# Example 22.1,PAGE NUMBER:270¶

In [1]:
# Variable declaration
T_d=37;# The dry bulb temperature of air in °C
H=24;# % saturation
n_s=75;# Saturation efficiency in %
h=62.67;# The entering enthalpy in kJ/kg

# Calculation
# By construction on the chart, or from tables, the ultimate saturation condition would be 21.5°C, and 75% of the drop from 37°C to 21.5°C gives a fi nal dry bulb of 25.4°C.
h_fg=2425;# The average latent heat of water over the working range in kJ/kg
q=(h_fg)**-1;# The amount of water to be evaporated in kg/(s kW)
print"The amount of water to be evaporated is %0.0e kg/(s kW)"%q

The amount of water to be evaporated is 4e-04 kg/(s kW)


## Example 22.2,PAGE NUMBER:271¶

In [2]:
# Variable declaration
T_d=37;# The dry bulb temperature of air in °C
T_w=25.4;# The cooling temperature of water in °C
cf=0.80;# Contact factor

# Calculation
T_df=T_d-(cf*(T_d-T_w));# The dry bulb temperature (final) in °C
print"\nThe dry bulb temperature (final)=%2.1f°C (point D , Figure 22.4b )"%T_df
print"\nThe wet bulb is now 18.9°C and the enthalpy is 53 kJ/kg."

The dry bulb temperature (final)=27.7°C (point D , Figure 22.4b )

The wet bulb is now 18.9°C and the enthalpy is 53 kJ/kg.


## Example 22.3,PAGE NUMBER:271¶

In [3]:
# Variable declaration
T_d=26;# The dry bulb temperature of air in °C
T_w=20;# The wet bulb temperature of water in °C
T_win=29;# The temperature of water at inlet in °C
T_wout=24;# The temperature of water at outlet in °C
C_pw=4.187;# The specific heat capacity of water in kJ/kg.K

# Calculation
Q=C_pw*(T_win-T_wout);# Heat from water in kJ/kg
h_ain=57.1;# Enthalpy of entering air in kJ/kg
h_aout=78.1;# Enthalpy of leaving air in kJ/kg
print"\nHeat from water=%2.0f kJ/kg \nEnthalpy of entering air=57.1 kJ/kg \nEnthalpy of leaving air=78.1 kJ/kg"%Q
print"From the chart, the air leaves at approximately 25.7°C dry bulb."

Heat from water=21 kJ/kg
Enthalpy of entering air=57.1 kJ/kg
Enthalpy of leaving air=78.1 kJ/kg
From the chart, the air leaves at approximately 25.7°C dry bulb.