**Unformatted text preview: **2AT
X 10 -% 2)/24S
T_. = 20"C
T1 = 400"C
-D= 0.003 m
Soldering
L- 0.01 m
Iron
qe = hco A AT = he (/D /4+ 7DL) (T,-T.) = q
q = 20 W/(m. K)
7 (0.003 m)?
4
-+ /(0.003m) (0.01m) (400.C - 20C)
q = 0.77 W
qr = A1 61 6(T)* - T.') - (TD + R DL ) eo(; - T.."
7 (0.003 m )?
4
-+ 7(0.003m) (0.01m) | (0.8) [5.67 x 10-$ W/(m' K')] [(673 K)* - (293 K)*]
qr = 0.91 W
The power requirement of the soldering iron, q, is equal to the total rate of heat loss from the tip. The
total heat loss is equal to the sum of the convectiond radiative losses
q = qe + qr = 0.77 W + 0.91 W = 1.68 W
(40 pnt) The cross-section of a spherical fuel element (Thorium) in a nuclear reactor is shown. Energy
generation occurs uniformly in the thorium fuel rod (k-60 W/m.K), which is of diameter D-20 mm. It is
proposed that, under steady-state conditions, the system operates with a generation rate of
e gen =5x10* W/m' and cooling system characteristics of T.-20"C and h=10 kW/m K. If the melting
temperature of the Thorium is 2023 K, will the fuel melt inside the nuclear reactor?
Coolant
T_, h
fuel rod
dr
1 1 B. C.
Ther tell Symmetry
two dTo =>4-0
T()--3+ca
dr
64
end gc.
T() - 9
TE (."-t) +30. +To
k dT(rero) . h. (T - Tea)
3h
dr
T()~Slo? (01013 2, S.1053.0)
-t.(-s.co. ) . h . (T(.)-Too
+ (204231)
=138, 83 - 1 282.885, 9cz
41666, 614 293
C2 - 36 40
(T( ) - 2038 , 56 -4385 804,5,2
may ~ T(r =0) = 2098, 56 k. 2 2023 1
14 Melts!
Time: 90 min.
Good Luck......

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- Spring '15
- Thermodynamics, Nuclear technology, Enrico Fermi, Sellafield