# Heating System Flow Rates

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

q = h / (c_{p}ρdt)(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 SI-Units

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

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*in

^{o}C*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

Heating systems - capacity and design of boilers, pipelines, heat exchangers, expansion systems and more.

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