# Chapter 23:Air-Conditioning Load Estimation¶

## Example 23.1,PAGE NUMBER:275¶

In :
#Variable declaration
R_si=0.3;# The inside resistance in (m**2 K)/W
R_1=0.040/0.09;# The thermal resistance of concrete panels in  (m**2 K)/W
R_2=0.050/0.037;# The thermal resistance of insulation in  (m**2 K)/W
R_3=0.012/0.16;# The thermal resistance of plaster board in  (m**2 K)/W
R_so=0.07;# The outside resistance in (m**2 K)/W

#Calculation
U=1/(R_si+R_1+R_2+R_3+R_so);# U factor in W/(m**2 K)
print"U factor=%0.2f W/(m**2 K)"%U

U factor=0.45 W/(m**2 K)


## Example 23.2,PAGE NUMBER:278¶

In :
#Variable declaration
T_d1=21;# The dry bulb temperature of air in °C
H=45;# % saturation
T_d2=27;# The dry bulb temperature of air in °C
T_wb1=20;# The wet bulb temperature of air in °C
m=1.35;# The mass flow rate of air in kg/s
C_pa=1.006;# 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
# 1.Total heat:
h_2=57.00;# Enthalpy at 27°C DB, 20°C WB in kJ/kg
h_1=39.08;# Enthalpy at 21°C DB, 45% sat in kJ/kg
dh=17.92;# Heat to be removed in kJ/kg
Q_t=dh*m;# Total heat in kW
print"Total heat,Q_t=%2.1f kW"% Q_t

# 2.Latent heat:
x_2=0.0117;# Moisture at 27°C DB, 20°C WB in kg/kg
x_1=0.0070;# Moisture at 21°C DB, 45% sat in kg/kg
dx=x_2-x_1;# Moisture to be removed in kg/kg
Q_l=dx*m*2440;# Latent heat in kW
print"Latent heat,Q_l=%2.1f kW"% Q_l

# 3.Sensible heat:
Q_s=(C_pa+((C_pw*x_2)))*(T_d2-T_d1)*m;# Sensible heat in kW
print"Sensible heat,Q_s=%1.1f kW"% Q_s

Total heat,Q_t=24.2 kW
Latent heat,Q_l=15.5 kW
Sensible heat,Q_s=8.5 kW


## Example 23.3,PAGE NUMBER:280¶

In :
#Variable declaration
Q_tl=15;# Total lighting load
P_ra=90;# % of load taken from return air
P_a=25;# % of load rejected to ambient

#Calculation
Q_ra=Q_tl*(P_ra*10**-2);# Picked up by return air in kW
Q_a=Q_ra*(P_a*10**-2);# Rejected to ambient in kW
Q_net=Q_tl-Q_a;# Net room load in kW
print"\nNet room load=%2.3f kW"%Q_net

Net room load=11.625 kW