Euro-C 4000-JLGBEN
29/31
1
1
1
2
+
−
×
=
V
V
x
x
V
V
tot
tot
HS
)
/
³
12371
(
/
³
.
3
1
1
.
2
.
5
.
383
200
.
5
44
67
00
60
00
h
m
s
m
V
HS
=
+
−
×
=
Note
In this case 2.
67
m³/s
(= 9600m³/h)
(see table
7) flows through the heat exchanger
enclosure.
The remaining 2.
33
m³/s
(=8400m³/h)
flows
through the by-pass duct
(5.
00
m³/s – 2.
67
m³/s = 2.
33
m³/s)
C) Pressure loss :
In this case the total pressure loss over the
EURO-C module with by-pass can be
extracted from figures 13 & 14. The airflow
through the heat exchanger enclosure must
be applied to determine the total pressure
drop.
Example :
- Airflow through the heat exchanger
enclosure = 2.
67
m
³
/s (9600 m
³/
h
)
- Total pressure drop = 40 Pa (approx.)
(see fig. 14)
11.3 Calculation of alternative by-pass dimension
Due to space restrictions it may not be
possible to use the above mentioned maximum
dimension by-pass
In this case a smaller by-pass can be applied.
Obviously, this smaller by-pass will result in
higher portion of the total air flow through the
heat exchanger enclosure and consequently in
a higher total pressure drop
A) To calculate this air flow through the heat
exchanger enclosure, following inter-
polation rule should be applied
:
With
:
V = Recommended air flow
(Table 7)
(m³/s)
V
tot
= Total air flow (m³/s)
x
1
= Recommended by-pass (dim
C or D)
(as calculated in section
11.2).
x
2
= Applied by-pass dim. (mm)
Then V
HS
(air flow through heat exchanger
enclosure can be calculated as :
As explained above, the air flow V
HS
through the heat exchanger enclosure
must be used in figure 13 & 14
to
extract the correct total pressure drop
over the RHC module with by-pass.
The air flow V
HS
may not exceed the
maximum airflow allowed according to
section 4.1.2 to avoid supercooling of
the flue gases.
B)
Example
.
EURO-C 4072
with :
- Total air flow : 5m³/s (18000m³/h)
- Expected min. inlet temp. : -5°C
- Recommended max. dim C = 383.
60
mm
(see above 2B)
- Applied by-pass dim. C = 200mm
Then :
x1
= 383.
60
mm
x2 = 200mm
Vtot = 5m³/s
V = 2.
67
m³/s (9600m³/h)
(see table 7)
Result
:
Consequently, 3.
44
m³/s (12371 m³/h) flows
through the heat exchanger enclosure and
the remaining 1.
56
m³/s
(=5629m³/h)
(=
5.
00
m³/s – 3.
44
m³/s) flows through the by-
pass channel.
From section 4.1.2 the maximum allowable
airflow through the heating module
enclosure is 16180 m³/h (for this example).
We notice the 12371 m³/h is still below
this maximum allowable figure.
Consequently there is no probability of
condensation forming within the tubes.
C) Pressure drop :
The total pressure drop can be derived from
fig. 13 & 14. Use the air flow through the
heat exhanger enclosure to determine the
total pressure drop. (@12371 m³/h)
Result : 70Pa (approx) (see fig. 14).
Summary of Contents for EURO-C 4000 DJL
Page 5: ...Euro C 4000 JLGBEN 5 31 Figure 1 Dimensions EURO C 4011 15 RJL DJL Fig 1 A Fig 1 B...
Page 6: ...Euro C 4000 JLGBEN 6 31 Fig 1 C...
Page 7: ...Euro C 4000 JLGBEN 7 31 Figure 2 Dimensions EURO C 4018 72 RJL DJL Fig 2 A Fig 2 B...
Page 8: ...Euro C 4000 JLGBEN 8 31 Fig 2 C...
Page 9: ...Euro C 4000 JLGBEN 9 31 Figure 3 Dimensions EURO C 4100 RJL DJL Fig 3 A Fig 3 B...
Page 10: ...Euro C 4000 JLGBEN 10 31 Fig 3 C...
Page 18: ...Euro C 4000 JLGBEN 18 31 Figure 8a models 4011 72 Figure 8b model 4100...
Page 19: ...Euro C 4000 JLGBEN 19 31 Figure 8c all models Figure 9a models 4011 72...
Page 22: ...Euro C 4000 JLGBEN 22 31 Figure 10b MODELS 4011 72 MODEL 4100...
Page 24: ...Euro C 4000 JLGBEN 24 31 Figure 12 Models 4011 72 Model 4100...
