# 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 Systems

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

## Related Documents

### District Heating - Heat Capacity vs. Temperature

Water temperature and heating capacity.

### Heating Systems - Safety Valves Size vs. Boiler Power

Safety valves with boilers ranging *275 to 1500 kW*.

### Heating Up Applications - Energy Required and Heat Transfer Rates

Energy required to heat up a substance.

### Hot Water Boiler - Circulation Rates

Boiler power and water flow - Imperial and SI-units.

### Hot Water Expansion Tanks - Sizing

Required hot water expansion volume in open, closed and diaphragm tanks.

### Hot Water Heating System Temperatures vs. Outdoor Temperatures

Hot water heating temperatures adapts to outdoor temperatures.

### Hot Water Heating Systems - Flow Temperatures vs. Outside Temperatures

Seasonal effects on hot-water heating systems flow temperatures.

### Hot Water Heating Systems - Online Design Application

Free online design tool for designing hot water heating systems - metric units.

### Hot Water Heating Systems - Online Design Application, Imperial Units

Online design tool for hot water heating systems.

### Hot Water Heating Systems - Pressure Loss in Steel Pipes

Pressure loss nomogram for hot water steel pipes.