Is sensible heat involved in the equivalent evaporation of boiler?

[Elz]

1.Is sensible heat involved in the equivalent evaporation of boiler?
While studying about the equivalent evaporation of boiler I got confused about the equation. The equation says,2. E=Total heat required to evaporate feed water/2257(latent heat of water)
The definition of E is the amount of dry saturated steam generated from feed water at 100°C to steam at 100°C at the saturation pressure corresponding to 100°C. But the equation shows that

E×2257= Total heat required to evaporate feed water

E×2257=m_e (h-h_w)

this h_w is the enthalpy of the feed water eg: 32°C, and this h is the enthalpy of the final steam generated which in the mathematical examples is normally shown at 10/11 bar. So it means on the right side of the equation sensible heat is involved but on the left side it is only latent heat. How does this happen?

Also how are we getting E from and at 100°C when on the right side we have feed water temp below 100°C and steam temp above 100°C??
3. Can anyone please answer?

Is it something like we are comparing the actual heat required with the latent heat? I'm confused...

 

[member 553083]

Yes, that is correct. The equation is comparing the total heat required to evaporate the feed water with the latent heat of evaporation of water at 100°C to steam at 100°C at the saturation pressure corresponding to 100°C. This is known as the equivalent evaporation of the boiler. The sensible heat of the feed water and steam is taken into account in the equation, but it is not explicitly shown. The difference between the enthalpy of the feed water and the enthalpy of the final steam generated accounts for the sensible heat.