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An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be 6100 m2. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in m2) would be ______
    Correct answer is between '8000,8040'. Can you explain this answer?
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    An electrostatic precipitator (ESP) with 5600 m2 of collector plate ar...
    Given Information:
    - Collector plate area of an electrostatic precipitator (ESP) = 5600 m2
    - Efficiency of ESP in treating flue gas = 96%
    - Flue gas flow rate = 185 m3/s
    - Power plant capacity = 200 MW

    Efficiency Calculation:
    Efficiency of an electrostatic precipitator (ESP) is given by the formula:
    Efficiency = (1 - (Emission rate after ESP / Emission rate before ESP)) * 100

    Given that the ESP is 96% efficient, we can calculate the emission rate after the ESP:
    Emission rate after ESP = (1 - (Efficiency / 100)) * Emission rate before ESP
    Emission rate after ESP = (1 - (96 / 100)) * 185 m3/s
    Emission rate after ESP = 7.4 m3/s

    Calculation for 97% Efficiency:
    To achieve 97% efficiency, we need to calculate the required collector plate area.

    Let's assume the required collector plate area is A m2.

    Using the formula for efficiency, we can write:
    97 = (1 - (Emission rate after ESP / Emission rate before ESP)) * 100

    Substituting the values, we get:
    97 = (1 - (7.4 / 185)) * 100
    97 = (1 - 0.04) * 100
    97 = 96 * 100 / 100
    97 = 96

    Since the efficiency is already 96% and we need to increase it to 97%, it is not possible to achieve 97% efficiency with the given ESP collector plate area of 5600 m2.

    Calculation for 99% Efficiency:
    To achieve 99% efficiency, we need to calculate the required collector plate area.

    Let's assume the required collector plate area is B m2.

    Using the formula for efficiency, we can write:
    99 = (1 - (Emission rate after ESP / Emission rate before ESP)) * 100

    Substituting the values, we get:
    99 = (1 - (Emission rate after ESP / 185)) * 100

    To calculate the emission rate after the ESP, we can use the formula:
    Emission rate after ESP = (1 - (Efficiency / 100)) * Emission rate before ESP

    Substituting the values, we get:
    Emission rate after ESP = (1 - (99 / 100)) * 185 m3/s
    Emission rate after ESP = 1.85 m3/s

    Now, we can substitute this value back into the equation for efficiency:
    99 = (1 - (1.85 / 185)) * 100
    99 = (1 - 0.01) * 100
    99 = 99 * 100 / 100
    99 = 99

    Since the efficiency is already 96% and we need to increase it to 99%, it is not possible to achieve 99% efficiency with the given ESP collector plate area of 5600 m2.

    Conclusion:
    To achieve a higher efficiency of 99%, the ESP collector plate area would need to be increased beyond the given value of 5600 m2. However, the exact required collector plate area cannot be determined from the
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    An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be 6100 m2. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in m2) would be ______Correct answer is between '8000,8040'. Can you explain this answer?
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    An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be 6100 m2. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in m2) would be ______Correct answer is between '8000,8040'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be 6100 m2. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in m2) would be ______Correct answer is between '8000,8040'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electrostatic precipitator (ESP) with 5600 m2 of collector plate area is 96 percent efficient in treating 185 m3/s of flue gas from a 200 MW thermal power plant. It was found that in order to achieve 97 percent efficiency, the collector plate area should be 6100 m2. In order to increase the efficiency to 99 percent, the ESP collector plate area (expressed in m2) would be ______Correct answer is between '8000,8040'. Can you explain this answer?.
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