# Heating System Flow Rates

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 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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