# Aerospace Finding The Hidden Inviscid Approximation In The General Thrust Equation

1. Sep 9, 2012

### Compressible

Everyone knows the general thrust equation:

$$T = {{\dot m}_i}\left[ {(1 + f){V_e} - {V_\infty }} \right] + ({P_e} - {P_\infty }){A_e}$$

Where mdot_i is the incoming mass flow rate, f is the fuel flow rate, and the subscripts ∞ and e represent free-stream and exit conditions, respectively.

For the past few years, every time that I have derived this equation, it has been through an inviscid approximation where shear stress has been set to zero everywhere. Recently, however, I was presented with a different method of deriving this equation that does not make the inviscid approximation, yet they managed to derive the final inviscid thrust equation through manipulation of the control volume. This confused me a bit because viscosity shouldn't (as far as my knowledge goes) be a function of the control volume, i.e. if there is shear stress in the flow, it should still be represented in the governing equations. My guess then is that the inviscid approximation is hidden somewhere in the derivation, but unfortunately, I have no been able to find it. I'm hoping that you guys can help me find where this approximation lies. In the following paragraph, I will briefly go through their method of derivation.

For the following control volume:

It can be shown that the fuel mass flow rate equals:

$${{\dot m}_e} - {{\dot m}_i} = {\rho _e}{V_e}{A_e} - {\rho _\infty }{V_\infty }{A_c} = {{\dot m}_f}$$

Where the integral over control surface As was set to zero due to the velocity being perpendicular to the normal vector. Next, applying the x-momentum equation across the control volume gives the following:

$${\rho _e}{V_e}^2{A_e} - {\rho _\infty }{V_\infty }^2{A_c} = T + \int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}}$$

where,

$${\tau _{xj}} = \mu \left( {\frac{{\partial {u_x}}}{{\partial {x_j}}} + \frac{{\partial {u_j}}}{{\partial {x_x}}}} \right)$$

Note that the above is given in index notation. Next, the mass flow and momentum equation can be combined to derive the following:

$${{\dot m}_i}\left[ {(1 + f){V_e} - {V_\infty }} \right] = T + \int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}}$$

The shear stress and pressure integral can be broken up in the following way:

$$\int {{\tau _{xj}}} d{S_j} = \int_{{A_c}} {{\tau _{xj}}} d{S_j} + \int_{{A_e}} {{\tau _{xj}}} d{S_j} + \int_{{A_S}} {{\tau _{xj}}} d{S_j} = \int_{{A_S}} {{\tau _{xj}}} d{S_j}$$

$$\int {Pd{S_x}} = {P_e}{A_e} - {P_\infty }{A_c} + \int_{{A_s}} P d{S_x}$$

Where shear stress across Ac and Ae where set to zero due to the fact that there is no velocity gradients on those two surfaces. Those equations can be combined with the equation before them to get:

$$T + \int_{{A_S}} {{\tau _{xj}}} d{S_j} - \int_{{A_s}} P d{S_x} = {{\dot m}_i}\left[ {(1 + f){V_e} - {V_\infty }} \right] + {P_e}{A_e} - {P_\infty }{A_c}$$

Now here is the part where they use CV manipulation to get rid of the shear stress. Utilizing the following control volume:

Where A1 excludes Ac and A2 excluded Ae. A3 is at R->∞. Noting that there will be a small radial component of velocity at A3 in order to balance the inflow and outflow mass flux of the control volume. Applying the continuity equation over the control volume gives the following for the mass flow across A3:

$$\int\limits_{A3} {\rho (u \cdot dS) = - {\rho _\infty }} {V_\infty }({A_2} - {A_1}) = - {\rho _\infty }{V_\infty }({A_c} - {A_e})$$

Noting that there is no internal force, T, the momentum equation can be applied to get:

$${\rho _\infty }{V_\infty }^2({A_2} - {A_1}) + \int\limits_{A3} {\rho (u \cdot dS)} = \int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}}$$

where plugging in the mass flow across A3 into the momentum equation gives:

$${\rho _\infty }{V_\infty }^2({A_c} - {A_e}) - {\rho _\infty }{V_\infty }^2({A_c} - {A_e}) = 0 = \int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}}$$

Evaluating the integrals for shear stress and pressure across all control surfaces, noting that the shear stress across A1, A2, and A3 equals zero because of zero velocity gradients and the pressure across A3 equals zero because it does not act in the x-direction, we can arrive at the following equation:

$$\int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}} = \int_{{A_s}} {{\tau _{xj}}} d{S_j} - \int_{{A_s}} P d{S_x} = {P_\infty }({A_c} - {A_e})$$

Noting that the surface As on this control volume is the opposite of the previous CV, we get the following for the first control volume:

$$\int_{{A_s}} {{\tau _{xj}}} d{S_j} - \int_{{A_s}} P d{S_x} = {P_\infty }({A_e} - {A_c})$$

Finally, combining the last equation with the final derived equation of the first control volume we can get the thrust equation:

$$T = {{\dot m}_i}\left[ {(1 + f){V_e} - {V_\infty }} \right] + ({P_e} - {P_\infty }){A_e}$$

I've checked this derivation several times and it seems to be correct, but no matter what I manipulate, I can't see where the inviscid approximation is (unless viscosity is a function of control volume which doesn't make sense to me). Sorry for the extremely long derivation, but I had to do it to make my point. I'd appreciate any suggestions.

2. Sep 9, 2012

### Compressible

Since there hasn't been any replies so far, I'll offer some of my thoughts to get the discussion going. Note that this isn't for any homework assignment; it's more for the sake of bettering my (and perhaps some of you guys') understanding of propulsion and thermodynamics.

We all know that control volume analysis are generally done in order to avoid analyzing the complex behavior of the internal mechanics of the system. In this way, we can avoid having to worry about the internal performance of the system and only worry about what comes in and what goes out of our control volume. In this case, the majority of the internal inefficiencies can be summed up in the value of the exit velocities; i.e. the viscosity and inefficient performance of the compressor, combustor, turbine, and other components would negatively affect the exit velocity. So essentially, the thrust equation isn't exactly "inviscid".

The other drag characteristics that are hidden in the equation is the ram drag given by mdot_i*V_inf, but I believe that has more to do with the pressure surface forces than the viscous surface forces. Another assumption in the derivation that may imply inviscid behavior, though it does not necessarily have to, is the assumption that the inflow and outflow are uniform across the control surface area. Real engines obviously do not have such uniform velocity distributions and would suffer from some shear stress as a result of it. But again, I do not think (though I have not mathematically worked it out yet) that a uniform velocity distribution necessarily has to imply that there is inviscid internal flow.

What I'm having trouble with the most is that the shear stress is cancelling out across the stream tube (As control surface). It seems that through the use of the two control volumes, the shear stress contribution is being replaced with an equivalent pressure term, but this does not make sense to me on a fundamental level. I'm sure there's something I've missed here, so I'm open to any suggestions!

3. Sep 9, 2012

As far as I can tell, you made your inviscid assumption here:
$$\int {{\tau _{xj}}} d{S_j} - \int {Pd{S_x}} = \int_{{A_s}} {{\tau _{xj}}} d{S_j} - \int_{{A_s}} P d{S_x} = {P_\infty }({A_c} - {A_e})$$

Somehow your shear stress integral over As was assumed zero. You mentioned your reasoning for doing so at A3 but not As.

The curved streamlines forming As in your original control voluem seem a bit odd to me. That would necessarily imply that the flow is expanding as it heads toward your nozzle (I assume that is a nozzle). The only way that would make sense is if your inflow conditions at Ac are at higher pressure than the surroundings coming in through A1. That would imply a velocity difference and therefore a shear layer, meaning your assumption of zero shear stress on the As boundary is necessarily an inviscid approximation.

Your uniform velocity at your exit plane A2 is suspect as well given the local surface curvature in your outer control volume, though it would be difficult to say just how large an effect that may be since you allow for outflow through A3 (and therefore your second control volume is not a streamtube).

4. Sep 9, 2012

### Compressible

In my attempt at reducing the overall post length, I skipped a lot of the intermediate steps in the derivation. If you look closely, the shear stress across As is not assumed to be zero. The equation simply states that the integral of shear stress over the entire control volume reduces down to only the integral across As. Similarly, the pressure integral reduces down to the integral over As and some pressure terms from A1 and A2 which have been moved to the RHS of the equation listed above.

The shape of the curved streamline, As, is arbitrary. The only thing that is assumed is that the flow velocity is always perpendicular to the surface normal vector, which is a valid assumption as long as you allow for the surface geometry to be arbitrary. The arbitrariness of the control surface does not affect the derivation in any way because the integrals of pressure and shear stress over As are never directly calculated. The derivation uses the convenience of the two control volumes to essentially cancel out the shear stress. As I said earlier, this is the part that bothers me the most. It almost seems as if the shear stress was replaced by an equivalent pressure term and that does not make sense to me.

The uniform velocity is an assumption that is generally made in all thrust equation derivations, and although I agree with you on the fact that it will contribute slightly to the shear stress, I do not think it is a major issue.

I'm not quite sure what you mean by saying that the second control volume is not a streamtube. There has to be a mass flow through A3 in order to ensure that the net flux is zero, i.e. conservation laws have to be satisfied. The only way A3 would not have a mass flow through it is when Ae = Ac (A1 = A2), but since this is being derived in the most general case, then A1 =/= A2. Regardless, mass flow through A3 should not have any effect on whether or not As is a streamtube because it is assumed that A3 is located at R->∞, so essentially it is the flow in the farfield.

5. Sep 9, 2012