A correlation for particulate removal efficiency in an electrostatic precipitato

Question

A correlation for particulate removal efficiency in an electrostatic precipitator (%u03B7) as a function of gas velocity (u) may be written in the form: %u03B7 = 1 %u2012 e%u2012k / u. An engineering student, recognizing that the "ideal gas law" requires gas velocity to be function of temperature, suggests that an easy way to increase efficiency would be to change the temperature of the exhaust gas being treated. If the removal efficiency in a precipitator is 93% at a temperature of 1200oC, calculate a value of temperature that could be used to increase the efficiency to 99%.

Explanation / Answer

v = sqrt (3RT/M)

so v is directly proportional to the square root of temp

n = 1 - e^(-k / u)

0.93=1-e^(-k/sqrt(1200))

k=102.06

now for n=0.99

n = 1 - e^(-k / u)

0.99=1 - e^(-102.06/ sqrt(T))

T=491.15 K

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