CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
1
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
CIH EXAM Equations simply explained and with
examples
by Dr. Daniel Farcas, CIH, CSP, CHMM
The author will like to thank the following reviewers for their insightful comments
and efforts towards improving this book:
Michael McCawley, Ph.D. – West Virginia University.
Lorenzo Cena, Ph.D. – West Chester University.
Clive Diaz, PE - Stratified, Inc.
2
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
CONTENT
Equations Sheet……………….………………………………………...4
Chapter 1: Ventilation………………………………………………...8
Chapter 2: Noise………………………………………………………..36
Chapter 3: General Science, Statistics, Standards………49
Chapter 4: Heat Stress……………………………………………….75
Chapter 5: Radiation………………………………………………….77
Chapter 6: Hood Airflow Equations……………………………95
3
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
VENTILATION
Q = VA
V1A1 = V2A2
TP = VP + SP
Page 8
Page 9
Page 11
V = 4005 V = 1.29
Page 14
C- = ⌊/
!
VP = "
#$$%
3
)VP = "
35
Page 19
OP-3' M?4 S
=-
Page 23
Q=
`$3
E
C^ =
WY
P = P "6 & "WY6 &
4
4
Page 32
NOISE
Pa-
dJ ef
h
gE
i
)Q =
A
4
% WY
4
t
Lm = 10 log "t &
s
Page 37
L-u = 10 log Lv ∑x
my L104s t m NN
w4
Wo = 10-12 watts
Page 38
Lq = L> -20logr-0.5 + DI + CF
Page 41
M
M
%D = 100 "v4 + v6 + ⋯ + vH &
6
Page 43
U
%
H
#/PVWGH $X +
YZM? TWA = 16.61 log "$$& + 90dBA
Page 45
f=
U WY
Q = Q "6 & "WY6 &
4
4
Page 31
FTP = TP<lT − TPm]
Page 34
Z
SPL = SPL + 20 log "4 &
L> = 10 log "Z &
6
s
Page 37
wzH
Page 38
Lv = 10 log L∑x
my 10 4s N
xWY
`$
Page 45
Page 41
TL = 10 log "{&
Page 40
~ Lq = L> − 20logr − 11 + DI + CF
Page 43
?4
Page 24
Page 39
v
w bw
L
N
A!
M?6
Page 34
Page 38
Tq =
t − t = − 3EJ ln LM N
FSP = SP<lT − SPm] − VPm]
Page 33
Page 36
|/0 |
Page 21
H
dJ ef
N
gE
L
Page 30
PWR = PWR "6 & "WY6 &
s
Q = 1.29 C- Page 27
C^ = C^ e
Page 29
4
bc
3J
SPL or Lq = 20 log " &
M
Page 26
Page 28
|/0 |
Page 24
#$U/PVWGH $X YZM? Page 25
N;\]^-_ =
QF = G
B@CDE
3' T6 -T4 E
hood entry loss = Fh x VPd
Page 18
3
Q ;<22 = Q =<>-2 /
Page 22
OP-3' M?6 S
.*
& df+
Page 20
/?@A
Q = 4005C- A:|SP |
Page 13
Page 17
Q = 4005 C- 3
35
Page 19
& df
Page 17
VP2 = " 4 & VP + " 6 & VP
0⌋
SP = − F + 1VP Page 12
Page 16
ln
SP + VP = SP + VP + ∑ losses
DI = 10 log Q
%
Page 42
TWA-u = 10 log "$$& + 85dBA
;
f=
Page 46
Page 44
f = 2f
Page 46
f; = :f f
Page 47
f = √2 f
5
Page 48
4
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
GENERAL SCIENCES, STATISTICS, STANDARDS
ppm =
R- =
@f
x10`
HE
Page 49
log
Page 53
ppm =
ts
t
A
x10`
f
Page 50
= abc
ppm =
pH = −log$ H ¡
Page 54
PT<T\= = X P + X P + ⋯ Xm Pm vapor/hazard ratio =
TLVGm =
°4
°6
±wg4 ±wg6
⋯
RF = x
°
±wg
Page 59
C\_¦ =
V¥
$$$³
GM = 10
µ
¹
¸µ ¸´
Page 65
GSD =
90%Conf Interval = X ± 1.645
HEAT STRESS
/
√]
WBGT = 0.7t ]>¦ + 0.2t ^ + 0.1t ¦
Page 75
SD = TLVGm =
M4
v¯4
4
4
4 6
§£¤ ¡¨£b ¡
§¡
+
M6
v¯6
+⋯+
M
v¯
Page 58
`²
C\_¦ =
M³ M´ ¥
$$$¥µ ³
Page 60
∑H 6
GM = :x x … x]
]
Page 64
SAE = 1.645CVT<T\=
Page 67
SDq<<=- = /@@BDÁ
Page 52
K¦ =
Page 63
4 -6
^ 6 O S
Page 55
Page 69
V¯v4 v6 ⋯ v Page 73
²
%$% Tm=- \=l-
/¥Vv4 6 M4 6 v6 6 M66 ⋯ v 6 M 6
Page 70
#$
RF = x
GSD = %.¿% Tm=- \=l-
Page 68
M
£¤ ¡¥b ¡
£¥¡
Page 66
t=
Vv/ =
Page 60
Page 65
E; = E + E + ⋯ + E]
LCL = V¯
-
#
`
Page 63
#.U% Tm=- \=l%$% Tm=- \=l-
6 6
]Wv6
Page 51
_\T.;<];-]T2\Tm<]
-q<_l2- ulm-=m]-
$.`
¥µ
=
Page 55
d = _m]º
Page 62
∑B@?¾
K\ =
Page 59
Em¦-2 -]_mT· =
Page 61
Page 57
4 4
]Wv4
G.>.
Page 50
Page 54
Page 56
?
#.#%
5
CV =
/
À
Page 67
]4 /4 6 ]6 /6 6
]4 ]6 Page 70
RWL = LCÂHMÂVMÂDMÂAMÂFMÂCM
Page 71
95%Conf Interval = X + 1.645
Page 74
WBGT = 0.7t ]>¦ + 0.3t ^
Page 75
¯
LI = WZ¯
Page 72
/
√]
∆S = M − W ± C ± R − E
Page 76
5
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
RADIATION
I =
I " 4 &
6
Rem = RADQF
Page 77
Page 77
I = 1/2¥ I$
I = 1/10§ I$
Page 80
v4/6Dµµ
v4/6EÁ
+
Page 84
r = "#ËV¯&
#Î
I =
t4
É
$ Êgw
Page 87
/
D_ = Ø "Ëv¯ -a &
spatial ave = L
Page 89
D ¯ = : a + Ø r Page 93
?
X = 3.32log " 4 & HVL
t
t6
V6
U¿¿$
#Î /
rx£Ì = ¦s "ËV¯&
6
∑¸
HÑ4 Ð/H
x
/
N
s
Page 90
Page 80
I = I< Bel
#
¥
Page 86
Î ;<_Ï /
&
ËV¯
Page 89
$.$$UÒ/;G6
VDµµ
Page 91
S=
Page 86
rx£Ì = "
bs.XÈ5f
±4/6
Page 83
S = 37.7H Page 88
t=
A = Am e
Page 79
Page 85
$.`*U
N
v4/6 m
Page 83
S=
v4/6EÁ v4/6´H@
/
#Î
Am =
Page 82
Ov4/6EÁ SOv4/6´H@ S
rx£Ì = Ø "ËV¯ − a &
Page 87
É
Êgw
Page 78
Page 85
/
P
t4
A = Am 0.5
Page 81
T/- =
v4/6´H@
Page 78
I =
Page 81
=
D=
f
±4/6
Ç¥
6
V¯
t = Y¯ x0.1h
O. D. = log
Page 92
Page 92
ts
t
G = 104s
Page 93
CONSTANT AND CONVERSIONS
0F=9/5(0C)+32
0R=0F+460
1 ft3=7.481 U.S. gal
0K=0C+273.15
1L=1.0566 qt
molar volume at 250C, 1 atm=24.45L
1 inch=2.54 cm
1 lb=453.6 grams
1ft3 =28.32L
1 gram=15.43 grains
1 atm=14.7 psi=760 mm Hg=29.92 inHg=33.93 ft water=1013.25 mbar=101,325 pascals
1 Curie=3.7x1010 disint/sec (Becquerel)=2.2x1012 dpm
1 Tesla=10,000 Gauss
1 Gray=100 Rad
1 BTU=1054.8 joules=0.293 watt hr
Speed of sound in the air at 680F (200C)=1130 fps (344 m/s)
Planck’s constant=6.626x10-27 erg sec
1 cal=4.184 joules
speed of light=3x108 m/s
Avogadro’s number=6.024x1023
gas constant, R=8.314 l/mole K=0.082 L atm/mole K
g=981 cm/sec2 =32ft/sec2
1 Sievert=100 Rem
density of air=1.29 g/L at 1 atm, 00C
Ac=385 mm2 for 25 mm filter
Af=0.00785 mm2
6
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
HOOD AIRFLOW EQUATIONS
HOOD TYPE
DESCRIPTION
ASPECT RATIO,
H/L
AIRFLOW
Q = 3.7 LVxX
Slot
0.2 or less
Page 95
Q = 2.6 LVxX
Flanged slot
0.2 or less
Page 96
Plain opening
Flanged opening
0.2 or greater and
round
0.2 or greater and
round
Q = Vx(10X2+Af)
Page 97
Q = 0.75Vx(10X2+Af)
Page 98
Q = VA = VfWH
Booth
To suit work
Page 99
Q = 1.4 PVX
Canopy
To suit work
Page 100
Plain multiple
slot opening (2)
or more slots
Flanged multiple
slot opening (2)
or more slots
Q = Vx(10X2+As)
0.2 or greater
Page 101
Q = 0.75Vx(10X2+As)
0.2 or greater
Page 102
7
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
VENTILATION
Q = VA (1)
Q = volumetric flow rate, in cubic feet minute, cfm, (ft3/min) or cubic
meters per second (m3/sec).
A = cross-sectional area of the duct, in square feet (ft2) or square
meters (m2).
V = duct velocity, in feet per minute (ft/min) or meters per second
(m/sec).
This equation correlates the volumetric flow rate (Q) in a duct to the crosssectional area of the duct (A) and duct velocity (V) of the air or gas flowing
through it. Most exams questions give the diameter or circumference of the
duct, and ask to calculate the duct area using these formulas:
×
Ø
The cross − sectional area of the duct or circle area = ÕÖ × = Õ
Ù
The circumference of a duct (or circle) = ×ÕÖ
Where:
r = radius of the duct
d = diameter of the duct
Ú = 3.14
Example: What is the flow (Q) of an 8-inch diameter duct with a 100 feet
per minute (ft/min) duct velocity?
8
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
First, identify the variables. You have 100 feet per minute, which is velocity
(V), and an 8-inch diameter value that you can use to calculate the duct area
(A).
Remember that 1 foot = 12 inches, and so 8 inches divided by 12-inch equals
0.66666… feet (round up to 0.667 feet).
Then,
Duct area = π
6
#
= 3.14 ∗
$.``¿ T6
#
= 0.349 ft2
In the end the flow, Q = VA = 100 ft2 * 0.349 ft/min = 34.9 ft3/min or fpm.
Example: An Industrial hygienist measures the circumference of a duct to
be 25.12 inches, and the velometer registers a value of 100 feet per minute
(ft/min). What is the flow (Q)?
First, identify the radius of the duct based on the circumference.
25.12 = 2πÝ
So, r = 4 inch, (diameter = 4 inch = 0.333 ft)
Then, duct area = πr = 3.14 ∗ 0.333 ft = 0.349 ft Again, Q = VA = 100 ft2 * 0.349 ft/min = 34.9 ft3/min.
V1A1 = V2A2 (2)
A1,2… = cross-sectional area for ducts or into hood openings, in square
feet (ft2) or square meters (m2).
V1,2… = duct velocity, in feet per minute (ft/min) or meters per second
(m/sec).
9
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
This equation is also known as the Constant Volumetric Flow, Conservation
of Mass, or the Continuity Equation. The equation assumes no friction loses
and states that the gas mass flow rate at one point in a stream equals the
mass flow rate at another location of the duct. Translated: the flow increases
when transitioning from the large area duct into the small area duct or
velocity is inverse proportional to the duct area.
Example: The duct area of an 8-inch diameter duct with a 100 feet per
minute duct velocity is reduced to a 4-inch diameter duct. What is the new
velocity?
Initial diameter = 8 inch/12 inch = 0.666 ft
Reduced diameter = 4 inch/12 inch = 0.333 ft
d
0.667 ft
Initial area = π = 3.14 ∗
= 0.349 ft 4
4
d
0.333 ft
Reduced area = π = 3.14 ∗
= 0.087 ft 4
4
Notice that the initial area is four times bigger than the reduced area.
Then, V1A1=V2A2
100 ft/min *0.394 ft2 = V2*0.087 ft2
10
CIH EXAM Equations 2020
V =
Dr. Daniel Farcas, CIH, CSP, CHMM
µf
∗$.U#* T6
H
$.$¿ T6
$$
= 400 fpm (flow increases 4 times).
TP = VP + SP
(3)
TP = total pressure within a duct, “head” or energy in the ventilation
flow stream and is measured in inches of water column (commonly
written as “in. wc.”)
VP = velocity pressure within a duct (not duct velocity which is V),
inches of water column (also written as “in. wc” and not ft/min.)
SP = static pressure within a duct, this is mainly manifested as
pressure on the duct walls, inches of water column (also written as “in.
wc.”)
This equation is known as Conservation of Energy (again assuming friction
losses). It states that TP (total pressure), which represents total energy is the
sum of VP (velocity pressure), which is movement pressure and always
positive and SP (static pressure) which represents the pressure of the air
exerted in all directions. Static Pressure (SP) is pushing inside the duct walls
while Velocity Pressure (VP) is moving the air in the direction of the flow,
because of pressure differentials (from high pressure to low pressure).
11
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
Example: What is the total pressure (TP) in a duct pre fan if static pressure
(SP) is -2.25 in. wc. and velocity pressure (VP) is 1.5 in. wc. Remember that
the velocity pressure (VP) is only be exerted in the direction of airflow is
always positive!
TP = VP + SP = 1.5 in. wc. + (-2.25) in. wc. = -0.75 in. wc.
Pre fan, the TP is always negative; it means that the air flows into the duct
system.
Þßà + áßà = Þß× + áß× + ∑ âãääåäà×
(4)
VP1,2 = velocity pressure within the duct at the location, in. wc.
SP1,2 = static pressure within the duct at the location, in. wc.
losses = head (energy) loss as air moves downstream from point 1 to
point 2, in. wc.
Energy loss in the duct from point 1 to point 2 due to frictions are directly
proportional to the pressures inside the duct.
12
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
Example: One measurement at the entry point of a section of ductwork has
a total pressure of 3 in. wc., and 1.5 in. wc. at the exit point. What is the head
(energy) loss across the ductwork?
Þßà + áßà = Þß× + áß× + ∑ âãääåäà× and,
because TP = VP +SP equation (3) then,
TP1 = TP2 +∑ æçèèéèà×
Head (energy) loss across the ductwork = 3 in. wc. – 1.5 in. wc. = 1.5 in.wc.
Þßê = − ëê + àáßØ (5)
SPh = hood static pressure, in. wc.
VPd = velocity pressure within the duct, in. wc.
Fh = hood entry loss factor or friction loss coefficient, dimensionless.
This equation is used to calculate the hood static pressure required to
overcome as air or gas enters a hood.
13
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
Example: Calculate the hood static pressure if the duct velocity pressure
(VPd) is 0.6 in. wc. and the hood entry loss factor (Fh) is 0.45.
SPh = - ( (0.45 + 1)*0.6 in. wc. = -0.87 in. wc.
In this example, -0.87 in. wc. is necessary to accelerate air into the hood
opening to duct velocity and overcome any associated turbulence losses.
á = Ùììí áß
Øî
(6)
V = average velocity of airflow, (fpm)
4005 = a constant and conversion factor based on air flowing at
standard temperature and pressure (STP)
df = air correction density factor equals 1 at standard temperature and
pressure (STP), unitless
VP = average velocity pressure, in. wc.
Since the velocity of the air in the duct can not be measured directly, we must
measure the velocity pressure first with a pitot tube and calculate the velocity
14
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
of the airflow. Velocity in a duct is directly proportional to the square root of
velocity pressure divided by the density factor.
NOTE: Pitot tubes require a velocity pressure of 4,005 fpm to show a 1 in.
wc. reading.
Example: The velocity pressure (VP) in a duct is 2.25 in. wc. What is the
velocity (V) if the density factor is 1 (at 70 °F and 14.7 psi)?
V = 4005
.%
Øî =
= 4005 ∗ 1.5 = 6,007.5 fpm
ß
×÷Ù °ø
ß
íðì°ò
∗
=
∗
àÙ. ô õäö
ó
àìà, ð×í ßù
ó
The formula above is not on the list “USEFUL EQUATIONS FOR THE ABIH
EXAMINATION,” but it may be necessary to solve problems.
df = air correction density factor, equals 1 at standard temperature and
pressure (STP), unitless
°R = Rankine degrees = 0F + 460
°K = Kelvin degrees = 0C + 273.15
°F = Fahrenheit degrees = 9/5 (0C) + 32
The standard temperature and pressure (STP) are 101,325 Pascals (Pa) or
14.7 pounds per square inch (psi), and 530 °R or 294 °K.
Example: The velocity pressure (VP) in a duct is 2.25 in. wc. What is the
velocity (V) if the density factor must be corrected for 100 0C and 14.5 psi?
°F = 9/5*(100) + 32 = 212
°R = 212 + 460 = 672
15
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
df =
530 °R 14.5 psi
∗
= 0.778
672 °R 14.7 psi
ú = 4005 .%
$.¿¿
= 4005 ∗ 1.7 = 6,810.8 fpm
The flow is higher than in the previous example because hotter and less
compressed air has a larger volume, although the same mass of air is
transported in the same unit of time.
á = à. ×÷
áß
Øî
!
(7)
V = velocity of the air flow, (m/sec)
1.29 = a constant and conversion facor based on air flowing at
standard temperature and pressure (STP),
df = air correction density factor, equals 1 at STP, unitless
VP = velocity pressure, Pascals (Pa)
This formula is the metric version of the equation (6). The velocity of the
airflow is measured in meters per second and the velocity pressure in
Pascals.
Example: The velocity pressure in a duct is 559.89 Pa (the equivalent of
2.25 in. wc.). What is the velocity in fpm if the density factor is 1, and the
conversion factor m to ft is 3.28?
16
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
559.89
V = 1.29 û
= 1.29 ∗ 23.66 =
1
= 30.524 m/ sec = 1831.43 m/ min ≌ 6,007.5 fpm
áß = "
á
Ùììí
×
& Øî
(8)
V = velocity of airflow, (fpm)
4005 = a constant and conversion factor based on air flowing at
standard temperature and pressure (STP), fpm/in.wc
df = air correction density factor equals 1 at standard temperature and
pressure (STP)
VP = velocity pressure, in. wc.
This formula is derived from equation (6), the velocity pressure is calculated
based on the velocity and air correction density factor.
Example: The velocity in a duct is 6,007.5 fpm. What is the velocity pressure
if the density factor is 1?
6,007.5 VP = L
N ∗ 1 = 2.25 in. wc.
4005
)áß = "
á
à.×÷
×
& Øî+
(9)
V = velocity of air flow, (m/sec)
1.29 = a constant and conversion factor based on air flowing at
standard temperature and pressure (STP)
17
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
df = air correction density factor, equals 1 at standard temperature and
pressure (STP)
VP = velocity pressure, Pascals (Pa)
This formula is the metric version of the equation (8). The velocity pressure
is calculated based on the velocity and air correction density factor.
Example: The velocity in a duct is 30.524 m/sec. What is the velocity
pressure at standard temperature and pressure?
30.524 VP = L
N ∗ 1 ≌ 559.89 Pa
1.29
hood entry loss = Fh x VPd
(10)
he = hood entry loss, in. wc.
VPd = velocity pressure in the duct, in. wc.
Fh = hood entry loss factor, dimensionless
The hood entry loss, commonly known as (he) calculates the energy it takes
to accelerate the air entering the hood to the duct velocity. The bell mouth
hood is the most efficient (Fh = 0.04).
Example: What is the hood entry loss if the hood entry loss factor is 0.95
and the velocity pressure in the duct is 0.25 in. wc.?
he (hood entry loss) = 0.95*0.25 in. wc. = 0.2375 in. wc
18
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
þå = ⌊
áß
Þßê ⌋
(11)
Ce = hood entry coefficient, unitless
SPh = value of hood static pressure, in. wc. (absolute value)
VP = velocity pressure, in. wc.
The hood entry coefficient (Ce) is related to the hood static pressure (SPh),
velocity pressure (VP), and measures how efficiently the air moves into the
system through the hood. Bell shape has the best (Ce = 0.98) hood entry
coefficient.
Example: Determine the coefficient of entry for a hood within a duct with a
velocity pressure of 0.4 in. wc. and a hood static pressure of 0.6 in. wc.
C- = û
0.4 in. wc.
= √0.666 ≌ 0.816
⌊0.6⌋ in. wc.
áßÖ = " à & áßà + " × & áß× (12)
ð
ð
VPr = resulting velocity pressure of the merged flows, in. wc.
Q3 = volumetric flow rate of the merged branches flows, ft3/min, cfm
Q1 = volumetric flow rate of duct/branch 1, ft3/min, cfm
VP1 = velocity pressure in duct 1, in. wc.
Q2 = volumetric flow rate of duct/branch 2, ft3/min, cfm
VP2 = velocity pressure in duct/branch 1, in. wc.
19
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
This formula answers the question: “What is the resultant velocity pressure
in a duct pulling air from two duct branches?”
Example: What is the resulting velocity pressure in a duct resulting from two
merged flow ducts, first with a volumetric flow rate of the duct of 180 ft3/min
and 1.8 in. wc. and the second with a volumetric flow rate of the duct of 220
ft3/min and 2.2 in. wc.?
The volumetric flow rate of the merged branches flows 220 + 180 = 400 cfm.
VP2 L
180
220
N ∗ 1.8 in. wc. L
N ∗ 2.2 in. wc.
400
400
0.81 in. wc. 1.21 in. wc. 2.02 in. wc.
|Þßê |
Ùììí þå Øî
(13)
Q = volumetric flow rate of duct, ft3/min
4005 = a constant and conversion factor based on air flowing at
standard temperature and pressure (STP)
Ce = hood entry coefficient, dimensionless
SPh = value of hood static pressure, in. wc.
df = air correction density factor, equals 1 at STP
A = cross-sectional area of the duct, (ft2)
20
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
This equation is used to determine the volumetric flow rate of the air flowing
into the duct through the hood with just the hood static pressure if the hood
entry coefficient is known.
Example: Calculate the volumetric flow rate (Q) of a 6-inch diameter (A =
0.196 ft2) duct from a hood, if the hood static pressure measurement is 2.0
in. wc. also, the hood entry coefficient is 0.98 (bell-shaped) with a density
factor of 1.
Q = 4005 ∗ 0.98 ∗ û
|2.0|
∗ 0.196 ≌ 1,088 ft U /min
1
= à. ×÷ þå |Þßê |
Øî
!
(14)
Q = volumetric flow rate of duct, m3/sec
4005 = a constant and conversion factor based on air flowing at
standard temperature and pressure (STP)
Ce = hood entry coefficient, dimensionless
SPh = value of hood static pressure, Pa (absolute value)
df = air correction density factor, equals 1 at STP
A = Cross-Sectional Area of the duct, in square meters (m2)
This formula is the metric variation of the equation (14).
Example: Calculate the volumetric flow rate in fpm of a 6-inch diameter, 1
inch = 2.54 cm (A = 0.0182 m2) duct from a hood, if the hood static pressure
21
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
measurement is 497.68 Pa (equivalent 2.0 in. wc.) and the hood entry
coefficient is 0.98 (bell-shaped) with density factor 1, and the conversion
factor m3 to ft3 is 35.28.
Q = 1.29 ∗ 0.98 ∗ û
|497.68|
∗ 0.0182 =
1
= 0.513 mU / sec = 30.8 mU / min ≌ 1,088 ft U /min
= Ùììíþå :|Þßê |
(15)
Q = volumetric flow rate of duct, m2/sec
4005 = a constant based on air flowing at standard temperature and
pressure (STP)
Ce = hood entry coefficient, dimensionless
SPh = value of hood static pressure, Pa (absolute value)
A = cross-sectional area of the duct, (ft2)
Essentially, this is formula (13), when density factor (df) equals one at
standard temperature and pressure (STP).
Example: Calculate the volumetric flow rate (Q) of a 6-inch diameter duct
(A = 0.196 ft2) from a hood, if the hood static pressure measurement is 2.0
in. wc. and the hood entry coefficient is 0.98 (bell-shaped).
Q = 4005 ∗ 0.98 ∗ 0.196 ∗ :|2| ≌ 1,088 ft U /min
22
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
ãÖÖ = âãåÖ Þßã
ÞßâãåÖ
(16)
Qcorr = corrected (new) flow rate, ft3/min
Qlower = design (existing) flow rate, ft3/min
SPgov = governing static pressure, in. wc
SPlower = (new duct) design static pressure, in. wc
This equation is used to calculate the resulting flow of two ducts that join
based on their static pressures. This happens because the static pressure
must be equivalent at the duct’s junction. The duct with lower static pressure
has its flow corrected (increased) to achieve a balanced junction. The ratio
of governing static pressure to the new duct static pressure must be 1.2 or
lower.
Example: Calculate the corrected flow for a ventilation branch converging
the main duct with a static pressure of 1.4 in. wc. The joining ventilation
branch has a volumetric flow rate of 1800 ft3/min and a static pressure of 1.2
in. wc.
The ratio of governing static pressure to the new duct static pressure 1.4/1.2
= 1.17
Q ;<22 = 1800 ∗ .#
.
≌ 1944 ft3/min
23
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
F =
ö
(17)
The effective rate of ventilation is a dilution ventilation quality factor. Simply
stated: an effective rate of ventilation tells us how well the airflow cleans the
contaminates in a room. The safety factor is used to simulate real-life
conditions where the air mixing is not ideal (ideal is 1).
Q’ = effective rate of ventilation, ft3/min
Q = actual airflow (rate of ventilation), ft3/min
mi = mixing factor, nondimensional (usually 1 to 10)
Example: What is the effective rate of ventilation in a room if the rate of
ventilation is 20 ft3/min, and the mixing factor is 10?
QF =
20
= 2 ft U /min
10
× − à = −
Jâ L
áÖ
þ×
þà
N
(18)
t2 = final time, min
t1 = initial time, min
ln = natural logarithm, 2.7182818
Vr = volume of enclosure, ft3
Cg1 = initial concentration of gas or vapor, parts-per-million (ppm)
Cg2 = final concentration of gas or vapor, parts-per-million (ppm)
Q’ = Q/mi = effective rate of ventilation, ft3/min
24
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
This equation calculates the time needed for the dilution ventilation to dilute
the contaminant in a room to a lower concentration when there is no
contaminant generation.
Example: Benzene spills in a room that measures 40’ x 50’ x 12’ (24,000
ft3). An initial concentration is measured at 50 ppm, and the benzene use is
stopped (no more benzene vapors are released). With the effective rate of
ventilation 4,000 ft3/min of dilution air, how long would it take to reach the
PEL level of 1 ppm?
24000 ft U
1ppm
t − t = −
∗
ln
L
N=
ft U
50 ppm
4000
min
= −6 min ∗ ln0.02 = 23.47 min
ln
O J þ× S
O J þà S
=−
J × à áÖ
(19)
ln = natural logarithm, 2.7182818
Vr = volume of the room enclosure), ft3
Cg1 = initial concentration of gas or vapor, percentage
Cg2 = final concentration of gas or vapor, percentage
G = rate of generation of contaminant, ft3
Q’ = Q/mi = effective rate of ventilation, ft3/min
t2 = final time, min
t1 = initial time, min
25
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
Just like the equation (18), this formula is a dilution ventilation formula but
with continuous contaminant generation in a room.
Example: Benzene evaporates at a rate of 4 ft3/min in a room that measures
40’ x 50’ x 12’ (24,000 ft3). If an initial concentration measured is 50 ppm,
what will the concentration be after 45 minutes of 4,000 ft3/min of dilution air
at?
4 − 4,000 ∗ C
4,000
ln L
N=−
∗ 45 − 0
4 − 4,000 ∗ 0.000050
24,000
#,$$$
4 − 4,000 ∗ C
"#,$$$∗#%$&
=e
4 − 4,000 ∗ 0.000050
4 − 4,000 ∗ C
= 0.000553084
3.8
4 − 4,000 ∗ C = 0.002101721
4,000 ∗ C = 3.997898279
C2 = 0.000999475 ≌ 1000 ppm
=ۿ
ÙìðÞ òö àì þ (20)
Q = volumetric flow required to limit contaminant concentration, ft3/min
403 = constant and conversion factor
SG = specific gravity, nondimensional
ER = evaporation rate, pints/min
mi = mixing ventilation (dilution) safety factor, nondimensional
26
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
106 = unit conversion (ppm to volume percent)
MW = molecular weight
Cg = contaminant concentration in air, ppm
This equation calculates the dilution ventilation flow required to retain an
evaporating contaminant under a dangerous concentration in the air.
Example: Benzene evaporates from an open bottle that must remain open
on a laboratory table at a rate of 0.1 pints/min. How many ft3/min dilution air
is required to maintain the concentration below the PEL? Assume the PEL
is 1 ppm and a ventilation safety factor of 2. The molecular weight is 78.11
g/mol, and the specific gravity is 0.876.
403 ∗ 0.876 ∗ 0.1 ∗ 2 ∗ 10`
Q=
≌ 903,925 ft U /min
78.11 ∗ 1
In this situation, local exhaust ventilation (LEV) is recommended instead of
general dilution.
) =
+
×ÙÞ òö àì þ (21)
Q = volumetric flow required to limit contaminant concentration, m3/sec
24 = constant and conversion factor
SG = specific gravity, nondimensional
ER = evaporation rate, liters/second
27
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
mi = mixing ventilation (dilution) safety factor, nondimensional
106 = unit conversion (ppm to volume percent)
MW = molecular weight
Cg = contaminant concentration in air, ppm
This equation also calculates required dilution ventilation to control
evaporation as in formula (20) but in metric units.
Example: Benzene evaporates from an open bottle that must remain open
on a laboratory table at a rate of 0.00079 liters/second. How many ft3/min
dilution air is required to maintain the concentration below the PEL? Assume
the PEL is 1 ppm and a ventilation safety factor of 2. The molecular weight
is 78.11 g/mol, the specific gravity is 0.876, and the conversion factor m3 to
ft3 is 35.28.
24 ∗ 0.876 ∗ 0.00079 ∗ 2 ∗ 10`
Q=
= 425.27 mU /sec
78.11 ∗ 1
= 25,516.26 mU / min ≌ 901,099 ft U /min
êù
åä
=
ì
áÖ
(22)
Nchanges = number of air changes per hour
60 = conversion factor from minutes to hours
Q = room ventilation rate, ft3/min
Vr = room volume, ft3
28
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
This equation calculates the room air changes per hour and is commonly
used in indoor air quality (IAQ).
Example: A room measures 40 x 40 x 10 feet (16000 ft3). How many air
changes per hour are done using a ventilation rate of 4000 ft3/min?
N;\]^-_ =
`$∗#$$$
þ× =
`$$$
= 15 air changes per hour
J bc á h
Ö i
aàå
J
(23)
Cg2 = final concentration of gas or vapor, ppm
G = rate of generation of contaminant ft3/min
e = natural logarithm, 2.7182818
Δt = amount of time, min
Q’ = effective room ventilation rate, ft3/min
Vr = room volume, ft3
When airborne contaminant starts being release with a known constant rate
of generation in a ventilated room, the final concentration of gas or vapor is
time-dependent.
Example: A room measures 40 x 40 x 10 feet (16000 ft3) and has a
ventilation rate of 4000 ft3/min. Benzene evaporates from an open bottle left
on a laboratory table at a rate of 0.1 ft3/min. What is the final concentration
of benzene after 15 min?
29
CIH EXAM Equations 2020
C^ =
Dr. Daniel Farcas, CIH, CSP, CHMM
$. c.¿
=
#$$$
b"
sss∗4
&
4Xsss h
$. $.$U%
#$$$
þ× = þà å
=
$. O.¿b5. S
= 0.00002 ppm
J L
N
áÖ
#$$$
=
(24)
Cg2 = final concentration of gas or vapor, ppm
Cg1 = initial concentration of gas or vapor, ppm
e = natural logarithm, 2.7182818
Δt = amount of time, min
Q’ = effective room ventilation rate, ft3/min
Vr = room volume, ft3
In this equation the final concentration of a contaminant is based on the initial
concentration of the contaminant in a ventilated room and is also timedependent (no contaminant generation).
Example: A room measures 40 x 40 x 10 feet (16000 ft3) and has a
ventilation rate of 4000 ft3/min. Benzene evaporated from an open bottle left
on a laboratory table and reached 100 ppm concentration before closing the
bottle with a rubber stopper. What is the final concentration of benzene after
15 min and 30 min?
C^/%Gm] = 100 ∗
C^/U$Gm] = 100 ∗
#$$$∗%
" `$$$ &
2.7182818
#$$$∗U$
" `$$$ &
2.7182818
= 2.35 ppm
= 0.055 ppm
30
CIH EXAM Equations 2020
= ×ۿ
Dr. Daniel Farcas, CIH, CSP, CHMM
Ø× ð òß ×
à " & "
&
Øà
òß à
(25)
Q2 = volumetric flow rate for the secondary condition, ft3/min
Q1 = volumetric flow rate for the initial condition, ft3/min
d2 = fan diameter for the secondary condition, inches
d1 = fan diameter for the initial condition, inches
RPM2 = fan speed for the secondary condition, rpm
RPM1 = fan speed for the initial condition, rpm
Fan Law 1. Ventilation engineers use this equation to calculate how the
change in fan speed (rotations per minute) and diameter will affect the flow
based on known initial values.
Example: A local exhaust ventilation system with a flow of 300 ft3/min has
its fan turning at 4500 rpm replaced by a 5500 rpm. If the fan diameter
remains the same, what is the new flow?
Since the fan diameter does not change, that part of the equation equals 1.
ft U
5500 rpm
Q = 300
∗ 1U ∗ L
N ≌ 367 ft U /min
min
4500 rpm
31
CIH EXAM Equations 2020
ß× =
Dr. Daniel Farcas, CIH, CSP, CHMM
Ø× × òß × ×
ßà " & "
&
Øà
òß à
(26)
P2 = system pressure for the secondary condition, in. wc.
P1 = system pressure for the initial condition, in. wc.
d2 = fan diameter for the secondary condition, inches
d1 = fan diameter for the initial condition, inches
RPM2 = fan speed for the secondary condition, rpm
RPM1 = fan speed for the initial condition, rpm
Fan Law 2. Ventilation engineers use this equation to calculate how the
change in fan speed (rotations per minute) and diameter will affect the fan
pressure based on known initial values.
Example: A local exhaust ventilation system with a fan pressure 3 of in. wc.
has its fan turning at 4500 rpm replaced by a 5500 rpm. If the fan diameter
remains the same, what is the new flow?
Since the fan diameter does not change, that part of the equation equals 1.
5500
rpm
P = 3 in. wc.∗ 1 ∗ L
N ≌ 4.48 in. wc.
4500 rpm
32
CIH EXAM Equations 2020
= × ܀܅۾
Dr. Daniel Farcas, CIH, CSP, CHMM
Ø× í òß × ð
ßò à " & "
&
Øà
òß à
(27)
PWR2 = system power for the secondary condition, horsepower, hp or
brake horsepower, bhp
PWR1 = system power for initial condition, horsepower, hp or brake
horsepower, bhp
d2 = fan diameter for the secondary condition, inches
d1 = fan diameter for the initial condition, inches
RPM2 = fan speed for the secondary condition, rpm
RPM1 = fan speed for the initial condition, rpm
Fan Law 3. Ventilation engineers use this equation to calculate how the
change in fan speed (rotations per minute) and diameter will affect the fan
power based on known initial values.
Example: To improve the local exhaust ventilation of a 30-horsepower fan
turning at 5000 rpm the fan speed is increased by 20%, what is the new
horsepower of the fan?
6000 rpm U
PWR = 30 hp ∗ 1 ∗ L
N = 51.84 hp
5000 rpm
%
33
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
ëÞß = Þßã − Þßö − áßö
(28)
FSP = fan static pressure, in. wc.
SPout = static pressure measured on the outlet side of the fan, in. wc.
SPin = static pressure measured on the inlet side of the fan, in. wc.
VPin = velocity pressure on the inlet side of the fan, in. wc.
Fan Law 4. Fan static pressure (FSP) represents the resistance pressure
that the fan has to blow against in order to move the air downstream.
Example: What is the fan static pressure (FSP) when the static pressure at
the inlet (SPin) is -3.3 in. wc. the static pressure at the outlet (SPout) is 5 in.
wc. and the velocity pressure on the inlet side of the fan (VPin) is 1 in. wc.?
FSP = 5 in. wc.- (-3.3 in. wc.) -1 in. wc.= 7.3 in. wc.
The fan in this example is blowing against a high resistance pressure, and
this will require more horsepower and will deliver less air.
ëóß = óßã − óßö
(29)
FTP = fan total pressure, in. wc.
TPout = total pressure measured on the outlet side of the fan, in. wc.
TPin = total pressure measured on the inlet side of the fan, in. wc.
34
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
Fan Law 5. The fan total pressure (FTP) is the difference between the total
pressures in the fan outlet and inlet openings.
Example: What is the fan total pressure (FTP) when the total pressure at
the inlet (TPin) is - 4.5 in. wc. the total pressure at the outlet (TPout) is 4 in.
wc.?
FTP = 4- in. wc. (-4.5 in. wc.) = 8.5 in. wc.
END OF DEMO COPY
VISIT:
https://www.amazon.com/dp/1700473417/
35
CIH EXAM Equations 2020
Dr. Daniel Farcas, CIH, CSP, CHMM
36
