Errata / Text
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| Chapter |
Page |
Line-Fig-Eq-Ex | Original | Corrected | Date posted |
|---|---|---|---|---|---|
| 2 | 051 | 23 |
[...] specific power is 36.7 $\mathrm{kW}/\mathrm{kg}_\mathrm{IHM}$. |
[...] specific power is 37.2 $\mathrm{kW}/\mathrm{kg}_\mathrm{IHM}$. |
Dec 2, 2011 |
| 2 | 052 | Figure 2.10 |
[The vertical axis of the figure is not printed correctly. The specific power should be 0 at the bottom of the plot and 100 at the top. The coordinates of the current plotted PWR operating condition are $T_{res}=4.5\:\mathrm{years}$ and $Q^{M}=37.2\:\mathrm{kW/kg_{IHM}}$] |
Nov 21, 2011 | |
| 3 | 127 | Ref. 3 |
[add at the end] An Erratum issued on Oct 19, 2011 changes the total uncertainty values listed for Examples 1 and 2 in Appendix B (page 32). |
Nov 7, 2011 | |
| 4 | 132 | Table 4.2 |
$h=u+p\upsilon$ |
$h=u+p\nu$ |
Nov 7, 2011 |
| 4 | 132 | Eq. 4.2 |
$\frac{dc}{dt}=\frac{dc}{dt}+\overrightarrow{\upsilon_{0}}\cdot\nabla c$ |
$\frac{dc}{dt}=\frac{\partial c}{\partial t}+\overrightarrow{\upsilon_{0}}\cdot\nabla c$ |
Nov 7, 2011 |
| 4 | 137 | Eq. 4.17a |
$\frac{Dm}{Dt}=\frac{D}{Dt}-\iiint_{V_{m}}\rho dV=0$ |
$\frac{Dm}{Dt}=\frac{D}{Dt}\iiint_{V_{m}}\rho dV=0$ |
Nov 7, 2011 |
| 4 | 138 | Eq. 4.17c |
$\frac{Dm}{Dt}=\frac{d}{dt}-\iiint_{V_{m}}\rho dV=\left(\frac{dm}{dt}\right)_{c.m.}=0$ |
$\frac{Dm}{Dt}=\frac{d}{dt}\iiint_{V_{m}}\rho dV=\left(\frac{dm}{dt}\right)_{c.m.}=0$ |
Nov 7, 2011 |
| 4 | 145 | 7 |
Hence, applying Equations 4.36 and 4.40a [...] |
Hence, applying Equations 4.36 and 4.40b [...] |
Nov 7, 2011 |
| 4 | 146 | 1 |
Realizing that for a stationary volume $\frac{dV}{dt}=\frac{\partial V}{\partial t}$ [...] |
Realizing that for a stationary volume $\frac{dW}{dt}=\frac{\partial W}{\partial t}$ [...] |
Nov 7, 2011 |
| 4 | 159 | Eq. 4.88 |
$-\nabla\cdot\left[\mu\nabla\cdot\overrightarrow{\upsilon}\right]$ |
$-\nabla\times\left[\mu\nabla\times\overrightarrow{\upsilon}\right]$ |
Nov 28, 2012 |
| 4 | 171 | Eq. 4.88 |
$-\nabla\cdot\left[\mu\nabla\cdot\overrightarrow{\upsilon}\right]$ |
$-\nabla\times\left[\mu\nabla\times\overrightarrow{\upsilon}\right]$ |
Nov 28, 2012 |
| 5 | 204 | Eq. 5.62 |
$\nu_{\ell z}$ [in the equation] |
$\upsilon_{\ell z}$ |
Nov 28, 2012 |
| 6 | 237 | Table 6.3 |
$R$ [it appears 3 times] |
$R_{sp}$ |
Oct 29, 2011 |
| 6 | 239 | Eq. above Eq. 6.5 |
$m_{v}c_{v_{c}}=(T_{e}-T_{1_{c}})=m_{f}c_{v_{f}}(T_{1_{f}}-T_{e})$ |
$m_{v}c_{v_{c}}(T_{e}-T_{1_{c}})=m_{f}c_{v_{f}}(T_{1_{f}}-T_{e})$ |
Nov 7, 2011 |
| 6 | 241 | Eq. 6.12 and Eq. 6.13 |
$R$ |
$R_{sp}$ |
Oct 29, 2011 |
| 6 | 243 | Second Equation |
$R$ |
$R_{sp}$ |
Oct 29, 2011 |
| 8 | 363 | Table 8.2B |
[Thermal conductivity of SiC] |
[Thermal conductivity of SiC] |
Feb 7, 2012 |
| 8 | 369 | Fig. 8.3 |
[replace existing figure] |
[Image available at the following link] |
Nov 28, 2012 |
| 8 | 371 | 9 |
The integral of Equation 8.22c is |
The integral of Equation 8.22c where T is in kelvin units is: |
Nov 28, 2012 |
| 9 | 442 | Figure 9.2 |
Incompressible flow [under left hand side arrow] |
Inviscid flow |
Apr 4, 2012 |
| 2 | 45 | First Equation |
$1000\mathrm{\: kg/m^{2}}$ |
$1000\mathrm{\: kg/m^{3}}$ |
Nov 28, 2012 |
| 2 | 45 | Table 2.4 |
[Operating cycle length of refueling interval (years), Defining Equation] 226 |
2.26 [the equation number should also be aligned in the same column as the other ones] |
Dec 17, 2012 |
| 9 | 453 | First Equation |
$[...]=\sqrt{\frac{2\left(9.81\,\mathrm{m/s^{2}}\right)\left(0.914\,\mathrm{m}\right)}{\left[\left(\frac{0.711}{0.686}\right)^{4}-1\right]}}=10.71\,\mathrm{m/s}$ |
$[...]=\sqrt{\frac{2\left(9.81\,\mathrm{m/s^{2}}\right)\left(0.924\,\mathrm{m}\right)}{\left[\left(\frac{0.711}{0.686}\right)^{4}-1\right]}}=10.85\,\mathrm{m/s}$ |
Dec 30, 2011 |
| 9 | 453 | Second Equation |
$[...]=\left(1000\,\mathrm{kg/m^{3}}\right)\left(10.71\,\mathrm{m/s}\right)\left[\frac{\pi}{4}\left(0.711\,\mathrm{m}\right)^{2}\right]$ |
$[...]=\left(1000\,\mathrm{kg/m^{3}}\right)\left(10.85\,\mathrm{m/s}\right)\left[\frac{\pi}{4}\left(0.711\,\mathrm{m}\right)^{2}\right]$ |
Dec 30, 2011 |
| 9 | 483 | 10 |
[...] Equation9.92a |
[...] Equation 9.92a |
Dec 30, 2011 |
| 10 | 558 | 3 |
Equations 10.74 and 10.75, when substituted into [...] |
Equations 10.74 and 9.60, when substituted into [...] |
Dec 18, 2012 |
| 10 | 568 | 18 |
Equation 10.97 |
Equation 10.96 |
Oct 29, 2011 |
| 11 | 641 | 16 |
[...] channl |
[...] channel |
Dec 30, 2011 |
| 11 | 641 | 19 |
The Homogeneous equilibrium model (HEM) model implies that [...] |
The Homogeneous Equilibrium Model (HEM) implies that [...] [correct capitalization and delete repeated word "model"] |
Dec 30, 2011 |
| 11 | 658 | 15 |
[...] high pre< |
[...] high pressure boiler [...] |
Dec 18, 2012 |
| 11 | 659 | 4th Equation |
$\left.\frac{\mathrm{d}p}{\mathrm{d}z}\right|_{fric}^{\ell o}=f\frac{L}{D_{e}}\frac{G_{m}^{2}}{2\rho_{f}}$ |
$\left.\frac{\mathrm{d}p}{\mathrm{d}z}\right|_{fric}^{\ell o}=f\frac{1}{D_{e}}\frac{G_{m}^{2}}{2\rho_{f}}$ [L replaced with 1] |
Nov 28, 2012 |
| 11 | 670 | Eq. 11.132c |
$\rho^{\prime\prime\prime}=\left\{ \left[\frac{x}{\rho_{g}}+\frac{\left(1-x\right)S}{\rho_{f}}\right]^{2}\left[x+\frac{1-x}{s^{2}}\right]\right\} ^{-1}$ |
$\rho^{\prime\prime\prime}=\left\{ \left[\frac{x}{\rho_{g}}+\frac{\left(1-x\right)S}{\rho_{f}}\right]^{2}\left[x+\frac{1-x}{s^{2}}\right]\right\} ^{-0.5}$ [exponent changed to -0.5] |
Nov 27, 2012 |
| 11 | 670 | Eq. 11.132b |
$G=\sqrt{\rho^{\prime\prime\prime}2\left[h_{o}-xh_{g}-\left(1-x\right)h_{f}\right]}$ |
$G=\rho^{\prime\prime\prime}\sqrt{2\left[h_{o}-xh_{g}-\left(1-x\right)h_{f}\right]}$ |
Nov 27, 2012 |
| 12 | 711 | Eq. 12.25c |
$\lambda_{D}=2\pi\left(\frac{3\sigma}{\rho_{l}-\rho_{\mathrm{v}}}\right)^{1/2}$ |
$\lambda_{D}=2\pi\left(\frac{3\sigma}{g\left(\rho_{l}-\rho_{\mathrm{v}}\right)}\right)^{1/2}$ |
Dec 2, 2011 |
| 12 | 711 | Eq. 12.25b |
$\mathrm{Nu_{\lambda}}=\frac{q^{\prime\prime}\lambda_{D}}{k_{\mathrm{v}}\left(T_{w}-T_{sat}\right)}$ |
$\mathrm{Nu_{\lambda_{D}}}=\frac{q^{\prime\prime}\lambda_{D}}{k_{\mathrm{v}}\left(T_{w}-T_{sat}\right)}$ |
Nov 7, 2011 |
| 12 | 711 | Eq. 12.25a |
$\mathrm{Nu_{\lambda}}=C\sqrt{3}R^{n}\mathrm{Pr}_{\mathrm{v}}^{1/3}f\left(\mathrm{Ja}\right)$ |
$\mathrm{Nu_{\lambda_{D}}}=C\sqrt{3}\left[\frac{\mathrm{Ra}_{\mathrm{m}}\nu^{2}_\mathrm{v}}{2}\right]^{n}\mathrm{Pr}_{\mathrm{v}}^{\frac{(1-3n)}{3}}f\left(\mathrm{Ja}\right)$ |
Nov 7, 2011 |
| 12 | 711 | Eq. 12.25e |
$\mathrm{Ja}=\frac{c_{p\mathrm{v}}\left(T_{w}-T_{sat}\right)}{h_{l\mathrm{v}}}$ |
$\mathrm{Ja}=\frac{c_{p\mathrm{v}}\left(T_{w}-T_{sat}\right)}{h_{fg}}$ |
Dec 2, 2011 |
| 12 | 711 | Eq. 12.25d |
$R=\frac{g\lambda_{D}^{3}}{\left(3\right)^{3/2}\left(\mu/\rho\right)_{\mathrm{v}}^{3/2}}\left(\frac{\rho_{l}-\rho_{\mathrm{v}}}{\rho_{\mathrm{v}}}\right)$ |
$\mathrm{Ra}_{\mathrm{m}}=\frac{g\lambda_{D}^{3}}{\left(3\right)^{3/2}\left(\mu/\rho\right)_{\mathrm{v}}^{2}}\mathrm{Pr}\left(\frac{\rho_{l}-\rho_{\mathrm{v}}}{\rho_{\mathrm{v}}}\right)$ |
Nov 7, 2011 |
| 12 | 711 | 8 |
$R>10^{8}$ |
$\frac{\mathrm{Ra}_{\mathrm{m}}\nu_\mathrm{v}^{2}}{2\mathrm{Pr}}>10^{8}$ |
Nov 7, 2011 |
| 12 | 712 | Table 12.2 title |
Parameters for Equation 12.25 |
Parameters for Equation 12.25a |
Dec 2, 2011 |
| 12 | 720 | 7 |
From Table 12.2 |
From Table 12.3 |
Dec 30, 2011 |
| 12 | 738 | Reference 25 |
Dhir, V. K. |
Dhir, V. J. |
Oct 29, 2011 |
| 14 | 846 | Eq. 14.41 |
\[ |
\[ |
Dec 16, 2012 |
