Fan Motor............................................................................................................................................14
System Control Switch ("WS" Models) ................................................................................................15
System Control Switch ("WE" & "WY" Models)....................................................................................15
Check V alv e.........................................................................................................................................19
Drain Pa n Valv e...................................................................................................................................20
Rev ersing V alv e ("WY" Models Only) ..................................................................................................20
Sealed Refrigeration System Repairs..................................................................................................21
In the application and sizing of room air conditioners for cooling, it is most important to give full consideration to all
factors which may contribute to the heat loss or gain of the space to be conditioned. It is therefore necessary to
make a survey of the space to be conditioned and calculate the load requirements before a selection of the size of
the equipment needed can be made.
The load requirement may be determined very easily by simply using the standard “AHAM” Load Calculating
Form, on Page 6. This form is very easy to use and is self explanatory. It is necessary only to insert the proper
measurements on the lines provided and multiply by the giv en factors, then add the result f or the total load requirements.
Cooling load requirements are generally based on the cooling load for comfortable air conditioning which does not
require specific conditions of inside temperature and humidity. The load calculation form is based on outside
design temperature of 95° FDB and 75° FWB. It can be used for areas in the Continental United States having
other outside design temperatures by applying a correction factor for the particular locality as determined from the
map shown on Page 6.
When sizing a TwinTemp unit for cooling and heating, we must remember that the heating capacity of any given
unit varies directly with the outdoor ambient temperature. Also, we must keep in mind the average low temperatures which might be experienced in the locality where the unit is to be installed. Theref ore, when sizing a T winTemp
unit, both cooling and heating requirements must be calculated. Do not oversize, or undersize, one phase of the
unit’s capacity at the expense of the other. In those cases where the unit will provide satisfactory cooling at all
times but will be inadequate for those few times that the outdoor temperature is below the maximum low for the
unit, additional auxiliary heating facilities must be provided to insure that adequate heat is available at all times.
5
INSTRUCTIONS FOR USING COOLING LOAD ESTIMATE
FORM FOR ROOM AIR CONDITIONERS
(AHAM PUB. NO. RAC-1)
A.This cooling load estimate form is suitable for estimating the cooling load for comfort air conditioning installations
which do not require specific conditions of inside temperature and humidity.
B.The form is based on an outside design temperature of 95°F dry bulb and 75°F wet bulb. It can be used for areas
in the continental United States having other outside design temperatures by applying a correction factor for the
particular locality as determined from the map.
C.The form includes "day" factors for calculating cooling loads in rooms where da ytime comfort is desired (such as
living rooms, offices, etc.)
D .The numbers of the following paragraphs refer to the corresponding numbered item on the f orm:
1.Multiply the square feet of window area f or each exposure b y the applicab le factor. The window area is the
area of the wall opening in which the window is installed. For windows shaded b y inside shades or venetian
blinds, use the factor for "Inside Shades." For windows shaded by outside awnings or by both outside
awnings and inside shades (or venetian blinds), use the factor for "Outside Awnings." "Single Glass"
includes all types of single thickness windows, and "Doub le Glass" includes sealed airspace types, storm
windows, and glass b lock. Only one n umber should be entered in the right hand column for Item 1, and this
number should represent only the exposure with the largest load.
2.Multiply the total square feet of all windows in the room b y the applicable factor.
3a.Multiply the total length (linear feet) of all walls e xposed to the outside by the applicab le factor . Doors should
be considered as being part of the wall . Outside walls facing due north should be calculated separately
from outside walls facing other directions. Walls which are permanently shaded by adjacent structures
should be considered “North Exposure.” Do not consider trees and shrubbery as providing permanent
shading. An uninsulated frame wall or a masonry wall 8 inches or less in thickness is considered "Light
Construction." An insulated wall or masonry wall over 8 inches in thickness is considered "Heavy Construction."
3b.Multiply the total length (linear feet) of all inside walls betw een the space to be conditioned and any uncon-
ditioned spaces by the given f actor . Do not include inside walls which separate other air conditioned rooms.
4.Multiply the total square feet of roof or ceiling area by the factor given for the type of construction most
nearly describing the particular application (use one line only.)
5.Multiply the total square feet of floor area by the factor giv en. Disregard this item if the floor is directly on the
ground or over a basement.
6.Multiply the number of people who normally occupy the space to be air conditioned by the f actor given. Use
a minimum of 2 people.
7.Determine the total number of watts f or light and electrical equipment, except the air conditioner itself , that
will be in use when the room air conditioning is operating. Multiply the total wattage by the factor given.
8.Multiply the total width (linear feet) of any doors or arches which are continually open to an unconditioned
space by the applicable factor.
NOTE: Where the width of the doors or arches is more than 5 feet, the actual load may exceed the
calculated value . In such cases , both adjoining rooms should be considered as a single large room, and the
room air conditioner unit or units should be selected according to a calculation made on this new basis.
9.T otal the loads estimated for the foregoing 8 items.
10.Multiply the subtotal obtained in item 9 by the proper correction factor, selected from the map, for the
particular locality. The result is the total estimated design cooling load in BTU per hour.
E.For best results, a room air conditioner unit or units having a cooling capacity r ating (determined in accordance
with the NEMA Standards Publication for Room Air Conditioners, CN 1-1960) as close as possible to the estimated load should be selected. In general, a greatly oversized unit which would operate intermittently will be
much less satisfactory than one which is slightly undersized and which w ould operate more nearly continuously.
F.Intermittent loads such as kitchen and laundry equipment are not included in this form.
6
COOLING LOAD ESTIMATE FORM
HEAT GAIN FROM
1.WINDOWS: Heat gain from the sun.
Northeast
East
Southeast
South
Southwest
West
Northwest
North
2.WINDOWS: Heat by conduction
(Total of all windows.)
Single glass
Double glass or glass block
3.WALLS: (Based on linear feet of wall)
a. Outside walls
North Exposure
Other than North exposure
b. Inside Walls (between conditioned and
unconditioned spaces only.)
4.ROOF OR CEILING: (Use one only)
a. Roof, uninsulated
b. Roof, 1 inch or more insulation
c. Ceiling, occupied space above
d. Ceiling, insulated, with attic space above
e. Ceiling, uninsulated, with attic space above
* These factors are for single glass
only. For glass block, multiply the
above factors by 0.5; for double glass
or storm windows, multiply the above
factors by 0.8.
QUANTITY
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____ ft.
____ ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
____sq. ft.
5.Floor: (Disregard if floor is directly on ground or over
a basement.
6.NUMBER OF PEOPLE
7.LIGHTS AND ELECTRICAL EQUIPMENT IN USE
8.DOORS AND ARCHES CONTINUOUSLY OPENED
TO UNCONDITIONED SPACE: (TOTAL LINEAR
FEET OF WIDTH.)
9.SUBTOTAL
10. TOTAL COOLING LOAD (BTU per hour to be used
for selection of room air conditioner(s).)
____ Total in Item 9 X ____(Factor from Map) =_______
____sq. ft.
____
____watts
____ft.
*****
3
600
3
300
*****
_____
_____
_____
_____
_____
7
HEAT LOAD FORM
The heat load form, Page 8, may be used by servicing
personnel to determine the heat loss of a conditioned
space and the ambient winter design temperatures in
which the unit will heat the calculated space.
The upper half of the form is for computing the heat loss
of the space to be conditioned. It is necessary only to
insert the proper measurements on the lines provided
and multiply by the given factors, then add this result for
the total heat loss in BTU/Hr./°F.
The BTU/Hr. per °F temperature difference is the 70°F
inside winter designed temperature minus the lowest
outdoor ambient winter temperature of the area where
the unit is installed. This temperature difference is used
as the multiplier when calculating the heat loss.
The graph shows the following:
Left Hand ScaleUnit capacity BTU/Hr . or heat loss
BTU/Hr.
Bottom ScaleOutdoor ambient temperature,
base point.
Heat Pump ModelBTU/Hr. capacity heat pump will
deliver at outdoor temperatures.
Balance PointMaximum BTU/Hr . heat pump will
deliver at indicated ambient
temperature.
Below is an example using the heat load form:
A space to be conditioned is part of a house
geographically located in an area where the lowest
outdoor ambient winter temperature is 40°F. The
calculated heat loss is 184 BTU/Hr./°F.
Subtract 40°F (lowest outdoor ambient temperature for
the geographical location) from 70°F (inside design
temperature of the unit) for a difference of 30°F . Multiply
184 by 30 for a 5500 BTU/Hr. total heat loss for the
calculated space.
On the graph, plot the base point (70°) and a point on the
40°F line where it intersects with the 5500 BTU/Hr. line
on the left scale. Dr aw a straight line from the base
point 70 through the point plotted at 40°F. This is the
total heat loss line.
Knowing that we ha ve a 5500 BTU/Hr. heat loss, and we
expect that our heat pump will maintain a 70°F inside
temperature at 40°F outdoor ambient, we plot the selected
unit capacity BTU/Hr. of the unit between 35° and 60° on
the graph and dr aw a straight line betw een these points .
Where the total heat loss line and the unit capacity line
intersect, read down to the outdoor ambient temperature
scale and find that this unit will deliver the required BTU/
Hr. capacity to approximately 30°F.
8
HEATING LOAD FORM
FRIEDRICH ROOM UNIT HEAT PUMPS
BTU/HR PER
WALLS:(Linear Feet)°F TEMP. DIFFERENCE
2" InsulationLin. Ft. x 1.6
AverageLin. Ft. x 2.6
WINDOWS & DOORS (Area, sq. ft.)
Single Glass:Sq. Ft. x 1.13
Double Glass:Sq. Ft. x 0.61
INFILTRATION - WINDOWS & DOORS: A VG.Lin. Ft. x 1.0
LooseLin. Ft. x 2.0
CEILING: (Area, Sq. Ft.)
Insulated (6")Sq. Ft. x 0.07
Insulated (2")Sq. Ft. x 0.10
Built-up Roof (2" insulatedSq. Ft. x 0.10
Built-up Roof (1/2" insulated)Sq. Ft. x 0.20
No InsulationSq. Ft. x 0.33
FLOOR: (Area, Sq. Ft.)
Above Vented Crawl space
Insulated (1:)Sq. Ft. x 0.20
UninsulatedSq. Ft. x 0.50
* Slab on GroundLin. Ft. x 1.70
1" Perimeter insulationLin. Ft. x 1.00
* Based on Linear Feet of outside wallTOTAL HEA T LOSS PER °F BTU/HR/°F
Multiply total BTU/HR/°F X 30 and plot on the graph below at 40°F. Draw a straight line from
the 70 base point thru the point plotted at 40°F. The intersection of this heat loss line with the
unit capacity line represents the winter design heating load.
* Heating Element comes on at 35°F outside ambient and compressor shuts off.
** AHAM Rating Conditions.
12
COMPONENTS OPERATION & TESTING
WARNING
DISCONNECT ELECTRICAL POWER TO
UNIT BEFORE SERVICING OR TESTING
COMPRESSORS
GROUND TEST
Use an ohmmeter set on its highest scale. Touch one
lead to the compressor body (clean point of contact as
a good connection is a must) and the other probe in
turn to each compressor terminal (see Figure 2.) If a
reading is obtained, the compressor is grounded and
must be replaced.
Compressors are single phase, 115 or 230/208 volt, depending on the model unit. All compressor motors are
permanent split capacitor type using only a running capacitor across the start and run terminal.
All compressors are internally spring mounted and externally mounted on rubber isolators.
COMPRESSOR WINDING TEST
Remove compressor terminal box cover and disconnect
wires from terminals. Using an ohmmeter, check continuity across the following:
(See Figure 1)
Figure 1:Compressor Winding Test
Figure 2:Typical Ground Test
CHECKING COMPRESSOR EFFICIENCY
The reason for compressor inefficiency is normally due
to broken or damaged suction and/or discharge v alves ,
reducing the ability of the compressor to pump refrigerant gas.
This condition can be checked as follows:
1. Install a piercing valve on the suction and discharge or liquid process tube.
1. Terminal “C” and “S” - no continuity - open winding - replace compressor.
2. Terminal “C” and “R” - no continuity - open winding - replace compressor.
3. Terminal “R” and “S” - no continuity - open winding - replace compressor.
2. Attach gauges to the high and low sides of the
system.
3. Start the system and run a “cooling or heating
perf ormance test.”
If test shows:
A. Below normal high side pressure.
B. Above normal low side pressure.
C. Low temperature difference across coil.
The compressor valves are faulty - replace the
compressor.
THERMAL OVERLOAD (External)
Some compressors are equipped with an e xternal overload which is located in the compressor terminal box
adjacent to the compressor body (see Figure 3.)
13
The overload is wired in series with the common motor
terminal. The overload senses both major amper age and
compressor temperature. High motor temperature or
amperage heats the disc causing it to open and break
the circuit to the common motor terminal.
Figure 3: External Overload
Should the internal temperature and/or current draw
become excessive, the contacts in the overload will open,
turning off the compressor . The ov erload will automatically
reset, but may require several hours before the heat is
dissipated.
CHECKING THE INTERNAL OVERLOAD (see Figure
4.)
Figure 4
Heat generated within the compressor shell is usually
due to:
1. High amperage.
2. Low refrigerant charge.
3. Frequent recycling.
4. Dirty condenser.
TERMINAL OVERLOAD - TEST
(Compressor - External Type)
1. Remove overload.
2. Allow time for ov erload to reset before attempting
to test.
3. Apply ohmmeter probes to terminals on overload
wires. There should be continuity through the
overload.
TERMINAL OVERLOAD (Internal)
Some model compressors are equipped with an internal
over load. The overload is embedded in the motor
windings to sense the winding temperature and/or
current draw. The overload is connected in series with
the common motor terminal.
1.With no power to unit, remove the leads from the
compressor terminals.
2. Using an ohmmeter, test continuity between
terminals C-S and C-R. If not continuous, the
compressor overload is open and the compressor
must be replaced.
F AN MOTOR
A single phase permanent split capacitor motor is used
to drive the evaporator b lower and condenser fan. A selfresetting overload is located inside the motor to protect
against high temperature and high amperage conditions.
Although fan motors are lubricated at the factor y and
sealed, oil ports are provided to lubricate to motor annually
after the first year of operation (see Figure 5.) To lubricate
(oil), remove the oil plugs on each end of the fan motor
and put up to 30 drops of SAE10W30 grade motor oil in
each hole, then replace oil plugs.
14
Figure 5: F an Motor
Figure 6: System Control Panel
F AN MOTOR - TEST
1.Determine that capacitor is serviceable.
2.Disconnect fan motor wires from fan speed s witch
or system switch.
3.Apply “live” test cord probes on bl ack wire and
common terminal of capacitor. Motor should run
at high speed.
4.Apply “live” test cord probes on red wire and
common terminal of capacitor. Motor should run
at low speed.
5. Apply “live” test cord probes on each of the
remaining wires from the speed switch or system
switch to test intermediate speeds.
SYSTEM CONTROL SWITCH ("WS" Models)
A five position control switch is used to regulate the
operation of the fan motor and compressor. The
compressor can be operated with the fan operating at
low, medium or high speed. The fan motor can also be
operated independently on medium speed. See switch
section as indicated on decorative control panel (see
Figure 7.)
1. “Off” Position - no continuity between terminals.
2. “Lo Cool” Position - between terminals “L1” and “C”,
“LO” and “MS”.
3. “Med Cool” Position - between terminals “L1” and
“C”, “M” and “MS”.
4. “Hi Cool” Position - between terminals “L1” and “C”,
“H” and “MS”.
5. “Fan Only” Position - between terminals “L1” and
Figure 7: System Control Switch
SYSTEM CONTROL SWITCH - TEST
Disconnect leads from control switch (see Figure 8.)
There must be continuity as follows:
SYSTEM CONTROL SWITCH
("WE" & "WY" Models)
An eight position switch is used to regulate the operation of the fan motor, compressor and electric heater.
15
The unit can be operated in cooling or heating mode
with the compressor or electric heater on and the fan
motor operating on low , medium or high speed.
The fan motor can also be operated independently on
medium speed. See s witch section as indicated on decorative control panel, in Figure 8.
Figure 8: System Control Panel
Figure 9: System Control Switch
(Heat Pump & Electric Heat Models)
“2”.
NOTE:
Units will operate in constant fan in the cooling mode and auto fan in the heating mode.
CAPACITOR, RUN
SYSTEM CONTROL SWITCH - TEST
Disconnect leads from control switch. Turn control to
position being tested (see Figure 8.) There must be continuity as follows:
1. "Off" Position-no contin uity between terminals.
2. "Lo Cool" Position-between terminals "C" and "3",
"C2" and "2", "LO" and "M/S", "AR" and "5".
3. "Med Cool" Position-between terminals "C" and
"3", "C2" and "2", "M" and "M/S", "AR" and "5".
4. "Hi Cool" Position-between terminals "C" and "3",
"C2" and "2", "H" and "M/S", "AR" and "5".
5. "Hi Heat" Position-betw een terminals "C" and "1",
"C2" and "4", "H" and "M/S", "AR" and "5".
6. "Med Heat" Position-between terminals "C" and
"1", "C2" and "4", "M" and "M/S", "AR" and "5".
7. "Lo Cool" Position-between terminals "C" and "1",
"C2" and "4", "LO" and "M/S", "AR" and "5".
8. "Fan Only" Position-between terminals "L1" and
"M".
A run capacitor is wired across the auxiliary and main
winding of a single phase permanent split capacitor motor such as the compressor and fan motor. A single capacitor can be used for each motor or a dual rated capacitor can be used for both.
Figure 10: Run Capacitor Hook-Up
The capacitor’s primar y function is to reduce the line
current while greatly improving the torque characteristics of a motor. The capacitor also reduces the line current to the motor by improving the power factor of the
load. The line side of the capacitor is marked with a red
dot and is wired to the line side of the circuit (see Figure
10.)
16
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