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

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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 :)  *

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

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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

*

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

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Eternalson17

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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

*

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

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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

*

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

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Eternalson17

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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

*

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

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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

***

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

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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

*

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

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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%.

*

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

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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

*

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

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Eternalson17

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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

*

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

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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

*

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

 

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 Eternalson17

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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 *

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

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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

*