I came up with this question when looking through the answer for this question, which states that a convergent duct(obstruction) compresses the exhaust so that its pressure matches the ambient pressure. In the comments section another question was answered like this:

Q:Doesn't the pressure decrease and the air accelerate when the diameter decreases ?

A:The pressure decrease happens in the obstruction and after it - it's increased by the obstruction, between the fan and the nozzle exit.

I am a bit confused by what's meant by "pressure decrease in the obstruction" and "pressure increased by the obstruction". What really is the effect of converging the duct, increase exhaust speed to decrease static pressure or compressing exhaust to increase static pressure?


2 Answers 2


It depends where in the system you look. Think a narrowing in a pipe.

That which moves the fluid must work harder to maintain mass flow. Pressure in the pipe before the narrowing increases. Velocity and pressure in the narrowing increase and decrease respectively.

The issue was that the apparatus in question gives less thrust per unit energy input than a free propeller under the conditions tested.


For a convergent nozzle the exit area will not change the jet pressure or velocity if the fan pressure ratio is the same. The major effect will be on the mass flow; a smaller nozzle area will let less air pass through.

For a complete system the fan pressure ratio and adiabatic efficiency will change with mass flow (the fan compressor map shows you the curves), so if we immagine to keep the same RPM of the fan and gradually shrink the nozzle aera, the pressure ratio will increase and so will the exit velocity.

A consideration on the thrust: Remember that thrust is, as a fist order approximation, the product of mass flow through the fan times the difference between velocity at the exit of the nozzle and free stream velocity (how fast the propeller is moving in the air). So reducing fan nozzle area will slightly increase the speed but greatly reduce mass flow

------ EXAMPLE ------

Here I will write a simple example for clarification

Before starting I want to remind that Static Pressure is what it's commonly known as pressure while Total Pressure is the pressure that the fluid would have if you brought it to a complete stop.

Let's imagine a ducted Fan moving at a speed of Mach 0.25 with a exit area of 0.01 m^2 at sea level condition, with a compression ratio of 1.3 and adiabatic efficiency of 0.6:

$M_0 = 0.25\;;\; A_{exit} = 0.01\: m^2 \;;\; T_0 = 228\: K \;;\; p_0 = 103205\: Pa\;;\; \pi_{fan} = 1.3 \;;\; \eta_{fan} = 0.6 $

For air ideal gas constant are $R = 287\; \frac{J}{kg\; K} \;;\; \gamma = 1.4 $

From the Mach number and pressure we can calculate the total pressure and temperature for free stream condition:

$p_{t0} = p_0 (1 + \frac{\gamma -1}{2} M_0^2)^{\frac{\gamma}{\gamma-1}} = 107791 \: Pa \\ T_{t0} = T_0 (1 + \frac{\gamma -1}{2} M_0^2) = 291.6 \: K$

From the fan pressure ratio and efficiency we can compute total pressure and temperature behind it

$p_{t2} = \pi_{fan}\cdot p_{t0} = 140128 \: Pa \\ T_{t2,iso} = \pi_{fan}^{\frac{\gamma-1}{\gamma}}\cdot T_{t0} = 314.3 \: K \\ T_{t2} = \frac{T_{t2,iso} - T_{t0}}{\eta_{fan}} + T_{t0} = 329.4 \: K$

Now we let expand the fluid up to atmospheric conditions

$p_{4} = p_{0}\\ T_{4} = (\frac{p_{t0}}{p_{t2}})^{\frac{\gamma-1}{\gamma}}\cdot T_{t2} = 301.8 \: K\\ M_4 = \sqrt{\frac{2}{\gamma -1} (\frac{T_{t2}}{T_{4}} -1)} = 0.676$

The speed of sound at the exit nozzle and jet velocity are

$a_4 = \sqrt{T_{4}\gamma R} = 348.2 \: m/s \\ V_4 = a_4\cdot M_4 = 235 \: m/s$

the mass flow of the fan is

$\dot{m} = \rho_4 \cdot V_4 \cdot A_{exit}\\ \rho_4 = \frac{p_4}{R T_4} = 1.191 \: kg/m^3 \\ \dot{m} = 2.799 \:kg/s$

The thrust of this fan is

$Thrust = \dot{m}(V_4 - V_0)\\ V_0 = \sqrt{T_{0}\gamma R}\cdot M_0 = 85 \:m/s \\ Thrust = 419.8 \: N$

You can see how only the mass flow depends on the exit area, if we half the area, the mass flow and thrust are halved. In reality the parameter $\pi_{fan}$ would also change, maybe going from $1.3$ to $1.45$. If $\pi_{fan}$ increases $V_4$ also increases, but it isn't enough to compensate for the reduction in mass flow.

  • $\begingroup$ Sorry it’s a bit hard for me to understand. A converging nozzle should keep the mass flow rate constant by increasing flow speed, and with the mass flow rate constant, this speed increase should also increase the velocity difference, thus increasing thrust. Is there anything wrong with my understanding? $\endgroup$
    – Frank
    Commented Jun 19, 2023 at 11:12
  • $\begingroup$ What you say is true only for very small variation of the exit area (5% max). I added an example for more clarity. $\endgroup$ Commented Jun 20, 2023 at 15:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .