22.313
THERMALHYDRAULICS IN NUCLEAR POWER TECHNOLOGY
Tuesday, May 17
th
, 2005, 9 a.m. – 12 p.m.
OPEN BOOK
FINAL (solutions)
3 HOURS
Problem 1 (30%) – Hydraulic analysis of the PWR primary system at cold zeropower
conditions
i) The momentum equation for the loop is:
2
2
sg
core
pump
A
2
m
)
K
K
(
P
dt
dm
A
L
l
ρ
+
−
Δ
=
⋅
(
1
)
where m is the mass flow rate, L=40 m is the total length of the loop, A=1.65 m
2
is the flow area,
K
core
=7 and K
sg
=4 are the form loss coefficients for the core and steam generator, respectively.
The acceleration and friction terms were neglected in Equation 1, as per the problem statement.
Moreover the gravity term is zero because the fluid is isothermal.
At steadystate
0
dt
dm
=
and Equation 1 can be easily solved for the steadystate mass flow rate,
m
ss
:
)
K
K
(
P
A
2
m
sg
core
pump
2
ss
+
Δ
ρ
=
l
≈
9,960 kg/s
(2)
ii) Equation 1 can be rewritten as follows:
2
2
ss
sg
core
m
m
dt
dm
)
K
K
(
L
A
2
−
=
⋅
+
⋅
ρ
l
(
3
)
Equation 3 can be integrated to find m(t) during startup.
Separating the variables, making use of
the hint in the problem statement, and setting the initial condition m(0)=0, one gets:
τ
−
τ
−
+
−
=
/
t
/
t
ss
e
1
e
1
m
)
t
(
m
(
4
)
where the time constant,
τ
, is defined as follows:
pump
sg
core
ss
sg
core
P
)
K
K
(
2
L
m
)
K
K
(
L
A
Δ
+
ρ
=
+
⋅
ρ
=
τ
l
l
≈
0.6 s
(5)
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View Full DocumentEquation 4 is plotted in Figure 1.
The time it takes to reach 50% of the steadystate value can be
calculated by setting m=0.5
⋅
m
ss
in Equation 4, and solving for t.
≈
0
.
6
6
s
(
6
)
)
3
ln(
50
⋅
τ
=
τ
Figure 1.
PWR primary system mass flow rate during cold zeropower startup.
iii) Equation 5 indicates that the time constant is proportional to the loop length and inversely
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 Spring '03
 RogerD.Kamm

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