Submittal Sheet

Data sheet Thermostatic expansion valves, type TUA/TUAE and TCAE
7DKRCC.PD.AG0.A4.22 / 520H8945
Determine required valve capacity.
Use the design evaporator capacity, Q
e
, to select
the required valve size at a given evaporating
temperature. If necessary, correct the evaporator
capacity for subcooling and pressure drop in
liquid line (pdw).
Subcooled liquid refrigerant entering the
evaporator increases evaporator capacity, so that
a smaller valve may be required.
In this example, the subcooling is: Δ
tsub
= t
c
− tl =
75 − 55 = 20°F
From the subcooling correction factor table, see
below, we nd the appropriate correction factor
“f
sub
” equals 1.06 for Δt
sub
= 20°F.
From the Distributer correction factor table, see
below, we nd the appropriate correction factor
“fp” equals 0.92 for pressure drop liquid = 25 psig
and t
evap
= +10°F
For optimum performance, it is important
to select a TU/TC valve according to system
conditions and application. Selecting an incorrect
valve will result in operational diculties or poor
system performance.
The following procedure will help you select the
correct valve for your needs.
Example:
Refrigerant = R407C
Evaporator capacity:
Qe = 30,000 Btu (2.5 TR)
Evaporating temperature: te = +10°F
Condensing temperature: tc = +75°F
Liquid refrigerant temperature: tl = +55°F
Pressure drop in Evaporator: dp
evap
: 3 psi
Presure drop in Condenser: dp
cond
: 3 psi
Pressure drop in distributer: 15 psi
Pressure drop in other components in liquid line:
dp line components: 2 psi
Determine the pressure drop in the liquid line.
Pressure drop in liquid line (dp
liquid
) = the
sum of other pressure drops in the liquid line,
evaporator, and distributor.
In this example, the total pressure drop in the
liquid line will be:
dp
liquid
= dp
evap
+ dp
cond
+ dp
distrib
+ dp
linecomp
dp
liquid
: 3 + 3 + 15 +2 = 23 psi
Now, determine the required valve capacity
by multiplying the evaporator capacity by the
correction factor for subcooling and pressure
drop
Required valve capacity
Qe × F
sub
x fp = 2.5 × 1.06 x 0.92 = 2.56 TR
Use the calculated valve capacity to select the
corresponding orice size from the capacity table
for R407C as indicated below.
Subcooling [°F] 2 7 10 20 30 40 50 60 70 80 90
Correction factor 0.95 0.98 1.00 1.06 1.12 1.19 1.25 1.31 1.37 1.43 1.49
Evaporating temperature [°F] -40 -30 -20 -10 0 10 20 30 40 50
"Pressure drop [psi]"
0 1 1 1 1 1 1 1 1 1 1
15 0.96 0.96 0.96 0.96 0.96 0.95 0.95 0.95 0.94 0.93
25 0.93 0.93 0.93 0.93 0.93 0.92 0.92 0.91 0.90 0.88
30 0.92 0.92 0.92 0.91 0.91 0.90 0.90 0.89 0.88 0.86
*calculated at 90°F condensing temperature
Subcooling correction factor 'fsub' ∆t
sub
Distributer correction factor 'fp' *
Step 2
Step 1
Selection