Measuring Efficiency



Efficiency can be measured either using a standard or as absolute efficiency.  Absolute efficiency is calculated as efficiency of removal, efficiency of transfer, or efficiency of addition.



Using a Standard

Efficiency ratings can be used to set goals, maintain efficient performance, and even break those goals.  In these cases, efficiency is often measured against a standard.  A standard is the most that can be achieved in a given set of circumstances.  

Let's consider a wastewater treatment plant.  When running at its most efficient, this plant requires $50 worth of energy to remove a ton of sewage from the water.  So the standard in this case is $50/ton.  

The following formula can be used to determine the percent efficiency of the plant compared to the standard:




What is the percent efficiency of the plant if it removes sewage at a cost of $50/ton?




In this case, the efficiency was 100% because the sewage was removed using the minimum amount of energy set by the standard.  

How about if the plant required $100 worth of energy to remove the same ton of B.O.D. from the wastewater?





This time, the efficiency was only 50%.  

In the example above, the goal is the standard - using only $50 worth of energy to remove a ton of B.O.D. from the water.  In the first example, the goal had been met and efficiency was 100%.  In the second example, the goal had not been met, since the efficiency was only 50%.  The second system would need to be made more efficient in order to meet the goal of 100% efficiency.  



Efficiency of Removal

There are several different types of efficiencies that play a role in water and wastewater treatment.  One of the most important in the wastewater field is the efficiency of removal, which involves the removal of a substance.

The efficiency of removal can be calculated using the following formula:



For example, the influent flowing into a wastewater treatment plant has a B.O.D. of 200 milligrams per liter.  Once the water has been treated, the B.O.D. of the effluent has been reduced to 10 milligrams per liter.  What is the efficiency of removal?

The initial amount is 200 mg/L and the reduced to amount is 10 mg/L.  So the efficiency would be calculated as follows:




So the efficiency of removal of the plant is 95%.

In addition to being important in the water and sewage treatment fields, the efficiency of removal works in other processes.



Efficiency of Transfer

The efficiency of transfer is a second type of efficiency which involves the ability to move a substance from one location to another.  In water and wastewater treatment, oxygen must be transferred from air to water.  Since aerating water is often a very expensive process, it is important that the efficiency of transfer be as great as possible.  

The efficiency of oxygen addition to water is influenced, in nature, by the air to surface area ratio, by the atmospheric pressure, and by the temperature.  In a wastewater treatment plant, the rate of oxygen addition is usually influenced most by the air to surface area ratio and by the method used to bring air in contact with water.

Large bubbles have less surface area for the mass of air than small bubbles do.  So these large bubbles are less efficient than small bubbles at the transfer of oxygen.  Despite this, most treatment plants use larger bubbles (greater than 1/8") anyway because less maintenance is required on the aerator nozzle.  

Since maintenance of the aerator nozzle often makes it impractical to pump small bubbles through water, we must look for a different way to improve aerator efficiency.  The two main types of aerators - air diffusion aerators and step aerators - differ greatly in efficiency.  

Energy is lost when compressed air is uncompressed.

Air diffusion aerators bring air in contact with water by pumping air through the water in bubbles, as discussed above. In order to pump air, the air must be compressed, a step which uses energy.  When the air is released into the water, the air becomes uncompressed and the energy used to compress the air is lost.  As a result of the energy lost in this way, pumping air is one of the most inefficient ways of putting oxygen in water.



Instead, oxygen can be added to water using step aerators.  In this case, water is pumped through air rather than vice versa. Water is non-compressible, so when it is pumped, no energy is used up compressing the water and energy of compression is not lost in the system.  

In a step aerator, water is pumped over a set of baffles, or steps.  This method of aeration is much more efficient and much cheaper than the compressed air method.  However, step aerators are limited by space - there must be enough space for the reaction to occur.  

At the Blountville, Tennessee, sewage plant, water is pumped over baffles and the average electricity cost per ton of B.O.D. removal is $89.  In contrast, the CNW plant at Coeburn, Virginia, pumps air and the average electricity cost per ton of B.O.D. removal is $350.  So the Coeburn plant is a much less efficient system.  (Most of the greater cost at the Coeburn plant is due to the type of aeration.  However, the Blountville plant also does not thicken sludge, which makes the treatment process even cheaper than it would have been otherwise. Instead, the Blountville plant mixes the sludge with lawn clippings, wood chips, leaves, etc. to make compost for land application.)



Efficiency of Addition

The third type of efficiency we will consider is the efficiency of addition, which is concerned with the addition of some substance.   The efficiency of addition is really just another way of looking at the efficiency of transfer.  The substance which is added has to be transferred from one location to another.

To illustrate, let's consider the process of raising animals.  In order for a farmer to make a profit from selling livestock, he must be concerned with the efficiency of addition.  In this case, that means the efficiency with which energy is transferred from the food to the animal.  A pig farmer would determine efficiency by pounds produced in meat versus pounds in grain fed:


 

To improve efficiency, the farmer would only breed pigs that consistently yielded an increase in weight with the least amount of food.  

A very efficient pig may gain 17.5 pounds of weight for every 50 pounds of grain fed.  So its efficiency would be:



The pig has an efficiency of addition of only 35%.  In contrast, a fish may be about 75% efficient in food to weight gain. Some of this difference in efficiency can be attributed to the warm-blooded nature of pigs and the cold-blooded nature of fish.  Pigs and other warm-blooded animals must use up some energy to keep themselves warm, while fish are dependent on the surrounding waters to maintain body temperature.  The energy pigs spend in keeping themselves warm is like the energy some aerators use to compress air - it is lost to the system and makes the system less efficient.