Basic Mechanical Engineering (3110006)

Question-2b

BE | Semester-1 Summer-2019 | 04-06-2023

Q2) (b)

Equivalent evaporation (E):

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 $2257 kJ / kg$ amount of latent heat to convert it to dry saturated steam.

So mathematically, equivalent evaporation from and at $100 ° C$ can be written as,

$E = \frac{m_{s} \times (h - h_{f})}{2257}$
Factor of evaporation ( f ):

Factor of evaporation is given by,

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