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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr ofsteam with enthalpy content of 2426 kJ/kg from feed water at temperature40°C (liquid enthalpy = 168 kJ/kg, enthalpy of vaporisation of water at100°C = 2258 kJ/kg) is
- a)2000
- b)2149
- c)1682
- d)1649
Correct answer is option 'A'. Can you explain this answer?
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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of...
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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of...
Rate of steam producing Ms= 2000kg/hr
specific enthalpy steam h =2426 Kg/hr
hf=168
hfg=2258
me=Ms(h-hf)/hfg = 2000(2426-168)/2258 =2000kg/hr
specific enthalpy steam h =2426 Kg/hr
hf=168
hfg=2258
me=Ms(h-hf)/hfg = 2000(2426-168)/2258 =2000kg/hr
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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr of...
Answer:
Given:
- Steam production rate = 2000 kg/hr
- Enthalpy content of steam = 2426 kJ/kg
- Feed water temperature = 40°C
- Liquid enthalpy = 168 kJ/kg
- Enthalpy of vaporization of water at 100°C = 2258 kJ/kg
To find the equivalent evaporation rate, we need to calculate the amount of water that is being converted into steam per unit time.
The enthalpy content of steam is the sum of the liquid enthalpy and the enthalpy of vaporization. So, we can write:
Enthalpy content of steam = Liquid enthalpy + Enthalpy of vaporization
2426 kJ/kg = 168 kJ/kg + 2258 kJ/kg
From this equation, we can find the enthalpy of vaporization at the given temperature.
Enthalpy of vaporization at 100°C = 2426 kJ/kg - 168 kJ/kg
= 2258 kJ/kg
Now, we can calculate the equivalent evaporation rate using the formula:
Equivalent evaporation rate = Steam production rate * (Enthalpy of vaporization at 100°C / Enthalpy of vaporization at feed water temperature)
Equivalent evaporation rate = 2000 kg/hr * (2258 kJ/kg / 2258 kJ/kg)
= 2000 kg/hr
Therefore, the equivalent evaporation rate of the boiler is 2000 kg/hr. This means that the boiler is converting 2000 kg of feed water into steam per hour.
Given:
- Steam production rate = 2000 kg/hr
- Enthalpy content of steam = 2426 kJ/kg
- Feed water temperature = 40°C
- Liquid enthalpy = 168 kJ/kg
- Enthalpy of vaporization of water at 100°C = 2258 kJ/kg
To find the equivalent evaporation rate, we need to calculate the amount of water that is being converted into steam per unit time.
The enthalpy content of steam is the sum of the liquid enthalpy and the enthalpy of vaporization. So, we can write:
Enthalpy content of steam = Liquid enthalpy + Enthalpy of vaporization
2426 kJ/kg = 168 kJ/kg + 2258 kJ/kg
From this equation, we can find the enthalpy of vaporization at the given temperature.
Enthalpy of vaporization at 100°C = 2426 kJ/kg - 168 kJ/kg
= 2258 kJ/kg
Now, we can calculate the equivalent evaporation rate using the formula:
Equivalent evaporation rate = Steam production rate * (Enthalpy of vaporization at 100°C / Enthalpy of vaporization at feed water temperature)
Equivalent evaporation rate = 2000 kg/hr * (2258 kJ/kg / 2258 kJ/kg)
= 2000 kg/hr
Therefore, the equivalent evaporation rate of the boiler is 2000 kg/hr. This means that the boiler is converting 2000 kg of feed water into steam per hour.
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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr ofsteam with enthalpy content of 2426 kJ/kg from feed water at temperature40°C (liquid enthalpy = 168 kJ/kg, enthalpy of vaporisation of water at100°C = 2258 kJ/kg) isa)2000b)2149c)1682d)1649Correct answer is option 'A'. Can you explain this answer?
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The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr ofsteam with enthalpy content of 2426 kJ/kg from feed water at temperature40°C (liquid enthalpy = 168 kJ/kg, enthalpy of vaporisation of water at100°C = 2258 kJ/kg) isa)2000b)2149c)1682d)1649Correct answer is option 'A'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr ofsteam with enthalpy content of 2426 kJ/kg from feed water at temperature40°C (liquid enthalpy = 168 kJ/kg, enthalpy of vaporisation of water at100°C = 2258 kJ/kg) isa)2000b)2149c)1682d)1649Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The equivalent evaporation (kg/hr) of a boiler producing 2000 kg/hr ofsteam with enthalpy content of 2426 kJ/kg from feed water at temperature40°C (liquid enthalpy = 168 kJ/kg, enthalpy of vaporisation of water at100°C = 2258 kJ/kg) isa)2000b)2149c)1682d)1649Correct answer is option 'A'. Can you explain this answer?.
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