Errata to: Powered Flight

The online version of the original book can be found under DOI 10.1007/978-1-4471-2485-6

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Errata to: D. R. Greatrix, Powered Flight, DOI 10.1007/978-1-4471-2485-6

• 1. There is a possibility of an extra page 333 and 334 in some copies of the book. Please ignore.

• 2. The original version of Chap. 6, Fig. 6.62 contained incorrect caption. The corrected caption is as follows: Schematic diagrams illustrating the nominal forward-thrust setup for the rear of a jet engine in flight (at left), and at right, the temporary deployment of external buckets at the rear of the engine in order to provide a reverse thrust capability for the airplane as it decelerates in a landing ground roll.

• 3. In Chap. 9, Eq. 9.29 should appear as

  $$ \frac{{A_{t} }}{{A_{e} }} = \frac{{Ma_{e} }}{{Ma_{t} }}\left[\frac{{2 + (\gamma - 1)Ma_{t}^{2} }}{{2 + (\gamma - 1)Ma_{e}^{2}}}\right]^{{\frac{\gamma + 1}{2(\gamma - 1)}}} $$

  (9.29)

• 4. Bottom of p. 294, should appear as:

  $$ \ldots ,{\text{air density}}\,\varDelta = {\text{ f}}\left( h \right)\, \ldots $$

• 5. In Chap. 10, Eq. (10.10) should appear as:

  $$ r_{e} = \frac{{h\left( {T_{F} - T_{S} } \right)}}{{\rho_{s} \left[ {C_{s} \left( {T_{S} - T_{i} } \right) - \Delta H_{s} } \right]}} $$

  (10.10)

• 6. Middle of p. 220, solution for Prob. 6.3, p ₀₅ value is shown as 193 kPa in the solution of the equation for finding p ₀₆, but that value should be 183 kPa, as per Prob. 6.2. Note that the value for p ₀₆ is correct as shown (172 kPa).

• 7. Near the bottom of p. 229 for the sample solution of Prob. 6.9, one sees the solution for the exiting mass flow, which should be shown as follows:

  $$ \dot{m}_{e} = \rho_{e} V_{e} A_{e} = 44.4\,{\text{kg/s}}$$

• 8. Near the bottom of p. 491 (Appendix III), should be as follows:

  $$ 1 \, {{\text{kg}} \mathord{\left/ {\vphantom {{\text{kg}} {{\text{m}}^{3} }}} \right. \kern-0pt} {{\text{m}}^{3} }} = \cdots = 3.61\,{\text{H}}\,10^{ - 5} {{\text{lbm}} \mathord{\left/ {\vphantom {{\text{lbm}} {{\text{in}}^{3} }}} \right. \kern-0pt} {{\text{in}}^{3} }} $$

• 9. Near the bottom of p. 432 (solution, Prob. 12.4), in the equation for flame zone thickness * _o , the wrong value was used for solid specific heat C _s (should be 2000, not 1100). As a result, the actual end solution for total burning rate r _b should be 0.00742 m/s (not 0.019 m/s) for that first iteration using the initial guess for r _b as 0.019 m/s. By repeated guesses for different values for r _b , one can eventually show that the converged value for r _b is 0.0173 m/s, and the axial mass-flux (base) burning rate at 500 g is 0.00703 m/s (as compared to 0.01165 m/s at 0 g).

In addition to the above corrections below text is revised content of Chap. 12:

Book Practice Problem Solution

12.4 (revised June 9, 2013)

Looking at a hybrid rocket fuel’s burning rate under mass flux and acceleration. Worst case is that there is sufficient oxidizer available for complete r _b augmentation due to a _n .

We will need to iterate between the two mechanisms of burning, given that each mechanism is dependent on the other mechanism as a base burning rate.

Effect of mass flux G:

From Prob. 12.3 (b),

$$ h^{*} = \frac{k}{d}\text{Re}_{d} {\text{Pr}}^{1/3} \frac{f}{8} = \frac{0.205}{0.08}4.926 \times 10^{6} \left( {0.73^{0.333} } \right)\frac{0.0118}{8} = 16766\,{{\text{W}} \mathord{\left/ {\vphantom {{\text{W}} {{\text{m}}^{ 2} }}} \right. \kern-0pt} {{\text{m}}^{ 2} }} \cong {\text{K}},{\text{ remains constant}}. $$

Begin iteration, guess r _b = 1.63 × 0.01165 m/s = 0.019 m/s:

$$ h = \frac{{\rho_{s} r_{b} C_{p} }}{{\exp \left( {\frac{{\rho_{s} r_{b} C_{p} }}{h^{*}}} \right) - 1}} = \frac{{1100\left( {2083} \right)r_{b} }}{{\exp \left( {\frac{{1100(2083)r_{b} }}{16766}} \right) - 1}} = \frac{{2.291 \times 10^{6} r_{b} }}{{\exp \left( {136.66r_{b} } \right) - 1}} = { 35}06\,{{\text{W}} \mathord{\left/ {\vphantom {{\text{W}} {{\text{m}}^{ 2} }}} \right. \kern-0pt} {{\text{m}}^{ 2} }} \cong {\text{K}} $$

$$ \begin{aligned} r_{b} &= r_{{o,a_{n}}} + \frac{{h(T_{F} - T_{S} )}}{{\rho_{s} C_{s} (T_{S} - T_{i} ) - \rho_{s} \Delta H_{S} }} = r_{{o,a_{n} }} + \frac{h(2725 - 800)}{1100(2000)(800 - 288) - 0} = r_{{o,a_{n} }} + 1.709 \times 10^{ - 6} h\\ & = r_{{o,a_{n} }} + 0.00 6 = \left( {0.0 1 9 - 0.00 6} \right) + 0.006 = 0.0 1 3 + 0.00 6 = 0.019\,{{\text{m}}/{\text{s}}}, {\text{ tentatively.}} \end{aligned}$$

Need to check via remaining equations, to bring convergence to the solution. For now, r _o,G = 0.006 m/s.

Effect of normal acceleration a _n :

$$ \begin{aligned} \delta_{0} & = \frac{k}{{\rho_{s} r_{o} C_{p} }}\ell n\left[ {1 + \frac{{C_{p} (T_{F} - T_{S} )}}{{C_{S} (T_{S} - T_{i} ) - \Delta H_{S} }}} \right] = \frac{0.205}{{1100\left( {r_{o,G} } \right)2083}}\ell n\left[ {1 + \frac{{2083\left( {1925} \right)}}{{2000\left( {512} \right) - 0}}} \right] = \frac{{1.4255 \times 10^{ - 7} }}{{r_{o,G} }}\\ & = { 2}. 3 7 6\times 10^{ - 5}\,{\text{m}}\\ G_{a} & = \frac{{a_{n} p}}{{r_{b} }}\frac{{\delta_{o} }}{{RT_{F} }}\frac{{r_{o} }}{{r_{b} }} = \frac{{ - 4905\left( {8.0 \times 10^{6} } \right)}}{{r_{b} }}\frac{{1.4255 \times 10^{ - 7} }}{{361.5\left( {2725} \right)r_{o,G} }}\frac{{r_{o,G} }}{{r_{b} }} = - \frac{{5.678 \times 10^{ - 3} }}{{r_{b}^{2} }} = - 1 5. 7 3\, {{\text{kg}}/{\text{s}}}\cong {\text{m}}^{ 2}\\ r_{b} & = \frac{{C_{p} \left( {T_{F} - T_{S} } \right)}}{{C_{S} \left( {T_{S} - T_{i} } \right) - \Delta H_{S} }} \cdot \frac{{r_{b} + {{G_{a} } / {\rho_{s} }}}}{{\exp \left[ {\frac{{C_{p} \delta_{o} \rho_{s} }}{k}\left( {{{r_{b} + G_{a} } / {\rho_{s} }}} \right)} \right] - 1}}\\ & = 3.92 \cdot \frac{{r_{b} - \frac{{5.162 \times 10^{ - 6} }}{{r_{b}^{2} }}}}{{\exp \left[ {\frac{{2083\left( {1.4255 \times 10^{ - 7} } \right)1100}}{{0.205r_{o,G} }}\left( {r_{b} - \frac{{5.162 \times 10^{ - 6} }}{{r_{b}^{2} }}} \right)} \right] - 1}}\\ & = 3.92 \cdot \frac{{r_{b} - \frac{{5.162 \times 10^{ - 6} }}{{r_{b}^{2} }}}}{{\exp \left[ {\frac{1.5934}{{r_{o,G} }}\left( {r_{b} - \frac{{5.162 \times 10^{ - 6} }}{{r_{b}^{2} }}} \right)} \right] - 1}} \cdot 0.00 7 4 2\,{{\text{m}} / {\text{s}}} ,\end{aligned}$$

so guessed incorrectly on r _b .

To potentially speed things up a bit, one can note that there is a lower limit value for r _b at a given a _n , as prescribed by the above equation:

$$ r_{b} - \frac{{5.162 \times 10^{ - 6} }}{{r_{b}^{2} }} = 0 $$

or in other words,

$$ r_{b,\lim } \approx \left\{ {\frac{{a_{n} p}}{{RT_{F} }}\frac{k}{{\rho_{s}^{2} \cdot C_{p} }}\ell n\left[ {1 + \frac{{C_{p} \left( {T_{F} - T_{S} } \right)}}{{C_{s} \left( {T_{S} - T_{i} } \right) - \Delta H_{s} }}} \right]} \right\}^{1/3} $$

Depending on the influence of other factors, the overall r _b may be at or above r _b,lim for a given value of a _n . In this case, it turns out that r _b is at the r _b,lim bound for 500 g, hence equal to 0.0173 m/s. The base burn rate (i.e., due to axial mass flux alone) for the 500 g case is 0.00703 m/s.

Thus, $\frac{{r_{b} }}{{r_{o} }} = \frac{0.0173}{0.00703} = 2.46$ as augmentation of burning rate due to radial vibration of 500 g.

Note how much mass-flux base burning was brought down by the vibration… from 0.01165 m/s down to 0.00703 m/s (about a 40 % decrease). Referencing the pre-vibration state, the augmentation ratio would in that case be 0.0173/0.01165 = 1.49.