Difficulty: Easy
Correct Answer: greater than unity
Explanation:
Introduction / Context:
The factor of evaporation (FoE) is used to convert actual steam generation under given conditions to the standardized “equivalent evaporation from and at 100°C.” It allows fair comparison of different boilers and operating states by normalizing to a common reference enthalpy rise.
Given Data / Assumptions:
Concept / Approach:
Because the actual boiler must raise feedwater from below 100°C to saturation at a higher temperature and perhaps superheat it, the enthalpy rise per kilogram is usually greater than the latent heat at 100°C. Therefore, when we divide this larger enthalpy rise by the 100°C latent heat, the resulting ratio (FoE) exceeds 1. Only in the special case where feedwater enters at 100°C and steam is produced as dry saturated at 100°C would FoE be exactly 1.
Step-by-Step Solution:
Define FoE = (h_steam_actual − h_feed_actual) / h_fg_100.For typical plants: h_steam_actual includes sensible heating to saturation at elevated pressure plus latent heat and perhaps superheat.Since h_feed_actual is below 100°C and h_steam_actual is at conditions hotter than 100°C, numerator > h_fg_100.Therefore, FoE > 1 in normal practice.
Verification / Alternative check:
Worked examples routinely yield FoE values like 1.05 to 1.3 or higher depending on pressure and feedwater heating. Only laboratory reference conditions make FoE approach unity.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing FoE with boiler efficiency; FoE is a normalization factor, not an efficiency measure by itself.
Final Answer:
greater than unity
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