Chapter 18:Refrigeration Load Estimation

EXAMPLE 18.1,PAGE NUMBER:228

In [1]:
# Variable declaration
T_1=15;# °C
T_2=0;# °C
C_pw=4.187;# The specific heat capacity of water in kJ/kg.k
m=20*10**3;# The mass flow rate of water in kg/day
h_l=334;# kJ/kg
t=24*3600;# The time available for cooling in s

# Calculation
Q=(m*((C_pw*T_1)+334))/t;# The cooling load in kW
print"The cooling load,Q=%2.0f kW."%Q

The cooling load,Q=92 kW.

EXAMPLE 18.2,PAGE NUMBER:228

In [2]:
# Variable declaration
T_1=22;# °C
T_2=1;# °C
C_p=3.1;# The specific heat capacity of meat in kJ/kg.K
m=8*10**3;# The mass of meat in kg
t=14*3600;# The time available for cooling in s

# Calculation
Q=(m*((C_p*(T_1-T_2))))/t;# The cooling load in kW
print"The cooling load,Q=%2.1f kW."%Q

The cooling load,Q=10.3 kW.

EXAMPLE 18.3,PAGE NUMBER:230

In [3]:
# Variable declaration
n=12;# The number of lighting fittings
P=280;# W
P_3f=660;# W
P_h=18;# kW
I=80;# A
V=24;# V

# Calculation
L=[1.12,3.36];# Lighting,12*280,8h/day [Average over 24 h,Peak]
F=[7.78,7.92];# Fan motors, 12*660 W [Average over 24 h,Peak]
Dh=[1.50,18.00];# Defrost heaters,72 kW,1/2 h/day [Average over 24 h,Peak]
Fl=[0.21,1.92];# Fork-lift,1.92 kW,(1/3)*8h [Average over 24 h,Peak]
Fld=[0,0.12];# Fork-lift driver,120 kW,(1/3)*8h [Average over 24 h,Peak]
P=[0,0.24];# Packers,240 W,(1/3)*8h [Average over 24 h,Peak]
Avg=L[0]+F[0]+Dh[0]+Fl[0]+Fld[0]+P[0];# Average over 24 h
Peak=L[1]+F[1]+Dh[1]+Fl[1]+Fld[1]+P[1];# Peak
print"\nAverage over 24 h=%2.2f \nPeak=%2.2f"%(Avg,Peak)

Average over 24 h=10.61 
Peak=31.56

EXAMPLE 18.6,PAGE NUMBER:231

In [4]:
# Variable declaration
m=1000;# The capacity of meat store in tonnes
m_l=50;# The amount of meat leaving the store in t/day
m_s=300;# The amount of meat arrives from the ships in t/day
t=24*3600;# Time in s

# Calculation
# Case(1)
m=90;# t/day
T_1=2;# °C
T_2=-12;# °C
C=3.2;# Specific heat capacity in kJ/(kg.K)
T_fp=-1;# Freezing point of meat in °C
h_fg=225;# Latent heat of freezing in kJ/kg
C_fm=1.63;# Specific heat of frozen meat in kJ/(kg.K)
Q_f=(m*1000*((C*3)+h_fg+(C_fm*11)))/(t);# Cooling load in kW
print"\nCase(1):Cooling load,Q_f=%3.0f kW"%Q_f
# Case(2)
Q_f=(m_s*10**3*(C_fm*T_1))/t;# Cooling load in kW
print"\nCase(2):Cooling load,Q_f=%2.0f kW"%Q_f
# Case(3)
Q_f=(m_l*10**3*((C*3)+h_fg+(C_fm*11)))/t;# Cooling load in kW
print"\nCase(3):Cooling load,Q_f=%3.0f kW"%Q_f

Case(1):Cooling load,Q_f=263 kW

Case(2):Cooling load,Q_f=11 kW

Case(3):Cooling load,Q_f=146 kW