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# Steam Heating Processes - 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)

where

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:

P = m cp dT / t                                       (2)

where

P = mean heat transfer rate or power (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 / P

= (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

P = cp dT m / t                                   (3)

where

P = 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 = P / he                                       (4)

where

ms = mass of steam (kg/s)

P = 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 with 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:

P = (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:

P = (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

## Related Topics

• Steam and Condensate - Steam & condensate systems- properties, capacities, pipe sizing, systems configuration and more.
• Thermodynamics - Thermodynamics of steam and condensate systems.
• Heat Loss and Insulation - Steam and condensate pipes - heat loss uninsulated and insulated pipes, insulation thickness and more.
• Pipe Sizing - Sizing of steam and condensate pipe lines - pressure loss, recommended velocity, capacity and more.

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