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# Errata / Text

## Primary tabs

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]
$70$

[Thermal conductivity of SiC]
$70^{\mathrm{b}}$
[add as a footnote below the table the following text:]
$^{\mathrm{b}}$ The value of SiC thermal conductivity in Table 8.2B is for the unirradiated material at room temperature. Both high temperature and irradiation degrade SiC thermal conductivity. At 700K, the thermal conductivity of unirradiated SiC is only 35 W/m K. Irradiation at LWR conditions for two months reduces the conductivity to between 5 and 10 W/mK, which seems to become stable even at longer irradiations.
See D. Carpenter, K. Ahn, S. Kao, P. Hejzlar, and M.S. Kazimi, "Assessment of Silicon Carbide for High Performance Light Water Reactors", MIT-NFC-TR-098, MIT, November 2007. and
J. D. Stempien, D. Carpenter, G. Kohse, and M. S. Kazimi, " Behavior of Triplex Silicon Carbide Fuel Cladding Designs Tested Under Simulated PWR Conditions", MIT-ANP-TR-134, MIT, 2011.

Feb 7, 2012
8 369 Fig. 8.3

[replace existing figure]

[Image available at the following link]
http://nuclearsystems.mit.edu/sites/default/files/images/Fig.%208.3.gif

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

$[...]=\left(1000\,\mathrm{kg/m^{3}}\right)\left(10.71\,\mathrm{m/s}\right)\left[\frac{\pi}{4}\left(0.711\,\mathrm{m}\right)^{2}\right]$
$=4252\,\mathrm{kg/s}$

$[...]=\left(1000\,\mathrm{kg/m^{3}}\right)\left(10.85\,\mathrm{m/s}\right)\left[\frac{\pi}{4}\left(0.711\,\mathrm{m}\right)^{2}\right]$
$=4308\,\mathrm{kg/s}$

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

$R=\frac{g\lambda_{D}^{3}}{\left(3\right)^{3/2}\left(\mu/\rho\right)_{\mathrm{v}}^{3/2}}\left(\frac{\rho_{l}-\rho_{\mathrm{v}}}{\rho_{\mathrm{v}}}\right)$

$\mathrm{Ra}_{\mathrm{m}}=\frac{g\lambda_{D}^{3}}{\left(3\right)^{3/2}\left(\mu/\rho\right)_{\mathrm{v}}^{2}}\mathrm{Pr}\left(\frac{\rho_{l}-\rho_{\mathrm{v}}}{\rho_{\mathrm{v}}}\right)$
(a modified Rayleigh Number)

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

$\mathrm{sin}\left(\frac{\pi z}{L}\right.$

$\mathrm{sin}\left(\frac{\pi z}{L}\right)$

Dec 16, 2012