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3.1 SELECTION EXAMPLE
Assume a process chiller with a remote condenser located below the processing unit; operation is also required in winter and consequently with a low conden-
sing temperature.
The evaporating temperature considered is the highest expected value corresponding to the condensing temperature in winter.
Design data
a.
Type of refrigerant
b1.
Condensing temperature
b2.
Evaporating temperature
c.
Cooling capacity
Using the Selection sheet relating to R410A refrigerant, proceed as follows:
Being initially unknown, use Table 1 to calculate the pressure head ΔPC corresponding to Tcond and Tevap.
1.
ΔPC (bar) - Pressure head according to the temperature
20
-40
12.7
-35
12.2
-30
11.7
-25
11.1
10.4
-20
-15
9.6
-10
8.7
-5
7.6
0
6.4
5
5.1
10
15
The value is calculated by interpolation.
Determine the pressure difference ΔPV across the valve using the formula:
the pressure exerted by the column of liquid is negative, as the condenser is installed below the valve.
N.B.:
2.
The temperature of the refrigerant at the valve inlet is not known; assume a subcooling value of 5 °C and consequently a temperature of the liquid
Tliq = Tcond – 5°C = 32 °C. Table 2 is used to determine the Correction Factor:
Tliq [°C]
-22
-16
CF
0.56
0.58
3.
The expansion valve must have an equivalent capacity RATING determined by the product of the cooling capacity CAP by the Correction Factor CF:
4.
In Table 3 identify the cell relating to the design saturated evaporating temperature Tevap. Determine, corresponding to the column with the pressure
difference nearest to the ΔPV calculated in point 3 above, the model of valve whose capacity is immediately higher than the required equivalent value. The
numbers in the table can be interpolated. In the case the model is: E2V18
R410A
Tcond
= 37 °C
Tevap
= 5 °C
CAP
= 9 kW
Tcond – Saturated condensing temperature (°C)
25
30
35
40
14.7
17.1
19.6
22.4
14.3
16.6
19.2
22
13.8
16.1
18.7
21.5
13.2
15.5
18.1
20.9
12.5
14.8
17.4
20.2
11.7
14
16.6
19.4
10.8
13.1
15.6
18.4
9.7
12.0
14.6
17.4
8.5
10.8
13.4
16.2
7.2
9.5
12
14.8
5.7
8
10.5
13.3
6.3
8.8
11.6
ΔPV = ΔPC − ΔPH − ΔPL + 0,1 × ΔH = 13,1 − 0,6 − 0,8 + 0,1 × (− 6) = 11,1 bar
CF – Correction factor for the temperature (°C) of the liquid at the valve inlet
-10
-4
2
8
0.61
0.64
0.67
0.71
RATING = CAP × CF = 9 × 0,92 = 8,3 kW
Tevap. 5°C
8
12
E2V09B
2.4
2.9
E2V11B
4.2
5.1
E2V14B
6.4
7.8
E2V18B
9.1
11.2
E2V24B
18.1
22.2
E2V35B
36.5
44.7
E4V55A
88.5
108.4
E4V65A
122
149
E4V85A
171
209
E4V95A
--
--
d1.
Pressure drop in high branch
d2.
Pressure drop in low branch
e.
Height of condenser above valve
f.
Temperature of the liquid
45
50
55
60
25.5
28.8
32.5
36.6
25
28.4
32.1
36.1
24.5
27.9
31.6
35.6
23.9
27.3
31
35
23.2
26.6
30.3
34.3
22.4
25.8
29.5
33.5
21.5
24.9
28.6
32.6
20.4
23.8
27.5
31.5
19.2
22.6
26.3
30.3
17.9
21.3
25
29
16.4
19.8
23.4
27.5
14.7
18.1
21.8
25.8
Table 1
ΔPC = 13,1 bar
CF = 0,92
14
20
26
32
0.75
0.80
0.86
0.92
Table 2
ΔPv [bar]
16
20
24
28
3.3
3.7
4.1
4.4
5.9
6.6
7.2
7.8
9.1
10.1
11.1
12
12.9
14.4
15.8
17
25.6
28.7
31.4
33.9
51.6
57.7
63.3
68.3
125.2
140
153
166
172
192
211
228
242
270
296
320
--
--
--
--
Table 3
6
ΔPH
= 0,6 bar
ΔPL
= 0,8 bar
ΔH
= — 6 m
Tliq
= unknown
65
41
40.5
40
39.4
38.7
37.9
37
35.9
34.7
33.4
31.9
30.2
38
44
50
56
1.00
1.10
1.22
1.39
32
4.7
8.4
12.8
18.2
36.3
73
177
243
342
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