Heating System Flow Rates
Calculate flow rates in heating systems.
The volumetric flow rate in a heating system can be expressed as
q = h / (c_{p} ρ dt) (1)
where
q = volumetric flow rate (m^{3} /s )
h = heat flow rate (kJ/s, kW)
c_{p} = specific heat (kJ/kg ^{o}C )
ρ = density (kg/m^{3} )
dt = temperature difference (^{o}C)
This generic equation can be modified for the actual units  SI or imperial  and the liquids in use.
Volumetric Water Flow Rate in Imperial Units
For water with temperature 60 ^{o}F flow rate can be expressed as
q = h (7.48 gal/ft^{3} ) / ((1 Btu/lb_{m} ^{o}F) (62.34 lb/ft^{3} ) (60 min/h) dt)
= h / (500 dt) (2)
where
q = water flow rate (gal/min)
h = heat flow rate (Btu/h)
ρ = density ( lb/ft^{3} )
dt = temperature difference ( ^{o}F)
For more exact volumetric flow rates the properties of hot water should be used.
Water Mass Flow Rate in Imperial Units
Water mass flow can be expressed as:
m = h / ((1.2 Btu/lbm. ^{o}F) dt)
= h / (1.2 dt) (3)
where
m = mass flow (lb_{m} /h)
Volumetric Water Flow Rate in SIUnits
Volumetric water flow in a heating system can be expressed with SIunits as
q = h / ((4.2 kJ/kg ^{o}C) (1000 kg/m^{3} ) dt)
= h / (4200 dt) (4)
where
q = water flow rate (m^{3} /s)
h = heat flow rate (kW or kJ/s)
dt = temperature difference (^{o}C)
For more exact volumetric flow rates the properties of hot water should be used.
Water Mass Flow Rate in SIunits
Mass flow of water can be expressed as:
m = h / ((4.2 kJ/kg ^{o}C) dt)
= h / (4.2 dt) (5)
where
m = mass flow rate (kg/s)
Example  Flow Rate in a Heating System
A water circulating heating systems delivers 230 kW with a temperature difference of 20 ^{o}C .
The volumetric flow can be calculated as:
q = (230 kW) / ((4.2 kJ/kg ^{o}C) (1000 kg/m^{3} ) (20 ^{o}C))
= 2.7 10 ^{3} m^{3} /s
The mass flow can be expressed as:
m = (230 kW) / ((4.2 kJ/kg ^{o}C) (20 ^{o}C))
= 2.7 kg/s
Example  Heating Water with Electricity
10 liters of water is heated from 10 ^{o}C to 100 ^{o}C in 30 minutes . The heat flow rate can be calculated as
h = (4.2 kJ/kg ^{o}C) (1000 kg/m^{3} ) (10 liter) (1/1000 m^{3} /liter) ((100 ^{o}C)  (10 ^{o}C)) / ((30 min) (60 s/min))
= 2.1 kJ/s (kW)
The 24V DC electric current required for the heating can be calculated as
I = (2.1 kW) (1000 W/kW)/ (24 V)
= 87.5 Amps
Related Topics

Heating Systems
Design of heating systems  capacities and design of boilers, pipelines, heat exchangers, expansion systems and more.
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