Force produced by flow in aircraft engine

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1. May 27, 2017

lampCable

1. The problem statement, all variables and given/known data
Consider an aircraft turbine as in the figure below. In the reference frame of the turbine, air of density $\rho_0$ comes into the turbine with speed $U_0$ through the entrance with cross sectional area $S_0$. In the combustion zone, air is mixed with the fuel and burned (contribution of the fuel to the gas mass may be neglected). The density of the burned gas is $\Theta$ times smaller than the initial air density (typically $\Theta \gg 1$). The burned gas leaves the turbine through the exit with cross sectional area $S_1$. Find the thrust (the force produced by the flow on the turbine). Assume for simplicity that the flow is stationary, uniform over any turbine cross-section and incompressible, and that the gas density changes only in the combustion zone. Disregard pressure modifications in the combustion zone.

2. Relevant equations
(1)
Mass conservation:
$$\rho_0 U_0S_0 = \frac{\rho_0}{\Theta} U_1S_1.$$

(2) Bernoulli equation along a streamline:
$$P_0 + \frac{1}{2}\rho_0U_0^2 + \rho_0 gh_0= P_1 + \frac{1}{2}\rho_1U_1^2 + \rho_1 gh_1.$$

(3) Momentum conservation (stationary flow):
$$0 = -\oint _S\rho\mathbf{u}(\mathbf{u}\cdot d\mathbf{S}) + \mathbf{F}.$$

3. The attempt at a solution
We begin by splitting the force term into two parts $\mathbf{F} = \mathbf{F}_p + \mathbf{F}_{ByWall}$, where $\mathbf{F}_p$ is the pressure force on the fluid at the openings and $\mathbf{F}_{ByWall}$ is the pressure force on the fluid from the turbine (walls and fans). We use Newton's third law to obtain the thrust as $\mathbf{F}_{thrust} = -\mathbf{F}_{ByWall}$. Using (3) with $\hat{\mathbf{x}}$ to the right in the figure, we get

$$\mathbf{F}_{thrust} = -\oint _S\rho\mathbf{u}(\mathbf{u}\cdot d\mathbf{S}) + \mathbf{F}_p = (\rho_0 U_0^2 S_0 - \frac{\rho_0}{\Theta}U_1^2 S_1)\hat{\mathbf{x}} + (P_{atm}S_0-P_1S_1)\hat{\mathbf{x}}.$$

Using now (1) and (2) with $h=0$ we find that

$$P_{atm} + \frac{1}{2}\rho_0 U_0^2 = P_1 + \frac{1}{2}\frac{\rho_0}{\Theta} U_1^2 \Rightarrow P_1 = P_{atm} + \frac{1}{2}\rho_0 U_0^2\Big(1-\frac{S_0^2\Theta}{S_1^2}\Big).$$

Thus,

$$\mathbf{F}_{thrust} = (\rho_0 U_0^2 S_0 - \frac{\rho_0}{\Theta}U_1^2 S_1)\hat{\mathbf{x}} + (P_{atm}S_0-(P_{atm} + \frac{1}{2}\rho_0 U_0^2\Big(1-\frac{S_0^2\Theta}{S_1^2}\Big))S_1)\hat{\mathbf{x}} = (\rho_0 U_0^2 S_0 - \frac{\rho_0}{\Theta}U_1^2 S_1 - \frac{1}{2}\rho_0 U_0^2\Big(1-\frac{S_0^2\Theta}{S_1^2}\Big)S_1)\hat{\mathbf{x}} + P_{atm}(S_0-S_1)\hat{\mathbf{x}}.$$

Now, there are quite alot of terms here and it can be simplified a bit, but what I don't understand is the last term in the expression for the thrust, $P_{atm}(S_0-S_1)$, which comes from considering the pressure forces on the fluid at the entrance and the exit. If we put the velocities to zero, we get thrust from nothing, which doesn't make sense. Is the way I split the force $\mathbf{F}$ incorrect?

Last edited: May 27, 2017
2. May 27, 2017

Nidum

I can't understand the problem statement . It appears to be mostly gibberish .

There are in any case some quite simple standard methods for estimating engine thrust .

Last edited: May 27, 2017