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Aerospace

Exit speed with a nozzle efficiency

Exhaust speed from a nozzle with efficiency ηn, the ratio of actual to ideal isentropic kinetic energy for the same inlet total state and exit static pressure pe.

Ref
AER-46
Rev
v1
Field
Aerospace
Result
Ve [m/s]

The ideal exhaust speed from isentropic expansion to pe is √{2cpTt[1 − (pe/pt)^((γ−1)/γ)]}. A nozzle efficiency ηn is the ratio of the actual to the ideal exit kinetic energy for the same inlet state and exit static pressure, so Ve = √(ηn) times the ideal speed. Real nozzles have ηn of 0.92 to 0.99. The exit pressure must be one that the nozzle can really reach.

Exhaust speed from a nozzle with efficiency ηn, the ratio of actual to ideal isentropic kinetic energy for the same inlet total state and exit static pressure pe. It needs 0 < pe ≤ pt, 0 < ηn ≤ 1 and constant cp.

Calculator

Solve for
m/s
0.81
J/(kg·K)
9001,300
K
2503,000
1.11.67
Pa
20,000200,000
Pa
200,0001,000,000
  • Ideal exit speed (ηn = 1)465.4 m/s
  • Exit kinetic energy per kg, ½Ve²102,900 J/kg
  • Pressure ratio pe/pt0.5

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.

Exit speed (m/s) current value 453.6 m/s
peExit static pressureKey driver
0743
TtInlet total temperature
2931,010
ptInlet total pressure
743
γHeat-capacity ratio
264527
cpSpecific heat at constant pressure
429516
ηnNozzle efficiency
416465
  • The gas leaves at 453.59 m/s, against an ideal speed of 465.37 m/s.
  • The expansion from 200 kPa to 100 kPa converts 17.07 % of the total enthalpy cpTt into kinetic energy.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (m/s)Local stepChange in result
pe Exit static pressure20,000 to 200,000 Pa0743743+10%−6.51 %
Tt Inlet total temperature250 to 3,000 K292.81,014721.5+10%+4.88 %
pt Inlet total pressure200,000 to 1,000,000 Pa453.6743289.4+10%+5.96 %
γ Heat-capacity ratio1.1 to 1.67 264.4527.3262.8+10%+9.59 %
cp Specific heat at constant pressure900 to 1,300 J/(kg·K)429.351686.66+10%+4.88 %
ηn Nozzle efficiency0.8 to 1 416.2465.449.13+10%+4.88 %

Sweep of Exit static pressure

Exit static pressure (Pa)Exit speed (m/s)
20,000743
42,500639.9
65,000560.8
87,500490.8
110,000424
132,500356.5
155,000283.6
177,500195.9
200,0000

Sweep one variable

Exit speed (m/s)Hover, touch, or use the arrow keys on the chart
0200400600800 50k100k150k200k Exit static pressure (Pa) now 453.6

Variables and how each one contributes

Ve

Exit speed

Result · m/s

What the equation gives you. Everything else feeds it.

ηn

Nozzle efficiency

· typical 0.8 to 1

The nozzle efficiency, actual over ideal exit kinetic energy. Raises the exit speed with its square root. Raising it by 10 % (to 1.045) moves the result by +4.88 %, from 453.6 to 475.7 m/s.

cp

Specific heat at constant pressure

J/(kg·K) · typical 900 to 1,300

The specific heat at constant pressure, in J/(kg·K). Raises the exit speed with its square root. Raising it by 10 % (to 1,105) moves the result by +4.88 %, from 453.6 to 475.7 m/s.

Tt

Inlet total temperature

K · typical 250 to 3,000

The inlet total temperature, in K. Raises the exit speed with its square root. Raising it by 10 % (to 660) moves the result by +4.88 %, from 453.6 to 475.7 m/s.

γ

Heat-capacity ratio

· typical 1.1 to 1.67

The heat-capacity ratio cp/cv of the gas, about 1.4 for air. A lower γ (hotter, more complex gas) changes the compressible-flow exponents and the sizes of the shock and expansion effects. Raising it by 10 % (to 1.54) moves the result by +9.59 %, from 453.6 to 497.1 m/s.

pe

Exit static pressure

Pa · typical 20,000 to 200,000

The exit static pressure, in Pa. A lower exit pressure (more expansion) raises the exit speed, with diminishing returns. Raising it by 10 % (to 110,000) moves the result by −6.51 %, from 453.6 to 424 m/s.

pt

Inlet total pressure

Pa · typical 200,000 to 1,000,000

The inlet total pressure, in Pa. A higher total pressure for the same exit pressure raises the exit speed. Raising it by 10 % (to 220,000) moves the result by +5.96 %, from 453.6 to 480.6 m/s.

Worked example with the values above

  1. Pressure term: 1 − (pe/pt)^((γ−1)/γ) = 1 − 0.5^0.28571 = 0.179665.
  2. Energy: 2ηn cp Tt × term = 2 × 0.95 × 1,005 × 600 × 0.179665 = 205,739.4 m²/s².
  3. Ve = 453.585 m/s.

Assumptions and limits

  • Steady, adiabatic flow with constant γ and cp, and a nozzle that is choked where the entry says so.
  • Uniform properties across the throat and exit plane, unless a survey is used.

Common mistakes

  • Using static instead of total pressure upstream.
  • Using the choked formula when the pressure ratio is below the critical value.
  • Forgetting that a convergent-divergent nozzle has different regimes from a convergent one.

Statistics and uncertainty

Ideal nozzle formulas overestimate real performance: discharge coefficients of 0.95 to 0.99 lower mass flow by 1 % to 5 %, and nozzle efficiency of 0.9 to 0.98 lowers exit speed by 1 % to 5 %. Total conditions are the key measurement, and a 1 % error in total pressure gives 1 % in mass flow.

Where it is used

  • Sizing throats and predicting thrust of jet and rocket nozzles.
  • Estimating mass flow through orifices, valves and venturis.

Maths you need

  • Critical pressure ratio: (2/(γ+1))^(γ/(γ−1)), about 0.528 for air
  • Momentum flux: ρu²A, the momentum carried by a stream
  • Nozzle efficiency: actual divided by ideal kinetic energy

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