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Aerospace

Corrected engine mass flow

Mass flow that the engine would pass at reference inlet conditions with the same Mach number in the compressor, from the actual flow, total temperature and total pressure at the inlet.

Ref: AER-36
Rev: v1
Field: Aerospace
Result: ṁ_corr [kg/s]

{\dot{m}}_{corr} = \dot{m} \frac{\sqrt{T_{t} / T_{ref}}}{p_{t} / p_{ref}}

Engine maps are drawn in corrected variables so that one map covers all inlet conditions. The corrected mass flow is the flow the engine would pass if the inlet were at the reference state with the same Mach numbers: ṁcorr = ṁ√(Tt/Tref)/(pt/pref). A hotter inlet raises the corrected flow and a higher inlet pressure lowers it. Use the same reference state as the map.

Mass flow that the engine would pass at reference inlet conditions with the same Mach number in the compressor, from the actual flow, total temperature and total pressure at the inlet. Maps are only comparable when the reference state is the same.

Calculator

Solve for

ṁ_corrCorrected mass flowComputed

kg/s

ṁActual mass flow

kg/s

1500

T_tInlet total temperature

200500

T_refReference temperature

280300

p_tInlet total pressure

20,000300,000

p_refReference pressure

95,000105,000

Temperature factor, √(Tt/Tref)1.054
Pressure factor, pt/pref0.8
Change from the actual flow31.76 %

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.

Corrected mass flow (kg/s) current value 26.35 kg/s

ṁActual mass flowKey driver

1.32659

p_tInlet total pressure

7.03105

T_tInlet total temperature

20.832.9

p_refReference pressure

2527.7

T_refReference temperature

25.826.7

The engine at the inlet conditions passes 20 kg/s, equivalent to 26.352 kg/s on the map, a change of 31.76 %.
The low inlet pressure raises the corrected flow and the high temperature raises it further, so the operating point sits higher on the map than the raw flow suggests.

Chart data as tables

Ranked variables

Variable	Typical range	Lowest result	Highest result	Swing (kg/s)	Local step	Change in result
ṁ Actual mass flow	1 to 500 kg/s	1.318	658.8	657.5	+10%	+10 %
p_t Inlet total pressure	20,000 to 300,000 Pa	7.027	105.4	98.38	+10%	−9.09 %
T_t Inlet total temperature	200 to 500 K	20.83	32.94	12.11	+10%	+4.88 %
p_ref Reference pressure	95,000 to 105,000 Pa	25.03	27.67	2.635	+10%	+10 %
T_ref Reference temperature	280 to 300 K	25.82	26.73	0.9062	+10%	−4.65 %

Sweep of Inlet total pressure

Inlet total pressure (Pa)	Corrected mass flow (kg/s)
20,000	105.4
55,000	38.33
90,000	23.42
125,000	16.87
160,000	13.18
195,000	10.81
230,000	9.166
265,000	7.955
300,000	7.027

Sweep one variable

Vary

Corrected mass flow (kg/s)Hover, touch, or use the arrow keys on the chart

Variables and how each one contributes

ṁ_corr

Corrected mass flow

Result · kg/s

What the equation gives you. Everything else feeds it.

ṁ

Actual mass flow

kg/s · typical 1 to 500

The actual inlet mass flow, in kg/s. Scales the corrected flow in direct proportion. Raising it by 10 % (to 22) moves the result by +10 %, from 26.35 to 28.99 kg/s.

T_t

Inlet total temperature

K · typical 200 to 500

The inlet total temperature, in K. Raises the corrected flow with the square root of the temperature. Raising it by 10 % (to 352) moves the result by +4.88 %, from 26.35 to 27.64 kg/s.

T_ref

Reference temperature

K · typical 280 to 300

The map reference temperature, in K. Lowers the corrected flow with the inverse square root. Raising it by 10 % (to 316.8) moves the result by −4.65 %, from 26.35 to 25.13 kg/s.

p_t

Inlet total pressure

Pa · typical 20,000 to 300,000

The inlet total pressure, in Pa. Lowers the corrected flow in inverse proportion. Raising it by 10 % (to 88,000) moves the result by −9.09 %, from 26.35 to 23.96 kg/s.

p_ref

Reference pressure

Pa · typical 95,000 to 105,000

The map reference pressure, in Pa. Raises the corrected flow in direct proportion. Raising it by 10 % (to 110,000) moves the result by +10 %, from 26.35 to 28.99 kg/s.

Worked example with the values above

Temperature factor: √(320/288) = 1.05409.
Pressure factor: pt/pref = 80,000/100,000 = 0.8.
ṁcorr = 20 × 1.05409 / 0.8 = 26.3523 kg/s.

Assumptions and limits

Steady, adiabatic, reversible flow of a calorically perfect gas, with no shocks or friction.
A constant heat-capacity ratio, and quasi-one-dimensional flow where an area enters.

Common mistakes

Using the subsonic branch of the area–Mach relation for a supersonic nozzle, or the reverse.
Entering static temperature where the formula needs stagnation temperature.
Applying the area-velocity relation at M = 1, where it is singular.

Statistics and uncertainty

Isentropic relations are exact for a perfect gas. Real nozzle flow departs from them: boundary layers and non-equilibrium chemistry make mass flow a few per cent lower, and for hot combustion gas γ is a function of temperature, not a constant.

Where it is used

Sizing nozzles, inlets and diffusers.
Referring engine data to standard conditions for comparison.

Maths you need

Stagnation conditions: the state reached if the flow is slowed to rest isentropically
Choking: the flow is sonic at the minimum area and the mass flow is maximal
Mach number: M = V/a

Related formulas

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