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Aerospace

Turbine exit temperature from shaft-power balance

Total temperature leaving a turbine that must supply shaft power Pload, including the connected compressor and accessory demand.

Ref: AER-50
Rev: v1
Field: Aerospace
Result: T_t5 [K]

T_{t5} = T_{t4} - \frac{P_{load}}{η_{m} {\dot{m}}_{g} c_{p,g}}

A turbine delivering shaft power P must extract it from the gas. An energy balance gives ηm ṁg cpg (Tt4 − Tt5) = P, so Tt5 = Tt4 − P/(ηm ṁg cpg). The gas flow ṁg includes the fuel added and excludes the bleed taken off. The mechanical efficiency ηm covers bearing and gearbox losses. Check that the exit temperature is above zero and physically sensible.

Total temperature leaving a turbine that must supply shaft power Pload, including the connected compressor and accessory demand. mg is the actual turbine gas flow after fuel addition and bleed, ηm the mechanical efficiency, and cp is constant. Check that the exit temperature is physically feasible and positive.

Calculator

Solve for

T_t5Turbine exit total temperatureComputed

T_t4Turbine inlet total temperature

9002,000

P_loadRequired shaft power

100,0001e7

η_mMechanical efficiency

0.91

ṁ_gTurbine gas mass flow

kg/s

2100

c_p,gSpecific heat of the gas

J/(kg·K)

1,0001,300

Temperature drop, Tt4 − Tt5177.5 K
Gas power extracted, ṁg cp ΔT2,041,000 W
Exit-to-inlet temperature ratio0.8732

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Turbine exit total temperature (K) current value 1,223 K

T_t4Turbine inlet total temperatureKey driver

7231,820

P_loadRequired shaft power

5131,390

ṁ_gTurbine gas mass flow

5131,380

c_p,gSpecific heat of the gas

1,2001,240

η_mMechanical efficiency

1,2101,230

The turbine takes the gas from 1,400 K to 1,222.5 K, a drop of 177.5 K, to deliver 2,000 kW.
The exit is 87.32 % of the inlet temperature on an absolute scale; the same power from half the gas flow would need twice the temperature drop.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (K)	Local step	Change in result
T_t4 Turbine inlet total temperature	900 to 2,000 K	722.5	1,823	1,100	+10%	+11.5 %
P_load Required shaft power	100,000 to 1e7 W	512.7	1,391	878.4	+10%	−1.45 %
ṁ_g Turbine gas mass flow	2 to 100 kg/s	512.7	1,382	869.6	+10%	+1.32 %
c_p,g Specific heat of the gas	1,000 to 1,300 J/(kg·K)	1,196	1,243	47.1	+10%	+1.32 %
η_m Mechanical efficiency	0.9 to 1	1,207	1,226	19.32	+10%	+1.32 %

Sweep of Required shaft power

Required shaft power (W)	Turbine exit total temperature (K)
100,000	1,391
1,338,000	1,281
2,575,000	1,172
3,813,000	1,062
5,050,000	951.9
6,288,000	842.1
7,525,000	732.3
8,763,000	622.5
1e7	512.7

Sweep one variable

Vary

Turbine exit total temperature (K)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

T_t5

Turbine exit total temperature

Result · K

What the equation gives you. Everything else feeds it.

T_t4

Turbine inlet total temperature

K · typical 900 to 2,000

The turbine inlet total temperature, in K. Raises the exit temperature one-for-one. Raising it by 10 % (to 1,540) moves the result by +11.5 %, from 1,223 to 1,363 K.

P_load

Required shaft power

W · typical 100,000 to 1e7

The required shaft power, including the compressor and accessory demand, in W. Lowers the exit temperature in direct proportion to P. Raising it by 10 % (to 2,200,000) moves the result by −1.45 %, from 1,223 to 1,205 K.

η_m

Mechanical efficiency

· typical 0.9 to 1

The mechanical efficiency. A lower efficiency means more gas power is needed, lowering the exit temperature. Raising it by 10 % (to 1.078) moves the result by +1.32 %, from 1,223 to 1,239 K.

ṁ_g

Turbine gas mass flow

kg/s · typical 2 to 100

The actual gas mass flow through the turbine, in kg/s. A larger flow spreads the work, so a smaller temperature drop. Raising it by 10 % (to 11) moves the result by +1.32 %, from 1,223 to 1,239 K.

c_p,g

Specific heat of the gas

J/(kg·K) · typical 1,000 to 1,300

The specific heat of the gas, in J/(kg·K). A larger specific heat gives a smaller temperature drop for the same power. Raising it by 10 % (to 1,265) moves the result by +1.32 %, from 1,223 to 1,239 K.

Worked example with the values above

Gas power needed: P/ηm = 2,000,000 / 0.98 = 2,040,816 W.
Temperature drop: 2,040,816 / (10 × 1,150) = 177.462 K.
Tt5 = 1,400 − 177.462 = 1,222.54 K.

Assumptions and limits

Constant specific heats in each stream, adiabatic components and total-to-total efficiencies.
Steady operation at one operating point, with no bleed and no leakage unless the page says so.

Common mistakes

Mixing total and static states.
Using a total-to-static efficiency where a total-to-total one is needed.
Forgetting the fuel added to the turbine mass flow.

Statistics and uncertainty

Constant-cp cycle calculations are rough: cp rises from 1005 J/(kg·K) in cold air to about 1150 J/(kg·K) in hot gas, which shifts results by 5 % to 10 %. Component efficiencies from maps are good to 1 % to 2 % at design and worse off design.

Where it is used

Matching compressors, combustors and turbines and finding fuel flow and power.
Sizing engines and checking cycle designs.

Maths you need

Isentropic efficiency: ideal work divided by actual work (or the reverse for a turbine)
Total temperature: the temperature the gas would reach at rest
Brayton cycle: compression, heat addition, expansion

Related formulas

Practise this formula