Flow Rates in Heating Systems

Calculating flow rates in heating systems

The volumetric flow rate in a heating system can be expressed as

q = h / (cp ρ dt)         (1)

where

q = volumetric flow rate

h = heat flow rate

cp = specific heat capacity

ρ = density

dt = temperature difference

The basic 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 oF flow rate can be expressed as

q = h (7.48 gal/ft3) / ((1 Btu/lbm.oF) (62.34 lb/ft3) (60 min/h) dt)        

    = h / (500 dt)         (2)

where

q = water flow rate (gal/min)

h = heat flow rate (Btu/h)

dt = temperature difference (oF)

For more exact volumetric flow rates for hot water 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.oF) dt)        

    =  h / (1.2 dt)     (3)

where

m = mass flow (lbm/h)

Volumetric Water Flow Rate in SI-Units

Volumetric water flow in a heating system can be expressed with SI-units as

q = h / ((4.2 kJ/kgoC) (1000 kg/m3) dt)        

    = h / (4200 dt)        (4)

where

q = water flow rate (m3/s)

h = heat flow rate (kW or kJ/s)

dt = temperature difference (oC)

For more exact volumetric flow rates for hot water the properties of hot water should be used.

Water Mass Flow Rate in SI-units

Mass flow of water can be expressed as:

m = h / ((4.2 kJ/kg.oC) 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 20oC.

The volumetric flow can be expressed as:

q = (230 kW) / ((4.2 kJ/kg oC) (1000 kg/m3) (20 oC))

    = 2.7 10-3 m3/s

The mass flow can be expressed as:

m = (230 kW) / ((4.2 kJ/kg oC) (20oC))

    = 2.7 kg/s

Example - Heating Water with Electricity

10 liters of water is heated from 10oC to 100oC in 30 minutes. The heat flow rate can be calculated as 

h = (4.2 kJ/kg oC) (1000 kg/m3) (10 liter) (1/1000 m3/liter) ((100oC) - (10oC)) / ((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 - Heating systems - capacity and design of boilers, pipelines, heat exchangers, expansion systems and more

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