Sunday, December 25, 2016

How to Design Walk-in Cooler - Heat Load, Electrical Calculations



How to Design Walk-in Cooler     
Heat Load Calculations
Electrical Load Calculations
Mechanical Installation
Electrical Wiring



Note:
I had errors putting pictures, charts, tables and other images.
Pls. see Youtube Video here:       

https://www.youtube.com/watch?v=74aUw3ZrFgE




References:
-          ASHRAE Tables
-          Carrier charts
-          Paragon Company
-          Sumit Products Inc
-          Eaton/Cutler Hammer
-          Penn/Johnson Controls
-          Keeprite Refrigeration
-          Encore Wire Corporation
-          Coldmatic Refrigeration
-          Thomson Delmar Learning
-          NATIONAL REFRIGERANTS INC
-          Tecumseh Products Company
-          Industrial Controls – Eriks company
-          Sporlan – Parker Hannifin Corporation
-          Modern Refrigeration and Air-conditioning – Althouse
-          … and many more (if I forget you, I don’t mean it…sorry)


                               Walk-in cooler Design, sizing and selection Process:
Ø      Heat load calculation
Ø      How to Size Evaporator
Ø      How to size compressor and condensing unit
Ø      How to select liquid receiver
Ø      How to select TXV (thermostatic expansion valve)
Ø      How to select filter/drier
Ø      How to select sight glass
Ø      How to select liquid line solenoid valve
Ø      How to size refrigerant lines and piping
Ø      How to calculate electrical load input
Ø      How to size copper wire conductors
Ø      How to wire refrigeration walk-in cooler
Ø      How to wire evaporator
Ø      How to wire condensing unit
Ø      How to wire compressor, condenser fan motor
Ø      How to wire defrost timer
Ø      How to wire thermostat
Ø      How to wire liquid line solenoid valve
Ø      How to wire low pressure control/switch
Ø      How to calculate and set low pressure switch settings
Ø      How to prepare estimate bill of materials – electrical
Ø      How to prepare estimate bill of materials – mechanical
Ø      How to prepare order information and specification





Given:
PROJECT WALK-IN COOLER:
Walk-in Cooler Room Temp = 35 F
Outside Temp = 75 F  (walk-in cooler is inside air-conditioned building)
TD = 75 F – 35 F
TD = 40 F
Two Evaporator/Cooler Fans =  1/8 HP  (100 watt) each
Lights = 40 watt Lamp working 8 hrs per day
Occupancy Load = 2 persons working 8 hrs per day

CABINET:
Cabinet Dimensions:  26 GA  White Stucco embossed finish int/exterior
       Length = 6.5 ft
       Width = 6.5 ft
       Height = 8.25 ft
Cabinet Material Characteristics:
       Outside air film factor, Fo = 6.0
       Inside air film factor, Fi = 1.65
Cabinet insulation characteristics:
       1/4  inch thick Polyurethane, K = 0.16 Btu in./sq ft/hr/F
       1  inch thick Particle Board, K = 0.94 Btu in./sq ft/hr/F
       1/2  inch thick Celotex, K = 0.31 Btu in./sq ft/hr/F

PRODUCT:
Product Load (meat) = 500 lbs  (storage quantity - newly added)
Product Stock Shift = 250 lbs  (regularly placed in walk-in cooler)
Total Product Load = 500  +  250 
Total Product Load =  750 lbs
Product (Meat, brined) specific heat above freezing = 0.75  Btu/lb-F
Product entering temp = 45 F
Walk-in Cooler Room Temp = 35 F
TD product = 45 – 35
TD product = 10 F


Other Design Factors/Considerations:
1.     Ventilation/Infiltration Loss Factor = Normal
                   (Note: Heavy Usage is 4 or more door openings per hour)
2.     Loading Percentage = 100 %  (design for Full Load scenario)

3.     Walk-in Cooler has No Windows

4.     Walk-in Cooler is not exposed to the sun (no solar load)

5.     No Defrost

6.     Split (TD between refrigerant & air) is  10 F  for wet produce & meat

7.     Total System Running Time is 20 hrs

8.     Add 20% Reserve Capacity



Cold Room Calculator (ALFA LAVAL) results:
Transmission losses = 264 watt
Ventilation losses = 299 watt
Other heat sources = 303 watt
Cooling down/congel = 23 watt   (Cool down/Congel. Time = 24 hr)

SUBTOTAL  = 889 watt
SUBTOTAL = 889 watt x 3.41 Btu/hr per watt
SUBTOTAL = 3031 Btu/hr

REQUIRED CAPACITY = 3031 Btu/hr + 20 % Reserve Capacity
REQUIRED CAPACITY = 3031 x 1.2
REQUIRED CAPACITY = 3638  BTU/HR
SPECIFIC CAPACITY = 108  Watt/cu. m
SPECIFIC  CAPACITY = 10.4  Btu/hr per cu. ft


Approximate Conversion:
1  Btu/hr  per cu. ft  ~  10  Watt/cu. m







Solution:
TOTAL HEAT LOAD = HEAT LEAKAGE (TRANSFER) LOAD  +
                                             HEAT USAGE (SERVICE) LOAD


HEAT LEAKAGE LOAD:
HEAT LEAKAGE LOAD = HEAT walls  + 
                                                    HEAT ceiling  + 
                                                    HEAT floor  + 
                                                    HEAT windows


HEAT LEAKAGE LOAD = U x A x TD
              where:
               U = overall (combined) heat transfer coefficient of walls, ceiling,
                       floor, windows of cabinet
               A = total outside area of cabinet/refrigerated space
               TD = temperature difference between outside and inside of
                          refrigerated space

                                        U = 1/R total
                                        R = 1/C
                                        R = t/K
                                        R total = R1  +  R2  +  R3  +  R4  +  R5
                                        R total = 1/C1  +  1/C2  +  t3/K3  +  t4/K4  +  t5/K5
                                        R total = 1/Fo  +  1/Fi  +  t3/K3  +  t4/K4  +  t5/K5    
             
                                   where:
                                   R = thermal resistance of a material
                                   R total = total thermal resistance of cabinet (composite,
                                                     made of different materials)   
                                   R1, R2, R3, R4, R5 = individual thermal resistances of
                                                                            materials
                                   C1, C2 = individual thermal conductance of materials
                                                    (given thickness)
                                   t3, t4, t5 = individual thickness of materials (inch)
                                   K3, K4, K5 = individual thermal conductivity of materials
                                                             (per inch of thickness)










Calculate U, the overall (combined) heat transfer coefficient:
U = 1/R total
R total = 1/Fo  +  1/Fi  +  t3/K3  +  t4/K4  +  t5/K5    

From the given Cabinet Material Characteristics:
       Outside air film factor, Fo = 6.0
       Inside air film factor, Fi = 1.65
From the Cabinet insulation characteristics:
       t3 = 1/4 inch thick Polyurethane, K3 = 0.16 Btu in./sq ft/hr/F
       t4 = 1 inch thick Particle Board, K4 = 0.94 Btu in./sq ft/hr/F
       t5 = 1/2 inch thick Celotex, K5 = 0.31 Btu in./sq ft/hr/F

Substituting to the formula,
R total = 1/6  +  1/1.65  +  0.25/0.16  +   1/0.94  +  0.5/0.31
R total = 0.167  +  0.606  +  1.562  +  1.064  +  1.613
R total = 5.012 
U = 1/R total   
U = 1/5.012
U = 0.2  Btu/sq ft/hr/F

HEAT walls:
HEAT walls = U x A x TD
HEAT walls =  0.2  Btu/sq ft/hr/F    x    A(walls)   x   40 F



Length = 6.5 ft                                                       H=8.25
Width = 6.5 ft
Height = 8.25 ft                                                          
                  L=6.5
                                                         W=6.5

Area of Walls:
A(walls)  =  ( 2 x W x H )   +   ( 2 x L x H )
A(walls)  =  ( 2 x 6.5 x 8.25 )   +   ( 2 x 6.5 x 8.25 ) 
A(walls)  =  214.5  sq ft

HEAT walls =  0.2  x  214.5  x  40
HEAT walls =  1716  Btu/day

HEAT ceiling and floor:
A(ceiling and floor) =  2  x  L  x  W
A(ceiling and floor) =  2  x  6.5  x  6.5
A(ceiling and floor) = 84.5  sq ft

HEAT ceiling and floor =  0.2  x  84.5  x  40
HEAT ceiling and floor =  676  Btu/day

HEAT windows:
HEAT windows = 0


HEAT LEAKAGE LOAD:
HEAT LEAKAGE LOAD = HEAT walls  + 
                                                    HEAT ceiling and floor  + 
                                                    HEAT windows

HEAT LEAKAGE LOAD  =  1716  +  676  +  0
HEAT LEAKAGE LOAD  =  2392  Btu/day


HEAT USAGE LOAD:
HEAT USAGE LOAD = HEAT product  + 
                                              HEAT air change (infiltration)  + 
                                              HEAT motors  + 
                                              HEAT lights  + 
                                              HEAT occupants  +
                                              HEAT defrost  +
                                              HEAT solar


HEAT product:
HEAT product = m  x  Cp  x  TD
Substituting values below,
      Total Product Load =  750 lbs
      Product (Meat, brined) specific heat above freezing = 0.75  Btu/lb-F
      TD product = 10 F
HEAT product =  750  x  0.75  x  10
HEAT product =  5625  Btu/day


HEAT air change (infiltration):
HEAT air change (infiltration)  =   Inside Volume  x  Air Change  x  Heat

Inside Volume:
Total thickness = 1  +  0.5  +  0.25
Total thickness =  1.75  inches  (0.146 ft)

Inside Length = 6.5  -  ( 2  x  0.146 )
Inside Length = 6.21  ft
Inside Width = 6.5  -  ( 2  x  0.146 )
Inside Width = 6.21  ft
Inside Height = 8.25  -  ( 2  x  0.146 )
Inside Height = 7.96  ft

Inside Volume = Inside Length  x  Inside Width  x  Inside Height
Inside Volume = 6.21  x  6.21  x  7.96
Inside Volume = 307  cu ft











Using the Inside Volume and ASHRAE tables,
corresponding to Inside Volume of 307  cu ft, we find
Air Changes per 24 hr = 34.5
For Normal Ventilation/Infiltration Loss Factor (Long Storage):  0.6
Air Changes per 24 hr = 34.5  x  0.6
Air Changes per 24 hr = 20.7






Properties of Air:
The given Outside Temperature is 75 F  (walk-in cooler is inside air-conditioned building), however, to be on the safe side, we will use:
Outside Air at:                          85 F  80% RH
Cold Storage Room Temp:   35 F  60% RH 






Heat to be Removed from Air:
From ASHRAE Table,
Heat to be removed from air at 85 F  80% RH  to  35 F  60% RH
Heat to be Removed = 1.70  Btu/cu ft 

HEAT air change (infiltration)  =   307  x  20.7  x  1.7
HEAT air change (infiltration)  =   10,803  Btu/day


HEAT motors :
HEAT motors = No. of Motors  x  Btu/motor  x  HP  x  24 hr

No. of Motors (Evaporator/Cooler Fans) =  2 
HP of Motors (each) =  1/8 HP 

From the Table, for 1/8 HP motor,
Btu/motor  =  4600  Btu/HP-hr

HEAT motors = 2  x  4600  x  1/8  x  24
HEAT motors = 27,600  Btu/day


HEAT lights:
HEAT lights = No. of Lamps  x  Watts/lamp  x  Operation hr  x  3.42 Btu/watt

No. of Lamps = 1
Lamp wattage = 40 watts
Lamp operation = 8 hrs per day
HEAT lights = 1  x  40  x  8  x  3.42
HEAT lights = 1094  Btu/day

 
HEAT occupants:
HEAT occupants =  No. of Occupants  x  Heat/occupant  x  Hours of Work

No. of Occupants = 2 persons
Hours of Work =  8 hrs per day

Heat/occupant:
From the Table, we can interpolate the Heat/occupant:
Heat/occupant =  ( 840  +  950 )/2
Heat/occupant = 895  Btu/hr
HEAT occupants =  No. of Occupants  x  Heat/occupant  x  Hours of Work
HEAT occupants =  2  x  895  x  8
HEAT occupants =  14,320  Btu/day


HEAT defrost:
HEAT defrost  = 0
(Note: 
  Heat of Defrost is typically for Storage Temperature of 32 F or lower, 
  in which case, 10% of Defrost Heat Input is added to the Heat Load.)

HEAT solar:
HEAT solar = 0
(Note:
  Illumination (solar) heat from the sun is typically 15 watt/sq m of surface.)


HEAT USAGE LOAD:
HEAT USAGE LOAD = HEAT product  + 
                                              HEAT air change (infiltration)  + 
                                              HEAT motors  + 
                                              HEAT lights  + 
                                              HEAT occupants  +
                                              HEAT defrost  +
                                              HEAT solar
HEAT USAGE LOAD = 5625  +  10,803  +  27,600  +  1094  +  14,320  +  0  +  0                                              
HEAT USAGE LOAD = 59,442  Btu/day                                              
                           
                  
TOTAL HEAT LOAD:                                             
TOTAL HEAT LOAD = HEAT LEAKAGE (TRANSFER) LOAD  +
                                             HEAT USAGE (SERVICE) LOAD
                                              
TOTAL HEAT LOAD =   2392  +  59,442 
TOTAL HEAT LOAD =  61,834  Btu/day                                       

Based on Total System Running Time of 20 hrs:
TOTAL HEAT LOAD/20 hr  =  61,834/20
TOTAL HEAT LOAD/20 hr  =  3092  Btu/hr

Adding 20% Reserve Capacity (R.C.):
TOTAL HEAT LOAD with R.C.  =  3092  x  1.2
TOTAL HEAT LOAD with R.C.  =  3710  Btu/hr

TONNAGE:
TONNAGE =  3710/12,000
TONNAGE =  0.31  TON
TONNAGE  ~  1/3  TON


Split (TD between refrigerant & air):
In a walkin cooler with mixed products, the product requiring the highest humidity will determine the split. 
Pork requires 85% to 90% RH and beans 90% to 95% RH. 
To prevent moisture loss and product damage, we will select  10 °F  split to maintain high humidity. 

SIZING THE EVAPORATOR:
Evaporator Size at  1 °F  TD =  TOTAL HEAT LOAD with R.C. /Split

TOTAL HEAT LOAD with R.C.  =  3710  Btu/hr
Split = 10 F
Evaporator Size at  1 °F  TD =  3710/10
Evaporator Size at  1 °F  TD =  371  Btu/hr-F

From KeepRite Refrigeration tables, evaporators are available at 10 F TD, therefore, we will multiply the Evaporator Size at  1 °F  TD  by  10 °F.
Evaporator Size @  10 °F  TD = Evaporator Size at  1 °F  TD  x  10 °F
Evaporator Size @  10 °F  TD =  371  x  10
Evaporator Size @  10 °F  TD =  3710  Btu/hr

Selecting Evaporator Capacity at Evaporator Temperature:
With Box Temperature of 35 F and Split of 10 F,
Evaporator Temperature (temperature of refrigerant entering evaporator) is calculated as,
Evaporator Temperature = 35 – 10
Evaporator Temperature = 25 F
Using the calculated Evaporator Size @  10 °F  TD of  3710  Btu/hr
and the calculated Evaporator Temperature of 25 F, we select
       KLP Model 104M with 4300 Btu/hr 
       Air Defrost model:  A
       Power supply is 115 Volts, 1-phase, 60 Hz:  S1
       Generation (2nd):  B
Evaporator Order Information:  KLP104MA-S1B



SIZING THE CONDENSING UNIT AND COMPRESSOR:
Based on good design engineering practice, we will select a condensing unit
that will match the capacity of the evaporator model that was chosen on the
previous step.

Using the nomenclature and the table
TECUMSEH AIR-COOLED CONDENSING UNITS,
the closest match we found is
CDU Model AKA9440ZXAXC with a capacity of 4000 Btu/hr,
having compressor model AKA9438ZXA with a capacity of 3800 Btu/hr, 
1/2 HP using Refrigerant R-404A.

Condensing Unit Order Information:  AKA9440ZXAXC


Evaporator Range and Condenser Ambient Range:



HOW TO CALCULATE REFRIGERANT LIQUID RECEIVER CAPACITY:
Specification from Tecumseh Cond. Unit:
       Liquid Receiver Tank capacity = 33  cubic inch
       Tank Design Pressure = 500 psig



Calculate mass flowrate of R404A refrigerant:
Q = m x Cp x TD
m = Q/Cp x TD

Q = 3710  Btu/hr
Cp = Specific Heat from NRI = 0.36 Btu/lb-F
TD = 40 F

m = 3710/(0.36 x 40)
m = 258  lb/hr

mass flowrate per minute = 258/60 = 4.3  lb/min


The general rule of thumb for R22 and R404a charge per ton is 2-4 lbs/ton.
Tonnage = 1/3 ton
Since we are using short lines, we will use Charge per Ton of 2 lbs/ton,

Calculate Refrigerant Charge:
Charge = 2 lbs/ton  x   1/3 ton
Charge = 0.67  lbs

Charge in ounce (oz) = 0.67  lbs  x  16  oz/lbs
Charge in oz =  11 oz.

Calculate Volume of R404A refrigerant charge:
D = m/v
v = m/D

D = Density from NRI = 66.37  lb/cu ft

v = 0.67/66.37
v = 0.01  cu ft

Convert cubic feet to cubic inches:
v = 0.01  x  1728
v = 17.28  cu in.

The liquid receiver capacity from Tecumseh Condensing Unit is 33 cu in.
However, it can only be filled up to 80% capacity according to code and for  safety reasons.

Max effective capacity = 33  x  0.80
Max effective capacity = 26.4  cu. in.

Since the calculated volume of refrigerant charge of  17.28  cu. in.  is LESSER than the Maximum Effective Capacity of 26.4  cu.in., the liquid receiver size is appropriate.


FILTER DRIER SIZE SELECTION:
From Sporlan Tables, we select the
1/4  inch  Catch-All  C-032  Flared filter drier




SELECTION OF SIGHT GLASS:
Using recommended Sporlan sight glass tables, we select
1/4  inch  SA-12FM  Female X Male Flare sight glass




SELECTING TEV/TXV – THERMOSTATIC EXPANSION VALVE:
Application:  Medium Temperature Refrigeration
Design Evaporator Temperature:    25  F
Design Ambient Temperature:          90 F  (from Tecumseh Condensing Unit)
Design TD (condenser & ambient):  35  F

Design Condenser Temperature  =  Design Ambient  +  Design TD
Design Condenser Temperature  =  90  +  35
Design Condenser Temperature  =  125  F

Design Condenser Sub-cooling:  10 F
Design System Capacity:                1/3  TON

Refrigerant Liquid Temperature  = Design Condenser Temperature  -
                                                                        Design Condenser Sub-cooling

Refrigerant Liquid Temperature  =  125  -  10
Refrigerant Liquid Temperature  =  115 F

Calculate Available Pressure Drop across TEV:
Design Condenser Temperature:    125  F
Design Evaporator Temperature:    25  F
    From  R404A  PT chart, determine:
         Condensing Pressure @ 125 F:  333  psig
         Evaporating Pressure @  25 F:    63  psig

        
Pressure drop due to friction, accessories, lines, and others:  50 psig

Available Pressure Drop across TEV =  333  -  63  -  50
Available Pressure Drop across TEV =  220  psig

Using the table with interpolation, we find the correction factor (CF)
CF Liquid Temperature @ 115 = ( 0.89 + 0.77 )/2
CF Liquid Temperature @ 115 =  0.83

Using the following information & using the table to find the closest match:
       Design Evaporator Temperature:          25  F
       Available Pressure Drop across TEV:  220  psig
CF Pressure Drop =  1.34

VALVE TYPE SELECTION:
The following tables will help us choose a TEV that is most suitable for the Walk-in Cooler project application:


Based on project requirements, given data, and recommendations from the tables, we decide to choose the following valve type and characteristics:
Valve Type:  G
Valve Body:  Forged Brass  (for superior strength and durability)
Connection Type:  SAE flare
Capacity Adjustment:  Externally adjustable valve
Thermostatic Element:  Replaceable
Strainer:  Removable 100 mesh strainer at inlet connection
                                                                

            Valve Type:   G
            Design System Capacity:   1/3  TON
            Design Evaporator Temperature:   25  F
Using these information, we can now determine from the R404A table,
TEV Rating = 0.59  TON

TEV Capacity = TEV Rating  x  CF  Liquid Temperature  x  CF Pressure Drop

CF Liquid Temperature =  0.83
CF Pressure Drop =  1.34

TEV Capacity = 0.59  x  0.83  x  1.34
TEV Capacity =  0.66  TONS


Inlet connection, Outlet connection and External line connection sizes are determined from the Type G valve body Specifications table for R404a

---



Sporlan Nomenclature/Order:  GSE-1/2-SC 1/4” x 1/2” x 1/4” SAE flare x 5’
Body Type:  G
Sporlan Refrigerant R404A Code:  S
External equalizer:  E
Nominal Capacity:  1/2  TON
Thermostatic Charge:  SC
Inlet Connection:  1/4  in.
Outlet Connection:  1/2  in.
External Equalizer Connection:  1/4  in. SAE flare
Capillary Tubing Length:  5 ft


SELECTION OF LIQUID LINE SOLENOID VALVE:
From Sporlan chart, we select the corresponding Liquid Line solenoid valve
model  A3F1 – normally close, 1/4  inch,  MKC-1 coil 120 Volts, 50-60 Hz



Bill of Materials Estimate:  Refrigerant Lines and Mechanical Components
ACR Copper line 1/2 in. x 25 ft  for suction line and drain line
ACR Copper line 1/4 in. x 15 ft for liquid line and equalizer line
Armaflex black insulation size 1/2 in
Armaflex black insulation size 1/4 in
1/2 in. vibration isolator
1/2 in. refrigerant line clamps x 10
1/4 in. refrigerant line clamps x 10
1/2 in. 90 degree elbow x 10
1/4 in. 90 degree elbow x 6
Unistrut  1-5/8  inch width  x  3 ft  length
2 in. width  flat bar  x  3 ft  length
1/2 in. x 8 ft threaded rod
1/2 in. nuts x 4
1/2 in. washers x 4
1/2 in. flares x 8
1/4 in. flares x 8


ELECTRICAL LOAD CALCULATION - WALK-IN COOLER:

Notes:
1. Electrical wiring is to be sized in accordance with minimum circuit ampacity (MCA) rating.

2. Use 75 Celsius wire or higher.

3. Overcurrent protection for evaporator fan motors and defrost heaters must not exceed maximum value shown on evaporator nameplate.

4. Size fuses used must not exceed the Maximum Fuse Size ratings.


CALCULATE TOTAL ELECTRICAL LOAD IN VOLT-AMPERES (VA) INPUT:

I. Lighting Load
  
   Watts Output: 40 watt lamp x 1
   Voltage:     120 Volts (Input Voltage)
   Amperage:   0.67 Amps (Amp draw or Input Current)
   LED Light Bulb efficiency: 50%

   Power Output = Volts x Amps x Efficiency
   40 watts = 120 volts x Amps x Efficiency
   Amps = 40/(120 x 0.5)
   Amps = 0.67 Amps
   
   Input Volt Ampere (VA) = 120 Volts x 0.67 Amps
   Input Volt Ampere (VA) = 80 VA


II. Motor Load - Evaporator Fan Motor

    Electrical Data for Evaporator Model KLP104MA-S1B:
       HP:     1/15 HP
       Watts:   100 watts PSC Motor x 1
       Voltage: 115 Volts
       MCA:     1.3 Amps
       Max. Fuse: 15 Amps
       Max. Overcurrent Protection (MOP): 15 Amps
       Full Load Amps (FLA): 1 Amp
       Motor Efficiency: 100 watts/115 watts = 87%   

       Input Volt Ampere (VA) = 115 Volts x 1 Amp
       Input Volt Ampere (VA) = 115 VA            


III. Motor Load - Condensing Unit Compressor & Fan Motor
    
     Electrical Data for Condensing Unit Model AKA9440ZXAXC:
         a. Compressor
            HP: 1/2 HP ( 373 watts )
            Rated Load Amperage (RLA): 9.2 Amps
            Locked Rotor Amps (LRA):  58.8 Amps
            Voltage: 115 Volts
            Compressor Efficiency: 373 watts/1058 watts = 35%
       
            Input Volt Ampere (VA) = Voltage x RLA
            Input Volt Ampere (VA) = 115 Volts x 9.2 Amps
            Input Volt Ampere (VA) = 1058 VA  

         b. Condenser Fan Motor
            Output Watts: 35 watts
            Voltage:     115 Volts
            Amperes:     1.4 Amps
            Motor efficiency: 35/161 = 22%
            
            Input Volt Ampere (VA) = 115 x 1.4
            Input Volt Ampere (VA) = 161 VA

         c. Condensing Unit
            Voltage: 115 Volts
            MCA: 12.9 Amps
            Max Fuse Size: 20 Amps
            Heating Air Conditioning Refrigeration (HACR) Breaker: 20 A


IV. Motor Load - Defrost Timer Motor
   
    From Paragon 9145 Defrost Timer specifications,
    Voltage: 120 Volts
    Power Consumption: 6 VA
    Input Amps: 0.05 Amps
   
    Input Volt Ampere (VA) = 6 VA


V. Controls Load - Thermostat

   From PENN/Johnson Controls A419 Thermostat specifications,
   Voltage: 120 Volts
   Power Consumption: 1.8 VA
   Input Amp: 0.015 A

   Input Volt Ampere (VA) = 1.8 VA


VI. Controls Load - Low Pressure Control

    From Ranco Low Pressure Control model O10-1483,
    Input Voltage: 120 Volts
    Input Current: 24 Amps

    Input Volt Ampere (VA) = 120 X 24
    Input Volt Ampere (VA) = 2880 VA
  

VII. Solenoid Valve Load - Liquid Line Solenoid Valve 

    From Sporlan A3F1 MKC-1 Coil model,
    Voltage: 120 Volts
    Watts: 10 Watts    
    Magnetic Coil Efficiency: 70%

    Power Output = Volts x Amps x Efficiency
    10 watts = 120 volts x Amps x Efficiency
    Amps = 10/(120 x 0.7)
    Amps = 0.12 Amps
   
    Input Volt Ampere (VA) = 120 Volts x 0.12 Amps
    Input Volt Ampere (VA) = 14.4 VA


TOTAL INPUT VA:
  80 VA - I. Lighting Load
 115 VA - II. Motor Load - Evaporator Fan Motor
1058 VA - III. Motor Load - Condensing Unit Compressor
    161 VA          Motor Load - Condensing Unit Condenser Fan Motor
       6 VA - IV. Motor Load - Defrost Timer Motor
    1.8 VA -  V. Controls Load - Thermostat
2880 VA - VI. Controls Load - Low Pressure Control
  14.4 VA - VII. Solenoid Valve Load - Liquid Line Solenoid Valve
--------------------------
4316.2  VA   TOTAL


ADDING 20% RESERVE ELECTRICAL LOAD

4316.2  x  1.20 = 5179.4 VA


DIVIDE TOTAL INPUT VA BY LINE VOLTAGE
5179.4 VA/120 V = 43.16 A


FIND SERVICE AMP: USE TABLE 310.15(B)(16) with Copper 75 C
50 Amp Service (disconnect switch)
AWG 8 Service Conductors (service wire)

Order Information:
60-Amp Eaton Cutler-Hammer 
Lockable Indoor/Outdoor Air Conditioner Disconnect


ALLOWABLE AMPACITY: not less than 83% of the service rating
50 Amp  x  0.83 = 41.5 A










COPPER CONDUCTOR WIRE SIZES FOR WIRING:

Use Max Fuse Amp recommended by Manufacturer

Low Pressure Control   30 A : AWG 10
Condensing Unit              20 A : AWG 12

Evaporator             15 A : AWG 14
Light Bulb               15 A : AWG 14

L.L. Solenoid Valve    15 A : AWG 14
Defrost Timer               15 A : AWG 14
Thermostat                    15 A : AWG 14


TABLE 310.15(B)(16):




 

Estimate Bill of Material -  Electrical: Condensing Unit, Fan Motor:
Defrost timer – Paragon model 9145-00
Thermostat – Johnson Controls model A419
Liquid line solenoid valve – Sporlan model A3F1
Low pressure control – Ranco model O10-1483
Disconnect Switch 60-Amp – Eaton Cutler-Hammer
Evaporator Drain Line Heater
Junction box, square x 2
AWG 8 Wire x 15 ft
AWG 10 Wire x 5 ft
AWG 12 Wire x 15 ft
AWG 14 Wire x 40 ft
1-1/2 in. x 15 ft flexible conduit
1-1/2 in. straight fittings for flexible conduit x 12
1-1/2 in. 90 degree fittings for flexible conduit x 8
1-1/2 in. flexible conduit clamps x 20
1/2 in. x 15 ft BX cable
1/2 in. pipe/tubing clamps for BX cable x 20
Wire connectors x 40
BX cable Anti-shorts x 10


Electrical data - Evaporator

Condenser fan motor input 115 Volt, 1.4 Amp

Paragon 9145 Defrost Timer

Johnson Controls A419 Thermostat



Evaporator Installation:
Ceiling mounted evaporators are best located on the upper corner of the cabinet as far as possible from the entrance door.

LOW PRESSURE CONTROL (LPC) CUT-IN, DIFFERENTIAL, CUT-OUT:
 (note: Cut-in is High Event, Cut-out is Low Event)
 LPC CUT-IN Temperature = Evaporator Temperature – 5 F
 LPC CUT-IN Temperature = 25 – 5
 LPC CUT-IN Temperature = 20 F
 From PT chart for R404a,
 Pressure @ 20 F = 57 psig
 LPC CUT-IN Pressure = 57 psig

 Differential (Diff.) = 10 psig

 LPC CUT-OUT Pressure = LPC CUT-IN Pressure – Differential
 LPC CUT-OUT Pressure = 57 – 10
 LPC CUT-OUT Pressure = 47 psig
 From P/T Chart for R404a,
 LPC CUT-OUT Temperature @ 47 psig = 12 F

  [Note:  Adjust/Fine tune the differential.
    If it short cycles (run time is too short), adjust differential to 15 psig]



TYPICAL WALKIN REFRIGERATOR (R-22) TEMPERATURES/PRESSURE:


Note:
I had errors putting pictures, charts, tables and other images.
Pls. see Youtube Video here: 

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