Operation Manual

<13. GENERAL SPECIFICATIONS>
13-18
IM 01F06A00-01EN
Reference
Table 13.11 Inner Diameter and Nominal value
Model Code
Inner
Diameter
mm
Nominal
K-Factor
Pulse/L
Nominal Pulse
Rate
Hz / m/s Hz / m
3
/h
DY015
DY025
/R1
DY040
/R2
14.6 376 62.7 104
DY025
DY040
/R1
DY050
/R2
25.7 68.6 35.5 19.1
DY040
DY050
/R1
DY080
/R2
39.7 18.7 23.1 5.19
DY050
DY080
/R1
DY100
/R2
51.1 8.95 18.3 2.49
DY080
DY100
/R1
DY150
/R2
71.0 3.33 13.2 0.925
DY100
DY150
/R1
DY200
/R2
93.8 1.43 9.88 0.397
DY150
DY200
/R1
138.8 0.441 6.67 0.123
DY200 185.6 0.185 5.00 0.0514
DY250 230.8 0.0966 4.04 0.0268
DY300 276.2 0.0563 3.37 0.0156
DY400 354.2 0.0265 2.61 0.00736
Pressure Loss
Calculation of pressure loss for standard type
obtained from the following equations.
P=108×10
5
×
f
×
2
······ (1)
or
P=135×
f
×
Q
f
2
D
4
········· (2)
where,
P: Pressure loss (kPa )
f
: Density at operating condition (kg/m
3
)
: Flow velocity (m/s)
Q
f
: Actual ow rate (m
3
/h)
D: Inner diameter of digitalYEWFLO (mm)
(Example)
DY050, hot water: 80°C, owrate: 30 m
3
/h
1. Since the density of water at 80°C is 972 kg/
m
3
, substitute this value in equation (2):
P = 135 × 972 × 30
2
/ 51.1
4
= 17.3 kPa
2. Obtain the pressure loss using equation (1).
The ow velocity when the ow rate is 30 m
3
/h
is given by:
= 354×Q
f
/D
2
=
354×30
51.1
2
= 4.07m/s
Therefore, substitute this value in equation (1):
P = 108 × 10
-5
× 972 × 4.07
2
= 17.3 kPa
Calculation of pressure loss for reduced bore
type (Option code: /R1)
obtained from the following equations.
P = 124 × 10
-5
×
f
×
2
··········· (3)
or
P = 155 ×
f
× Q
f
2
/ D
4
············ (4)
(Example)
DY040/R1, hot water: 50 °C, owrate: 10 m
3
/h
1. Since the density of water at 50 °C is 992 kg/
m
3
, substitute this value in equation (4):
P = 155 × 992 × 10
2
/ 25.7
4
= 35.3 kPa
2. Obtain by using equation (3). The ow velocity
when the ow rate is 10 m
3
/h is given by:
= 354 × Q
f
× /D
2
= 354 × 10 × 25.7
2
= 5.4m/s
Therefore, substitute this value in equation (3):
P = 124 × 10
-5
× 992 × 5.4
2
= 35.3 kPa
Calculation of pressure loss for reduced bore
type (Option code: /R2)
obtained from the following equations.
P = 138 × 10
-5
×
f
×
2
··········· (5)
or
P = 173 ×
f
×
Q
f
2
D
4
················ (6)
(Example)
DY050-/R2, hot water: 50 °C, owrate: 15 m
3
/h
1. Since the density of water at 50 °C is 992 kg/
m
3
, substitute this value in equation (6):
P = 173 × 992 × 15
2
/ 25.7
4
= 88.5 kPa
2. Obtain by using equation (5). The ow velocity
when the ow rate is 15m
3
/h is given by:
= 354×Q
f
/D
2
=
354×15
25.7
2
= 8.0m/s
Therefore, substitute this value in equation (5):
P = 138 × 10
-5
× 992 × 8.0
2
= 88.5 kPa