Steam Heating Process - Load Calculating

Calculating the amount of steam in non-flow batch and continuous flow heating processes

In general steam heating is used to

  • change a product or fluid temperature
  • maintain a product or fluid temperature

A benefit with steam is the large amount of heat energy that can be transferred. The energy released when steam condenses to water is in the range 2000 - 2250 kJ/kg (depending on the pressure) - compared to water with 80 - 120 kJ/kg (with temperature difference 20 - 30 oC). 

Changing Product Temperature - Heating up the Product with Steam

The amount of heat required to raise the temperature of a substance can be expressed as:

Q = m cp dT                                       (1)


Q = quantity of energy or heat (kJ)

m = mass of substance (kg)

cp = specific heat of substance (kJ/kg oC ) - Material properties and heat capacities common materials

dT = temperature rise of substance (oC)

Imperial Units? - Check the Units Converter!

This equation can be used to determine a total amount of heat energy for the whole process, but it does not take into account the rate of heat transfer which is:

  • amount of heat energy transferred per unit time

In non-flow type applications a fixed mass or a single batch of product is heated. In flow type applications the product or fluid is heated when it constantly flows over a heat transfer surface.

Non-flow or Batch Heating

In non-flow type applications the process fluid is kept as a single batch within a tank or vessel. A steam coil or a steam jacket heats the fluid from a low to a high temperature.

The mean rate of heat transfer for such applications can be expressed as:

q = m cp dT / t                                       (2)


q = mean heat transfer rate (kW (kJ/s))

m = mass of the product (kg)

cp = specific heat of the product (kJ/kg.oC) - Material properties and heat capacities common materials

dT = Change in temperature of the fluid (oC)

t = total time over which the heating process occurs (seconds)

Example - Time required to Heat up Water with direct Injection of Steam

The time required to heat 75 kg of water (cp = 4.2 kJ/kgoC) from  temperature 20oC to 75oC with steam produced from a boiler with capacity 200 kW (kJ/s) can be calculated by transforming eq. 2 to

t = m cp dT / q

 = (75 kg) (4.2 kJ/kgoC) ((75 oC) - (20 oC)) / (200 kJ/s) 

 = 86 s

Note! - when steam is injected directly to the water all the steam condenses to water and all the energy from the steam is transferred instantly.

When heating through a heat exchanger - the heat transfer coefficient and temperature difference between the steam and the heated fluid matters. Increasing steam pressure increases temperature - and increases heat transfer. Heat up time is decreased.

Overall steam consumption may increase - due to higher heat loss, or decrease - due to to shorter heat up time, depending on the configuration of the actual system.  

Flow or Continuous Heating Processes

In heat exchangers the product or fluid flow is continuously heated.

A benefit with steam is homogeneous heat surface temperatures since temperatures on heat surfaces depends on steam pressure.

The mean heat transfer can be expressed as

q = cp dT m / t                                   (3)


q = mean heat transfer rate (kW (kJ/s))

m / t = mass flow rate of the product (kg/s)

cp = specific heat of the product (kJ/kg.oC)

dT = change in fluid temperature (oC)

Calculating the Amount of Steam

If we know the heat transfer rate - the amount of steam can be calculated:

ms = q / he                                       (4)


ms = mass of steam (kg/s)

q = calculated heat transfer (kW)

he = evaporation energy of the steam (kJ/kg)

The evaporation energy at different steam pressures can be found in the Steam Table with SI Units or in the Steam Table with Imperial Units.

Example - Batch Heating by Steam

A quantity of water is heated with steam of 5 bar (6 bar abs) from a temperature of 35 oC to 100 oC over a period of 20 minutes (1200 seconds). The mass of water is 50 kg and the specific heat of water is 4.19 kJ/kg.oC.

Heat transfer rate:

q = (50 kg) (4.19 kJ/kg oC) ((100 oC) - (35 oC)) / (1200 s)

    = 11.35 kW

Amount of steam:

ms = (11.35 kW) / (2085 kJ/kg)

    = 0.0055 kg/s

    = 19.6 kg/h

Example - Continuously Heating by Steam

Water flowing at a constant rate of 3 l/s is heated from 10 oC to 60 oC with steam at 8 bar (9 bar abs).

The heat flow rate can be expressed as:

q = (4.19 kJ/kg.oC) ((60 oC) - (10 oC)) (3 l/s) (1 kg/l)

    = 628.5 kW

The steam flow rate can be expressed as:

ms = (628.5 kW) / (2030 kJ/kg)

    = 0.31 kg/s

    = 1115 kg/h

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