Direct steam injection involves the discharge of steam bubbles into a liquid at a lower temperature to transfer heat. This tutorial explains the process and the methods used, including the relevant heat transfer calculations.
Direct steam injection involves the discharge of a series of steam bubbles into a liquid at a lower temperature. The steam bubbles condense and give up their heat to the surrounding liquid.
Heat is transferred by direct contact between the steam and the liquid, consequently this method is only used when dilution and an increase in liquid mass is acceptable. Therefore, the liquid being heated is usually water. Direct steam injection is seldom used to heat solutions in which a chemical reaction takes place, as the dilution of the solution would reduce the reaction rate and lower the productivity.
Direct steam injection is the most widely used method for boiler feedtank heating throughout industry. This method is often chosen because of its simplicity. No heat transfer surface or steam trap set is required, and there is no need to consider the condensate return system.
During direct steam injection, heat is transferred in a different manner to indirect heat exchange. As the heat is not transferred across a surface, and the steam mixes freely with the process fluid being heated, the amount of usable heat in the steam must be calculated in a different way. This can be found using Equation 2.11.1:
Equation 2.11.1 shows that steam injection utilises all of the enthalpy of evaporation (or latent heat) and a proportion of the liquid enthalpy contained in the steam. The actual proportion of the liquid enthalpy used will depend on the temperature of the water at the end of the injection process.
One major difference between indirect heating and direct steam injection, is that the volume (and mass) of the process fluid is increased as steam is added, by the amount of steam injected.
Another difference is that, when calculating the steam flowrate to a steam coil, the pressure in the coil is considered, but for steam injection, the pressure before the control valve is considered.
In some cases (where the liquid surface is not at the overflow pipe level), this will increase the head of liquid over the injector as time progresses. However, this increase is likely to be small and is rarely taken into account in calculations.
In Equation 2.11.1, the steam consumption rate is directly related to the heat requirement. Unless the steam injection system is designed so that all conditions are conducive to maximum heat transfer, the steam bubbles may simply break the surface of the liquid and escape to the atmosphere; some of the heat contained in the steam will be lost to atmosphere and the actual heat transfer rate to the water will be less than anticipated.
In the case of a submerged coil, the maximum heat transfer rate at the start of the warm-up period will depend on the maximum steam flowrate allowed through the control valve and its associated pipework, and the maximum heat output allowed by the coil surface area.
During direct steam injection, it might be expected that the maximum heat transfer rate at the very start of the warm-up period is dependent on the maximum flowrate through the control valve, and the pipe or injector itself. However, as implied above, it will also depend on other factors such as:
It is always advisable to ensure that the temperature of the liquid is correctly controlled and is kept to the minimum required for the application, so that the maximum heat transfer rate is maintained and there is no wastage of energy.
This is simply a pipe mounted inside the tank, with the holes drilled at regular positions (typically 4 o’clock and 8 o’clock) when viewed from the end, equally spaced along the length of the pipe, and with the end blanked off. The steam exits the pipe through the holes as small bubbles, which will either condense as intended or reach the surface of the liquid (see Figure 2.11.1).
Sparge pipes are inexpensive to make and easy to install, but are prone to cause high levels of vibration and noise. A much more effective method is to use a properly designed steam injector.
These calculations (steps 1 to 5) are based on Examples 2.9.1 and 2.10.1 as far as heat losses are concerned, but with the tank containing water (cp = 4.19 kJ/kg °C), instead of weak acid solution and the water being heated by steam injection rather than a steam coil.
Step 1 - find the energy required to heat up 12 000 kg of water from 8°C to 60°C in 2 hours by using Equation 2.6.1:
Steam is supplied to the control valve at 2.6 bar g. In order to calculate the mean steam flowrate, it is necessary to determine the total enthalpy in the steam (hg) at this pressure. It can be seen from Table 2.11.1 (an extract from steam tables) that the total enthalpy of steam (hg) at 2.6 bar g is 2733.89 kJ/kg.
Table 2.11.1 Extract from steam tables
|Pressure bar g||Saturation temperature °C||Specific enthalpy (energy) in kJ/kg||Specific volume of dry saturated steam m3/kg
|2.4||138.011||580.741||2 150.53||2 731.27||0.536766|
|2.5||139.023||585.085||2 147.51||2 732.60||0.522409|
|2.6||140.013||589.333||2 144.55||2 733.89||0.50882|
|2.7||140.98||593.49||2 141.65||2 735.14||0.495939|
Step 3 - find the mean steam flowrate to heat the tank material (steel).
From Example 2.9.1, the mean heat transfer rate for the tank material = Q̇(tank) = 14 kW
The mean steam flowrate to heat the tank material is calculated by again using Equation 2.11.1:
Whilst it is reasonable to accept that the steam’s liquid enthalpy will contribute to the rise in temperature of the water and the tank material, it is more difficult to accept how the steam’s liquid enthalpy would add to the heat lost from the tank due to radiation. Therefore, the equation to calculate the steam used for heat losses (Equation 2.11.2) considers only the enthalpy of evaporation in the steam at atmospheric pressure.
It is important to remember with steam injection systems that the final mass of liquid is equal to the mass of cold liquid, plus the mass of steam added.
In this example, the process started with 12 000 kg of water. During the required heat-up period of 2 hours steam has been injected at the rate of 569 kg/h. The mass of liquid has therefore, increased by 2 h x 569 kg/h = 1 138 kg.
The final mass of the liquid is: 12 000 kg + 1 138 kg = 13 138 kg
The additional 1138 kg of condensate has a volume of about 1 138 litres (1.138 m³) and will also have increased the water level by:
Clearly, the process tank needs to have sufficient space above the starting water level to allow for this increase. For safety, an overflow should always be included in the tank construction where steam injection is involved.
Alternatively, if the process requirement had been to finish with a mass of 12 000 kg, the mass of water at the beginning of the process would be:
A more effective alternative to the sparge pipe is the steam injector as shown in Figure 2.11.3. The injector draws in cold liquid and mixes it with steam inside the injector, distributing heated liquid to the tank.
The engineered design of the injector body is more sophisticated than the simple sparge pipe, and allows steam at higher pressures to be used. A turbulent zone is created within the body of the injector, which ensures that thorough mixing of the steam and liquid occurs, even at relatively high pressures. This has the effect of agitating and circulating the liquid so that a constant temperature is maintained throughout the tank, without temperature stratification or cold spots.
These injectors are more compact than sparge pipes, consequently any interference with objects that may be dipped in the tank can be avoided. They are more robust and generally quieter than sparge pipes, although noise problems may still be encountered if not installed correctly.
When using high pressure steam injectors three distinct noise levels are produced under the following conditions:
Based on data from Example 2.11.1, propose a steam injection system.
Required steam injection rate = 569 kg/h
The steam injection pressure = 1.0 bar
Table 2.11.2 Typical steam injector capacity chart
|Steam pressure at inlet of injector (bar g)||Saturated steam capacity kg/h|
|14||264||1 045||2 710|
|15||282||1 095||2 815|
|16||300||1 170||3 065|
|17||318||1 225||3 200|
The largest injector (IN40M) has a capacity of 400 kg/h at 1.0 bar, so this application will require:
Ideally, because of the low pressures involved, the injectors would be installed at opposite ends of the tank to give good mixing.
An alternative would be to use higher pressure steam. This would allow the use of just one, smaller injector, reducing costs and still providing good mixing.
The previous method used in this Module to calculate the mean steam flowrate requires the mean heat load to be calculated first. This is depicted by Equation 2.11.1:
If the mean heat transfer rate is not known, another method can be used to determine the mean steam flowrate. This requires the use of a heat balance as described below.
It should be noted that both methods return exactly the same result, so whichever is used depends upon the user’s choice.
Calculating the mean steam flowrate by means of a heat balance
A heat balance is considered where the initial heat content in the water plus the heat added by the steam equals the final heat content. The heat balance equation for the water in the tank is shown in Equation 2.11.3:
Mass of steam to be injected
The mass of steam to be injected can be determined more directly from Equation 2.11.4, which is developed from Equation 2.11.3.
Consider the same conditions as that in Example 2.11.1.
Conducting a heat balance on the water in the tank by using Equation 2.11.4:
Conducting a heat balance on the tank material
The heat losses from the sides of the tank and the water surface are the same as previously calculated, that is 24 kg/h.
This is the same result as that obtained previously in this Module from Equations 2.11.1 and 2.11.2, and proves that either method can be used to calculate the mean steam flowrate to heat the tank and its contents.