# Chapter 15:Cold storage¶

## Example 15.1,PAGE NUMBER:188¶

In [1]:
# Variable declaration
n=2;# The number of two pellet truck doors
m_n=300;#The number of traffic movements per day
t=30;# seconds

# Calculation
T=n*m_n*t;# The time for the door openings seconds per day
A=2.2*3.2;# The cross sectional area in m**2
v=1;# m/s
I=A*T*v;# The air infiltration in m**3/d
V=50*70*10;# The store volume in m**3
R=I/V;# The rate of air change per day
print"\nThe store volume is %5.0f m**3. \nThe rate of air change is %1.1f per day."%(V,R)

The store volume is 35000 m**3.
The rate of air change is 3.6 per day.


## Example 15.2,PAGE NUMBER:188¶

In [2]:
# Variable declaration
T=5;#  The dry bulb temperature in
R=3.6;# The rate of air change per day
V=35000;# The store volume in m**3
v_spa=0.8;# The specific volume in m**3/kg
q=600;# m**3/h
n=2;# The number of two pellet truck doors
h_1=15.9;# kJ/kg
h_2=-24.3;# kJ/kg
T_1=20;# °C
T_2=-25;# °C
t=24;# Time duration for one day in hours
t_s=24*60*60;# Time duration for one day in seconds

# Calculation
R_woh=V*R/v_spa;# The rate of air change without dehumidification in kg/day
Q_woh=R_woh*(h_1-h_2)/t_s;# The cooling load without dehumidification in kW
R_wh=q*n*t/v_spa;# The rate of air change with dehumidification in kg/day
Q_wh=R_wh*(T_1-T_2)/t_s;# The cooling load with dehumidification in kW
print"\nThe rate of air change without dehumidification is %5.0f kg/day. \nThe cooling load without dehumidification %2.1f kW(calculation error)."%(R_woh,Q_woh)
print"\nThe rate of air change with dehumidification is %5.0f kg/day. \nThe cooling load with dehumidification %2.2f kW."%(R_wh,Q_wh)

The rate of air change without dehumidification is 157500 kg/day.
The cooling load without dehumidification 73.3 kW(calculation error).

The rate of air change with dehumidification is 36000 kg/day.
The cooling load with dehumidification 18.75 kW.