Carel SA0AALI000, Руководство по конструкции

Страница: 19 / 36

19
ENG
“ultimateSAM - design” +03U0071EF - rel. 2.0 - 20150301
Sample calculation using Example 1 in Sec.4.1: assume that a site has the
following conditions:
•
Humidification load: 90 kg/hr (200 lb/hr)
•
Distributor: SABEBLI300
•
Inlet adapter: SAKIT40200
•
40mm (1.6”) steam hose: 2 pieces, 3m (10’) long, 45 kg/hr (100 lb/hr) per
hose.
1.
Use Table 4.c to determine constant “A.”
For width code “E” and height code “B,” A=0.610 kPa (0.51 in H2O).
2. Calculate PB1.
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
3.
Use Table 4.f to determine constant “B.” For SAKIT40200, B=0.21 kPa (0.17
in H2O).
4.
Calculate PB2.
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
5.
Use Table 4.f to determine constant “C.” For 40mm (1.6”) steam hose,
C=0.36 kPa/m (0.091 inH2O per ft).
6.
Calculate PB3.
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
P
B1
= (0.610) = 0.49kPA
90
100
( (
2
P
B1
= (0.51) = 2.0 in H
2
O
200
100
( (
2
P
B2
= (0.21) = 0.17kPA
90
100
( (
2
P
B2
= (0.17) = 0.68 in H
2
O
200
100
( (
2
P
B3
= (0.36) (3) = 0.22kPA
45
100
( (
2
P
B3
= (0.091)(10) = 0...91 in H
2
O
100
100
( (
2
7.
P
total
= 0.49 + 0.17 + 0.22 = 0.88kPa (PTOTAL=2.0 + 0.68 + 0.91 = 3.6 in H2O)
Note: The static pressure of the duct must be less than 1.12kPa
(4.4 in H2O) to keep the outlet pressure acting on the UE090X****
under 2kPa (8” H2O).
4.5 Air Flow Resistance
The static pressure drop created due to the ultimateSAM distributor in the
duct or AHU is shown in Table 4.h e 4.i. A distributor that is properly sized to
the duct or AHU will minimize the pressure drop. The data table can only be
used to determine the flow resistance of air passing through the effective
area of the distributor. It does not include pressure losses due to facing off
areas of the duct for valves, drains, or other connections.
Pressure drop, Pa (in H
2
O) (SAB* / SAT* models)
Air Velocity,
m/s (fpm)
Upright configuration
S L H
3 (600) 0 (0.002) 1 (0.006) 5 (0.022)
6 (1200) 2 (0.008) 6 (0.024) 22 (0.088)
10 (2000) 5 (0.022) 17 (0.067) 61 (0.245)
Tab. 4.i
Pressure drop, Pa (in H
2
O) (SA0* models)
Air Velocity,
m/s (fpm)
Upright length, mm (in)
358 (14) 1270 (50) 2030 (80)
3 (600) 5 (0.020) 5 (0.020) 6 (0.024)
6 (1200) 18 (0.072) 20 (0.080) 24 (0.096)
10 (2000) 48 (0.193) 54 (0.217) 66 (0.265)
Tab. 4.j
4.6 Steam Losses
When designing an ultimateSAM Direct Steam Humidification System,
allowance must be made for steam that condenses within the system
before the steam mixes with the air in the duct. There are two areas in
which steam losses occur:
•
Within the ultimateSAM distributor itself;
•
Within the piping between the humidifier and the ultimateSAM distributor.
To achieve maximum operating efficiency, the ultimateSAM distributor is
insulated to minimize steam loss due to condensation. The design includes
a header wrapped with stainless-steel-clad insulating foam and uprights
with stainless-steel insulating shields.
Table 4.h provides information on the estimated steam loss, expressed as
a percentage of steam capacity. Values can be used to compare the effect
of different configurations on steam loss, given the same distributor size
(width code: “J”, height code: “J”). It is important to allow for this loss when
selecting a distributor configuration.
Nominal Steam Loss @ 15C (59F) (% of max. capacity)
Air velocity m/s (fpm)
Configurazione 3 (600) 6 (1200) 10 (2000)
SATJJSI*** 5 6 6
SABJJSI*** 9 12 14
SATJJLI*** 3 4 4
SABJJLI*** 6 8 9
SATJJHI*** 3 4 5
SABJJHI*** 8 10 11
SATJJSN*** 7 8 9
SABJJSN*** 13 15 18
SATJJLN*** 4 5 6
SABJJLN*** 8 10 12
SATJJHN*** 5 6 7
SABJJHN*** 11 13 15
Tab. 4.k
Note:
1.
Compared to a top-feed distributor with comparable configuration,
width and height, a bottom-feed distributor has 2 times the steam loss
as a percentage of capacity because the bottom-feed has 1/3 of the
maximum capacity of the top-feed.
2. Compared to insulated distributors, uninsulated distributors have 40%
more steam loss. For example, at 3 m/s (600 fpm) an SABJJSI*** has a
steam loss of 9% of 110 kg/hr (240 lb/hr), that is 10 kg/hr (22 lb/hr). The
uninsulated version, SABJJSN***, has 40% greater steam loss, that is 14
kg/hr (31 lb/hr), or 13% of capacity.
Note: Besides increased steam loss, uninsulated distributors are
likely to inject condensate into the airway because the
distributors do not have nozzle inserts. See section 4.8.)
To calculate estimated steam loss for specific width and height codes,
tables 4.i and 4.j provide steam loss per length of upright and header.
Note
: As shown, steam loss increases as ambient temperature
deceases. To calculate estimated steam loss at other ambient
temperatures (Ta), adjust the values by the ratio (Ta-100)/85 or (Ta-100)/75
for tables 4.i and 4.j, respectively.
To calculate the total steam loss,
1. Calculate the steam loss for the uprights
2.
Calculate the steam loss of the header(s)
Steam Loss @ 15C (59F) kg/hr/m (lb/hr/ft)
Air velocity m/s (fpm)
Configuration
3 ( 600) 6 (1200) 10 (2000)
Uprights
“SA***SI***
SA***HI***” 0.34 (0.23) 0.42 (0.28) 0.48 (0.32)
“SA***SN***
SA***HN***” 0.48 (0.32) 0.59 (0.39) 0.67 (0.45)
SA***LI*** 0.39 (0.26) 0.49 (0.33) 0.56 (0.38)
SA***LN*** 0.55 (0.37) 0.69 (0.46) 0.78 (0.53)
Headers
SAB***I*** 2.0 (1.4) 2.5 (1.7) 2.9 (1.9)
SAB***N*** 2.5 (1.7) 3.1 (2.1) 3.5 (2.4)
SAT***I*** 4.5 (3.0) 5.6 (3.8) 6.4 (4.3)
SAT***N*** 7.0 (4.7) 8.7 (5.8) 9.9 (6.7)
Tab. 4.l