12
Limit of elevation
difference between
the 2 units
No additional charge of compressor oil is necessary.
*
1
If total tubing length becomes 30 to 50 m, charge additional refrigerant by 40 g/m.
Models
GRFP255L
GRFP255R
GRFP365L
GRFP365R
GRFP485L
GRFP485R
Tubing Data
Tubing size
Narrow tube
mm (in.)
9.52 (3/8)
9.52 (3/8)
9.52 (3/8)
outer diameter
Wide tube
mm (in.)
15.88 (5/8)
15.88 (5/8)
15.88 (5/8)
Limit of tubing length
(m)
50
50
50
Outdoor unit is placed
30
30
30
higher
(m)
Outdoor unit is placed
15
15
15
lower
(m)
Max. allowable tubing length at shipment (m)
3 – 30
3 – 30
5 – 30
Required additional refrigerant *
1
(g/m)
a) 40
b) 40
b) 40
Refrigerant charged at shipment
(kg)
1.2
1.2
1.2
Table 1-6 Tubing Data for Models (Single, Twin, Triple, Double-Twin)
25 – 48
25 – 48
12 – 18
φ
15.88
φ
15.88
φ
15.88
φ
9.52
φ
9.52
φ
9.52
φ
6.35
φ
12.7
( 1, 2, 3, 4)
20 g
40 g
40 g
40 g
Wide tube
Narrow tube
Main tubing (L)
25
Double-Twin distribution tube (L1, L2)
Total type capacity of indoor units connected after
the branch
Indoor unit connection tube
Type capacity of indoor units
Amount of additional charge per 1 m
Table 1-7 List of Connection Tube Sizes
Charge with an amount of additional refrigerant calculated using the formula below, based on the values in Table 1-7
and the size and length of the liquid (narrow) tubing.
Amount of additional refrigerant charge (g)
Do not remove refrigerant from the system, even if the result of the calculation is negative. (Use with the current
refrigerant charge.)
(a) Actual
length (m)
of main
tubing
(
φ
9.52)
Refrigerant charge per 1 m of actual length = 40 g/m (25, 36 and 48)
(b) Total length of distribution tubing (
φ
9.52)
Refrigerant charge per 1 m of actual length = 40 g/m
(c) Total length of distribution tubing (
φ
6.35)
Refrigerant charge per 1 m of actual length = 20 g/m
Example
●
Sample tubing lengths
L = 35 m
1 = 5 m
L1 = 10 m
2 = 5 m
L2 = 5 m
3 = 5 m
4 = 10 m
●
Find the liquid (narrow) tube size from Table 1-7.
L :
φ
9.52 (48 type)
L1 :
φ
9.52 (Total type capacity of indoor units 25)
L2 :
φ
9.52 (Total type capacity of indoor units 25)
1 – 4
:
φ
6.35
●
The amount of additional on-site refrigerant charge is found by subtracting the outdoor unit
charge-less refrigerant amount from the total charge amount for all tube sizes.
φ
9.52
→
L
: 35 m
×
40 g / m = 1400
φ
9.52
→
L1 + L2
: (10 + 5) m
×
40 g / m = 600
φ
6.35
→
1 –
4 : (5 + 5 + 5 + 10) m
×
20 g / m = 500
Outdoor unit charge-less refrigerant amount
–1200
Total +1300
●
The amount of additional on-site refrigerant charge is 1,300 g.
Outdoor unit
Indoor unit (12 type
×
4)
L
L1
L2
4
3
2
1
Additional refrigerant amount (g) = Additional refrigerant for main tubing (g) + Additional refrigerant for distribution tubing (g)
– Outdoor unit charge-less refrigerant amount (g)
= 40X (a) + 40X (b) + 20X (c) – 1200 (25, 36 and 48)
04-393 DC INV Tech p a-44 12/15/04 3:04 PM Page 12
