Q3:A- a cylindrical fuel rod consist of a fuel region of diameter (10mm). If the temperature at thecenter line of the fuel is (1800°C) and the heat generation rate is (q//=4x10 W/m').1- Find the temperature distribution along the radial direction of the fuel meat. Assume steadystate one dimensional heat flowd?Tr1 d+= 0dr?r dr2- Calculate the temperature at the surface of the fuel. Take the average thermal conductivity ofthe oxide fuel to be (1.9 W/m °C).3- Calculate the temperature distribution along the radius of the fuel meat4- Draw the temperature distribution along the radius of the fuel meat

Given data as Diameter(d)=2r=10mmr=5 mm=5×10-3Centre line temperature (Tc)=1800°CHeat generation…

Answered: Q3:A- a cylindrical fuel rod consist of…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Q3:A- a cylindrical fuel rod consist of a fuel region of diameter (10mm). If the temperature at the
center line of the fuel is (1800°C) and the heat generation rate is (q//=4x10 W/m').
1- Find the temperature distribution along the radial direction of the fuel meat. Assume steady
state one dimensional heat flow
d?Tr
1 d
+
= 0
dr?
r dr
2- Calculate the temperature at the surface of the fuel. Take the average thermal conductivity of
the oxide fuel to be (1.9 W/m °C).
3- Calculate the temperature distribution along the radius of the fuel meat
4- Draw the temperature distribution along the radius of the fuel meat

Expert Solution

Step 1

Given data as

$D i a m e t e r (d) = 2 r = 10 m m r = 5 m m = 5 \times 10^{- 3} C e n t r e l i n e t e m p e r a t u r e (T_{c}) = 1800 ° C H e a t g e n e r a t i o n r a t e (q''') = 4 \times 10^{8} W / m^{3}$

The conduction equation for steady-state one-directional heat conduction in the cylinder is as

$\frac{1}{r} \frac{d}{d r} (r \frac{d T}{d r}) + \frac{q'''}{k_{f}} = 0 \frac{d}{d r} (r \frac{d T}{d r}) = - \frac{q''' r}{k_{f}} I n t e g r a t e t h e a b o v e e q u a t i o n r \frac{d T}{d r} = - \frac{q''' r^{2}}{2 k_{f}} + C_{1} A s w e k n o w m a x i m u m t e m p e r a t u r e i n t h e f u e l m e a t i s a t c e n t r e l i n e S o, S l o p e o f t h e t e m p e r a t u r e c u r v e w i l l b e z e r o a t c e n t r e (r = 0) . T o s a t i s f y t h i s c o n d i t i o n C_{1} m u s t b e e q u a l t o z e r o C_{1} = 0 A b o v e e q u a t i o n c a n b e w r i t t e n a s \frac{d T}{d r} = - \frac{q''' r}{2 k_{f}} I n t e g r a t i n g t h e a b o v e e q u a t i o n a g a i n T = - \frac{q''' r^{2}}{4 k_{f}} + C_{2} A s g i v e n C e n t r e l i n e t e m p e r a t u r e (r = 0) 1800 ° C S o, p u t r = 0, T = 1800 i n a b o v e e q u a t i o n S o, C_{2} = 1800$

$S o a b o v e e q u a t i o n o f t e m p e r a t u r e d i s t r i b u t i o n w i l l b e T = - \frac{q''' r^{2}}{4 k_{f}} + 1800$

Step 2

2. Given

$k_{f} = 1.9 W / m . ° C$

For finding the temperature at the surface (T_s) put the values in the above equation

$r = 5 \times 10^{- 3} i n t h e a b o v e e q u a t i o n o f t e m p e r a t u r e d i s t r i b u t i o n T_{s} = - \frac{4 \times 10^{8} \times (5 \times 10^{- 3})}{4 \times 1.9} + 1800 T_{s} = 484.21 ° C$