Heat balance calculation method for refrigerated trucks

How to calculate the thermal balance of the refrigerated truck's insulated cargo compartment, now we will use the commonly used 4-meter refrigerated truck as an example to analyze: 4 meters refrigerated truck insulation compartment wall panels are insulated, due to the goods There is a certain temperature difference between the inside and the outside of the insulation wall of the compartment, so there are three processes in which the heat is transferred from one side of the insulation wall to the other side of the cooling fluid: the heat from the high temperature side is transferred to the high temperature side. The surface of the wall, the thermal conductivity in the insulating wall, that is, the heat is transferred from the high temperature side surface of the heat insulating wall to the low temperature side surface, and the heat is transmitted from the low temperature side surface of the heat insulating wall to the low temperature side air. Take the insulated cargo compartment as an example. In general, the outside temperature is higher than the inside temperature. The heat is transferred from the air outside the cabin to the outer skin surface of the heat insulation wall, and then transmitted to the inner skin surface through the outer skin heat insulation layer and the inner skin material, and then transmitted from the inner skin surface to the air in the cabin. go with.

1) =K. F(-)=0.4×13.83×10=55.32 (2-1)

In the formula: k-insulation compartment total heat transfer coefficient, W/(.K)

F- heat transfer area of ​​the insulated cabin,

- Air temperature outside the insulated cabin, K

- Air temperature in the insulated cabin, K

2) Heat flow into the compartment due to air leakage caused by gaps in the cargo compartment of the 4m refrigerated truck

The following empirical formulas are commonly used in actual calculations:

= (0.1 to 0.2) = 0.1 × 55.325.532 (2-2)

Under normal circumstances, take = 0.1 serious cargo leakage, take = 0.2

3) Heat flux of solar radiation

=K(-)=0.4×6.915×20×0.5=27.66 (2-3)

In the formula: - The area where the cargo compartment is exposed to solar radiation is generally taken from 35% to 50% of the heat transfer area.

- The average surface temperature of the cargo compartment affected by solar radiation, taken as = +20, K

The T-cargo is exposed to the sun every day for 12 to 14 hours.

4) Unload the incoming heat from the door

=f(+)=0.75×170.72=128.04 (2-4)

In the formula: f-opening frequency coefficient. Do not switch during transportation, take f = 0.25, open the door 1 ~ 5 times, take f = 0.5, open the door 6 ~ 10 times, take f = 0.75, open the door 11 ~ 15 times, to f = 1.

5) Heating value of goods loaded in the cargo compartment

= =1000×15=15000 (2-5)

In the formula: - The quality of a certain cargo in the cargo compartment, Kg

- Heat of a certain cargo in the cargo compartment

6) Heating capacity of fans and lights in the cargo compartment

=(+)=2900×=120.8 (2-6)

Where: - The power of the lamp, W

- Average daily lighting time, h.

7) Power consumed when the cargo compartment body is precooled

===22500 (2-7)

In the formula: -The quality of the cooling part of the cargo compartment, Kg

- The average specific heat of the cooling section of the cargo compartment, J/(Kg.K)

- The difference between the outside temperature and the inside temperature, K,

t - time for precooling cargo bed, h.

8) Heat flow taken from pre-cooling cargo in cargo compartment

=()=(10001500+502500)=12500(2-8)

In the formula: - Quality of goods in cargo compartment, Kg

- The quality of the carton container, K

- The specific heat of the packaging container, J/(Kg.h), j = 2500 J/(Kg.h)

When a refrigerated automobile transports fruits, vegetables and fresh eggs, the refrigeration unit is required to pre-cool the insulated cargo compartment and cargo at the same time. At this time, the heat load of the refrigeration unit is:

Q=+++++++=23kw (2-9)

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