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#### **Appendix A. Derivation of the Blowing Correction**

Close to the wall, *u* ≈ 0 and *∂u*/*∂x* ≈ 0, so that continuity becomes *ρv* = *ρ*w*v*w and, in the absence of an axial pressure gradient, the momentum equation can be written as

$$
\pi = \pi\_W + \rho\_W v\_W u. \tag{A1}
$$

Substitution of *B* = <sup>2</sup>(*ρ*w*v*w)/(*ρ*e*u*e*Cf*) and *τ*w = 12*Cf <sup>ρ</sup>*e*<sup>u</sup>*2e into Equation (A1) leads to

$$
\tau = \tau\_W \left( 1 + B\phi \right),
\tag{A2}
$$

where *φ* ≡ *<sup>u</sup>*/*<sup>u</sup>*e. Next, we equate Equation (A2) to the definition of shear stress for a turbulent boundary layer; namely,

$$
\pi = \left(\mu + \rho\epsilon\right) \frac{\partial u}{\partial y'} \tag{A3}
$$

where  represents the turbulent eddy diffusivity. This produces

$$
\pi\_W \left( 1 + B\phi \right) = \left( \mu + \rho \epsilon \right) \frac{\partial u}{\partial y}. \tag{A4}
$$

At this juncture, switching to non-dimensional forms may be achieved using Re*δ* ≡ *ρ*e*u*e*δ*/*μ*, *η* = *y*/*δ*, *φ*, and *τ*w. We ge<sup>t</sup>

$$\frac{1}{2}\mathcal{C}\_f \left(1 + B\phi\right) = \text{Re}\_\delta^{-1} \left(1 + \frac{\rho\epsilon}{\mu}\right) \frac{\partial\phi}{\partial\eta}.\tag{A5}$$

According to Prandtl's mixing length concept,  ∝ *<sup>η</sup>*<sup>2</sup>*∂φ*/*∂η*, where, for the purpose of estimating , a power law profile (such as *φ* = *η<sup>n</sup>*) may be used for the range of Reynolds numbers typical of hybrid motors [15]. Then, *∂φ*/*∂η* may be approximated as *nη<sup>n</sup>*−1. Furthermore, since *n* is usually small (∼ 1/7), the eddy diffusivity becomes

$$
\epsilon \approx c \eta^{1+n} \approx c \eta,\tag{A6}
$$

where *c* represents a constant that is proportional to the mixing length—its exact value proves immaterial to this analysis. Substituting Equation (A6) into Equation (A5) yields

$$\frac{1}{2}\mathcal{C}\_f \left(1 + B\phi\right) = \text{Re}\_\delta^{-1} \left(1 + \frac{\rho}{\mu} c\eta\right) \frac{\partial\phi}{\partial\eta}.\tag{A7}$$

Note that integration from the wall to the boundary layer edge (i.e., from 0 to 1 for both *η* and *φ*), leads to an expression for the skin friction coefficient as a sole function of Re*δ* and *B*:

$$\frac{1}{2}\mathbb{C}\_f = \lg\left(\text{Re}\_\delta\right) \frac{\ln\left(1+B\right)}{B}.\tag{A8}$$

Marxman [7] argued that, since the Reynolds number and blowing parameter dependencies can be distinctly separated in the expression for *Cf* , the functional form of *g* (Re*δ*) may be obtained by comparing Equation (A8) to known results in the absence of blowing; namely,

$$\frac{1}{2}\mathcal{C}\_{f\_0} = g \left( \mathrm{Re}\_{\delta} \right) = 0.0225 \mathrm{Re}\_{\delta}^{-0.25} \text{\textdegree} \tag{A9}$$

where a suitable empirical expression from Schlichting [10] is used. To generalize, we have

$$\frac{1}{2}\mathbf{C}\_f = 0.0225 \text{Re}\_\delta^{-0.25} \frac{\ln\left(1+B\right)}{B}.\tag{A10}$$

Evaluating Equation (A10) for the case of no blowing enables us to write

$$\frac{C\_f}{C\_{f\_0}} = \left(\frac{\delta\_0}{\delta}\right)^{0.25} \frac{\ln\left(1+B\right)}{B},\tag{A11}$$

where (*δ*0/*δ*) accounts for the thickening effect of mass addition and differentiates Marxman's expression from that of Lees in Equation (17).

At this point, one is left with the task of determining a relation to describe *δ* in terms of *B* in Equation (A11). Such a relation may be obtained by performing a control volume analysis of the boundary layer to deduce an appropriate momentum integral expression. A sketch of a suitable control volume is provided in Figure A1. In this context, a statement of mass conservation (per unit thickness) across the control volume leads to

**Figure A1.** Sketch of the control volume for the momentum integral analysis.

$$
\rho \Delta \dot{m} + \int\_0^{\delta\_1} \rho u\_1 \mathbf{d}y - \int\_0^{\delta\_2} \rho u\_2 \mathbf{d}y + \rho\_\mathbf{W} v\_\mathbf{W} \Delta \mathbf{x} = 0. \tag{A12}
$$

Similarly, a statement of axial momentum conservation yields

$$
\rho u\_0 \Delta \dot{m} + \int\_0^{\delta\_1} \rho u\_1^2 \mathbf{d}y - \int\_0^{\delta\_2} \rho u\_2^2 \mathbf{d}y - \pi\_\mathbf{W} \Delta x = 0. \tag{A13}
$$

Re-arranging and non-dimensionalizing different terms may be used to produce

$$\begin{aligned} \text{Mass:} \qquad \Delta \dot{m} + \frac{1}{2} \mathbb{C}\_f \rho u\_\text{e} B \Delta \mathbf{x} &= \rho u\_\text{e} \delta\_2 \int\_0^1 \phi \mathbf{d} \eta - \rho u\_\text{e} \delta\_1 \int\_0^1 \phi \mathbf{d} \eta \\\ \frac{\Delta \dot{m}}{\rho u\_\text{e} \Delta \mathbf{x}} + \frac{1}{2} \mathbb{C}\_f B &= \frac{\delta\_2 - \delta\_1}{\Delta \mathbf{x}} \int\_0^1 \phi \mathbf{d} \eta. \end{aligned} \tag{A14}$$

Momentum:

$$\begin{split} \mathbf{u} \cdot \quad \quad \mu\_{\mathbf{e}} \Delta \dot{\boldsymbol{m}} + \frac{1}{2} \mathbb{C}\_{f} \rho \mu\_{\mathbf{e}}^{2} \Delta \mathbf{x} &= \rho \mu\_{\mathbf{e}}^{2} \delta\_{2} \int\_{0}^{1} \phi^{2} \mathbf{d} \eta - \rho u\_{\mathbf{e}}^{2} \delta\_{1} \int\_{0}^{1} \phi^{2} \mathbf{d} \eta \\ \quad \quad \frac{\Delta \dot{m}}{\rho u\_{\mathbf{e}} \Delta \mathbf{x}} - \frac{1}{2} \mathbb{C}\_{f} &= \frac{\delta\_{2} - \delta\_{1}}{\Delta \mathbf{x}} \int\_{0}^{1} \phi^{2} \mathbf{d} \eta. \end{split} \tag{A15}$$

By combining the expressions for conservation of mass, Equation (A14), and momentum, Equation (A15), we collect

$$\frac{\delta\_2 - \delta\_1}{\Delta x} = \frac{1}{2} \mathbb{C}\_f \frac{1+B}{\int\_0^1 \phi \left(1-\phi\right) \mathrm{d}\eta} = \frac{1}{2} \mathbb{C}\_f \frac{1+B}{\beta},\tag{A16}$$

where *β* = / 10 *φ* (1 − *φ*) d*η* is related to the momentum thickness of the boundary layer by *β* = *θ*/*δ*. In the limit as Δ*x* → 0, the control volume becomes infinitesimal, such that

$$\frac{\mathbf{d}\delta}{\mathbf{d}\mathbf{x}} = \frac{1}{2}\mathbf{C}\_f \frac{1+B}{\beta}.\tag{A17}$$

Substituting the expression for the skin friction, found earlier as Equation (A10), into Equation (A17), we ge<sup>t</sup>

$$\frac{d\delta}{dx} = 0.0225 \text{Re}\_{\delta}^{-0.25} \frac{(1+B)\ln\left(1+B\right)}{\beta B} = 0.0225 \frac{(1+B)\ln\left(1+B\right)}{\beta B} \left(\frac{\rho u\_{\text{e}}}{\mu}\right)^{-0.25} \delta^{-0.25}.\tag{A18}$$

Subsequent integration over the boundary layer yields

$$\frac{\delta}{\chi} = \left[ 0.02813 \frac{(1+B)\ln\left(1+B\right)}{\beta B} \right]^{0.8} \text{Re}\_x^{-0.2} \text{.} \tag{A19}$$

and so

$$\frac{\delta\_0}{\delta} = \left[ \frac{\beta}{\beta\_0} \frac{B}{(1+B)\ln\left(1+B\right)} \right]^{0.8}.\tag{A20}$$

Finally, combining Equation (A20) with Equation (A11) leads to an expression for the blowing correction in terms of the blowing parameter, specifically

$$\frac{C\_f}{C\_{f0}} = \left[\frac{\beta}{\beta\_0} \frac{B}{(1+B)\ln\left(1+B\right)}\right]^{0.2} \frac{\ln\left(1+B\right)}{B} = \left[\frac{\beta}{\beta\_0} \frac{1}{\left(1+B\right)}\right]^{0.2} \left[\frac{\ln\left(1+B\right)}{B}\right]^{0.8},\tag{A21}$$

where *β* depends on the velocity profile *φ*, which may be approximated relatively easily. Starting from Equation (A4), it is trivial to show that

$$\frac{\partial \phi}{\partial \eta} = \frac{\tau\_{\text{W}} \delta}{\mu\_{\text{e}} \left(\mu + \rho \epsilon\right)} \left(1 + B\phi\right) = f\left(y\_{\text{}}B\right) \left(1 + B\phi\right). \tag{A.22}$$

When *β* = 0, we may again assume a power law profile *φ* = *ηn* such that *∂φ*/*∂η* = *nη<sup>n</sup>*−1, which simplifies the unknown function *f* (*y*, *B*) to the assumed form *f* (*B*) *nη<sup>n</sup>*−<sup>1</sup> with the requirement that *f* (0) = 1. For *n* = 1/7, Equation (A22) becomes

$$\frac{\partial \Phi}{\partial \eta} = f\left(B\right)\eta^{-6/7} \left(1 + B\eta^{1/7}\right). \tag{A23}$$

Integrating Equation (A23) and evaluating the result at the boundaries, as with Equation (A7), leads to

$$\phi = \frac{\eta^{1/7} \left( 1 + \frac{1}{2} B \eta^{1/7} \right)}{1 + \frac{1}{2} B}. \tag{A24}$$

Finally, the expression for *φ* allows for the evaluation of *β* from its definition:

$$\beta = \frac{7\left(1 + \frac{13}{10}B + \frac{4}{11}B^2\right)}{72\left(1 + \frac{1}{2}B\right)^2}.\tag{A25}$$

It is clear that, in the case with no blowing (*B* = 0), *β* reduces to *β*0 = 7/72, such that

$$\frac{\beta}{\beta\_0} = \frac{\left(1 + \frac{13}{10}B + \frac{4}{11}B^2\right)}{\left(1 + \frac{1}{2}B\right)^2}.\tag{A26}$$
