发一个低温冷库的负荷计算

Let’s suppose we need to design the refrigeration system for the freezing room
Size: 10M*10M*3M
Room temp:-18℃ evaporating temp:-23℃
Stock: meat ρ=650KG/M^3
Intake temp:35℃ outtake temp:-15℃
Freezing time: 20H
Panel: 1 aluminium panel +150 polyurethane panel+1 aluminium panel
designing cond. temp:45℃，sub cool:5℃, cond. pressure:P=1.5MPA
evaporation temp:-25℃，over heat:20℃, evaperation pressure: P=0.18MPA.



Room capacity :G=Vρη/1000=300*650*0.4/1000=78T
V-volume M^3
ρ-mass of unit volume KG/M^3
η-efficiency 0.4



1.1 enclosure head load:Q1a=KFΔT=0.17*220*50=1.87 KW
K-heat transfer coefficient, 0.17W/M^2
F-area M^2
ΔT- temp difference for indoor and outdoor ,indoor:-15℃，outdoor:35℃

1.2 ground heat load Q1b=KFΔT=0.3*100*50=1.5 KW
K- heat transfer coefficient,0.3W/M^2
F- room area M^2
ΔT- temp difference for indoor and outdoor ,indoor:-15℃，outdoor:35℃

Total enclosure head load:Q1=(Q1a+Q1b)*α=(1.87+1.5)*1.2=4.044KW

2.1 stock freezing heat load:Q2a=G’(h1-h2)/T/3.6=78*1000*5%*（350-15）/20/3.6=18.15KW
G’-stock mass, 5% of the total capacity , KG
h1，h2- enthalpy of the stock KJ/KG，
T-freezing time, 20 H
3.6-referance for 1KJ/H, exchanging to W.

2.2package freezing heat load:Q2b=G’B(T1-T2)C/T=78*1000*5%*0.1*(35+25)*2.5/20/3.6=0.8KW
G’- stock mass, 5% of the total capacity , KG
B-mass reference, 0.1
T1，T2-package temp for indoor and outdoor
C-specific heat of the package, KJ/KG 度，2.5 KJ/KG ℃
T-freezing time，20H

Total stock heat load:Q2=Q2a+Q2b=18.95KW

3.1ventilation heat load:Q3a=（hw-hn）*n*V*ρ/24/3.6=（120+12）*2*300*1.36/24/3.6=1.2KW
hw-hn-enthalpy of the inside and the outside of the freezing room KJ/KG
n- times of ventilation ,2
V-volume of the freezing room M^3
ρ-density of the air of the inside of the freezing room KG/M^3

3.2ventilation for people in the freezing room:Q3b=30* n*ρ*（hw-hn）/3.6=30*3*1.36*（120+12）/3.6=4.5KW
30-air volume for each person30M^3/H
n-quantity of the persons in the freezing room ,3
ρ- density of the air of the inside of the freezing room ,1.36 KG/M^3

total ventilation heat load:Q3=Q3a+Q3b=5.7KW

4.1heat load of the motor of the eva.
150W per unit
Q4=150*5=0.75KW

5.1light heat load Q5 a=q*F=3*100=0.3KW
q-light heat road per unit square，3W/M^2

5.2door opening heat load:Q5 b=V n（hw-hn）ρ/3.6/24=300*5*132*1.36/24/3.6=3.1KW
n-times of opening the door ,5

5.2heat load of the person in the freezing room:Q5 c=3nq/24=3*3*410/24=0.15KW
3/24, three hours per day
n- persons in the freezing room,3
q-heat load for each person ,410W/per one

total heat load of manipulating:Q5=Q5 a+Q5 b+Q5 c=0.3+3.1+0.15=3.55KW

total heat load:Q= Q1+ Q2+ Q3+ Q4+ Q5=33.29 KW





to chose the comp. according to the total heat load:
We chose BITZER-HSN5353-25，
eva. temp.:-25 ℃，cond. Temp.:50 ℃，cooling capacity:Q=35.8KW
Power input:P=22.4KW

To chose the eva. According to the heat load
Let’s chose 7 units, capacity of each :Q=5KW

To chose the cond. According to the sum of the heat load and the comp. power input:
Q=22.4+35.8=58.2KW

To chose the expansion valve according to the heat load:
The pressure drop is 0.6MP，cooling capacity is 5KW，evaperation temp:-25℃

Let’s set the defrost times:4 times per day.
The time during defrost:25 min.
We can see whether the frost is cleaned by watching the eva. When defrosting. and then chose the time during defrost.

Let’s chose the copper pipe to connect the system.
Suction pipe diameterφ65,
Discharge pipe diameterφ28

Let’s chose 150MM thickness insulation material


Accumulator:V=Gv/0.8=50*0.0008/0.8=0.5M^3
G-charging quantity of the freon KG
V- M^3/KG

Separator:D=（4q/πω） ^0.5=0.2M
D-diameter
q-gas displacement,M^3/S
ω-speed of the gas M/S