Consider the adiabatic saturator Moist air enters the satura

Consider the adiabatic saturator. Moist air enters the saturator at 30 ^oC, 1 bar, 10% relative humidity, with a volumetric flow rate of 120 m³/min. The system operates at a constant pressure of 1 bar.

a)   Evaluate humidity ratio, dew point temperature at the inlet.

b)   Determine the mass flow rates of dry air, water vapor, and the mixture at the inlet.

c)   Starting with the most general equations, simplify the equations for the conservation of mass, first law of thermodynamics, and second law of thermodynamic as applied to the saturator. Neglect the kinetic and potential energy effects Expand the resulting equations in terms of properties of dry air an water vapor.

d)   Derive equation (12.48) by using your results of the first law of thermodynamics in part (c).

e)   Evaluate the adiabatic saturation temperature

f)   Determine the rate of evaporation of liquid water in kg/s.

g)   Determine the rate of entropy generation in KW/K

Solution

Solution:

a)

using psychrometric chart for air at 1 bar pressure

first we need to plot the point on graph using given data that is 10% relative humidity & 30 C temp

streching that point horizontally, it will value for humidity ratio that is 3g per kg.

Dew point temp is given by relation

DWT = 4030(DBT + 235)/(4030-(DBT+235)(ln relative humidity) - 235

= -4.48 C

b)

for mass flow first we can find secific volume from graph which comes out = 0.86 m3/kg

mass flow of mixture = 1/0.86 (120)

= 139.53 kg/min

=2.33 kg/s

humidity ratio is 3 which means 3 g of water vapour in 1 kg dry air

total mixture is 2.33 kg/s

mass flow of water vapour is 0.003(2.33) = 0.0069 kg/s

mass flow of dry air is 2.323 kg/s

c)

First law of thermodynamic based on conservation of engery states that energy entering the system is equal to energy leaving the system, mathematically it can be written as

h1 + KE1 + PE1 + Q = h2 + KE2 + PE2 + W

here system is adiabatic therefore Q becomes zero

& KE & PE is also neglected

h1 = h2 + W

above equation is itself equation in dry air.

d) equation 12.48 is not given to derive.

e)

adiabatic temp is exist temp for air

this can be calculated drawing diagonal line towards right at 100% relative humidity it will give 13C

which is adiabatic saturation temp at this temp complete mixture will get saturated.

f)

rate of evaporation of liquid is equal to mass of water vaopur as complete water will get evaporated in saturator.

0.0069 kg/s

g)

entrpohy is given by change in enthaply/(temp)

= (w2-w1)h /(13+273)

= (9-3)36/(286)

= 0.755 kW/K