Difficulty: Easy
Correct Answer: the amount of water evaporated from and at 100° C into dry and saturated steam
Explanation:
Introduction:Because boilers operate at different pressures and feedwater temperatures, engineers normalize performance to a common reference called “equivalent evaporation.” This allows fair comparison of capacities and efficiencies across diverse conditions.Given Data / Assumptions:
Concept / Approach:Equivalent evaporation expresses the mass of water that would be evaporated per unit time at the reference condition if the same heat input were applied. It is a conversion of the actual evaporation at operating conditions to the standardized basis “from and at 100°C.” This is not an efficiency or a fixed constant; it is a normalized mass flow measure.
Step-by-Step Solution:
Identify the need: normalize for differing pressures and feedwater temperatures.Choose the reference: 100°C water to dry saturated steam at 100°C.Definition: equivalent evaporation = kg/s (or kg/h) of water evaporated from and at 100°C that is thermally equivalent to actual steaming rate.Verification / Alternative check:Handbooks show conversions using enthalpy differences: m_eq = (Q_actual) / (h_g,100 − h_f,100). This matches the verbal definition.
Why Other Options Are Wrong:
Common Pitfalls:Confusing equivalent evaporation with efficiency or boiler horsepower; forgetting the “dry saturated at 100°C” reference.
Final Answer:
the amount of water evaporated from and at 100° C into dry and saturated steam
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