The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of steam with enthalpy content of 2426 kJ/kg from feed water at temperature 40ºC (liquid enthalpy at 40ºC = 168 kJ/kg, enthalpy of vaporization of water at 100ºC = 2258 kJ/kg) will be

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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of steam with enthalpy content of 2426 kJ/kg from feed water at temperature 40ºC (liquid enthalpy at 40ºC = 168 kJ/kg, enthalpy of vaporization of water at 100ºC = 2258 kJ/kg) will be

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Answer 1

Concept:

We use the principles of boiler thermodynamics to determine the equivalent evaporation, which represents the amount of water that would be evaporated under standard conditions (100°C and 1 atm) using the same heat input as the actual boiler operation.

Given:

Steam production rate = \( 2000 \, \text{kg/hr} \)
Enthalpy of steam, \( h_{steam} = 2426 \, \text{kJ/kg} \)
Feedwater temperature = \( 40°C \)
Enthalpy of feedwater, \( h_{fw} = 168 \, \text{kJ/kg} \)
Enthalpy of vaporization at 100°C, \( h_{fg} = 2258 \, \text{kJ/kg} \)

Step 1: Calculate heat added per kg of steam

\( \text{Heat added} = h_{steam} - h_{fw} = 2426 - 168 = 2258 \, \text{kJ/kg} \)

Step 2: Compute equivalent evaporation

\( \text{Equivalent Evaporation} = \frac{\text{Steam production} \times \text{Heat added}}{h_{fg}} = \frac{2000 \times 2258}{2258} = 2000 \, \text{kg/hr} \)

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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of steam with enthalpy content of 2426 kJ/kg from feed water at temperature 40ºC (liquid enthalpy at 40ºC = 168 kJ/kg, enthalpy of vaporization of water at 100ºC = 2258 kJ/kg) will be

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