Table of Contents
Advertisement
Table 2.a: Cooling capacity of the correction factors in function of the condensing temperature.
Using table 1 and the suggested correction factors it is possible to choose the proper diameter in accordance to the
requested cooling capacity and the evaluated pressure drops.
Once the choice is taken a check shall be made by calculating the real pressure drop using the following formula.
In order to verify if the selected diameter could be correct the following maximum pressure drops shall be considered
according to ASHRAE:
Liquid line: ΔPmax = 0,5÷1 K
Suction line: ΔPmax = 1 K
Discharge line: ΔPmax = 1 K
Example:
How to choose the proper tube diameter.
Input data:
Equivalent length calculated: Leq=60 m
Saturated suction temperature: Tas=5°C
Condensing temperature: Tc=50°C
Cooling capacity at the evaporator: Pf=30 kW
Since the values of cooling capacity indicated in Table 1 are relative to pressure drop values equal to 0.02 K / m , first
check the pressure drops that would occur on the line using the diameter corresponding to the cooling capacity of
the unit. The equivalent length is about 60 meters , then with a pressure drops of 0.02 K / m the total pressure
drops are equal to: T = 60x0 , 02 = 1.2 K
These pressure drops are higher than the maximum specified by ASHRAE. To comply with the recommended value ( K
1 ) it is necessary to reduce the pressure drops length down to 0.016 K / m , using the formula [1 ]. Refer to Table
1 and to search in the discharge line column ( with suction temperature equal to 5 ° C) a cooling capacity close to
the unit one.
P = 29.7 kW at line with Δext = 28 mm .
This capacity is relative to a condensing temperature of 40 ° C. To obtain the corresponding 50 ° C capacity, this value
shall be multiplied to a factor shown in the table 2a: (discharge line , Tc = 50 ° C) factor = 1.11. The cooling
capacity delivered by the line with Δext = 28 mm ( corresponding to pressure drops of 0.02 k / m) is therefore:
P = 29,7 x1, 11 = 32.97 kW.
In conclusion, the data we need to use the formula [1] are the following:
ΔT=0,016x60=0,96 K
ΔTtab=0,02 K
Le=1m
Leff=60m
P=32,97 kW
As a consequence:
Advertisement
Table of Contents
