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Section 1.0 – Most Common Units

This table shows a list of the popular unit conversions applicable for HVAC & Refrigeration

1 Btu/h = 8.33 x10-5 tons 1 ton = 12,000 Btu/h

1 Btu/h = 2.93 x10-4 kW 1 kW = 3,412 Btu/h

1 Btu/h = 0.293 W 1 W = 3.412 Btu/h

1 EER = 0.083 kW/ton 1 kW/ton = 12 EER

1 SEER = 0.083 kW/ton 1 kW/ton = 12 SEER

1 feet = 03048 meter 1 meter = 3.2808 feet

1 lb = 0.4536 kg 1 kg = 2.205 lb

1 ft2 = 0.093 m2 1 m2 = 10.7639 ft2

1 gal = 3.79 liters 1 liters = 0.227 gal

1 ft3 = 0.1337 gal 1 ft3 = 7.481 gal

1 fpm = 5.08x10-3 m/s 1 m/s = 195.85 fpm

1 gpm = 0.063 l/s 1 l/s = 15.85 gpm

1 cfm = 0.472 l/s 1 l/s = 2.119 cfm

1 in. wg = 249.09 Pa 1 Pa = 4.01 x10-3 in. wg

1 ft. hd = 2.99 Pa 1 Pa = 3.34 x10-4 ft. hd

1 psi = 6.89 x10-3 MPa 1 MPa = 145.04 psi

1 psi = 27.68 in. wg 1 in. wg = 0.036 psi

1 psi = 2.31 ft. hd 1 ft. hd = 0.43 psi

1 lb/ft3 = 16.018 kg/m3 1 kg/m3 = 0.062 lb/ft3

1 Btu = 0.293 watthour 1 watthour = 3.412 Btu

1 Btu/h = 3.93x10-4 HP 1 HP = 2,545 Btu/h

1 HP = 0.7457 kW 1 kW = 1.34 HP

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 lbm kg kg lbm
1 3
=16.02 3 1 3
=0.0624 3
ft m m ft
kg lbm g lbm
Density (ρ) 1 3
=0.0624 3 1 3
=62.4 3
m ft cm ft
lbm g g lbm
1 3
=27.7 3 1 3
=0.0361 3
¿ cm cm ¿
1∈¿ 25.4 mm 1 mm=0.0394∈¿

Length (l) 1∈¿ 2.54 cm 1 cm=0.394 ∈¿

1 ft =0.3048 m 1 m=3.28 ft
2 2 2 2
1 f t =0.093 m 1 m =10.76 f t
Area ( A)
2 2 2 2
1 in =6.452 c m 1 c m =0.1550i n
3 3 3 3
1 f t =0.0283 m 1 m =35.32 f t
Volume (V )
3 3 3 3
1 in =16.39 c m 1 c m =0.0610i n
−4
1 BTU =1054 J 1 J =9.48 x 10 BTU
1 lbf −ft=1.356 J 1 J =0.738 lbf −ft
Energy (Q)
2
kg−m
1 BTU =778 lbf −ft 1 J =1 2
=1 N−m
s

1 BTUH =0.293 W 1 W =3.414 BTUH

BTU J
Power (P , q) 1 BTUH =1 1 W =1
HR S

12,000 BTUH =1 Ton

9 5
T ( ℉ ) = ( T ( ° K )−273 ) +32 T ( ° C )= ( T (° F )−32 )
5 9
Temperature (T )
9
T ( ℉ ) = ( T ( ° C )) + 32 T ( ° K )=T ( ° C ) +273
5

Pressure (P) 1∈wg=0.0360912 psi 1 psi=27.7076∈wg

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 Section 2.0 – Basic Engineering Practice

Conversion Formula Factor Value

Present Value to FV =PV x ¿

Multiply PV by (F/P, i, n)
Future Value

FV
Future Value to PV =
¿¿ Multiply FV by (P/F, i, n)
Present Value

Present Value to A=PV∗¿

Multiply PV by (A/P, i, n)
Annual Value

Annual Value to PV = A∗¿

Multiply AV by (P/A, i, n)
Present Value

Future Value to A=FV ¿

Multiply FV by (A/F, i, n)
Annual Value

Annual Value to FV = A∗¿

Multiply A by (F/A, i, n)
Future Value

I =current [amps]
Ohm’s Law V
I= V =voltage [volts ]
R
R=resistantce[amps]

Resistors in series
Req ,series =R 1+ R 2+ R 3+ R n

Resistors in parallel 1 1 1 1 1
= + + +
R eq R1 R2 R3 R n

P=I ∗V
2
Power Equations V
P=
R
2
P=I ∗R

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 h ft∗Q gpm∗( SG )
Pmech work , pump[ HP]= ;
3956
Q=volumetric flow rate [ gpm ]
p ∗Q ∗( SG )
Pmech work , pump,[ HP ]= psi gpm ; h=pressure [feet of head ]
Pump Water 1,714
Horsepower
P= power [horesepower ]
Equations
SG=1.0 for water at 39 F
SG=specific gravity
kg lbm
pwater =1,000 3
; 62.4 3 p= pressure [ psi]
m ft

Qcfm=volumetric flow rate of air[cubic feet per minute]

Fan Mechanical
Horsepower Qcfm∗TP¿wg
Pmech work ,fan[ HP ]= ; TP¿ wg=total pressure[inches water gauge]
Equation 6356
Pmech work ,fan[ HP ]=fan mechanical horsepower

Pump or Fan Brake Pfan/ pump[ HP]=P mechwork[ HP ]¿ ¿ ;

Horsepower ε fan/ pump
Equation

Motor Horsepower
Equation Pmotor =Pmech work[ HP] ¿ ¿ ;
ε motor

Electrical Power Pmotor [ HP ]

Supplied to Motor Psupplied ¿ motor [HP ]¿=
PF

Single Phase → P=IV∗PF

Real Power Three Phase → P=√ ❑ PF=power factor ; units → kW , W , MW

Single Phase → S=IV

Apparent Power Three Phase → S=√ ❑ PF=power factor ; units → KVA ,VA , MVA

Apparent Power P=IV PF=power factor ; units → KVA ,VA , MVA

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 Section 3.0 – Thermodynamics

Term Equation Description

Pressure P|¿|[ psi]=P gauge [ psi]+14.7 psi¿

Absolute
Temperature ° R=℉ + 460 temperature

Water Boilts at 212° F∧1 atm

BTU
0.18505
Enthalpy lbm
h=u+ pv∗( 3
)
1 psi∗f t

Water has a specific heat of 1.0 ( lbmBtu℉ )

Btu
c p , air =0.240
Heat Capacity cp( Btu
lbm ℉ ) lbm ° R
Btu
c v , air =0.171
lbm ° R

( )
Isentropic transition υ1
k K = 1.4 for air
from Pressure State p2= p1∗
υ2
1 to Pressure State 2 Q= ṁ∗(h2−h1 )

( )
k
T2 k−1 Q= ṁ∗c p∗( T 2−T 1 )
❑
p2= p1∗
T1

( )
k−1
υ1
T 2=T 1∗
υ2

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 ( )
k−1
P2 k
T 2=T 1∗
P1

K = 1.4 for air

( )
❑
υ2
T 2=T 1∗
Isobaric transition υ1 Q= ṁ∗(h2−h1 )
with heat gain or
❑

( )
❑
heat rejection T2 Q= ṁ∗c p∗( T 2−T 1 )
v 2=v 1∗
T1

Transition Description

Other transitions Adiabatic No change in energy

Isothermal No change in temperature

Isochoric No change in volume

Relation between h g=h f +h fg h g=enthalpy of saturated vapor ; h f =enthalpy of saturated

the Enthalpy of
Saturated Vapor and h fg =enthalpy of vaporization
Liquid

h mix=h f + x∗hfg x=vapor quality

Enthalpy of Mixed
Vapor-Liquid

s g=sf +s fg
Relationship of
Entropy of s g=entropy of saturated vapor
Vaporization,
Entropy of s f =entropy of saturated liquid
Saturated Vapor
and Liquid Water s fg=entropy of vaporization

smix =s f + x∗sfg
Entropy of Wet Steam
(Mixed Region) as a smix =entropy of wet steam(mix of liquid ∧vapor )
Function of Steam Quality
x=steam quality , drnyess fraction, % vapor

Heat Available from Condensing Steam Q= ṁ∗h fg

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 lbm
ṁ=mass flow rate [ ]
hr
Btu
Q=energy [ ]
hr

Throttling: Irreversible Adiabatic [Constant hinitial =h final

Enthalpy] or Isenthalpic

v initial=v final
Tank Heating/Cooling: Isometric [Constant 3
Volume] ft
v=specific volume [ ]
lb

Turbine Expansion: Isentropic [Constant sinitial =s final

Entropy] or Reversible Adiabatic

Compressor: Isentropic [Constant Entropy] sinitial =s final

or Reversible Adiabatic

Pinitial =P final
Boiler Heating: Isobaric [Constant Pressure]
P= pressure [ psia]

Simplified Steam Heating Coil: Steam to ṁsteam∗h fg =500∗GP M water∗∆ T

Water Heat Transfer

Simplified Steam Heating Coil: Steam to Air

Heat Transfer ṁsteam∗h fg =1.08∗CF M air∗∆ T

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 Wor k pump ,compressor =ṁ fluid∗( h leaving−hentering )
Turbines, Pumps and
Compressors Energy Balances
ṁfluid∗( hleaving −h entering )=Wor k turbine

Qboiler / evap + ṁliquid∗hliquid =ṁgas∗hgas

Boilers, Condensers and
Evaporators Energy Balances
Qcondenser + ṁliquid∗hliquid =ṁgas∗h gas

Nozzles, Diffusers Energy 2 2

v enter v exit
Balances h enter + =h exit +
2 2

Heat Exchangers Energy ṁcold∗c p ,cold∗¿

Balances

ṁcold∗c p ,cold∗T cold + ṁhot∗c p , hot∗T hot =ṁmix∗c p ,mix∗T mix

Mixing Energy Balances
ṁcold∗hcold + ṁhot∗hhot = ṁmix∗h mix

A∗C x H y + B∗(O ¿ ¿ 2+3.76 N 2 )→ Heat +C∗C O2 + D∗H 2 O ¿

where A , B , C∧D are numerical values that

describe theratio of each compound∈the equation

Combustion
Air=(O2 +3.76 N 2)

Actual Air
% Excess Air= −100 %
Theoretical Air

Coefficient of Performance EER

(COP) COP=
3.412
W out
COP= (both values must be same units )
W¿
Btu
Q net refrigeration effect (
)
hr
COP=
Btu
W compressor ( )
hr

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 Evaporator Net Refrigeration Effect

Qnet refrigeration effect [ Btuh ] =( H 1−H 4 )

[ ](
Btu
lb
∗ Refrig Flow Rate )
lb
min [ ] [ ]
∗( 60 )
min
hr

H 1=leaving evaporator enthalpy

[ ]
Btu
lb
; H 4=entering evaporator enthalpy [
Btu
lb
]

Compressor Work

W compressor [ Btuh ] =( H 2−H 1 )

[ ]
Btu
lb
∗( Refrig Flow Rate )
lb
min
∗( 60 )
hr[ ] [ ]
min

Refrigeration Cycle
H 2=leaving compressor enthalpy
[ ]
Btu
lb
; H 1 =entering condneser enthalpy [
Btu
lb
]

Net Condenser Effect

Qnet condenser effect [ Btuh ] =( H 2−H 4 )

[ ]
Btu
lb
∗( Refrig Flow Rate )
lb
min [ ] [ ]
∗( 60 )
min
hr

H 2=entering condenser enthalpy

[ ]Btu
lb
; H 4=leaving condneser enthalpy [
Btu
lb
]

Net Condenser Effect

Qnet condenser effect [ Btuh ]

¿ W compressor [ Btuh ] +Q net refrigeration effect [ Btuh ]

Rankine Cycle Pump → Wor k pump =ṁ∗(h2−h1 )

ṁ∗P 2−P1
Pump → Wor k pump =
ρ

where P2∧P1= [ psi ] , ρ=

[ ]
lbs
ft
3
; ṁ=[
lbs
hr
]; Work =
Btu
hr [ ]
Boiler → Q¿= ṁ∗(h3−h 2)

Condenser →Q out =ṁ∗(h 4−h1 )

Turbine →W turb= ṁ∗(h4−h3)

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 Wor k turbine −Wor k pump
Efficiency=
Q¿

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 Section 4.0 – Psychrometrics

Dry bulb temperature indicates the amount of energy independent of the amount of water
Dry Bulb
in the air. Measured with a thermometer.
Temperature
Units=[℉ ]
Wet bulb temperature indicates the amount of water in the air. Measured with a sling
Wet Bulb
psychrometer or hygrometer.
Temperature
Units=[℉ ]
The temperature at which moist air must be cooled to, in order for water to condense out of
Dew Point the air.
Units=[℉ ]
Humidity ratio or specific humidity is the measure of the amount of water in air.
Humidity Ratio lb of Water Varpor
Units=[ ]
lb of Dry Air
Relative Humidity indicates the amount of water in the air relative to the total amount of
Relative Humidity water the air can hold.
Units = [%]

Sensible heat indicates the amount of dry heat. It indicates the amount of energy either
absorbed or released to change the dry bulb temperature of the air.
Sensible Heat Btu
Units=[ ]
lb of air
Latent heat indicates the amount of energy in the air due to moisture. It is the amount of
heat released when water in the air condenses out or the amount of heat absorbed by water
Latent Heat in order to vaporize the water.
Btu
Units=[ ]
lb of air
Enthalpy is an indication of the total amount of energy in the air, both sensible and latent.
Enthalpy Btu
Units=[ ]
lb of air
Mixing of Air T mix , DB=T 1 , DB∗%1 +T 2 ,DB∗% 2
Streams
T 1 ,DB∗CFM 1 +T 2 , DB∗CFM 2
T mix , DB=
CFM 1 +CFM 2

W mix , DB=W 1 , DB∗% 1+W 2 , DB∗% 2

W 1 , DB∗CFM 1+W 2 , DB∗CFM 2

W mix , DB=
CFM 1+CFM 2

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 h mix , DB=h 1, DB∗% 1+ h2 , DB∗% 2

h1 , DB∗CFM 1+ h2 , DB∗CFM 2
h mix , DB=
CFM 1+CFM 2

Qlatent =0.68∗∆ W GR∗CFM

Qlatent =ṁ∗HV

HV =heat of vaporization
Btu
HV =1,060 at ideal conditions
hr
Latent Heat (Air at
Sea Level and Ideal

[ ]
Conditions) grain of H 20
∆ W GR =change∈ specific humidity
lb of dry air

Qlatent =4,734∗∆ W LB∗CFM

∆ W LB
[ lb of H 20
lb of dry air ]
=change∈ specific humidity

lbm
3
∗W grains ,h 20
ft
∗lbm H 20
min lbm dry air
ṁ=CFM∗60 ∗0.075
hr 7000 grains H 20
Latent Heat
or
lbm
3
∗W lbs, h 20
min ft
ṁ=CFM∗60 ∗0.075
hr lbm dry air

Sensible Heat (Air at Btu

Sea Level and Ideal Qsensible [ ]=ṁ c p ∆ T
h
Conditions)
Btu
where c p=0.240 ;
lbm−℉
Cubic Ft 60 Minutes .075 lbm
ṁ = ( )*( )*( )
Minute 1 Hour Cubic Ft .

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 Btu
Q sensible [ ]=1.08∗∆ T∗CFM
h

[ ]
Sensible Heat
Btu 5.2559
Adjusted for Qsensible =1.08∗∆ T∗CFM∗[ 1−elev∗6.8754 x 10−6 ] ; elev ∈feet
Different Elevations h

Qtotal=4.5∗( ∆ h )∗CFM

Btu
Qtotal=total heat [ ]
Total Heat hr
Equation
∆ h=change ∈enthalpy between entering∧leaving
CFM =volumetric flow rate ,cubic feet per minute

Total Heat
Equation
Qtotal
[ ]
Btu
h
=4.5∗CFM∗( ∆ h )∗[ 1−elev∗6.8754 x 10−6 ]
5.2559
; elev∈ feet

pw Ww
RH = x 100 % ≈ x 100 %
p SAT W SAT

RH =relative humidity
Relative Humidity
as a Function of pw =∂ pressure of water vapor ∈the air stream
Humidity Ratio and
Partial Pressures pSAT =saturated vapor pressure of water at the temperature∈question

W w =humidity ratio of the air stream

W SAT =humidity ratio of the air stream at saturation at the temperature∈question

Btu Btu
sensible heat ( ) sensible heat ( )
h h
SHR SHR= =
Btu Btu
total heat ( ) sensibleheat +latent heat ( )
h h

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 Section 5.0 – Heat Transfer

Term Equation Description

Btu
U =heat transfer coefficient [ 2
]
1 hr∗f t ∗℉
Convert U-Factor U=
to R-Value R 2
hr∗f t ∗℉
R=thermal resistance [ ]
Btu

Addition of R- Rtotal=R 1+ R 2+ …+ Rn
Values Layers∈series

1 1 1 1
Addition of U- = + +…+
Factors U total U 1 U 2 Un Layers∈series

Btu
Thermal k=
Conductivity Units hr∗ft∗℉

t=thickness of material [ft ]

[ ]
Convert Thermal t Btu
Conductivity to R- R= k =thermal conductivity
k hr∗ft∗℉
Value and U-Factor
k
U=
t

Btu
U =overall heat transfer coefficient [ 2
]
Heat Transfer Q=U∗A∗∆ T hr∗f t ∗℉
Equation 2
A=area of heat transfer [f t ]
∆ T =temp .difference between hot∧cold areas [℉ ]

∆T a −∆ T b
Log Mean LMTD=
∆Ta ∆ T a=temperature difference at entrance
Temperature ln ⁡( )
Difference (LMTD) ∆Tb
∆ T b=temperature difference at exit

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Counter-flow Heat
Exchanger

Parallel-flow Heat
Exchanger

Q=quantity of heat transferred

[ ]
Btu
hr

Conduction Heat
Transfer Equation, Q=
k∗A∗( T hot −T cold )
t
k =thermal conductivity of material
[ Btu
hr∗ft∗℉ ]
Through Wall
T hot −T cold =temp . diff . between indoors/ outdoors[℉ ]

t=thickness of material [ ft ]

A=area of heat transfer [ ft 2 ]

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 Btu
h=convective heat transfer coefficient [ 2
]
hr∗f t ∗℉
2
Convection Heat Q=h∗A∗∆T A=area of heat transfer [f t ]
Transfer Equation
∆ T =temp .diff . between hot∧cold areas of heat transfer [℉ ]

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 Btu
hrad =radiationheat transfer coefficient [ 2
]
hr∗f t ∗℉
Radiative Heat Q=hrad∗A∗∆ T 2
Transfer Equation A=area of heat transfer [f t ]
∆ T =temp .diff . between hot∧cold areas of heat transfer [℉ ]

Q=radiationheat ( Btuh ) ; A=area ( f t 2 ) ;

Radiative Heat
Transfer Between
Two Objects
Q=ε∗σ∗A∗(T obj1−T obj 2)
4 4
σ =0.1713 x 10
−8
( h∗fBtut ∗R );
2 4

T =surface temperature (° R ) ; ε =emissivity

2 πL∗( T inner −T outter )

Qcond +conv =
ln ln
( ) r2
r inner
+
+ln ln
( )
r outter
rn
+
1
+
1
ki km r inner h inner r outter houtter

Q=quantity of heat transferred

[ ]
Btu
hr

Heat Transfer
Equation, Through
Pipe
k =thermal conductivity of material
[ Btu
hr∗ft∗℉ ]
T inner =Temperature at the inner pipe wall[℉ ]

T outter =Temperature at the pipe exterior [℉ ]

r inner =inner radius of pipe [ft ]

r outter =outter radius of pipe[ft ]

L= pipe length [ ft ]

h=convective heat transfer coefficient ¿

μ∗c p
( lbm∗℉ );
Prandtl Number Btu
Pr= c p=specific heat
k

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 ( )
Btu
k =thermal conductivity ;
t 2∗℉
h∗f
ft

lbm
μ=dynamic viscosity ( )
hr−ft
D∗hconv
Nusselt Number Nu= =C D=diameter of pipe ( ft ) ,
k
k =thermal conductivity of the fluid

D∗hconv .8 .4
Turbulent flow inside circular pipe Nu= =.023∗R e ∗Pr
(heating) Nusselt Number k

D∗hconv .8 .3
Turbulent flow inside circular pipe Nu= =.023∗R e ∗Pr
(cooling) Nusselt Number k

D∗hconv
Laminar flow inside circular pipe Nu= =4.36
(heating) Nusselt Number k

D∗hconv
Laminar flow inside circular pipe Nu= =3.66
(cooling) Nusselt Number k

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 Section 6.0 – Fluid Mechanics

Term Equation Description

lb
μ[ ] 2 where ρ=density , μ=dynamic viscosity ;
Kinematic and ft∗s ft
v= =[ ]
Dynamic Viscosity lb s v=kinematic viscosity
ρ[ 3 ]
ft
ρfluid
Specific Gravity SG= lb
ρwater ρwater =62.4 @ 60℉
3
ft

u
M=
Mach Number c M =Mach number [unitless]; u=velocity
[ ]
ft
sec
; c=speed of sound

M =Mach number ; Rs =Specific gas constant ;

T =temperature∈ Rankine
cp
k =ratio of specific heats= ,
cv

c p=specific heat of gas at constant pressure∧¿

Speed of Sound c= √ ❑
c v =specific heat of gas at constant volume

Air Properties Units

Rs
[ ]
1,716 ft−lb
slug−° R

k 1.4 N/A

q ¿/out =any heat entering∨exitingthe system

1 2
q ¿ +h ¿ + v ¿
2 h¿/ out =enthalpy at the entrance/exit of the nozzle
Nozzles & Diffusers
1 2 v ¿/out =vel at the entrance∧exit of the nozzle
¿ q out +hout + v out
2

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 ∆ m=0 A¿/ out =area at the entrance/exit of the nozzle
Conservation of
mass ∆ ṁ=0 ρ¿/ out =velocity at the entrance/exit of the nozzle
ρ¿∗A¿∗v ¿= ρout∗A out∗v out

p= pressure
[ ]
lbf
ft
2
;

R=Individual Gas Constant

Ideal gas law
p= ρ∗R∗T ft−lbf
¿ 53.35 for air ;
lbm−R
T =temperature∈ Rankine

Darcy Weisbach
Equation
2
fL v
where h=ft of head ; f =Darcy friction factor ; v=velocity
[ ]
ft
sec
,
h=
2 Dg ft
D=inner diameter [ ft ] , g=gravity [32.2 2
]
sec

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 3
Convert GPM to 1 ft
Multiply GPM by ¿ get .
Cubic Feet 448.83 sec

( )
2
ft
ν=kinematic viscosity ;
sec
V ∗D
Reynolds number=
Reynolds Number ν
V =velocity of fluid ( secft );
D=diameter of pipeor hydraulic diameter [ft ]

C L∗ρ∗A∗v
2 where C L =coefficient of lift ; ρ fluid=fluid density ;
Lift F L=
2 A=area of object ; v =velocity of fluid

C D∗ρ∗A∗v
2 where C D =coefficient of drag ; ρ fluid=fluid density ;
Drag FD=
2 A=area of object ; v =velocity of fluid

B=buoyancy force ; (lbf ∨N )

kg lbm
ρ fluid=fluid density ;( 3
∨ 3)
m ft
Buoyancy
B=ρ fluid∗V fluid displaced∗[ g]
V fluid displaced=volume of displaced fluid

[ g=9.81
m
s
2
for SI
]

Fluid Power
p=
F
A
p= pressure
( )
lbs
in
2
; F=force ( lbs ) ;

2 2
A=π r =area(i n )

( ∆∆V P/V ) ; psi

Bulk Modulus
β=

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 2
−5 ft
Viscosity Centistokes=1.0764 x 10
s
lb
Poise=0.067197
ft−s
2 2
p v 1 pg z 1 p v 2 pg z 2
P1 + + + hmech =P 2+ +
2 gc gc 2 gc gc

Bernouli’s ft−lbm ft
gc =32.2 2;
g=32.2 2
lbf −s s
ft 2 2
v= ; z=ft ; P=lb/f t ; h mech=lbm /f t
s

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 Section 7.0 – Energy/Mass Balance

Term Equation Description

Dehumidification/
W 1∨2
[ lb of H 20
lb of dry air ]
=specific humidity

ṁwater =m˙air∗(W 1−W 2)

Humidification m˙air =mass flow rate of air ;

˙ =mass flow rate of water

mwater

Ṁ liquid− A , enter + Ṁ liquid−B , enter= Ṁ liquid−mix , leaving

Mixing General
Ṁ gas− A , enter + Ṁ liquid−B , enter= Ṁ gas−mix ,leaving + Ṁ liquid−mix , leaving
Equations
Ṁ gas− A , enter + Ṁ gas−B , enter= Ṁ gas−mix ,leaving

T mix , DB=T 1 , DB∗%1 +T 2 ,DB∗% 2

Mixing Air based T 1 ,DB∗CFM 1 +T 2 , DB∗CFM 2
on Temperature T mix , DB=
CFM 1 +CFM 2

h mix , DB=h 1, DB∗% 1+ h2 , DB∗% 2

Mixing Air based
on Enthalpy h1 , DB∗CFM 1+ h2 , DB∗CFM 2
h mix , DB=
CFM 1+CFM 2

W mix , DB=W 1 , DB∗% 1+W 2 , DB∗% 2

Mixing Air based W 1 , DB∗CFM 1+W 2 , DB∗CFM 2
on Humidity Ratio W mix , DB=
CFM 1+CFM 2

Latent Heat of
970 Btu/lbm
Vaporization

26http://www.engproguides.com
 Section 8.0 – Heating/Cooling Loads

Term Equation Description

Btu
1 U =heat transfer coefficient [ 2
]
Convert U-Factor to
U= hr∗f t ∗℉
R
R-Value 2
hr∗f t ∗℉
R=thermal resistance [ ]
Btu

Addition of R-Values
Rtotal=R 1+ R 2+ R 3 …+ R n

1 1 1 1 1
= + + …+
Addition of U-Factors U total U 1 U 2 U 3 Un

Btu
Thermal Conductivity
k=
hr∗ft∗℉
Units

t
R=
Convert Thermal k t=thickness of material [ft ]
Conductivity to R-
k k =thermal conductivity ¿
Value and U-Factor U=
t

Btu
U =overall heat transfer coefficient [ 2
]
hr∗f t ∗℉
2
Q=U∗A∗∆ T A=area of heat transfer [f t ]
Heat Transfer
Equation ∆ T =temperature difference between
hot∧cold areas of heat transfer [℉ ]

Roof/Wall Conductive Loads → Q=U∗A∗CLTD

Skylight/Window Conductive Loads → Q=U∗A∗CLTD

Or

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 Conductive Loads → Q=U∗A∗∆ T
Radiative Loads → Q= A∗SC∗SCL
SC=shading coefficient
SCL=solar cooling load factor

Q=N∗SHG∗CLF
N=number of people
People
Sensible loads SHG=sensible heat gain, activity dependent
[ ]
Btu
hr

CLF =cooling load factor

Q=N∗LHG∗CLF
N=number of people

[ ]
People Latent loads Btu
Latent=latent heat gain , activity dependent
hr

CLF =cooling load factor

Btu
N∗Watts∗3.412
hr
Q= ∗UF∗SAF∗SF
watts
wher N=number of light type .
Lighting
UF =usage factor
SAF=special allowance factor
SF=space fraction

Miscellaneous Btuh
Equipment ∗P ( HP)
HP
Total Heat =¿ Q=2545
Efficiency

Motor Heat Loss=¿ Total Heat∗(1−Efficiency )

Equipment Heat Loss=¿ Total Heat∗(Efficiency )

28http://www.engproguides.com
 Qmotor heat loss=Total Heat∗(1−ε motor )∗F U∗F L

Qequipment heat loss=Total Heat∗(ε motor )∗F U∗F L

P=horespower of motor
ε motor=efficiency of motor

F U =usage factor of the motor

F L =load factor of the motor

Q=qinput∗F U∗F R

Btu
q input=input ¿ the equipment
h
F R =fraction of thetotal heat that is radiated ¿ the space

F U =usage factor

Btu
Q total =4.5∗CFM∗∆ h[ ]
lb

Infiltration
Qsensible =1.08∗CFM∗(T outdoor −T indoor)

Qlatent =4,840∗CFM∗(W outdoor −W indoor )

W =humidity ratio ¿]

29http://www.engproguides.com
 Section 9.0 – Equipment & Components

Term Equation Description

Simplified Sensible
Heat Equation
Q
[ ]
Btu
h
=1.08∗CFM∗∆ T [℉ ]
¿ air conditions at 70 ℉∧1 atm .

Q1 D1
= ; if speed is held constant ,
Q2 D2

where Q=flow∧D=diameter

Q1 N1
= ; if diameter is held constant
Q2 N2
2
H 1 D1
Affinity laws = ; if speed is held constant ,
H 2 D22
(For when given a
system curve and
diameter change is where H= pressure , D=diameter
less than 5%) 2
H1 N1
= ; if diameter is held constant
H 2 N 22
3
P 1 D1
= ; if speed is held constant
P 2 D32

where P= power∧D=diameter
3
P1 N1
= ; if diameter is held constant
P 2 N 32

hentering coil−hleavingcoil where h is equal to the

Contact Factor Contact Factor= enthalpy
Equation for Coils hentering coil −happaratus dew point

T entering coil −T leaving coil where T is equal to the

Contact Factor Contact Factor= dry bulb temperature
Equation for Coils T entering coil −T apparatus dew point

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 W entering coil−W leavingcoil
Contact Factor Contact Factor=
Equation for Coils W entering coil −W apparatus dew point

Bypass Factor Bypass Factor=1−Contact Factor

H=60∗ρ∗Q∗(W exit−W enter )

[ lb of water ]
W =the humidity ratio entering∨leaving the system
[ lbof dry air ]
lb
ρ=density of air [ ]
Moisture Transfer ft
3
Equation
3
ft
Q=air flow rate[ ]
min
lb
H=moisture transferred [ ]
hr

Range=T water ,∈¿ [ ℉ ] −T water ,out [℉ ] ¿

Approach=T water ,out [℉ ]−T air∈,WB [℉ ]

Cooling Tower Range

Effectiveness=
Range+ Approach

Q [ Btuh ] =500∗flow rate

[ ]gal
min
∗¿

Cooling Tower
Evaporation Rate
.000943∗cooling tower flow rate
[ ]
gal
min
∗¿

Fans CFM∗TSP[¿ . wg]

Fan mechanical horsepower= 1
6,356
Brake horsepower =MHP∗( )
fan efficiency

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 Motor HP=BHP∗ ( motor efficiency
1
)

( )
2
p∗v lbm
VP= = 2
2∗gc ft

ft−lbm
gc =32.2 2
lbf −s
ft 2
v= ; z=ft ; VP=lb /f t
Fans s

Shortcut Equation
FPM
Fan velocity pressure= [ ¿ . wg ]
4005
where FPM =velocity feet per minute

Affinity Laws for all

( )
other cases (fans Q1 D1 3 N 1
and pumps) = ( )
Q2 D2 N 2
N = speed, Q = flow

( )( )
2 2
rate, H = head or H1 D1 N 1 p1
= ( )
pressure, H2 D2 N 2 p2
p = density,

( )( )
5
P1 D1 N 1 3 p1
= ( )
D = diameter, P2 D2 N 2 p2

( )( ) ( )
P = power Q1 D1
2
H1
.5
p2
.5

=
Q2 D2 H2 p1

( )( )
.5 .5
N1 D2 H 1 p2
= ( )
N2 D1 H 2 p1

( )( ) ( )
2 1.5 .5
P1 D1 H1 p2
=
P2 D2 H2 p1

32http://www.engproguides.com
 ( )
3
N1 D2 Q1
= ( )
N2 D1 Q2

( )( )
4
H1 D2 Q1 2 p1
= ( )
H2 D1 Q2 p2

( )( )
4 3
P1 D2 Q 1 p1
= ( )
P2 D1 Q 2 p2

SCFM= ACFM∗ ( P14.7 )∗( T519 )

actual

actual

ACFM vs. SCFM Pactual =actual pressure∈absolute terms ( psia ) ;

T actual=actual temperature∈Rankine (° R)

R
>10 → thin walled pressure vessel assumption
t
PR
for cylindrical thin walled pressure vessels → σ =
t
PR
for spherical thin walled pressure vessels → σ=
Pressure vessel 2T
σ =stress ( psi ) ; P= pressure ( psi ) ;

R=radius ( ¿ )

t=thickness(¿)

W ∗√ ❑
Gas pressure relief valve sizing → A= A=valve effective orifice area ( in ) ;
Pressure relief 2

❑
C=coefficient of gas ( 315 for air ) ;

K=coefficient of discharge

( typically 0.975 ) ;
K b =correction factor for back pressure

M =molecular weight of gas

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 ( see periodic table ) ;

P1=relieving pressure ( psia ) ;

T =absolute temperature ( ° R ) ;

W =relieving capacity ( lbshr ) ;

Z=compressibility factor

Q=volumetric flow rate ( gpm ) ;

Control valve liquid

Q=C v √ ❑ C v =valve coefficienct ;

∆ P=pressure drop ( psi )

SG=specific gravity of fluid

Q=volumetric flow rate

( scfh−standard cubic feet per hour ) ;

C v =valve coefficienct ;

Control valve gas

Q=59.64∗C v ∗Poperating √ ❑ ∆ P=pressure drop ( psi ) ;

T =absolute temperature(° R)

SG=specific gravity of gas ( air=1.0 ) ;

P1=operating pressure ( psi)

Control valve critical Q=C g∗P1∗√ ❑ where C g=gas valve coefficient

point

Btu
HHV =higher heating value of fuel [ ]
lb
Boiler fuel energy Qfuel = ṁ∗HHV
ṁ=mass flow rate of fuel
[ ]
lb
hr

Boiler efficiency ε boiler=( ṁ¿¿ feedwater )∗¿¿ ¿

Boiler energy for Q=500∗GP M water∗∆ T =ṁwater∗c p∗∆T

water temperature

34http://www.engproguides.com
 ṁwater =lbs/hr

increase c p=heat capacity of water

[ Btu
lbm∗℉ ]
=1.00

∆ T =change∈temperature of air [℉ ]

Q= ṁwater∗hfg

h fg =heat of vaporization of water ( lbm

Btu
)
ṁwater =mass flow rate∈lbs /hr

[ ]
Boiler energy for
3
vaporization of 1f t
water
∗62.4 lb
7.48 gallon
3
∗60 minute
gallon of water ft
1 ∗
minute hour

Q=500∗GP M water∗hfg =ṁwater∗h fg

Desuperheating Region
1 st Step :Q=ṁsteam∗c p ,steam∗( T initial−T sat )

Condensing Region
Feedwater heater 2 nd Step: Q=ṁsteam∗h fg

Subcooling Region
3 rd Step:Q=ṁsteam∗c p , steam∗( T sat −T final )

35http://www.engproguides.com
36http://www.engproguides.com
 Section 10.0 – Systems & Components

Term Equation Description

0.625
1.30∗( a∗b )
De = 0.250 where a∧b are the width[ft ]∧height [ft ]of the duct
Rectangular Duct ( a+ b )

1.55∗( A∗Area )0.625

De = 0.250
( Perimeter ) where A is themajor axis∧a is the minor axis
Oval Duct
Perimeter=π∗a+2∗( A−a)

FPM =air velocity∈feet per minute

[ ]
2
Velocity Pressure FPM
as a Function of Air VP= [¿ . wg]
4005 VP=velocity pressure[¿ . wg]
Velocity

¿ . wg
Friction loss due to F duct [¿ . wg]=L [ft ]∗f [ ]
length of duct 100 ft

Friction loss due to F duct [ ¿ . wg ] =C∗VP

duct fitting

2
0.1025 Lq
f= 5.31
rd
Where, f = pressure drop ( psi )
L= pipe length ( ft )
Compressed air
piping loss q=airflow ( cfm )

r =ratio ofcompression ( dischargefree

pressure @ pipeentrance
air pressure )
d=internal pipe diamter (¿)

37http://www.engproguides.com
 h=ft of head ;
f =Darcy friction factor ;

Darcy Weisbach
h=
fL v
2 Dg
2 v=velocity
[ ]
ft
sec
,

D=inner diameter [ ft ] ,
ft
g=gravity [32.2 2
]
sec
Pressure
1 psi is equal to 2.31 feet of head (water)

NPSH =Suction Hea d inlet of pump−Vapor Pressur e water

NPSHA=(P atm ± P elev −Pfric )−V P water

Net positive
suction head
NPSHA=P¿ −Pfric −P vapor ¿

or
NPSHA=Pgauge + Pvelocity −Pvapor

Velocity pressure 2 ft ft
V velocity∈ ; gravity=32.2
(Pumps) [ ft of head ] ; sec sec ⁡
2
2g

Air to air energy q total ,actual =4.5∗CF M outdoor∗(h ¿ ¿ outdoor−hsupply )¿

recovery devices
q stotal ,actual =4.5∗CF M return∗(h ¿ ¿ return−h exhaust )¿

q sensible , actual=1.08∗CF M outdoor∗(T ¿ ¿ outdoor−T supply )¿

q sensible , actual=1.08∗CF M return∗(T ¿ ¿ return−T exhaust )¿

q latent , actual=4,770∗CF M outdoor∗(W ¿ ¿ outdoor−W supply )¿

q latent , actual=4,770∗CF M return∗(W ¿ ¿ return−W exhaust )¿

38http://www.engproguides.com
 q sensible , actual
ε sensible=
q sensible , max

q latent , actual
ε latent=
q latent , max

qtotal , actual
ε total=
qtotal ,max

39http://www.engproguides.com
 Section 11.0 – Supportive Knowledge

Term Equation Description

Fresh air flow

based on ASHRAE
Fresh air flow rate=R p∗( ¿ of people )+ R A∗( Area)
62.1

Refrigeration Room
Q [ CFM ] =100 X G , where G = lbs of refrigerant.
0.5
Ventilation Rate

Vibration Control:
Natural Frequency f natural (Hz)=3.13∗√ ❑
of Spring

Vibration Control: 1 Transmissibility %=¿

T=

( )
2
Transmissibility f forcing frequency
and Vibration −1 100 %−Vibration Isolation Efficiency %
f naturalfrequency
Isolation Efficiency

Combining the D B1 DB2 D B3 D B4 DB5

10 10 10 10 10
Sound Levels of L A =10∗(10 +10 +10 + 10 +10 +..)
Multiple Sources

Ldb=Lequip −20∗x−1

Ldb=Sound level at a distance of x [ DB]

Sound Level at a Distance from a Point Source
(Spherical Propagation) Lequip =Sound level of equipment [DB ]
'
x=distance ¿ equipment [f t ]

Sound Level at a Distance from a Point Source Ldb=Lequip −20∗x+ 2

(Half-Spherical Propagation)

Sound Level at a Distance from a Point Source Ldb=Lequip −20∗x+ 5

(Quarter-Spherical Propagation)

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Sound Level at a Distance from a Point Source Ldb=Lequip −20∗x+ 8
(Eighth-Spherical Propagation)

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