The removal efficiency of an electrostatic precipitator can be expressed mathema
ID: 115148 • Letter: T
Question
The removal efficiency of an electrostatic precipitator can be expressed mathematically as:
E = 100 * (1-e-wA/Q)
Where E = percent removal efficiency
e = base of natural logarithm, equal to 2.718
w= effective drift velocity
A=total surface area of plates
Q=gas flow rate through the unit
From the equation above, which of the following statements can be directly concluded?
The gas flow rate must be slow and uniform across each plate to obtain high removal efficiencies.
The removal efficiency will increase proportionately with increasing surface area of the electrostatic plates
Doubling the drift velocity will cut the efficiency by half
Increasing drift velocity will increase efficiency, but there is a physical limit to this as we need to maintain constant drift velocity
The gas flow rate must be slow and uniform across each plate to obtain high removal efficiencies.
The removal efficiency will increase proportionately with increasing surface area of the electrostatic plates
Doubling the drift velocity will cut the efficiency by half
Increasing drift velocity will increase efficiency, but there is a physical limit to this as we need to maintain constant drift velocity
Explanation / Answer
The gas flow rate must be slow and uniform across each plate to obtain high removal efficiencies. this is because according to the formula the Q is in denominator so as Q decreases the Efficiency increases,
Increasing drift velocity will increase efficiency, but there is a physical limit to this as we need to maintain constant drift velocity. This is also true as the w increased so 1/ewa/Q decreases thus increasing the overall efficiency.