# 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 ^{o}C*).

### 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 c_{p}dT(1)

where

Q= quantity of energy or heat (kJ)

m= mass of substance (kg)

c_{p}= specific heat of substance (kJ/kg^{o}C ) - Material properties and heat capacities common materials

dT= temperature rise of substance (^{o}C)

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 c_{p}dT / t(2)

where

P = mean heat transfer rate or power (kW (kJ/s))

m= mass of the product (kg)

c_{p}= specific heat of the product (kJ/kg.^{o}C) -Material properties and heat capacities common materials

dT= Change in temperature of the fluid (^{o}C)

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 *(c _{p} = 4.2 kJ/kg^{o}C)* from temperature

*20*to

^{o}C*75*with steam produced from a boiler with capacity

^{o}C*200 kW (kJ/s)*can be calculated by transforming eq. 2 to

*t = m c _{p} dT / P*

* = (75 kg) (4.2 kJ/kg^{o}C) ((75 ^{o}C) - (20 ^{o}C)) / (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 = c_{p}dT m / t(3)

where

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

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

c_{p}= specific heat of the product (kJ/kg.^{o}C)

dT= change in fluid temperature (^{o}C)

### Calculating the Amount of Steam

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

m_{s}= P / h_{e}(4)

where

m_{s}= mass of steam (kg/s)

P= calculated heat transfer (kW)

h_{e}= 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 ^{ o}C to 100^{ o}C* 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.*.

^{o}CHeat transfer rate:

P = (50 kg) (4.19 kJ/kg^{o}C) ((100^{ o}C) - (35^{ o}C)) / (1200 s)

= 11.35 kW

Amount of steam:

m_{s}= (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 ^{ o}C to 60^{ o}C* with steam at

*8 bar (9 bar abs)*.

The heat flow rate can be expressed as:

P = (4.19 kJ/kg.^{o}C) ((60^{ o}C) - (10^{ o}C)) (3 l/s) (1 kg/l)

= 628.5 kW

The steam flow rate can be expressed as:

m_{s}= (628.5 kW) / (2030 kJ/kg)

= 0.31 kg/s

= 1115 kg/h