TTC Timmler Technology TTC Silent Gravitiy Cooling – Modultherm ...

TTC Capacity Diagrams
Category 1
Specific cooling capacity: category 1 Specific pressure drop heat exchanger: category 1
1700
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12.1
12
Category 1 (cooling)
2,2 m
6 7 8 9 10 11 12 13
Median temperature difference ∆ϑ [K] m
6,0 m
5,0 m
4,0 m
3,5 m
3,0 m
2,5 m
2,0 m
1,5 m
Formula to calculate the median temperature difference ∆ϑ m
1
<
Specific cooling capacity q spec. [W/m]
.
<
<
t [°C] + t [°C]
W1 W2 ∆ϑ [K] = t – for cooling mode
m R
2
∆ϑ m [K] = median temperature difference between cooling
medium and room temperature
t W1 [°C] = cold water inlet
t W2 [°C] = cold water outlet
t R [°C] = room temperature
Issued 02/2005. We reserve the right to make technical changes.
<
15
10
8
6
5
4
3
2
1,0
0,8
0,6
0,5
0,4
0,3
0,2
0,1
0,08
0,06
kg/h 10
kg/s
12.3
.
< >
Specific pressure drop ∆p w specif. [kPa/m] finned
0,003
15
20
25 30 40 50 60 80 100 150 200 300 400 500 700
0,005 0,007 0,01 0,015 0,02 0,03 0,04 0,06 0,08 0,1 0,15 0,20
.
< Water mass flow mw >
Formula to calculate the overall sensitive cooling capacity of
the TTC Modultherm cooling unit
3
. .
Q [W] = q [W/m] · L [m]
K(tot.) K(spec.) finned
·
QK.(tot.) .
qspecif. [W]
[W/m]
= total cooling capacity of a cooling unit
= specific cooling capacity of a cooling unit
Lfinned [m] = finned length of the cooling unit
.
Formula to calculate the specific water mass flow mw · q [kW] · L [m]
·
(specif.) finned
4 m
·
[kg/h] = 860 ·
W t - t [K]
W2 W1
Formula to calculate the overall pressure drop ∆p of the
w(tot.)
cooling unit on the water side
.
.
·
5 ∆p [kPa] = ∆p [kPa/m] • L [m]
w(tot.) w(spec.) finned
∆p. [kPa/m] = see figure 12.3
w(specif.)