{\mathrm {8s} })^{{\gamma }_{t}/({\gamma }_{t}-1)}$
The nozzle throat velocity (squared) is calculated using the Steady Flow Energy Equation:
$V_{8}^{2}=2gJC_{pt}(T_{8}-t_{\mathrm {8s} })$
The density of the gases at the nozzle throat is given by:
${\rho }_{\mathrm {8s} }=p_{\mathrm {8s} }/(R\cdot t_{\mathrm {8s} })$
Nozzle throat effective area is estimated as follows:
$A_{8}=w_{8}/({\rho }_{\mathrm {8s} }\cdot V_{8})$

#### Gross thrust

There are two terms in the nozzle gross thrust equation; ideal momentum thrust and ideal pressure thrust. The latter term is only non-zero if the nozzle is choked:
$F_{g}=C_{\mathrm {x} }((w_{8}\cdot V_{8}/g)+A_{8}(p_{\mathrm {8s} }-p_{0}))\,$

#### Unchoked nozzle

The following special calculation is required, if the nozzle happens to be unchoked.
Once unchoked, the nozzle static pressure is equal to ambient pressure:
$p_{\mathrm {8s} }=p_{0}\,$
The nozzle static temperature is calculated from the nozzle total/static pressure ratio:
$t_{\mathrm {8s} }=T_{8}/(P_{8}/p_{\mathrm {8s} })^{{(\gamma }_{t}-1)/{\gamma }_{t}}$
The nozzle throat velocity (squared) is calculated, as before, using the steady flow energy equation:
$V_{8}^{2}=2gJC_{pt}(T_{8}-t_{\mathrm {8s} })$

#### Gross thrust

The nozzle pressure thrust term is zero if the nozzle is unchoked, so only the Momentum Thrust needs to be calculated:
$F_{g}=C_{\mathrm {x} }((w_{8}\cdot V_{8}/g)\,$

#### Ram drag

In general, there is a ram drag penalty for taking air onboard via the intake:
$F_{r}=w_{0}\cdot V_{0}/g$

#### Net thrust

The ram drag must be deducted from the nozzle gross thrust:
$F_{n}=F_{g}-F_{r}\,$
The calculation of the combustor fuel flow is beyond the scope of this text, but is basically proportional to the combustor entry airflow and a function of the combustor temperature rise.
Note that mass flow is the sizing parameter: doubling the airflow, doubles the thrust and the fuel flow. However, the specific fuel consumption (fuel flow/net thrust) is unaffected, assuming scale effects are neglected.
Similar design point calculations can be done for other types of jet engine e.g. turbofan, turboprop, ramjet, etc.
The method of calculation shown above is fairly crude, but is useful for gaining a basic understanding of aeroengine performance. Most engine manufacturers use a more exact method, known as True Specific Heat. High pressures and temperatures at elevated levels of supersonic speeds would invoke the use of even more exotic calculations: i.e. Frozen Chemistry and Equilibrium Chemistry.

### Worked example

Question
Calculate the net thrust of the following single spool turbojet cycle at Sea Level Static, ISA, using Imperial units for illustration purposes:
Key design parameters:
Intake air mass flow, $w_{2}=100\ \mathrm {lb/s} \,$
(use 45.359 kg/s if working in SI units)
Assume the gasflow is constant throughout the engine.
Overall pressure ratio, $P_{3}/P_{2}=10.0\,$
Turbine rotor inlet temperature, $T_{4}=\mathrm {RIT} =1400\ \mathrm {K} \,$
(factor-up by 1.8, if working with degrees Rankine)
Design component performance assumptions:
Intake pressure recovery factor, $\mathrm {prf} =1.0\,$
Compressor polytropic efficiency, ${\eta }pc=0.89\ (i.e.89\%)\,$
Turbine polytropic efficiency, ${\eta }pt=0.90\ (i.e.90\%)\,$
Combustor pressure loss 5%, so the combustor pressure ratio $P_{4}/P_{3}=0.95\,$
Jetpipe pressure loss 1%, so the jetpipe pressure ratio $P_{8}/P_{5}=0.99\,$
Nozzle thrust coefficient, $C_{\mathrm {x} }=0.995\,$
Constants:
Ratio of specific heats for air, ${\gamma }_{c}=1.4\,$
Ratio of specific heats for combustion products, ${\gamma }_{t}=1.333\,$
Specific heat at constant pressure for air, $C_{\mathrm {pc} }=0.6111\ {\frac {\mathrm {hp} \cdot \mathrm {s} }{\mathrm {lb} \cdot \mathrm {K} }}\,$
(use 1.004646 kW·s/(kg·K) when working with SI units and use 0.3395 hp·s/(lb·°R) if working with American units)
Specific heat at constant pressure for combustion products , $C_{\mathrm {pt} }=0.697255\ {\frac {\mathrm {hp} \cdot \mathrm {s} }{\mathrm {lb} \cdot \mathrm {K} }}\,$ (use 1.1462 kW·s/(kg·K) when working with SI units and use 0.387363889 hp·s/(lb·°R) if working with American units)
Acceleration of gravity, $g=32.174\ \mathrm {ft} /\mathrm {s} ^{2}\,$ (use 1000 when working with SI units)
Mechanical equivalent of heat, $J=550\ \mathrm {ft} \cdot \mathrm {lb} /(\mathrm {s} \cdot \mathrm {hp} )\,$ (use 1 when working with SI units)
Gas constant, $R=96.034\ \mathrm {ft} \cdot \mathrm {lbf} /(\mathrm {lb} \cdot \mathrm {K} )\,$ (use 0.287052 kN·m/(kg·K) when working with SI units and use 53.3522222 ft·lbf/(lb·°R) if working with American units including degrees Rankine)
Answer
Ambient conditions
A sea level pressure altitude implies the following:
Ambient pressure, $p_{0}=14.696\ \mathrm {psia} \,$ (assume 101.325 kN/m² if working in SI units)
Sea level, ISA conditions (i.e. Standard Day) imply the following:
Ambient temperature, $t_{0}=288.15\ \mathrm {K} \,$
(Note: this is an absolute temperature i.e. $15\ ^{\circ }\mathrm {C} +273.15\ ^{\circ }\mathrm {C} \,$)
(Use 518.67 °R, if working with American units)
Freestream
Since the engine is static, both the flight velocity, $V_{0}\,$ and the flight Mach number, $M\,$are zero
So:
$T_{1}=t_{0}=288.15\ \mathrm {K} \,$
$P_{1}=p_{0}=14.696\ \mathrm {psia} \,$
Intake
$T_{2}=T_{1}=288.15\ \mathrm {K} \,$
$P_{2}=P_{1}\cdot \mathrm {prf} \,$
$P_{2}=14.696*1.0=14.696\ \mathrm {psia} \,$
Compressor
$T_{3}=T_{2}\cdot ((P_{3}/P_{2})^{({\gamma }_{c}-1)/({\gamma }_{c}\cdot {\eta }pc)}=288.15*10^{(1.4-1)/(1.4*0.89)}=603.456\ \mathrm {K}$
$P_{3}=P_{2}\cdot (P_{3}/P_{2})\,$
$P_{3}=14.696*10=146.96\ \mathrm {psia} \,$
Combustor
$T_{4}=\mathrm {RIT} =1400\ \mathrm {K} \,$
$P_{4}=P_{3}\cdot (P_{4}/P_{3})=146.96*0.95=139.612\ \mathrm {psia} \,$
Turbine
$w_{4}\cdot C_{\mathrm {pt} }(T_{4}-T_{5})=w_{2}\cdot C_{\mathrm {pc} }(T_{3}-T_{2})\,$
$100*0.697255*(1400-T_{5})=100*0.6111*(603.456-288.15)\,$
$T_{5}=1123.65419\ \mathrm {K} \,$
$P4/P5=(T4/T5)^{{\gamma }_{t}/(({\gamma }_{t}-1).{\eta }_{\mathrm {pt} })}\,$
$P4/P5=(1400/1123.65419)^{1.333/((1.333-1)*0.9)}\,$
$P4/P5=2.65914769\,$
Jetpipe
$T_{8}=T_{5}=1123.65419\ \mathrm {K} \,$
$P_{5}=P_{4}/(P_{4}/P_{5})\,$
$P_{5}=139.612/2.65914769=52.502537\ \mathrm {psia} \,$
$P_{8}=P_{5}\cdot (P_{8}/P_{5})\,$
$P_{8}=52.502537*0.99=51.9775116\ \mathrm {psia} \,$
Nozzle
$P_{8}/p_{0}=51.9775116/14.696=3.53684755\,$
$(P_{8}/p_{\mathrm {8s} })crit=(({\gamma }_{t}+1)/2))^{{\gamma }_{t}/(({\gamma }_{t}-1)}\,$
$(P_{8}/p_{\mathrm {8s} })crit=((1.333+1)/2)^{1.333/(1.333-1)}=1.85242156\,$
Since $P_{8}/p_{0}>P_{8}/p_{\mathrm {8s} }\,$ , the nozzle is CHOKED
Choked Nozzle
$t_{\mathrm {8s} }=T_{8}/(({\gamma }_{t}+1)/2)\,$
$t_{\mathrm {8s} }=1123.65419/((1.333+1)/2)\,$
$t_{\mathrm {8s} }=963.269773\ \mathrm {K} \,$
$p_{\mathrm {8s} }=P_{8}/((T_{8}/t_{\mathrm {8s} })^{{\gamma }_{t}/({\gamma }_{t}-1)})$
$p_{\mathrm {8s} }=51.9775116/(1123.65419/963.269773))^{1.333/((1.333-1))}\,$
$p_{\mathrm {8s} }=28.059224\ \mathrm {psia} \,$
$V_{8}^{2}=2gJC_{pt}(T_{8}-t_{\mathrm {8s} })\,$
$V_{8}^{2}=2*32.174*550*0.697255*(1123.65419-963.269773)=3957779.09\,$
$V_{8}=3957779.09^{0.5}=1989.41677\ \mathrm {ft} /\mathrm {s} \,$
${\rho }_{\mathrm {8s} }=p_{\mathrm {8s} }/(R\cdot t_{\mathrm {8s} })\,$
${\rho }_{\mathrm {8s} }=(28.059224*144)/(96.034*963.269773)=0.0436782467\ \mathrm {lb} /\mathrm {ft} ^{3}\,$
NOTE: inclusion of 144 in²/ft² to obtain density in lb/ft³.
$A_{8}=w_{8}/({\rho }_{\mathrm {8s} }\cdot V_{8})\,$
$A_{8}=(100*144)/(0.0436782467*1989.41677)=165.718701in^{2}\,$
NOTE: inclusion of 144 in²/ft² to obtain area in in².
Gross Thrust
$F_{g}=C_{\mathrm {x} }((w_{8}\cdot V_{8}/g)+A_{8}(p_{\mathrm {8s} }-p_{0}))\,$
$F_{g}=0.995(((100*1989.41677)/32.174)+(165.718701*(28.059224-14.696)))\,$
$F_{g}=6152.38915+2203.46344\,$
The first term is the momentum thrust which contributes most of the nozzle gross thrust. Because the nozzle is choked (which is the norm on a turbojet), the second term, the pressure thrust, is non-zero.
$F_{g}=8355.85259\ \mathrm {lbf} \,$
Ram Drag
$F_{r}=w_{0}\cdot V_{0}/g\,$
$F_{r}=(100*0)/32.174=0\,$
The ram drag in this particular example is zero, because the engine is stationary and the flight velocity is therefore zero.
Net thrust
$F_{n}=F_{g}-F_{r}\,$
$F_{n}=8355.85259-0=8356\ \mathrm {lbf} \,$
To retain accuracy, only the final answer should be rounded-off.

### Cooling Bleeds

The above calculations assume that the fuel flow added in the combustor completely offsets the bleed air extracted at compressor delivery to cool the turbine system. This is pessimistic, since the bleed air is assumed to be dumped directly overboard (thereby bypassing the propulsion nozzle) and unable to contribute to the thrust of the engine.
In a more sophisticated performance model, the cooling air for the first row of (static) turbine nozzle guide vanes (immeditely downstream of the combustor) can be safely disregarded, since for a given (HP) rotor inlet temperature it has no effect upon either the combustor fuel flow or the net thrust of the engine. However, the turbine rotor cooling air must be included in such a model. The rotor cooling bleed air is extracted from compressor delivery and passes along narrow passage ways before being injected into the base of the rotating blades. The bleed air negotiates a complex set of passageways within the aerofoil extracting heat before being dumped into the gas stream adjacent to the blade surface. In a sophisticated model, the turbine rotor cooling air is assumed to quench the main gas stream emerging from turbine, reducing its temperature, but also increasing its mass flow:
i.e.
$w_{\mathrm {rotorexit} }\cdot C_{\mathrm {pt} }\cdot T_{\mathrm {rotorexit} }=w_{\mathrm {rotorbleed} }\cdot C_{\mathrm {pc} }\cdot T_{\mathrm {rotorbleed} }+w_{\mathrm {rotorentry} }\cdot C_{\mathrm {pt} }\cdot T_{\mathrm {rotorentry} }\,$
$w_{\mathrm {rotorexit} }=w_{\mathrm {rotorbleed} }+w_{\mathrm {rotorentry} }\,$
The bleed air cooling the turbine discs is treated in a similar manner. The usual assumption is that the low energy disc cooling air cannot contribute to the engine cycle until it has passed through one row of blades or vanes.
Naturally any bleed air returned to the cycle (or dumped overboard) must also be deducted from the main air flow at the point it is bled from the compressor. If the some of the cooling air is bled from part way along the compressor (i.e. interstage), the power absorbed by the unit must be adjusted accordingly.

## Cycle improvements

Increasing the design overall pressure ratio of the compression system raises the combustor entry temperature. Therefore, at a fixed fuel flow and airflow, there is an increase in turbine inlet temperature. Although the higher temperature rise across the compression system implies a larger temperature drop over the turbine system, the nozzle temperature is unaffected, because the same amount of heat is being added to the total system. There is, however, a rise in nozzle pressure, because turbine expansion ratio increases more slowly than the overall pressure ratio (which is inferred by the divergence of the constant pressure lines on the TS diagram). Consequently, net thrust increases, implying a specific fuel consumption (fuel flow/net thrust) decrease.
So turbojets can be made more fuel efficient by raising overall pressure ratio and turbine inlet temperature in unison.
However, better turbine materials and/or improved vane/blade cooling are required to cope with increases in both turbine inlet temperature and compressor delivery temperature. Increasing the latter may also require better compressor materials. Also, higher combustion temperatures can potentially lead to greater emissions of nitrogen oxides, associated with acid rain.
Adding a rear stage to the compressor, to raise overall pressure ratio, does not require a shaft speed increase, but it reduces core size and requires a smaller flow size turbine, which is expensive to change.
Alternatively, adding a zero (i.e. front) stage to the compressor, to increase overall pressure ratio, will require an increase in shaft speed (to maintain the same blade tip Mach number on each of the original compressor stages, since the delivery temperature of each of these stages will be higher than datum). The increase in shaft speed raises the centrifugal stresses in both the turbine blade and disc. This together with increases in the hot gas and cooling air(from the compressor) temperatures implies a decrease in component lives and/or an upgrade in component materials. Adding a zero stage also induces more airflow into the engine, thereby increasing net thrust.
If the increase overall pressure ratio is obtained aerodynamically (i.e. without adding stage/s), an increase in shaft speed will still probably be required, which has an impact on blade/disc stresses and component lives/material.

## Other gas turbine engine types

Design point calculations for other gas turbine engine types are similar in format to that given above for a single spool turbojet.
The design point calculation for a two spool turbojet, has two compression calculations; one for the Low Pressure (LP) Compressor, the other for the High Pressure (HP) Compressor. There is also two turbine calculations; one for the HP Turbine, the other for the LP Turbine.
In a two spool unmixed turbofan, the LP Compressor calculation is usually replaced by Fan Inner (i.e. hub) and Fan Outer (i.e. tip) compression calculations. The power absorbed by these two "components" is taken as the load on the LP turbine. After the Fan Outer compression calculation, there is a Bypass Duct pressure loss/Bypass Nozzle expansion calculation. Net thrust is obtained by deducting the intake ram drag from the sum of the Core Nozzle and Bypass Nozzle gross thrusts.
A two spool mixed turbofan design point calculation is very similar to that for an unmixed engine, except the Bypass Nozzle calculation is replaced by a Mixer calculation (where the static pressures of the core and bypass streams at the mixing plane are usually assumed to be equal) followed by a Final (Mixed) Nozzle calculation.

## Nomenclature

* $A\,$ flow area
* $A_{\mathrm {8calc} }\,$ calculated nozzle effective throat area
* $A_{\mathrm {8despt} }\,$ design point nozzle effective throat area
* $A_{\mathrm {8geometricdesign} }\,$ nozzle geometric throat area
* ${\alpha }\,$ shaft angular acceleration
* ${\beta }\,$ arbitrary lines which dissect the corrected speed lines on a compressor characteristic
* $C_{\mathrm {pc} }\,$ specific heat at constant pressure for air
* $C_{\mathrm {pt} }\,$ specific heat at constant pressure for combustion products
* $C_{\mathrm {dcalc} }\,$ calculated nozzle discharge coefficient
* $C_{x}\,$ thrust coefficient
* ${\delta }\,$ ambient pressure/Sea Level ambient pressure
* $({\delta }H/T)_{\mathrm {turb} }\,$ turbine enthalpy drop/inlet temperature
* ${\delta }N\,$ change in mechanical shaft speed
* ${\delta }P_{w}\,$ excess shaft power
* ${\delta }\,{\tau }\,$ excess shaft torque
* ${\eta }_{\mathrm {pc} }\,$ compressor polytropic efficiency
* ${\eta }_{\mathrm {pt} }\,$ turbine polytropic efficiency
* $g\,$ acceleration of gravity
* $F_{g}\,$ gross thrust
* $F_{n}\,$ net thrust
* $F_{r}\,$ ram drag
* ${\gamma }_{\mathrm {c} }\,$ ratio of specific heats for air
* ${\gamma }_{\mathrm {t} }\,$ ratio of specific heats for combustion products
* $I\,$ spool inertia
* $J\,$ mechanical equivalent of heat
* $K\,$ constant
* $K_{1}\,$ constant
* $K_{2}\,$ constant
* $M\,$ flight Mach number
* $N\,$ compressor mechanical shaft speed
* $N_{\mathrm {cor} }\,$ compressor corrected shaft speed
* $N_{\mathrm {turbcor} }\,$ turbine corrected shaft speed
* $p\,$ static pressure
* $P\,$ stagnation (or total) pressure
* $P_{3}/P_{2}\,$ compressor pressure ratio
* $prf\,$ intake pressure recovery factor
* $R\,$ gas constant
* ${\rho }\,$ density
* $SFC\,$ specific fuel consumption
* $RIT\,$ (turbine) rotor inlet temperature
* $t\,$ static temperature or time
* $T\,$ stagnation (or total) temperature
* $T_{1}\,$ intake stagnation temperature
* $T_{3}\,$ compressor delivery total temperature
* ${\theta }\,$ ambient temperature/Sea Level, Standard Day, ambient temperature
* ${\theta }_{T}\,$ total temperature/Sea Level, Standard Day, ambient temperature
* $V\,$ velocity
* $w\,$ mass flow
* $w_{\mathrm {4corcalc} }\,$ calculated turbine entry corrected flow
* $w_{\mathrm {2cor} }\,$ compressor corrected inlet flow
* $w_{\mathrm {4cordespt} }\,$ design point turbine entry corrected flow
* $w_{\mathrm {4corturbchar} }\,$ corrected entry flow from turbine characteristic (or map)
* $w_{\mathrm {fe} }\,$ combustor fuel flow