Danfoss AVPL Data sheet

Data sheet

Differential pressure controller (PN 16)

AVPL - return mounting, adjustable setting

Description

Ordering

Example:
Differential pressure controller,
return mounting, DN 15, kvs 1.0,
PN 16, setting range 0.05 - 0.25 bar,
t

120 °C, ext. thread

max

- 1× AVPL DN 15 controller
Code No: 003L5030

Option:

- 1× Weld-on tailpieces
Code No: 003H6908

It can be used on primary side of house
substations for smaller systems such as one and
two-family houses.

The controller could be used to control the
differential pressure across radiator systems and
similar systems to keep a constant differential
pressure even with a variable system resistance
kva and/or supply pressure ∆p0.

Main data:

• DN 15

• kvs 1.0, 1.6 m3/h

• PN 16

• Setting range: 0.05-0.25 bar

(factory setting 0.1 bar)
AVPL is a self-acting differential pressure
controller primarily for use in district heating
systems. The controller closes on rising
differential pressure.

• Temperature:

- Circulation water / glycolic water up to 30%:
2 … 120 °C

• Connections:

- Ext thread (weld-on and thread tailpieces)

The controller has a control valve and an actuator
with one control diaphragm.

AVPL Controller

Picture

* Controller inc l. impulse tube set AH (1.5 m at kvs 1.0 and 2.5 m at kvs 1.6) and nipple G⁄ - R⁄ fo r impulse tube connecti on to pipe

DN

(mm)

15

kVS

(m3h)

1.0

1.6 003L5031

Ext. thread acc. to ISO 228/1 G ¾ A 0.05-0.25

Connection

∆p setting range

(bar)

Code No. *

003L5030

Accessories

Picture Type designation DN Connection Code No.

© Danfoss | 2020.03

Weld-on tailpieces

15

External thread tailpieces

Description:

Impulse tube set AH

Fitting for impulse tube connection to pipe

EPP insulation box

10 EPDM o-rings for impulse tube 003L8175

1)

The material fo r the insulation box is approve d according to the fire hazard classif ication B2, DIN 4102.

1)

- 1× copper tube Ø 3 × 1 mm

- 2× fitting for imp. tube connec tion
to actuator and pipe G /

Conical ex t. thread acc. to

EN 10226-1

- 003H6908

AI180286474027en-000902 | 1

R / 003H6902

1.5 m 003L3561

2.5 m 003L5043
5 m 003L3562

G / - R / 003L5042

G / - R ¼ 003L8151

00 3L817 0

Data sheet Differential pressure controller AVPL (PN 16)

Technical data

Application principles

The controller AVPL could be
installed in the return pipeline only.

Nominal diameter DN

k

value m3/h 1.0 1.6

vs

Cavitation factor z 0.5
Nominal pressure PN 16
Max. differential pressure bar 4.5
Medium Circulation water / glycolic water up to 30%
Medium pH Min. 7, max. 10
Medium temperature

Connections

Materials

Valve body, etc. Dezincing free brass CuZn36Pb2As
Cone, seat, spindle and spring Stainless steel
Diaphragm and O-ring EPDM

Impulse tube

o

C 2 … 120

valve External thread

tailpieces Weld-on and external thread

Copper tube Ø 3 × 1 mm

Stainless steel tube Ø 0.8 × 0.2 × 800 mm

15

Installation positions

Direct-connected heating system Indirectly connected heating system

The controllers can be installed in any position.

2 | © Danfoss | 2020.03

AI180286474027en-000902

Data sheet Differential pressure controller AVPL (PN 16)

Sizing

Considering the correlation between the
capacity of the system kva, the system flow Q
and the differential pressure ∆pa, the controller
setting ∆pi is determined by:

∆pi = ∆pa = (Q/kva)

2

Based on the stated differential pressure of
the district heating ∆po and the calculated
differential pressure of the system ∆pa, the
differential pressure across the controller valve is
expressed as:

∆pv = ∆po - ∆p

a

Finally, a check is required to ensure that the
actual capacity of the controller k
than its max. capacity k

kvv = Q / √∆pv ≤ k

vs

vs

is smaller

vv

Example:

A heating system with a number of parallel hot
surfaces.
Required flow:
Q = 0.24 m3/h
Total capacity of the system determined to be kva
= 0.6 m3/h.

Calculation of the differential pressure across the
system:

∆pa = (Q/ kva )2 = (0.24 /0.6)2 = 0.16 bar (16 kPa)

The differential pressure from the district heating
is stated to be:

∆po = 0.5 bar (50 kPa) min

Calculation of the differential pressure across the
controller valve:

∆pv = ∆po - ∆pa = 0.5 bar - 0.16 bar = 0.34 bar (34 kPa)

In this example the capacity of the controller
valve is:

kvv = Q/√∆pv = 0.24/√0.34 = 0.412 m3/h

which is less that the max. capacity of the
controller = kvs = 1.0 m3/h.

Design

1. Spindle for differential

pressure setting

2. Bushing

3. Actuator

4. Control diaphragm

5. Setting spring for diff.

pressure control

6. Connection for impulse tube

7. O-ring

8. Pressure relieved valve cone

9. Seat

10. Valve body

AI180286474027en-000902

© Danfoss | 2020.03 | 3