Everyone knows the general thrust equation:(adsbygoogle = window.adsbygoogle || []).push({});

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

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:

[tex]{{\dot m}_e} - {{\dot m}_i} = {\rho _e}{V_e}{A_e} - {\rho _\infty }{V_\infty }{A_c} = {{\dot m}_f}[/tex]

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:

[tex]{\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}} [/tex]

where,

[tex]{\tau _{xj}} = \mu \left( {\frac{{\partial {u_x}}}{{\partial {x_j}}} + \frac{{\partial {u_j}}}{{\partial {x_x}}}} \right)[/tex]

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

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

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

[tex]\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}[/tex]

[tex]\int {Pd{S_x}} = {P_e}{A_e} - {P_\infty }{A_c} + \int_{{A_s}} P d{S_x}[/tex]

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:

[tex]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}[/tex]

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:

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

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

[tex]{\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}} [/tex]

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

[tex]{\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}} [/tex]

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:

[tex]\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})[/tex]

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

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

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

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

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.

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

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