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23-Chem-B1 Transport Phenomena · December 2016

Question 4 of 6: B2 — Maximum temperature in a nuclear fuel rod

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. EGBC 04-CHEM-B1 Transport Phenomena, December 2016, 3 hours, open-book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth (four of six at 25 marks each, only the first four in the answer book are marked). All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species tables) in rectangular, cylindrical and spherical coordinates — these are quoted rather than re-derived.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, internal-flow heat transfer and film mass transfer; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the log-mean-temperature-difference and effectiveness–NTU methods; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — transport properties and the Colebrook friction correlation.

Question 4: B2 — Maximum temperature in a nuclear fuel rod (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A long cylindrical fuel rod with radially-varying volumetric generation $P(r)=P_0[1+b(r/R_1)^2]$ (equal to $P_0$ on the centreline; for $b>0$ it grows toward the fuel surface), surrounded by cladding of outer radius $R_c$, then convection to coolant at $T_w$. Steady state, constant conductivities $k_F,k_c$, one-dimensional radial conduction, no axial loss.

Find. The maximum temperature $T_{\max}$ (which occurs at the centreline $r=0$).

fuel kF cladding kc coolant Tw, hL R1 Rc T is maximum at the centre r = 0
Fig. B2: Cross-section of the fuel rod. Heat generated in the fuel ($0\le r\le R_1$) flows radially outward through the cladding ($R_1\le r\le R_c$) and is removed by convection to the coolant. Because all the heat flows outward, the temperature peaks on the centreline.

Approach. Integrate the radial conduction equation with the given source in the fuel region to get the heat flux and the fuel temperature profile; the cladding (no source) gives a logarithmic drop, and the interface gives a convective drop. Build $T_{\max}=T_w+(\text{three temperature rises})$.

  1. Heat flux inside the fuel. The steady radial balance $\dfrac1r\dfrac{d}{dr}(r\,q_r)=P(r)$ integrates (with $q_r$ finite at $r=0$) to $$q_r^{F}(r)=P_0\Big(\frac r2+\frac{b\,r^3}{4R_1^2}\Big).$$
  2. Total heat generated per unit length. $Q'=\displaystyle\int_0^{R_1}P(r)\,2\pi r\,dr=\pi P_0 R_1^2\Big(1+\frac b2\Big)$, and indeed $2\pi R_1\,q_r^{F}(R_1)=Q'$ — all the generated heat crosses the fuel surface. Define the convenient group $A_q\equiv\dfrac{Q'}{2\pi}=\dfrac{P_0 R_1^2(2+b)}{4}$.
  3. Convective rise (coolant → cladding surface). $Q'=h_L\,2\pi R_c\big(T(R_c)-T_w\big)$ gives $$T(R_c)-T_w=\frac{A_q}{R_c\,h_L}.$$
  4. Conductive rise across the cladding. With no source, $q_r r=A_q$ constant, so $-k_c\,dT/dr=A_q/r$ integrates to $$T(R_1)-T(R_c)=\frac{A_q}{k_c}\ln\!\frac{R_c}{R_1}.$$
  5. Rise from fuel surface to centre. Fourier $q_r^{F}=-k_F\,dT/dr$ integrates to $T^{F}(r)=T_{\max}-\dfrac{P_0}{k_F}\Big(\dfrac{r^2}{4}+\dfrac{b\,r^4}{16R_1^2}\Big)$, so $$T(0)-T(R_1)=\frac{P_0 R_1^2}{16k_F}(4+b).$$ (For uniform generation, $b=0$, this reduces to the classic $P_0R_1^2/4k_F$.)
  6. Assemble the maximum temperature. Adding the three rises to the coolant temperature, $$\boxed{\,T_{\max}=T_w+\frac{A_q}{R_c\,h_L}+\frac{A_q}{k_c}\ln\!\frac{R_c}{R_1}+\frac{P_0 R_1^2(4+b)}{16k_F}\,},\qquad A_q=\frac{P_0 R_1^2(2+b)}{4}.$$
ResultExpression
Surface heat rate per length$Q'=\pi P_0 R_1^2(1+b/2)=2\pi A_q$
Fuel temperature profile$T^{F}(r)=T_{\max}-\dfrac{P_0}{k_F}\big(\tfrac{r^2}{4}+\tfrac{b r^4}{16R_1^2}\big)$
Maximum (centreline) temperature$T_{\max}=T_w+\dfrac{A_q}{R_c h_L}+\dfrac{A_q}{k_c}\ln\dfrac{R_c}{R_1}+\dfrac{P_0 R_1^2(4+b)}{16k_F}$
Check — the source-term form and where the maximum sits

The paper prints $P=P_0[1+b(r/R_1)^2]$ (a plus sign), the form used in Bird–Stewart–Lightfoot §10.3; the cladding radius and conductivity are written $R_c,\ k_c$ and the coolant temperature $T_w$ in this solution. Even when $b>0$ makes the generation largest at the fuel surface, the local flux $q_r^F=P_0(r/2+br^3/4R_1^2)$ is outward at every radius, so $dT/dr<0$ throughout and the maximum temperature is still on the centreline. Setting $b=0$ recovers the uniform-generation result $T_{\max}-T(R_1)=P_0R_1^2/4k_F$; the result also holds for a negative $b$ provided $1+b\ge0$ (generation non-negative everywhere).