Shop at Amazon.ca

Friday, November 4, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - mixed air temperature, outdoor, return

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - mixed air temperature 


PROBLEM:


An air-conditioning system draws 10% outdoor air at a dry bulb temperature of 42 C / 108 F. If the return air is at 28 C / 82 F, determine the dry bulb temperature of the mixed air.



Given

10% outdoor air


42C/108F outdoor air temperature 


28C/82F return air temperature 



Unknown 

dry bulb temperature of mixed air



SOLUTION:


(% Outdoor Air  x  Temperature)  +  

(% Indoor Return Air  x Temperature)  = 100%  x  Air Mixture Temperature


using Celcius temperatures and substituting given values, we have


( 10% x 42 )  +  

( 90% x 28 ) = 100% x Air Mixture Temp


Mixed air temp 

= 4.2 + 25.2 

= 29.4 C / 85 F



*  Cheers :)  *




Wednesday, November 2, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - compressor discharge, suction pressures

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - compressor  discharge, suction pressures


PROBLEM 

(from sample questions on): https://www.red-seal.ca/

The compression ratio of a refrigeration system is 7:1. If the discharge pressure is 300 psig, what is the suction pressure?


Given 

7:1 compression ratio


300 psig discharge pressure (gage)



Unknown 

suction pressure 



SOLUTION 


Cr = Pd abs/Ps abs


Ps abs = Pd abs/Cr


where: 

Cr = compression ratio

       

Pd abs = absolute discharge pressure 


Ps abs = absolute suction pressure 


Absolute Pressure 

= gage pressure 

+  14.7 


Pd abs

= 300 psig + 14.7

= 314.7 psia


substituting the values, we get, 


Ps abs = 314.7/7


Ps abs = 45 psia


Ps gage = 45 - 14.7

Ps gage = 30.3 psig



Checking


Cr = 314.7/45

Cr = 7



*

Tuesday, November 1, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - fan speed, static pressure, power, amperage

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - fan speed, static pressure, power, amperage 


PROBLEM 


The rpm of a 3 HP ducted fan moving 1,200 CFM is increased from 900 rpm to 1800 rpm. If the initial static pressure is 0.08 in. w.c., what is the resulting CFM, static pressure, and HP?



Given


cfm1 = 1200 cfm

sp1 = 0.08 in. w.c.

HP1 = 3 HP

rpm1 = 900 rpm

rpm2 = 1800 rpm



Unknown 


cfm2

sp2

HP2



SOLUTION 


Using the Fan Laws,


cfm2 =

cfm1 x (rpm2/rpm1)


cfm2 

= 1200 x 1800/900

= 1200 x 2

= 2400 cfm


when the fan speed rpm is doubled, the cfm is also doubled 



sp2 = sp1 x 

         (rpm2/rpm1)^2


sp2 = 0.08 x 

          (1800/900)^2


sp2 = 0.08 x (2)^2


sp2 = 0.08 x 4


sp2 = 0.32 in. w.c.


when the fan speed rpm is doubled, the static pressure is increased 4 times



HP2 = HP1 x 

         (rpm2/rpm1)^3


HP2 = 3 x 

          (1800/900)^3


HP2 = 3 x (2)^3


HP2 = 3 x 8


HP2 = 24 HP


when the fan speed rpm is doubled, the Power needed to drive the fan is increased 8 times


if the supply voltage is constant, 

then the current draw (amperage) 

is also increased 8 times




*

Monday, October 31, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - wattage of duct heater

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - wattage of duct heater 


PROBLEM 


(from sample questions on): 

https://www.red-seal.ca/



A duct heater is supplied with 238 VAC and drawing 19.9 A. What is the heater's wattage? 


Given


E = 238 VAC

I = 19.9 A


Unknown 

duct heater wattage 



SOLUTION 


P = I x E


P = 19.9 x 238


P = 4,736 watts



*

Sunday, October 30, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - resistance of compressor motor windings

Watch this video in my YouTube channel 

Eternalson17

YouTube link



REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - resistance of compressor motor windings


PROBLEM 


(from sample questions on): 

https://www.red-seal.ca/


When diagnosing the motor windings of a hermetic compressor, the following values are noted: 


Common-to-start: 

9 ohms 


Start-to-run: 

12 ohms 


What is the common-to-run value? 


Given

resistance of 

hermetic compressor windings


Common-to-start: 

9 ohms 


Start-to-run: 

12 ohms 



Unknown 

common-to-run value


SOLUTION 


common-to-run value


= Start-to-run value - Common-to-start value


= 12 - 9


=  3 ohms





*

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - minimum sleeve size for insulated pipes

Watch this video in my YouTube channel 




REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - minimum sleeve size for pipes with insulation 


PROBLEM:
The suction line of a walk-in cooler is 5/8 inch and has an insulation of 1/2 inch. The discharge line is 3/8 inch and has 1/4 inch of  insulation. Determine the minimum required sleeve size of pipe through a concrete block exterior wall.

Given
suction line = 5/8 in.
suction line insulation = 1/2 inch

discharge line = 3/8
discharge line insulation = 1/4 inch

Unknown 
Size of sleeve

SOLUTION 

Sleeve size of pipe
= suction line 
   + ( 2 * insulation )
   + discharge line
   + ( 2 * insulation)

=  5/8 + (2 * 1/2) 
   + 3/8 + (2 * 1/4)

= 5/8 + 1 + 3/8 + 1/2

= 2  1/2  inch



***

Saturday, October 29, 2022

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - capacitance of series capacitors

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - capacitance of series capacitors 


PROBLEM:


Two capacitors are wired in series. The capacitance of the first capacitor is 50 micro Farads (uF). If the second capacitor has 150 uF, solve for the total capacitance of the circuit.



Given 


wiring = series 


capacitance1 = 50 uF


capacitance2 = 150 uF



Unknown 

Total capacitance of circuit



SOLUTION:


Series Capacitors:


Total Capacitance Ct =  1/ [ (1/C1) + (1/C2) ]


Ct = 1/ [ (1/50) + (1/150) ]


Ct = 1/(0.02 + 0.0067)


Ct = 1/0.0267


Ct = 37.5 uF


Alternate formula:


Ct = C1 x C2 / ( C1 + C2 )


Ct = 50 x 150 / (50 + 150)


Ct = 7500/200


Ct = 37.5 uF



*

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - capacitance of parallel capacitors

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - capacitance of parallel capacitors 


PROBLEM:


Two capacitors having 50 uF and 150 uF respectively are wired in parallel. What is the value of the total capacitance?


Given 

capacitance1 = 50 uF


capacitance2 = 150 uF



Unknown 

Total capacitance of circuit 



SOLUTION:


Parallel Capacitors: additive


Total Capacitance, Ct = C1 + C2


Ct = 50 + 150


Ct = 200 uF


Note:

If these same two capacitors were wired in series, the total capacitance would be 37.5 uF, reduced to almost 20%.



*

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - cooler/reefer tonnage capacity

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - cooler/reefer  tonnage capacity 



PROBLEM:


An 8 x 20 ft sea container reefer walkin cooler has an hourly load of 22,500 Btu/hr and operates 24 hours a day. The defrost system runs for 15 minutes every hour. Calculate the tonnage capacity in Tons of Refrigeration and in Kilowatts.


Given

22,500 Btu/hr hourly load

24 hours operation 

15 minutes defrost per hour


Unknown 

Reefer capacity in tons and KW



SOLUTION:


Total defrost time

= 15 minutes x 24

= 360 minutes

= 6 hours


Total Run-Time

= 24 - 6

= 18 hours


Q = hourly load 

       x 24/run time 


Walk in Cooler Condensing Unit capacity

= ( Hourly Load x 24 hr ) / 18 hr Run-Time

= ( 22,500 x 24 ) / 18

= 30,000 Btu/hr


Tonnage capacity = 30,000/12,000 = 2.5 tons


Capacity in Kilowatts = 2.5 x 3.5 = 8.8 KW



*

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - water flow rate of chiller

Watch this video in my YouTube channel 

Eternalson17

YouTube link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - water flow rate of chiller


PROBLEM:

A chiller has a capacity of 75 tons with a heat rejection ratio (HRR) of 1.3. If the change in temperature is 10F, solve for the water flow rate in US gallons per minute and imperial gallons per minute.


Given

chiller capacity = 75 tons

heat rejection ratio = 1.3

delta T = 10F


Unknown

US gpm

imperial gpm


SOLUTION:


Converting tons to Btu/hr 

Q = 75 tons x 12,000 Btu/hr per ton = 900,000 Btu/hr


Incorporating the HRR and rearranging the main equation,


US gpm = Q x HRR/(500 x delta T)


US gpm = 900,000 x 1.3/(500 x 10)


US gpm = 234 US gallons


To solve for imperial gallons, change the constant to 600


imperial gpm = Q x HRR/(600 x delta T)


imperial gpm = 900,000 x 1.3/(600 x 10)


imperial gpm = 195 imperial gpm




*

REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - water temperature entering chiller

Watch this video in my YouTube channel 

Eternalson17

YouTube Link


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - water temperature entering chiller


PROBLEM:


A 60 ton chiller circulates 160 US gallons of water. What is the temperature rise? If the water supply temperature is 45F, determine the temperature of the water entering the chiller.


Given

Cooling capacity = 60 tons 

gpm = 160 US gallons

suppy temperature = 45F


Unknown

Temperature rise = ? or change in temperature (delta T)

Temperature of water entering chiller = ? or return temperature


SOLUTION:


Converting tons to Btu/hr 

Q = 60 tons x 12,000 Btu/hr per ton = 720,000 Btu/hr


Rearranging the main equation Q = 500 x gpm x delta T 


delta T = Q/(500 x gpm)


delta T = 720,000/(500 x 160)


delta T = 9 F


return temperature = delta T + supply temperature


return temperature = 9 + 45


return temperature = 54 F



*






REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - chiller cooling capacity (tonnage)

 

Watch this video in my YouTube channel

 Eternalson17

YouTube Link 


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - chiller cooling capacity (tonnage) PROBLEM: Calculate the tonnage of a water chiller having a water flow rate of 240 US gallons with a supply temperature of 45F and a return temperature of 55F.

Given gpm = 240 US gallons supply temperature = 45F return temperature = 55F Unknown Q = ? cooling capacity in tons of refrigeration

SOLUTION: Q = 500 × gpm x delta T Q = 500 × 240 x (55 - 45) Q = 500 × 240 x 10 Q = 1,200,000 Btu/hr

The conversion is 1 Ton = 12,000 Btu/hr, so the final answer is, Tonnage = 1,200,000/12,000 Tonnage = 100 tons *







REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - walk in freezer hourly load

Watch this video in my YouTube channel Eternalson17:

 YouTube Link 


REFRIGERATION HVAC MECHANIC RED SEAL LICENSE EXAM REVIEWER - walk in freezer hourly load


PROBLEM:


A 3 ton walkin freezer operates 24 hours a day and the defrosting system runs for 10 minutes every hour. Determine the hourly load in Btu/hr.


SOLUTION:


Total defrost time

= 10 minutes x 24

= 240 minutes

= 4 hours


Total Run-Time

= 24 - 4

= 20 hours


Walk in Freezer Condensing Unit capacity

= ( Hourly Load x 24 hr ) / 20 hr Run-Time


3 x 12,000 = Hourly Load x 24/20


Hourly Load = 36,000/1.2


Hourly Load = 30,000 Btu/hr



*