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where L is obtained by a summation of different portion lengths, weighted by numerical coefficients
that can be approximated to be constant in the time:
ρσ v
(15)
i
L = ∑ li
= ∑li
ρ σ v
i
i
i
i
It is worthwhile to note that in pool cooling systems a unique definition of the loop length is not
easy to give. This matter rises because we are supposing a mono-dimensional problem, which actually
is at least two-dimensional. In any case, if we are able to evaluate the “effective length” L, equation
(14) leads to:
dQ
L
dt
m
dv
= σ ( ∆Pg
− ∆P)
i. e ρL
= ∆Pg − ∆P0
dt
(16)
Finally, if we remember the pressure drop formulation (9) and the initial steady state condition,
we can write:
ρL
dv
dt
= ∆P
g
− ∆P
g
ρ
0.75
v
1.75
0
ρ v µ
0.75 1.75
0 0
µ
0.25
0.25
0
(17)
that can be solved numerically in the following way:
∆v
k
1 ⎛
= P
L
⎜∆
ρ ⎝
gk
− ∆P
g 0
ρ
ρ
0.75
k−1
0.75
0
v
v
1.75
k−1
1.75
0
µ
µ
0.25
k−1
0.25
0
⎞
⎟ ∆t
⎠
k
(18)
i.e. by adding and subtracting ∆P g0
∆v
k
1 ⎡
= ⎢
ρL
⎢⎣
⎛
⎜
⎝ ρ
0
0.75 1.75 0.25
ρ
k −1
vk
−1
µ
k −1
( ∆Pgk
− ∆Pgo
) − ∆P
⎜
⎟
g 0
−1
∆tk
v
0.75 1.75
0
µ
0.25
0
⎞⎤
⎟⎥
⎠ ⎥ ⎦
where k indicates the generic time step relevant to the transient thermal-hydraulic solution.
In order to avoid numerical oscillations, we could also impose the limit of the asymptotic solution
to the evaluation of the resistant strength (per unit of volume), i.e. to the value of the pressure drop per
length unit. We will impose that the resistant strength will never exceed the gravitational pull. In
particular, a reduction of velocity will never occur as a consequence of a gravitational pull increasing
or an increase of velocity will never occur as a consequence of a gravitational pull decrease (as a
consequence of bad estimations of the resistant strength):
Practically, if ((∆P g k
>∆P g k-1
and ∆ ν k