Boiler Performance — Factor of Evaporation (Unit Basis) If the total heat of steam is h kJ/kg and the sensible heat of feedwater is hf1 kJ/kg, what is the factor of evaporation (based on conversion to dry saturated steam at 100°C)?

Introduction / Context:The factor of evaporation (FoE) normalizes boiler performance to a common reference: the energy required to generate 1 kg of dry saturated steam at 100°C from water at 100°C. This lets engineers compare different boilers and operating conditions fairly, regardless of actual pressure or feedwater temperature.

Given Data / Assumptions:

• Total heat of steam produced at working pressure: h kJ/kg.
• Sensible heat of feedwater entering the boiler: hf1 kJ/kg.
• Reference latent heat at 100°C ≈ 2257 kJ/kg.

Concept / Approach:FoE is defined as (actual heat added per kg of water) divided by the latent heat at the reference condition. Actual heat added equals h - hf1. Dividing by 2257 converts the duty to “equivalent kilograms of evaporation from and at 100°C.”

Step-by-Step Solution:Compute heat added: q_in = h - hf1.Use reference latent heat: h_fg@100°C ≈ 2257 kJ/kg.Factor of evaporation: FoE = q_in / 2257 = (h - hf1) / 2257.

Verification / Alternative check:If feedwater already at 100°C and steam is dry saturated at 100°C, then h - hf1 = 2257 → FoE = 1, confirming the definition.

Why Other Options Are Wrong:(h - hf1) / h: incorrect normalization; denominator must be the reference latent heat.(h - hf1) / 100: arbitrary denominator with no thermodynamic meaning.none of these: incorrect, since the standard formula is known.

Common Pitfalls:Using latent heat at working pressure; FoE is defined with the 100°C reference to allow comparisons.

Final Answer:(h - hf1) / 2257