With makeup conductivity of 390 micromhos, feedwater conductivity 180, and condensate conductivity 45, what is the condensate return percentage using CR = (MC - FC) / (MC - RC) × 100?

Unlock all questions

This demo includes only 20 questions. Upgrade to access hundreds of questions, flashcards, exam simulations, and disable ads.

Full question bankExam simulationsFlashcards

From $24.99Unlock all

Study for the Washington DC 1st Class Boiler Test. Prepare with comprehensive questions, each with explanations and insights. Equip yourself thoroughly for success!

Multiple Choice

With makeup conductivity of 390 micromhos, feedwater conductivity 180, and condensate conductivity 45, what is the condensate return percentage using CR = (MC - FC) / (MC - RC) × 100?

This question tests applying the condensate return percentage formula CR = (MC - FC) / (MC - RC) × 100 using conductivity values.

Plugging in MC = 390, FC = 180, RC = 45:

MC - FC = 390 - 180 = 210

MC - RC = 390 - 45 = 345

CR = 210 / 345 × 100 ≈ 0.6087 × 100 = 60.9%

So about 60.9% of makeup water is returned as condensate. The result follows directly from the given values; with these numbers the calculation yields 60.9%, not the other percentages.

With makeup conductivity of 390 micromhos, feedwater conductivity 180, and condensate conductivity 45, what is the condensate return percentage using CR = (MC - FC) / (MC - RC) × 100?

Study for the Washington DC 1st Class Boiler Test. Prepare with comprehensive questions, each with explanations and insights. Equip yourself thoroughly for success!

With makeup conductivity of 390 micromhos, feedwater conductivity 180, and condensate conductivity 45, what is the condensate return percentage using CR = (MC - FC) / (MC - RC) × 100?

Get the latest from Examzify