UFC 3-240-13FN
25 May 2005
EQUATION
B = 37.8 l/sec (600 gpm) (4-1) = 12.6 l/sec (200 gpm)
(23)
a) This formula is derived using data from previously presented equations:
(1)
M
=
B+E
from paragraph 4-2.1.1 (19)
(2)
C
=
MB
from paragraph 4-2.1.2.1 (20)
(3)
C
=
(B + E) B
from equation (2) and (1) above
(4)
C
=
1 + (E B)
rearranging equation (3)
(5)
(C - 1)
=
EB
rearranging equation (4)
E (C - 1)
(6)
B
=
rearranging equation (5)
b) If you know the quantity of evaporation, you can calculate the blowdown
required for a given value of COC. You can estimate the evaporation
using simple "rule of thumb" estimates:
b.1) For a typical recirculating cooling tower water system, approximately 1%
of the recirculating rate (R) of the cooling water is evaporated for every 5.5
C (10 oF) temperature drop in the cooling water as it passes through the
o
tower; therefore, you may calculate the evaporation rate (E) this way:
E(l/sec)=0.01 x R(l/sec) x ∆T drop in oC 5.5 oC
EQUATION
(24)
Since 0.01 5.5 = 0.0018, this can be condensed to:
E = R x ∆T x 0.0018
EQUATION
(25)
or
E = R x ∆T 550
EQUATION
(26)
NOTE: Newer cooling towers can have 0.75% of the recirculation rate evaporated for
every 5.5 C (10 F) drop.
EXAMPLE 4-2:
b.2) A cooling system operates at 315 liters per second (5000 gallons per
minute). The temperature drop through the tower is 7.8 oC (14 oF). The
evaporation estimate is represented by this equation:
E = 0.01 x 315 l/sec (5000 gpm) x 7.8 oC (14 oF)
EQUATION
5.5 oC (10 oF) = 4.5 l/sec (70 gpm)
(27)
or
EQUATION
E = 315 x 7.8 550 = 4.5 l/sec (70 gpm)
(28)
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