# Chapter 21:Air Treatment Fundamentals¶

## Example 21.1,PAGE NUMBER:251¶

In [1]:
# Variable declaration
m_a=68;# The mass flow rate of air in kg/s
T_1=16;# The temperature of air at inlet in °C
T_2=34;# The temperature of air at outlet in °C
T_win=85;# The temperature of hot water at inlet in °C
T_wout=74;# The temperature of hot water at outlet in °C
C_pa=1.02;# The specific heat capacity of air in kJ/kg.K
C_pw=4.187;# The specific heat capacity of water in kJ/kg.K

# Calculation
Q=m_a*C_pa*(T_2-T_1);# Heat input in kW
m_w=Q/(C_pw*(T_win-T_wout));# The mass flow rate of water in kg/s
print"\nHeat input,Q=%4.0f kW \nThe mass flow rate of water,Q=%2.0f kg/s"%(Q,m_w)

Heat input,Q=1248 kW
The mass flow rate of water,Q=27 kg/s


## Example 21.2,PAGE NUMBER:251¶

In [2]:
# Variable declaration
Q=500;# The amount of heat required for the building in kW
T=19;# The temperature at which air enters the heater coil in °C
m_a=68;# # The mass flow rate of air in kg/s
C_pa=1.02;# The specific heat capacity of air in kJ/kg.K

# Calculation
t=T+(Q/(m_a*C_pa));# The air supply temperature in °C
print"The air-supply temperature,t=%2.1f°C"%t

The air-supply temperature,t=26.2°C


## Example 21.3,PAGE NUMBER:254¶

In [3]:
# Variable declaration
T_ra=21;# The temperature of the returning air
H=50;# % saturation
T_d=28;# The dry bulb temperature in °C
T_w=20;# The wet bulb temperature in °C
m_a=20;# The mass flow rate of returning air in kg/s
m_b=3;# The mass flow rate of outside air in kg/s
x_ra=0.0079;# The moisture content in kg/kg
x_oa=0.0111;# The moisture content in kg/kg
h_a=41.8;# The enthalpy in kJ/kg
h_b=56.6;# The enthalpy in kJ/kg

# Calculation
# Method (b)
t_c=((T_ra*m_a)+(T_d*m_b))/(m_a+m_b);# °C
g_c=((x_ra*m_a)+(x_oa*m_b))/(m_a+m_b);# kg/kg
h_c=((h_a*m_a)+(h_a*m_b))/(m_a+m_b);# kJ/kg dry air
print"\nThe condition of the mixture,t_c=%2.1f°C"%t_c
print"\n                             g_c=%0.4f kg/kg"%g_c
print"\n                             h_c=%2.1f kJ/kg dry air"%h_c

The condition of the mixture,t_c=21.0°C

g_c=0.0083 kg/kg

h_c=41.8 kJ/kg dry air


## Example 21.5,PAGE NUMBER:257¶

In [4]:
# Variable declaration
T_s=100;# The temperature of steam in °C
T_d=21;# The dry bulb temperature in °C
H=50;#  % saturation
x_ab=0.0079;# Moisture content of air before in kg/kg
C_ps=1.972;# The specific heat capacity of the steam in kJ/kg°C
C_pa=1.006;# The specific heat capacity of air in kJ/kg.K

# Calculation
x=x_ab+x_a;# Final moisture content in kg/kg
t=((x_a*C_ps*T_s)+(C_pa*T_d))/(((x_a*C_ps)+(C_pa)));# The final dry bulb temperature in °C
print"\nFinal moisture content=%0.4f kg/kg \nThe final dry bulb temperature,t=%2.2f°C"%(x,t)

Final moisture content=0.0146 kg/kg
The final dry bulb temperature,t=22.02°C


## Example 21.6,PAGE NUMBER:259¶

In [5]:
# Variable declaration
T_d1=23;# The dry bulb temperature in °C
T_w=5;# The temperature of water in °C
H=50;#  % saturation
n_s=0.7;# Saturation efficiency in %
x_a=0.0089;# Moisture content in kg/kg
x_b=0.0054;# Moisture content in kg/kg

# Calculation
#(a)
print"(a) By construction on the chart ( Figure 21.7 ), the final condition is 10.4°C dry bulb,82% saturation"
#(b)
T_d2=T_d1-(n_s*(T_d1-T_w));# The final dry bulb temperature in °C
x_f=x_a-(n_s*(x_a-x_b));# kg/kg
print"\n(b)The final condition,\n   The final dry bulb temperature=%2.1f°C \n   The moisture content=%0.5f kg/kg"%(T_d2,x_f)

(a) By construction on the chart ( Figure 21.7 ), the final condition is 10.4°C dry bulb,82% saturation

(b)The final condition,
The final dry bulb temperature=10.4°C
The moisture content=0.00645 kg/kg


## Example 21.7,PAGE NUMBER:259¶

In [6]:
# Variable declaration
m_w=4;# The mass of water in kg
m_a=1;# The mass of air in kg
h_ab=45.79;# Enthalpy of air before in kJ/kg
h_aa=26.7;# Enthalpy of air after in kJ/kg
C_pw=4.187;# The specific heat capacity of water in kJ/kg.K

# Calculation
Q_l=h_ab-h_aa;# Heat lost per kilogram air in kJ
Q_g=Q_l/m_w;# Heat gain per kilogram water in kJ
dT=Q_g/C_pw;# Temperature rise of water in K
print"Temperature rise of water=%1.0f K"%dT

Temperature rise of water=1 K


## Example 21.8,PAGE NUMBER:261¶

In [7]:
# Variable declaration
T_d1=24;# The dry bulb temperature in °C
T_d2=7;# The dry bulb temperature in °C
H=45;#  % saturation
cf=0.78;# Contact factor
h_1=45.85;# The enthalpy in kJ/kg
h_2=22.72;# The enthalpy in kJ/kg

# Calculation
#(a) By construction on the chart ( Figure 21.9 ), 10.7°C dry bulb, 85% saturation.
#(b) By calculation, the dry bulb will drop 78% of 24 to 7°C:
dT=T_d1-(cf*(T_d1-T_d2));# The drop in dry bulb temperature in °C
dh=h_1-(cf*(h_1-h_2));# The drop in enthalpy in kJ/kg
print"\nThe drop in dry bulb temperature=%2.1f°C \nThe drop in enthlpy=%2.2f kJ/kg"%(dT,dh)

The drop in dry bulb temperature=10.7°C
The drop in enthlpy=27.81 kJ/kg


## Example 21.10,PAGE NUMBER:262¶

In [8]:
# Variable declaration
T_d=23;# The dry bulb temperature in °C
H=40;# % saturation
SH=36;# The sensible heat to be removed in kW
LH=14;# The latent heat in kW

# Calculation
# Plotting on the chart ( Figure 21.10 ) from 23°C/40% and using the ratio
R=SH/(SH+LH);
print"The process line meets the saturation curve at - 1°C, giving the ADP (which meansthat condensate will collect on the fins as frost)."
print"Taking the ‘ off ’ condition at 5°C dry bulb and measuring the proportion along theprocess line gives a coil contact factor of 75%."

The process line meets the saturation curve at - 1°C, giving the ADP (which meansthat condensate will collect on the fins as frost).
Taking the ‘ off ’ condition at 5°C dry bulb and measuring the proportion along theprocess line gives a coil contact factor of 75%.