Basic Mechanical Engineering (3110006)

Question-2b

BE | Semester-1 Summer-2020 | 11-06-2020

Q2) (b)

It is the amount of water evaporated from feed water at $100 ° C$ and formed into dry saturated steam at $100 ° C$ at normal atmospheric pressure.
Let,

Q =

total heat required to evaporate feed water

h_{f} =

enthalpy of feed water

h =

enthalpy of steam

m_{s} =

mass of water actually evaporated or steam produced

Then total heat required to evaporate the water, $Q = m_{s} \times (h - h_{f})$
Now if the water is already at boiling temperature then at $1.013$ bar, it requires $2257 \frac{kJ}{kg}$ amount of latent heat to convert it to dry saturated steam.
So mathematically, equivalent evaporation from and at 1000C can be written as,

E = \frac{m_{s} \times (h - h_{f})}{2257}

Factor of evaporation is given by,

f = \frac{h - h_{f}}{2257}