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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
*
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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
*
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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
*
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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%.
*
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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
*
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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
*
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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
*
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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 *
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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
*
