Bosch UT-L 1 Technical Manual page 54

54 | System examples
Boiler circuit pump flow rate V
The boiler circuit pump, also known as a shunt pump, is
required to control the return temperature (flow past
the sensor). The control characteristics can also be
optimised using the boiler circuit pump. This makes it
possible to minimise switching during the heat-up
phase. This results in lower emissions.
V =
PK
F. 3 Calculating the flow rate of the boiler circuit
pump
c Specific heat capacity
-3
c = 1.16 × 10 kWh/(l × K) = 1/860 kWh/(l × K)
Δϑ
Temperature differential for sizing the boiler circuit
K
pump 30 to 50 K (30 K for optimised heat-up
characteristics)
Rated output in kW
Q
K
V Flow rate of the boiler circuit pump in l/h
PK
Heating circuit flow rate V
----------------------------------------- -
V =
(
HK
F. 4 Calculating the flow rate of the heating circuits
c Specific heat capacity
-3
c = 1.16 × 10 kWh/(l × K) = 1/860 kWh/(l × K)
ϑ
/ϑ Return/flow temperature of the heating circuits
R V
in °C
Q Heat demand of the heating circuits in kW
HK
V Flow rate of the heating circuits in l/h
HK
6 720 807 794 (2013/04)
PK
Q
K
----------------------- -
Δυ ×
c
K
HK
Q
HK
ϑ ϑ ) c ×
–
V R
Total boiler flow rate V
Kges
The head of the boiler circuit pump is the result of:
• The pressure loss of the boiler at the selected flow
rate V
PK
• The pressure loss in the pipework and
• All individual pressure loss levels in the boiler circuit
(path: A–C–D–B,  Fig. 35).
Due to the pump and system curves, the total flow rate
via the boiler cannot be calculated simply by adding up
the individual flow rates. However, as an initial estimate,
the simple addition is adequate for a rough calculation.
Base the sizing of the pipework in the boiler
circuit on a flow velocity of 1 m/s to 2.3 m/s.
≤
V V
Kges PK
F. 5 Calculating the total flow rate of the
boiler
V Flow rate of the heating circuits in l/h
HK
Maximum total flow rate through the boiler in l/
V
Kges
h (estimate)
V Flow rate of the boiler circuit pump in l/h
PK
Example
Given
• Rated output Q = 2500 kW
K
• Heating system flow temperature ϑ
• Heating system return temperature ϑ
• Temperature differential (selected) Δϑ
Result
= 43000 l/h (path: C–D,  Fig. 35)
• V
PK
• V = 107500 l/h
HK
(paths: C–F, D–G and E–H,  Fig. 35)
≈ 150500 l/h
• V
Kges
(paths: A–C and B–D,  Fig. 35)
+ V
HK
= 90 °C
V
= 70 °C
R
= 50 K
K
UNIMAT