# CHAPTER 25:Air-Conditioning Methods¶

## EXAMPLE 25.1,PAGE NUMBER:305¶

In :
import math
# Variable Declaration
T_d=21;# The dry bulb temperature in °C
Q=14;# Internal load in kW
H=50;# % saturation
Q_l=1.5;# Latent heat gain in kW
T_ain=12;# The inlet air temperature in °C
C_p=1.02;# The specific heat capacity of air in kJ/kg.K

# Calculation
deltaT=T_d-T_ain;# Air temperature rise through room in K
m=Q/(deltaT*C_p);# Air flow for sensible heat in kg/s
x=0.007857;# Moisture content of room air, 21, 50%
x_p=Q_l/(2440*m);# Moisture to pick up
x_ain=x-x_p;# Moisture content of entering air
print"\nAir flow for sensible heat=%1.3f kg/s \nMoisture content of entering air=%0.5f"%(m,x_ain)

Air flow for sensible heat=1.525 kg/s
Moisture content of entering air=0.00745


## EXAMPLE 25.2,PAGE NUMBER:305¶

In :
import math
# Variable declaration
# From example 25.1
Q_i=14;# Internal load in kW
Q_l=1.5;# Latent heat gain in kW
Q_f=0.9;# The fan motor power in kW
T_win=5;# The temperature of water at inlet in °C
T_wout=10.5;# The temperature of water at outlet in °C
C_pw=4.19;# The specific heat capacity in kJ/kg.K

# Calculation
Q=Q_i+Q_l+Q_f;# Total cooling load in kW
m_w=Q/(C_pw*(T_wout-T_win));# Mass water flow in kg/s
print"\nMass water flow=%0.2f kg/s"%m_w

Mass water flow=0.71 kg/s


## EXAMPLE 25.3,PAGE NUMBER:305¶

In :
import math
# Variable declaration
# From example 25.2
Q=16.4;# Total load in kW
T_in=33;# The temperature at liquid R134a enters the expansion valve in °C
T_out=9;# The temperature at liquid  R134a leaves the cooler in °C
T_e=5;#  The temperature at which liquid  R134a evaporates in °C

# Calculation
h_v=405.23;# Enthalpy of R134a,superheated to 9 C in kJ/kg
h_f=246.71;# Enthalpy of liquid R134a at 33 C in kJ/kg
Re=h_v-h_f;# Refrigerating effect in kJ/kg
m_r=Q/Re;# Required refrigerant mass flow in kg/s
print"Required refrigerant mass flow=%0.3f kg/s"%m_r

Required refrigerant mass flow=0.103 kg/s


## EXAMPLE 25.4,PAGE NUMBER:306¶

In :
import math
# Variable declaration
T_d1=13;# The dry bulb temperature in °C
m_a=0.4;# The flow rate of primary air in kg/s
T_win=12;# The temperature of water at inlet in °C
T_wout=16;# The temperature of water at outlet in °C
H=72;# % saturation
T_d2=21;# The dry bulb temperature in °C
# From example 25.1
Q_i=14;# Internal load in kW
Q_l=1.5;# Latent heat gain in kW
C_pw=4.19;# The specific heat capacity in kJ/kg.K
C_pa=1.02;# The specific heat capacity of air in kJ/kg.K

# Calculation
x_a=0.006744;# Moisture in primary air, 13 C DB, 72% sat
x_r=Q_l/(2440*m_a);# Moisture removed in kg/kg
x_rise=x_a+x_r;# Moisture in room air will rise to in kg/kg
# which corresponds to a room condition of 21°C dry bulb, 53% saturation
Q_a=m_a*C_pa*(T_d2-T_d1);# Sensible heat removed by primary air in kW
Q_w=Q_i-Q_a;# Heat to be removed by water in kW
m_w=Q_w/(C_pw*(T_wout-T_win));# Mass water flow in kg/s
print"\nMass water flow=%0.2f kg/s"%m_w

Mass water flow=0.64 kg/s