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

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Eternalson17

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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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Eternalson17

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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

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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

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