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An electrostatic precipitator (ESP) with $5600 \mathrm{~m}^{2}$ of collector plate area is 96 percent efficient in treating $185 \mathrm{~m}^{3} / \mathrm{s}$ of flue gas from a $200 \mathrm{MW}$ thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be $6100 \mathrm{~m}^{2}$. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in $\mathrm{m}^{2}$ ) would be



Question ID - 156331 | SaraNextGen Top Answer

An electrostatic precipitator (ESP) with $5600 \mathrm{~m}^{2}$ of collector plate area is 96 percent efficient in treating $185 \mathrm{~m}^{3} / \mathrm{s}$ of flue gas from a $200 \mathrm{MW}$ thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be $6100 \mathrm{~m}^{2}$. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in $\mathrm{m}^{2}$ ) would be

1 Answer
127 votes
Answer Key / Explanation : (8000-8040) -

8000-8040

127 votes


127