2020 ALL-003 Hydraulics of Pumped Systems (20-05991)
33
Sidney Innerebner, PhD, PE, CWP Indigo Water Group Disclaimers 2 Many of the images and videos that we use in our training materials were collected from the internet over the span of many years. We have made every effort to reference the original source. When possible, we’ve contacted webmasters for permission to use these materials. If you are the owner of an image or video and don’t want us to use it, please let us know and we’ll remove it from our training materials immediately. Indigo Water Group, LLC has made attempts to verify the information provided in this web-based training against respected reference materials including Manual of Practice 11 by Water Environment Federation and the Sacramento Manuals. Neither the Author nor the Publisher, Indigo Water Group, assumes any responsibility for errors, omissions, or contrary interpretation of the subject matter herein. A minimum passing score of 70% is now required for each quiz in our on-line training classes. You must pass each quiz to receive a training unit certificate. If you don’t pass a quiz, the class will return to the slide following the previous quiz. You’ll be able to move freely through the slides to view material you might have missed before retaking the quiz. Speaker Info. Marker Tools Outline & Notes Click on each of the tabs along the top bar to access learning resources. Resources include the course handout and useful links. Resources Call 303 03- 3 - 489 89- 9 -9226 for technical assistance with this course.
Transcript of 2020 ALL-003 Hydraulics of Pumped Systems (20-05991)
Sidn
ey In
nere
bner
, PhD
, PE,
CW
PIn
digo
Wat
er G
roup
Disc
laim
ers
2
Man
y of
the
imag
es a
nd v
ideo
s th
at w
e us
e in
our
trai
ning
m
ater
ials
wer
e co
llect
ed fr
om th
e in
tern
et o
ver t
he s
pan
of m
any
year
s. W
e ha
ve m
ade
ever
y ef
fort
to re
fere
nce
the
orig
inal
so
urce
. W
hen
poss
ible
, we’
ve c
onta
cted
web
mas
ters
for
perm
issi
on to
use
thes
e m
ater
ials
. If
you
are
the
owne
r of a
n im
age
or v
ideo
and
don
’t w
ant u
s to
use
it, p
leas
e le
t us
know
and
w
e’ll
rem
ove
it fr
om o
ur tr
aini
ng m
ater
ials
imm
edia
tely
.
Indi
go W
ater
Gro
up, L
LC h
as m
ade
atte
mpt
s to
veri
fy th
e in
form
atio
n pr
ovid
ed in
this
web
-bas
ed tr
aini
ng a
gain
st re
spec
ted
refe
renc
e m
ater
ials
incl
udin
g M
anua
l of P
ract
ice
11 by
Wat
er
Envi
ronm
ent F
eder
atio
n an
d th
e Sa
cram
ento
Man
uals
. N
eith
er th
e A
utho
r nor
the
Publ
ishe
r, In
digo
Wat
er G
roup
, ass
umes
any
re
spon
sibi
lity
for e
rror
s, o
mis
sion
s, o
r con
trar
y in
terp
reta
tion
of th
e su
bjec
t mat
ter h
erei
n.
A m
inim
um p
assi
ng sc
ore
of 7
0% is
now
requ
ired
for
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qui
z in
our
on-
line
trai
ning
cla
sses
.
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t pas
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iz to
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uiz,
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iew
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re re
taki
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e qu
iz.
Spea
ker
Info
.M
arke
rTo
ols
Out
line
&N
otes
Clic
k on
eac
h of
the
tabs
alo
ng th
e to
p ba
r to
acce
ss le
arni
ng re
sour
ces.
Reso
urce
s in
clud
e th
e co
urse
ha
ndou
t and
use
ful l
inks
.
Res
ourc
es
Cal
l 30303
-3-48
989-9-92
26 fo
r tec
hnic
al a
ssis
tanc
e w
ith th
is c
ours
e.
Agen
daH
ydra
ulic
Con
side
rati
ons o
f the
Pum
p Pi
ping
Sys
tem
yW
ork,
Hor
sepo
wer
, and
Ene
rgy
pC
ost t
o Ru
n a
Pum
ppPe
riph
eral
Vel
ocit
y, D
iam
eter
, and
Spe
edp
yPu
mp
Aff
init
y La
ws
pSy
stem
y
mmH
ead
Curv
esy Pu
mp
Perf
orm
ance
Cur
ves
pD
iscu
ssio
n
Pum
p Ha
ve T
wo
Ope
ratin
g Co
nditi
ons
Stat
ic a
nd D
ynam
icy
Stat
ic m
eans
the
pum
p is
turn
ed o
ffp
pD
ynam
ic m
eans
the
pum
p is
runn
ing
STAT
IC
HEA
D
Stat
ic Co
nditi
ons
The
elev
atio
n di
ffere
nce
betw
een
two
wat
er s
urfa
ces
–pu
mpi
ng fr
om a
nd to
.
Onl
y Ve
rtica
l Dist
ance
Mat
ters
Stat
ic h
ead
is th
e SA
ME
for b
oth
pum
ps.
Suct
ion
Head
–Po
sitiv
e He
ad a
t th
e Pu
mp
Inta
ke (i
mpe
ller e
ye)
Pum
p in
take
lin
e is
bel
ow
the
wat
er le
vel
Sour
ce:
ACR
P Tr
aine
r Kit
Suct
ion
Lift –
Nega
tive
Head
at
the
Pum
p In
take
(im
pelle
r eye
)
Pum
p in
take
lin
e is
abo
ve
the
wat
er le
vel
Sour
ce:
ACR
P Tr
aine
r Kit
Theo
retic
ally,
wha
t is
the
max
imum
suc
tion
lift f
or a
pum
p op
erat
ing
at s
ea le
vel?
Why
?
Max
imum
Suc
tion
Lift
Atm
osph
eric
pre
ssur
e at
sea
leve
l is
14.7
psi
ap
pC
onve
rt to
feet
of h
ead
14.7
psi
0.43
3 ps
i1
foot
H2O
= 34
feet
Theo
reti
cal m
axim
um su
ctio
n lif
t is 3
4 fe
et fo
r WAT
ER
Max
imum
Suc
tion
Lift
Atm
osph
eric
pre
ssur
e at
sea
leve
l is
14.7
psi
ap
pC
onve
rt to
feet
of h
ead
Theo
reti
cal m
axim
um su
ctio
n lif
t is 3
4 fe
et fo
r WAT
ER
Will
it in
crea
se o
r dec
reas
e w
hen
pum
ping
som
ethi
ng
Will
it in
crea
se o
r dec
reas
e w
hen
pum
W light
er th
an w
ater
…..
Say
gaso
line?
Quiz
Click
the Q
uizbu
tton t
o ed
it th
is ob
ject
Dyna
mic
Cond
ition
sTh
e pu
mp
is O
Np
pTh
ings
get
mor
e co
mpl
icat
edg
gp
Stat
ic h
ead
is g
ravi
ty w
orki
ng a
gain
st th
e pu
mp
Mor
e w
ork
to b
e do
ne!
Tota
l Dyn
amic
Hea
d =
Stat
ic +
Fri
ctio
n +
Velo
city
Head
Loss
Def
ined
Loss
of e
nerg
y du
e to
fric
tion
of p
ipe
Loss
of e
nerg
y du
e to
fric
tion
of p
ipe
L mat
eria
ls, v
alve
s, fi
ttin
gs, e
ntra
nce
mat
eria
ls, v
alve
s, fi
ttin
gs, e
and
exit
from
con
tain
ers
Loss
es h
appe
n w
hen
wat
er c
hang
es
Loss
es h
apL di
rect
ion
Lost
as asas
heat
Expr
esse
d as
feet
of h
ead
pN
ot re
ally
feet
of h
ead,
it’s
the
Not
real
ly fe
N equi
vale
ntee
t of h
ead,
iy
fentnt
ener
gy lo
ss
Phot
o So
urce
: By
Joh
ann
Jarit
z(O
wn
wor
k) [G
FDL
(http
://w
ww
.gnu
.org
/cop
ylef
t/fdl
.htm
l), C
C-B
Y-SA
-3.0
(http
://cr
eativ
ecom
mon
s.or
g/lic
ense
s/by
-sa/
3.0/
), C
C
BY-S
A 2.
5-2.
0-1.
0 (h
ttp://
crea
tivec
omm
ons.
org/
licen
ses/
by-s
a/2.
5-2.
0-1.
0), G
FDL
(http
://w
ww
.gnu
.org
/cop
ylef
t/fdl
.htm
l) or
CC
BY-
SA 3
.0
(http
://cr
eativ
ecom
mon
s.or
g/lic
ense
s/by
-sa/
3.0)
], vi
a W
ikim
edia
Com
mon
s
Tota
l Dyn
amic
Head
Stat
ic H
ead d
––Th
e di
ffere
nce
in v
ertic
al h
eigh
t in
wat
er
Stat
ic H
ead
Sd
The
diffe
renc
e in
ver
tica
lTh
e di
ffere
nce
in v
erti
cT
leve
ls, p
umpi
ng fr
om to
pum
ping
to.
pp
gVe
loci
ty H
ead
gdd
–p
pg
g –Th
e am
ount
of e
nerg
y re
quir
ed to
Ve
loci
ty H
ead
Vd
The
amou
nt o
f ene
rT
mov
e th
e w
ater
at a
giv
en v
eloc
ity.
Hea
d Lo
ssM
ajor
Los
ses
jM
inor
Los
ses
Ever
ythi
ng th
e pu
mp
mus
t wor
k to
ove
rcom
e.
18
STAT
IC
HEA
D
Stat
ic Co
nditi
ons
The
elev
atio
n di
ffere
nce
betw
een
two
wat
er s
urfa
ces
–pu
mpi
ng fr
om a
nd to
.
Velo
city
Head
V =
the
velo
city
of t
he b
ody
in ft
/sec
yy
G =
the
acce
lera
tion
due
to g
ravi
ty @
32.
2 ft
/secc
2g
H =
the
dist
ance
the
body
falls
Dir
ectl
yVe
rsus
Inve
rsel
yR
elat
ed
or
Maj
or v
ersu
s Min
or Lo
sses
Maj
or L
osse
s
Fric
tion
loss
es in
pip
ep
pC
alcu
late
d w
ith
Dar
cycyycy-
Cal
cula
ted
wC W
eisb
ach
wit
h D
arc
edw
hhor
Haz
eny
Dar
ccyenen
-W
eisb
ach
Will
iam
sor
Haz
eo
enchh
om
sm
ssequ
atio
ns
Rec
orde
d by
Indi
go W
ater
Gro
up a
t LAC
SD
Maj
or v
ersu
s Min
or Lo
sses
Min
or L
osse
sEn
tran
ce to
ves
sel
Exit
from
ves
sel
Valv
esFi
ttin
gs
Sour
ce: I
ndig
o W
ater
Gro
up
Befo
re w
e go
any
furt
her…
.
Engi
neer
s ver
sus
Ope
rato
rs
Maj
or Lo
sses
by
Darc
y-W
eisb
ach
Equa
tion
L =
Leng
th o
f Pip
e in
Fee
tg
pD
= In
side
Dia
met
er o
f Pip
e in
Fee
tp
v =
Velo
city
in ft
/sec
ond
yg
= A
ccel
erat
ion
of G
ravi
ty (3
2.2
ft/s
ecc2 c2 )
gg F =
Fric
tion
Fac
tor
Dir
ectl
yVe
rsus
Inve
rsel
yR
elat
ed
Maj
or Lo
sses
by
Haze
n-W
illia
ms E
quat
ion
Whe
re e HH
fHHHH
f=
Hea
dlos
s in
Feet
et
per
100
feet
of p
ipe
f
Q =
Flo
w in
gpm
Qgp
Q D =
Dia
met
er o
f Pip
e in
Inch
esp
C =
Fri
ctio
n Fa
ctor
(com
mon
ly 6
0 to
160)
Haze
n W
illia
ms C
Fac
tors
Pip
e M
ater
ial
Low
Hig
hA
sbes
tos-
cem
ent
140
140
Cas
t iro
n10
014
0C
emen
t-M
orta
r Lin
edD
ucti
le Ir
on P
ipe
140
140
Con
cret
e10
014
0C
oppe
r13
014
0St
eel
90110
Gal
vani
zed
iron
120
120
Poly
ethy
lene
(PE)
140
140
Poly
viny
l chl
orid
e (P
VC)
130
130
Fibe
r-re
info
rced
pla
stic
(FRP
)15
015
0So
urce
: htt
p://
en.w
ikip
edia
.org
/wik
i/H
azen
%E2
%80
%93
Will
iam
s_eq
uati
on
Let’s
Pra
ctice
!H
azenen
-n-W
illia
ms
is ty
pica
lly u
sed
for d
esig
nyp
Find
Hea
dlos
s giv
eng
Leng
th o
f 150
0 fe
etg
5D
iam
eter
of 1
5 in
ches
600
gpm
gpC
fact
or o
f 100
Not
e: H
fis
per 1
00 fe
et o
f pip
e.
Calcu
latin
g M
inor
Loss
esM
etho
d 1:
Cal
cula
te th
e K
fact
or fo
r the
fitt
ing
Tabl
es in
hyd
raul
ic te
xts
yM
anuf
actu
rer p
rovi
ded
data
Met
hod
2: E
quiv
alen
t len
gth
tabl
esq
gA
dd e
quiv
alen
t len
gths
to e
ithe
r Dar
cycy-yyyy-W
eisb
ach
or th
e A
dd e
quA H
azeneq
uenen
-iv
alen
t len
gths
to e
ituiqu nn-W
illia
ms e
quat
ions
Dir
ectl
yVe
rsus
Inve
rsel
yR
elat
ed
Sour
ce:
http
://w
ww
.wes
tern
dyna
mic
s.co
m/D
ownl
oad/
fric
loss
fittin
gs.p
df
Tota
l Dyn
amic
Head
Stat
ic H
ead d
––Th
e di
ffere
nce
in v
ertic
al h
eigh
t in
wat
er
Stat
ic H
ead
Sd
The
diffe
renc
e in
ver
tica
lTh
e di
ffere
nce
in v
ertic
Tle
vels
, pum
ping
from
to p
umpi
ng to
.p
pH
ead
Loss
Maj
or L
osse
sj
Min
or L
osse
sVe
loci
ty H
ead dd
––Th
e am
ount
of e
nerg
y re
quir
ed to
Ve
loci
ty H
ead
Vd
The
amou
nt o
f ene
rT
mov
e th
e w
ater
at a
giv
en v
eloc
ity.
Ever
ythi
ng th
e pu
mp
mus
t wor
k to
ove
rcom
e.
31
Putti
ng it
all
toge
ther
….
10 fe
et o
f 6” p
ipe
Static head = 80 ft
Find
vel
ocity
Find
vel
ocity
hea
dFi
nd m
ajor
loss
esFi
nd m
inor
loss
esA
dd ‘e
m u
p
Equa
tions
Nee
ded
If w
e ha
d K
tabl
es, w
e w
ould
als
o lo
ok u
p lo
sses
for e
lbow
s,
entr
ance
, and
exi
t lo
sses
.
Find
Vel
ocity
Firs
t, co
nver
t flo
w in
gpm
to fp
s
200
gallo
nsm
inut
e1
min
ute
60 s
econ
ds7.
48 g
allo
ns1
cf=
0.44
6 cf
s
Find
the
Area
of t
he P
ipe
Area
= (0
.785
)(dia
met
er)2
Area
= (0
.785
)(0.3
33 ft
)(0.3
33 ft
)Ar
ea =
0.0
87 ft
2
Con
vert
the
diam
eter
of t
he p
ipe
from
Inch
es to
feet
.
4 in
ches
12 in
ches
1 fo
ot=
0.33
3 ft
Find
Vel
ocity
Find
Vel
ocity
Hea
dv2 2g
=h f
(5.1
3 ft/
sec)
2
(2)(3
2.2
ft/se
c2)=
h f
26.3
2 ft2
/sec
2
64.4
ft/s
ec2
=h f
0.4
ft=
h f
Step
s for
Fin
ding
Vel
ocity
Hea
dCo
nver
t flo
w ra
te to
o cf
sC
onve
rt p
ipe
diam
eter
to fe
etp
pFi
nd c
rossp
pssss
-p
p ss-se
ctio
nal a
rea
of p
ipe
in sq
uare
feet
Cal
cula
te v
eloc
ityy
Find
vel
ocit
y he
ad
Find
Maj
or a
nd M
inor
Loss
esFo
r thi
s exa
mpl
e, w
e’ll
just
find
the
maj
or lo
ss in
the
p,
For t
his e
xam
ple,
we
llF pu
mp
disc
harg
e lin
ep
pU
se H
azenenen
-g
nn-W
illia
ms
ms
mse
quat
ion
Maj
or Lo
ss fo
r 399
ftof
Pip
eW
e fo
und
that
at
Hf
Hff
or 1,
000
feet
of p
ipe
was
44 44
ftp
pM
ust r
atio
this
to a
ctua
l len
gth
of p
ipe
44ft
1,00
0 ft
=x
399
ft
0.04
4 =
x
399
ft17
.6 ft
= x
Putti
ng it
All
Toge
ther Va
lue
Tota
l Sta
tic
Hea
d80
feet
Maj
or L
osse
s17
.6 fe
et
Min
or L
osse
sN
ot c
alcu
late
d
Velo
city
Hea
d0.
4 fe
et
Tota
l Dyn
amic
Hea
d98
.4 fe
et
42
Total Dynamic Head
Dis
char
ge F
low
Rat
e, g
pm
Pum
p Cu
rves
and
Thr
ottli
ng
32-i
nch
30-i
nch
28-i
nch
n =
3500
rpm
Quiz
Click
the Q
uizbu
tton t
o ed
it th
is ob
ject
Wor
k an
d En
ergy
Both
exp
ress
ed a
s foo
tot-t-po
unds
(ftft-
ft-l
b)p
ppIf
a 2
00 lb
man
clim
bs a
flig
ht o
f sta
irs u
p 40
feet
, the
n If
a 2
00 lb
man
clim
bsI he
has
don
e 8,
000
ftmbs ftft-
a fl
ight
of
s am
bs ftft-l
b of
wor
k
Sour
ce: h
ttps:
//ww
w.fl
ickr
.com
/pho
tos/
perry
goh/
6800
0547
18/
Mar
ked
for r
euse
.
Pow
er is
the
Tim
e Ra
te
of D
oing
the
Wor
kIn
a le
isur
ely y
5 m
inut
eecl
imb,
the
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ld h
ave
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ely
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min
ut55
ew
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the
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of:
And
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cond
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the
stai
rs:
Pum
ps d
o W
ork
by Li
fting
Wat
erIf
a p
ump
lifts
330
lb o
f wat
er u
p a
vert
ical
dis
tanc
e of
If
a p
ump
lifts
330
lb o
f wat
er u
p a
vert
ical
I 10 fe
et, t
hen
the
pum
p ha
s don
e 3,
300
ftical
ftft-
dist
ance
ol
cal
ftft-l
b of
wor
kp
p3
3If
thos
e 33
00 p
ound
s wer
e lif
ted
in 6
sec
onds
, the
n th
e If
thos
e 33
00 p
ound
s wer
e lif
ted
in 6
se
I POW
ER g
ener
ated
wou
ld h
ave
been
:
Hors
epow
er H
istor
yTe
rm “H
orse
pow
er” c
oine
d by
Jam
es W
att (
173636
-6-18
19)
py
J733
Mar
keti
ng h
is n
ew st
eam
eng
ine
to b
rew
erie
sg
gBr
ewer
ies u
sed
hors
es to
gri
nd g
rain
for m
ash
1 hor
se242424
———fo
ot d
iam
eter
cir
cle
4 Pulls
wit
h a
forc
e of
abo
ut 18
0 po
unds
pAv
erag
e sp
eed
of 18
0.96
ft/m
in
(181
ft/m
in)(
180
lbs)
= 3
2,58
0 ftft
-ft
-lbs
/min
Sour
ce: h
ttps:
//ww
w.fl
ickr
.com
/pho
tos/
ivou
tin/1
4038
1684
5/in
/pho
tost
ream
/
Calcu
latin
g Ho
rsep
ower
One
hor
sepo
wer
is 3
3,00
0 ftftft
-ftft
-lb
of w
ork/
min
p33
Wat
er H
orse
pow
erp
Brak
e H
orse
pow
er
er
er––
incl
udes
pum
p ef
ficie
ncy
pM
otor
or W
ire
Hor
sepo
wer
p
er
er–
pp
yp
–in
clud
es p
ump
and
Mot
or o
r Wir
e H
oM m
otor
eff
icie
ncy
All
Cer
tifie
d W
ater
Pro
fess
iona
ls n
eed
to k
now
how
to
All
Cer
tifie
d W
ater
Pro
fess
ion
A calc
ulat
e ea
ch o
ne o
f the
se.
49
HP
= (g
pm)(
TDH
, ft)
(8.3
4 lb
/gal
)(3
3,00
0 ft
-lb/
min
)
HP
= (g
pm)(
TDH
, ft)
(1 lb
/gal
)(3
,960
ft-l
b/m
in)
Sim
plify
ing
the
equa
tion
give
s us
som
ethi
ng fa
mili
ar
50
Are
n’t u
nits
am
azin
g?
gal
min
ftlb ga
lm
inft-
lb
HPw
ater
= (g
pm)(
TDH
, ft)
(3,9
60)
HPb
rake
= (g
pm)(
TDH
, ft)
(3,9
60)(
Ep)
HPm
otor
= (g
pm)(
TDH
, ft)
(3,9
60)(
Ep)(
Em)
51
Spec
ific G
ravi
ty a
nd P
umpi
ngW
ater
Tem
p, o C
Spec
ific
Gra
vity
BH
P4
1.010
0.0
600.
983
98.3
100
0.95
895
.812
50.
939
93.9
150
0.91
791
.7
As s
peci
fic g
ravi
ty d
ecre
ases
, so
does
the
BHP.
(Wat
er H
P at
4o C
)(spe
cific
gra
vity
) = B
rake
HP
Sam
ple
Que
stio
nsA
pum
p is
use
d to
fill
a w
ater
tank
250
feet
abo
ve th
e 0
fe0
feet
abo
ve th
e fe
et a
boveve
A
pum
p is
use
d to
fill
a w
ater
tA
um
is u
sed
to fi
ll a
wat
er ta
nk 2
50A pu
mp
stat
ion.
The
wat
er e
nter
s the
tank
at 1
.5 c
fs.
The
pum
p st
atio
n. T
he w
ater
ent
ers t
he ta
nk a
t 1.5
cfs
. Th
e ta
nk a
t 1.5
cf
mot
or is
85%
eff
icie
nt a
nd th
e pu
mp
is 9
0% e
ffic
ient
. m
otor
is 8
5% e
ffic
ient
and
thFi
nd th
e br
ake
hors
epow
er.
HPb
rake
= (g
pm)(
TDH
, ft)
(3,9
60)(
Ep)
47 H
Pbra
ke=
(673
.2 g
pm)(
250
ft)
(3,9
60)(
0.90
)53
1.5
cfse
c60
sec
1 m
in1
cf7.
48 g
al=
673.
2 gp
m
The
plan
t inf
luen
t pum
ps d
isch
arge
thro
ugh
an 18188
-Th
e pl
ant i
nflu
ent p
umps
dis
char
grge
thro
ugh
an 1
TT18
inch
line
. Th
e fl
ow v
eloc
ity
is 5
fps.
Fin
d th
e w
ater
in
ch li
ne.
The
flow
vel
ocit
y is
5 f
hors
epow
er if
the
lift i
s 30
feet
.
wH
P=
(gpm
)(TD
H,f
t)39
60
The
plan
t inf
luen
t pum
ps d
isch
arge
thro
ugh
an 1818
-Th
e pl
ant i
nflu
ent p
umps
dis
char
grge
thro
ugh
an 1
TT18
inch
line
. Th
e fl
ow v
eloc
ity
is 5
fps.
Fin
d th
e w
ater
in
ch li
ne.
The
flow
vel
ocit
y is
5 f
hors
epow
er if
the
lift i
s 30
feet
.
Area
= 0
.785
d2Ar
ea =
(0.7
85)(1
.5 ft
)(1.5
ft)
Area
= 1
.8 ft
2
Velo
city
=Fl
owAr
ea
5 fp
s =
Flow
1.8
ft2
9cf
s=
Flow
9cf s
1cf
7.48
gal
60 s
ec1
min
= 40
39 g
pm
The
plan
t inf
luen
t pum
ps d
isch
arge
thro
ugh
an 18188
-Th
e pl
ant i
nflu
ent p
umps
dis
char
grge
thro
ugh
an 1
TT18
inch
line
. Th
e fl
ow v
eloc
ity
is 5
fps.
Fin
d th
e w
ater
in
ch li
ne.
The
flow
vel
ocit
y is
5 f
hors
epow
er if
the
lift i
s 30
feet
.
wH
P=
(gpm
)(TD
H,f
t)39
60
wH
P=
(403
9gp
m)(3
0ft)
3960
wH
P=
30.6
Find
the
wat
er to
wir
e ho
rsep
ower
giv
en th
e fo
llow
ing
Find
the
wat
erF in
form
atio
n:In
flue
nt f
low
= 0
.8 8 cf
sSt
atic
hea
d =
20 fe
etM
ajor
loss
es =
1.2
feet
jM
inor
loss
es =
0.5
feet
5M
otor
eff
icie
ncy
= 95
%y
95Pu
mp
effic
ienc
y =
80%
Conv
ert c
fsto
gpm
0.8
cfs
1cf
7.48
gal
60 s
ec1
min
= 35
9.04
gpm
HPm
otor
= (g
pm)(
TDH
, ft)
(3,9
60)(
Ep)(
E m)
Plug
in W
hat Y
ou K
now
HPm
otor
= (g
pm)(
TDH
, ft)
(3,9
60)(
Ep)(
E m)
HPm
otor
= (3
59 g
pm)(
20 +
1.2
+ 0.
5, ft
)(3
,960
)(0.
80)(
0.95
)H
Pmot
or=
(359
gpm
)(21
.7, f
t)(3
,960
)(0.
80)(
0.95
)
HPm
otor
= 2.
6
Pow
er a
nd E
nerg
yTh
e ut
ility
com
pany
send
s you
a
yp
yTh
e ut
ility
com
pany
send
s T “P
ower
” bill
eac
h m
onth
.Bu
t…..
They
are
cha
rgin
g fo
r Bu
t…..
They
are
cB ki
low
att h
ours
.Th
at’s
not h
orse
pow
er o
r or ftft
-ft
-lbsbsb
/min
Wha
t giv
es?
Take
n by
Indi
go W
ater
Gro
up
Ener
gy
But…
.. I j
ust c
alcu
late
d Po
wer
in
But…
.. I j
ust c
aB H
orse
pow
erp
1 HP
= 0.
746
kw74
1 kw
= 1,
000
wat
ts
Sour
ce: B
y M
ike1
024
at e
n.w
ikip
edia
[Pub
lic d
omai
n], f
rom
Wik
imed
ia C
omm
ons
Pow
er C
osts
It is
is
CRI
TIC
ALL
for o
pera
tors
to u
nder
stan
d th
e en
ergy
It
iI
is
RITI
CA
LC
RL
por
ope
rfo
bill
for t
heir
faci
litie
sEn
ergy
can
be
50%
or m
ore
of o
pera
ting
cos
tsgy
You
ygyuu
MU
ST5
pg
TTbe
abl
e to
cal
cula
te th
e co
st to
run
a pu
mp,
Yo
uY
u M
UST
MT
be a
ble
to c
alcu
late
the
cos
bbl
ower
, or o
ther
pie
ce o
f equ
ipm
ent
pq
Back
to u
nit c
onve
rsio
ns!
This
stu
ff d
oesn
’t gr
ow o
n tr
ees!
Goin
g fro
m H
P to
$$$
A 3
5 H
P pu
mp
runs
8 h
ours
a d
ay, 7
day
s a w
eek.
How
A
35
HP
pum
p ru
ns 8
hou
rs a
day
, 7 d
ays a
wee
k. H
ow
A 3
5 H
P pu
mp
runs
8 h
ours
a d
ay, 7
day
s am
uch
will
is c
ost t
o ru
n th
e pu
mp
for a
wee
k if
elec
tric
ity
muc
h w
ill is
cos
t to
rco
sts $
0.09
a k
wh?
35 H
P1
HP
0.74
6 kw
= $1
31.5
9/w
k
= 26
.11
kw
(26.
11 k
w)(5
6 hr
s)=
1462
.16
kwh
1462
.16
kwh
1kh
w$0
.09
Prac
tice
Mak
es P
erfe
ctA
30
HP
pum
p is
use
d to
dew
ater
an
aera
tion
bas
in.
A 3
0 H
P pu
mp
is u
sed
to d
ewat
er a
n ae
rati
on b
asin
. A
30
HP
pum
p is
use
d to
dew
ater
an
aera
tA Th
e pu
mp
runs
24
hour
s a d
ay fo
r thr
ee d
ays.
Ene
rgy
The
pum
p ru
ns 2
4 ho
urs a
day
for t
hree
day
s. E
nerg
y Th
epu
mp
runs
24ho
ursa
day
fort
hree
days
.En
ergy
is $
0.07
/ k
wh.
How
muc
h di
d it
cost
to ru
n th
e pu
mp?
A s
ubm
ersi
ble
pum
p is
use
d to
tran
sfer
wat
er fr
om o
ne
A s
ubm
ersi
ble
pum
p is
use
d to
tran
sfer
wat
er fr
om
A dige
ster
to a
noth
er.
The
tota
l dyn
amic
hea
d is
15
one
om
5fe
etdi
gest
er to
ano
ther
. Th
e to
tal d
ynam
ic h
ead
is 1
dige
ster
toan
othe
r.Th
eto
tald
ynam
iche
adis
15 5
eet
eet
fefean
d th
e pu
mp
rate
is 2
5 gp
m.
The
pum
p an
d m
otor
an
d th
e pu
mp
rate
is 2
5 gp
m.
The
pum
p an
d m
otor
an
dth
epu
mp
rate
is25
gpm
.Th
epu
mp
and
moot
orot
orha
ve a
com
bine
d ef
ficie
ncy
of 7
6%.
If th
e pu
mp
runs
ha
ve a
com
bine
d ef
ficie
ncy
of 7
6%.
If th
e pu
mp
runs
ha
vea
com
bine
def
ficie
ncy
of76
%.
Ifth
epu
mp
runs
16 h
ours
a w
eek
and
elec
tric
ity
cost
s $0.
095
/ kw
h, h
ow
16 h
ours
a w
eek
and
elec
tric
ity
cost
s $0.
095
/ kw
h16
hour
saw
eek
and
elec
tric
ity
cost
s$0.
095
/kw
hm
uch
does
it c
ost t
o ru
n th
e pu
mp
ever
y w
eek?
Peak
and
Dem
and
Char
ges
Ener
gy c
osts
var
y du
ring
a 2
4424-44-ho
ur p
erio
dgy
Peak
Part
ial P
eak
Off
Pea
kD
eman
d ch
arge
s are
the
high
est e
nerg
y co
nsum
ptio
n D
eman
d c
D over
a 15nd
c1515
-ha
rges
are
the
hch
d c
55-m
inut
e pe
riod
How
can
an
oper
ator
avo
id th
ese
char
ges?
Pow
er, W
atts
, and
Am
pera
geD
eter
min
ing
pow
er c
onsu
mpt
ion
in k
w is
diff
icul
t to
gp
Det
erm
inin
g po
wer
D asse
ss in
the
field
Find
ing
ampe
rage
is e
asy!
Wat
ts L
aw fo
r Sin
gle
Phas
e Po
wer
Wat
ts L
aw fo
r Thr
ee P
hase
Pow
er:
Exam
ple
A m
otor
runn
ing
on 4
60 V
, thr
eeee-e-ph
ase
pow
er, i
s A
mot
or ru
nnin
g on
460
V, t
hre
A m
otor
runn
inA
eeeha
se p
ower
, is
phpdr
awin
g 9.
37 a
mps
. Th
e po
wer
fact
or w
as m
easu
red
at
draw
ing
9.37
am
ps.
The
pow
er fa
ctor
80%
. Fi
nd th
e po
wer
dra
w in
wat
ts
Pow
er =
(V)(A
)(1.7
3)(P
F)Po
wer
= (4
60 v
)(9.3
7 am
p)(1
.73)
(0.8
0)Po
wer
= 5
,968
wat
ts
5,96
8 w
atts
1 H
P0.
746
kw10
00 w
atts
1 kw
= 8
HP
Quiz
Click
the Q
uizbu
tton t
o ed
it th
is ob
ject
Agen
da –
Part
2Pe
riph
eral
Vel
ocit
y, D
iam
eter
, and
Spe
edp
yPu
mp
Aff
init
y La
ws
pSy
stem
y
mmH
ead
Curv
esy Pu
mp
Perf
orm
ance
Cur
ves
Let’s
Talk
Abo
ut G
ravi
tyC
entr
ifuga
l pum
p m
ust o
verc
ome
Cen
trifu
gal p
ump
mus
t ove
rco
C grav
ity
(sta
tic
head
) to
wor
kg
yIf
I dr
op a
rock
from
the
Chr
ysle
r If
I dr
op a
rock
from
the
Chr
ysle
r I bu
ildin
g, g
ravi
ty w
ill p
ull i
t dow
nIt
s g
gy
tstsV
ELO
CIT
Yy
py
YYw
ill in
crea
se a
t a
ItIts V
ELO
Vra
te o
f O
CIT
ELO
ofof32
.2
CIT 22ft
Yw
TYY
IT ftft/s
ecl i
ncre
ase
at a
w
illw ccc
for e
very
sec
ond
rate
oof
2.2
322
ftth
at it
falls
Igno
re w
ind
resi
stan
ceg Cal
cula
te th
e ve
loci
ty w
hen
the
Cal
cula
te th
e ve
loci
ty w
C rock
hit
s the
gro
und
Sour
ce: A
lex
Proi
mos
from
Syd
ney,
Aus
tral
ia
Velo
city,
Heig
ht, a
nd G
ravi
ty
V =
the
velo
city
of t
he b
ody
in n ftft
/sec
yy
G =
the
acce
lera
tion
due
to g
ravi
ty @
32.
2 2 ftft
/secc
2g
H =
the
dist
ance
the
body
falls
Hey
… is
n’t t
his
the
sam
e eq
uatio
n as
vel
ocity
hea
d?
That
falli
ng st
one
Ass
ume
that
our
ston
e is
dro
pped
from
a b
uild
ing
100
Ass
ume
thA fe
et h
ighg
Als
o as
sum
e no
win
d re
sist
ance
or d
rag
Take
the
squa
re ro
ot o
f bo
th s
ides
Who
Car
es A
bout
Falli
ng R
ocks
?Sa
me
equa
tion
cal
cula
tes
qSa
me
equa
tion
cal
cula
tes
S init
ial v
eloc
ity
need
ed to
in
itial
vel
ocity
ne
thro
w th
e ro
ck ee
dey
ne k kU
Ped
to
ede
PPth
e sa
me
thro
w th
e di
stan
ce
Now
…. L
et’s
thro
w w
ater
N
ow…
. Let
s thr
owN in
stea
d of
rock
s
Use
a c
entr
ifuga
l pum
p!Ta
ken
by In
digo
Wat
er G
roup
at B
uffa
lo, W
Y
Cent
rifug
al F
orce
Rem
embe
r erer––
cent
rifu
gal
Rem
embe
Rer
en
trifu
gal
cpu
mps
wor
k by
“thr
owin
g”
pum
ps w
ork
by t
hrow
ing
w
ater
to th
e ou
tsid
e of
the
wat
er to
volu
teW
ater
ent
ers a
t the
eye
of
Wat
er e
nter
s aW th
e im
pelle
rp
Just
like
swin
ging
a b
ucke
t Ju
st li
ke sw
ing
J full
of w
ater
Cent
rifug
al F
orce
, Pre
ssur
e, a
ndPe
riphe
ral V
eloc
ityN
ow…
. Dri
ll a
hole
in th
e N
ow…
. Dri
ll a
hole
in th
N bott
om o
f the
buc
ket
Wha
t hap
pens
?
Wha
t hap
pens
as t
he
Wha
t hap
pens
as t
hW bu
cket
spee
ds u
p?
Sour
ce: h
ttps:
//ww
w.yo
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atch
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t1Pg
V0
Perip
hera
l Vel
ocity
Als
o ca
lled
Rim
Vel
ocit
yyD
ista
nce
trav
eled
by
a po
int o
n th
e im
pelle
r rim
ove
r D
ista
nce
trav
eled
D on
e fu
ll ro
tati
onD
ista
nce
trav
eled
is e
qual
to th
e e C
IRC
UM
FER
ENC
ED
ista
nce
trav
eled
D of th
e im
pelle
r
A Ta
le o
f Tw
o Im
pelle
rs6-
inch
Dia
met
er12
-inc
h D
iam
eter
The
poin
t on
the
12-in
ch d
iam
eter
impe
ller i
s tra
velin
g tw
ice
as fa
r with
eac
h re
volu
tion!
Use
the
Trac
k to
Fin
d To
tal
Dist
ance
Trav
eled
per
Min
ute
1.57
feet
revo
lutio
n1 m
inut
e35
00 re
volu
tions
= 5,
497
ft/m
in
3.14
feet
revo
lutio
n1 m
inut
e35
00 re
volu
tions
= 10
,990
ft/m
in
Chan
ging
the
Impe
ller S
ize Is
n’t t
he
hang
ing
the
Impe
ller S
ize Is
nCh
angi
ng th
e Im
pelle
r Size
Isn
ize Is
nO
nly
Way
to C
hang
e Ve
locit
yIf
I do
uble
the ee
DIS
TAN
CEEE
trav
eled
, I d
oubl
e th
e If
I do
uble
thI
e D
ISTA
Dpe
riph
eral
vel
ocit
y.p
py
If th
is re
latio
nshi
p is
true
:
Wha
t hap
pens
to P
erip
hera
l Vel
ocit
y w
hen
I cha
nge
Wha
tW th
e t h
appe
nW
hat
e eSP
EED
ns to
Per
iphe
rape
nDD
of th
e m
otor
?
Dir
ectl
yVe
rsus
Inve
rsel
yR
elat
ed
Two
Impe
llers
, One
Dist
ance
6-in
ch D
iam
eter
12-i
nch
Dia
met
erRo
tate
at 3
500
rpm
35p
Dis
tanc
e is
1.57
p7 7
ftp ftft/r
evol
utio
n57
Spee
d is
5,4
97 577 7 77ft7 ftftft
/min
Rota
te a
t 175
0 rp
m75
pD
ista
nce
is 3
.14 p 4 4ftp ftft
/rev
olut
ion
34
Spee
d is
5,4
97 44 7 7ft4 ftftft
/min
Now
bot
h pu
mps
hav
e th
e sa
me
rim s
peed
.
A ph
iloso
phica
l que
stio
n….
Perip
hera
l Vel
ocity
and
Hea
d Ar
e Cl
osel
y Re
late
d
H =
Hea
d in
Fee
tD
= Im
pelle
r Dia
met
er in
Inch
esp
N =
rpm
or
The
Prac
tical
App
licat
ion
Prob
lem
: Fo
r an
1800
rpm
pum
p, fi
nd th
e im
pelle
r di
amet
er n
eces
sary
to d
evel
op a
hea
d of
200
feet
:
Step
1 –
Find
the
impe
ller p
erip
hera
l vel
ocit
y ne
eded
.
Step
2 –
Find
the
impe
ller s
peed
in re
volu
tion
s pe
r sec
ond.
1800
revo
luti
ons
min
ute
1 min
ute
60 s
econ
ds=
30 re
v/se
c
Step
3 –
Cal
cula
te th
e nu
mbe
r of f
eet a
poi
nt
on th
e im
pelle
r’s ri
m tr
avel
s in
a s
ingl
e ro
tati
on:
113 fe
etse
cond
1 sec
ond
30 re
vs=
3.77
feet
/rev
olut
ion
Step
4 –
Find
the
impe
ller d
iam
eter
The
dist
ance
trav
eled
by
the
edge
of t
he
impe
ller i
s equ
al to
the
peri
met
er o
r ci
rcum
fere
nce
of th
e im
pelle
r.
Peri
met
er =
(di
amet
er)
Peri
met
er =
(3.14
)(di
amet
er)
3.77
feet
= (3
.14)(
diam
eter
)3.
143.
141.2
feet
= d
iam
eter
Rem
embe
r: Y
ou c
an
do a
nyth
ing
you
wan
t to
an e
quat
ion
as lo
ng a
syou
do
it to
BO
TH s
ides
.
Pum
p Af
finity
Law
sU
se to
com
pare
new
ope
ratin
g co
nditi
on to
old
pp
gp
pg
Allo
ws a
n op
erat
or to
det
erm
ine
pIf
cha
ngin
g m
otor
spe
ed w
ill s
olve
a c
apac
ity
prob
lem
gg
pp
yp
If c
hang
ing
impe
llers
will
sol
ve a
cap
acit
y pr
oble
mg
gp
py
pW
hat h
appe
ns to
the
pum
p di
scha
rge
head
whe
n sp
eed
Wha
t hap
pens
to th
e pu
mp
disc
hW or
impe
ller d
iam
eter
is c
hang
edp
gIf
mak
ing
a ch
ange
will
cau
se a
n un
acce
ptab
le c
hang
e in
If
mak
ing
a I am
p dr
aw
Pre
tty
Use
ful S
tuff
!
The
follo
win
gpu
mp
affin
ityla
ws
appl
yto
CEN
TRIF
UG
AL P
UM
PS O
NLY
Pum
p Af
finity
Law
#1
FLO
WW(Q
) var
ies d
irec
tly w
ith
the
chan
ge in
the
FLO
WF
WSP
EED
Q)
Q) v
arie
s dir
WQQ(Q(Q
DD(N
) or t
he
yre
ctly
wit
h th
s dir
e eD
IAM
ETER
he c
hh
thRR
(D):
If sp
eed
doub
les,
flo
w d
oubl
es.
pIf
dia
met
er d
oubl
es, f
low
dou
bles
.
89
and
Pum
p Af
finity
Law
#2
The e
HEA
DD(H
) var
ies w
ith
the
squa
re o
f the
e SP
EED
ThTe
HEA
DH
(N) o
r the
)
H) v
arie
s wi
((HA
DD((
e eD
IAM
ETER
its w
iRR
:
90
Pum
p Af
finity
Law
#3
The e
BR
AK
E H
OR
SEP
OW
ER
R(
R(b
hphp) v
arie
s wit
h th
e ThT
eR
AK
E H
Bcu
be o
f the
H
OR
SEK
E H
e eSP
EED
EPO
WER
RSE DD
(N) o
r the
hbhb
phpva
ries
wva
r)) v
RRR ((b e eeD
IAM
ETER
wit
h t
s w RR(D
):
91
Exam
ple
Q1 =
200
gpm
Qgp
Q H1 =
100
feet
of h
ead
BHp1
= 7
.2 2 2H
pp
7p
Am
ps =
7.9
Am
ps
Imag
ine
a pu
mp
wit
h an
8I 888-
agin
e a
pum
p w
agin
e a
pum
pa
ImI 888-in
ch d
iam
eter
88
nch
diam
eter
nin
impe
ller.
It h
as th
e im
pelle
r. It
has
the
char
acte
rist
ics o
n th
e ch
arac
tri
ght.
Wha
t hap
pens
if th
e W
hat h
appe
ns if
the
W impe
ller i
s cha
nged
to a
im
p10im
p1010
-el
ler i
s cha
nged
pem
p00-
inch
dia
met
er?
Ratio
to C
reat
e a
Fact
orTh
e m
ath
invo
lved
is…
wel
l…. L
ess t
han
easy
Mak
e it
sim
pler
by
crea
ting
a fa
ctor
Appl
y Fa
ctor
to A
ffini
ty La
ws
Appl
y Fa
ctor
to A
ffini
ty La
ws
BHp
and
Ampe
rage
are
Dire
ctly
Re
late
d; U
se F
cube
d
Qui
ck C
ompa
re fo
r Exa
mpl
e8-
inch
Impe
ller
10-i
nch
Impe
ller
Flow
= 2
00 g
pmgpH
ead
= 10
0 fe
etBH
ppp=
7.2
ppA
mps
= 7
.9
Flow
= 2
50 g
pm5
gpH
ead
= 15
6 fe
etBH
p5
pp=
14pp
4A
mps
= 15
.4
Wha
t if t
he m
otor
is ra
ted
for 1
1 Am
ps?
Will
the
brea
ker s
hort
out?
How
abo
ut th
e th
erm
al o
verlo
ads?
Grou
p Ex
ercis
eQ
1 = 3
25 g
pmQ
35
gpQ H
1 = 15
0 fe
et o
f hea
d5
BHp1
= 12
2 2H
pp
pA
mps
= 15
Am
ps
Imag
ine
a pu
mp
wit
h han
I 15Ima
I 15151-
gp
pgi
ne a
pum
p w
igag
ma
55-in
ch d
iam
eter
1155
nch
diam
eter
nin
impe
ller.
It h
as th
e im
pelle
r. It
has
the
char
acte
rist
ics o
n th
e ch
arac
tri
ght.
Wha
t hap
pens
if th
e W
hat h
appe
ns if
the
W impe
ller i
s cha
nged
to a
im
pe17
.5mpe .55-
ler i
s cha
nged
toll
pe 55-in
ch d
iam
eter
?
Ratio
to C
reat
e a
Fact
orTh
e m
ath
invo
lved
is…
wel
l…. L
ess t
han
easy
Mak
e it
sim
pler
by
crea
ting
a fa
ctor
Appl
y Fa
ctor
to A
ffini
ty La
ws
Appl
y Fa
ctor
to A
ffini
ty La
ws
BHp
and
Ampe
rage
are
Dire
ctly
Re
late
d; U
se F
cube
d
Grou
p Ex
ercis
eQ
1 = 3
25 g
pmQ
35
gpQ H
1 = 15
0 fe
et o
f hea
d5
BHp1
= 12
2 2H
pp
pA
mps
= 15
Am
psp
5RP
M =
1800
Imag
ine
a pu
mp
wit
h h an
I 10Ima
I 10101-
gp
pgi
ne a
pum
p w
igag
ma
00-in
ch d
iam
eter
1100
nch
diam
eter
nin
impe
ller.
It h
as th
e im
pelle
r. It
has
the
char
acte
rist
ics o
n th
e ch
arac
tri
ght.
Wha
t hap
pens
if th
e W
hat h
appe
ns if
the
W mot
or sp
eed
is in
crea
sed
mot
or sp
eed
is
to 2
200
rpm
?
Ratio
to C
reat
e a
Fact
orTh
e m
ath
invo
lved
is…
wel
l…. L
ess t
han
easy
Mak
e it
sim
pler
by
crea
ting
a fa
ctor
Appl
y Fa
ctor
to A
ffini
ty La
ws
Appl
y Fa
ctor
to A
ffini
ty La
ws
BHp
and
Ampe
rage
are
Dire
ctly
Re
late
d; U
se F
cube
d
Quiz
Click
the Q
uizbu
tton t
o ed
it th
is ob
ject
Syst
em C
urve
Gag
e A
Gag
e B
85 p
si
Step
1 –
Plot
Sta
tic
Hea
d
Sour
ce: A
CRP
Pum
ps B
ook
Step
2 –
Calcu
late
Maj
or Lo
sses
w
ith H
azen
Will
iam
s Equ
atio
nSt
ep 2
–Ca
lcula
te M
ajor
Loss
es
with
Haz
en-W
illia
ms E
quat
ion
Q,
gpm
Hea
dlos
sft
/100
ftTo
tal
Hea
dlos
sH
ead
at
gage
BH
ead
at
gage
A0
0.00
020
020
070
00.
2222
200
222
1400
0.78
7820
027
822
001.8
118
120
038
1
Step
3 –
Plot
Sys
tem
Cur
ves
Cur
ves
chan
ge b
ased
on
wet
wel
l lev
el.
Sour
ce: A
CRP
Pum
ps B
ook
Wha
t hap
pens
to th
e sy
stem
cu
rve
if he
adlo
ss ch
ange
s?C
urve
s st
eepe
n (g
o up
fast
er) w
hen n
head
losss
incr
ease
sp
(gPi
pe d
eter
iora
tes
pA
ddin
g va
lves
gPa
rtia
lly c
lose
d va
lve
Curv
es fl
atte
n w
hen n
head
lossss
decr
ease
sPi
pe c
lean
edp
Pipe
e esl
iplin
edp
pVa
lves
/ o
bstr
ucti
ons r
emov
ed
Pum
p Cu
rves
Onl
y lo
ok c
ompl
icat
edy
py
pTO
NS
of in
form
atio
n on
one
gra
phg
pTh
e ke
y is
to re
ad th
em c
orre
ctly
Rew
ard:
Kno
win
g ho
w c
hang
es in
hea
d, sp
eed,
Re
war
d: K
now
ing
how
cha
nges
in h
ead,
sR im
pelle
r dia
met
er, a
nd s
ucti
on li
ft a
ffect
p,
Hor
sepo
wer
dem
and
pFl
owH
ead
Effic
ienc
y
116
The
pum
p AL
WAY
S op
erat
es o
n its
cu
rve
Thre
e Ba
sic T
ypes
of C
urve
s for
Ce
ntrif
ugal
Pum
psH
ead
capa
city
cur
vep
yEf
ficie
ncy
curv
ey
Hor
sepo
wer
dem
and
curv
e
Oft
en, a
ll ar
e pl
otte
d on
the
sam
e gr
aph
May
als
o ha
ve a
n N
PSH
cur
ve
118
Head
Cap
acity
Cur
ve
119
Head
Cap
acity
Cur
veIm
pose
the
Syst
em C
urve
and
the
Pum
p Cu
rve
to F
ind
Ope
ratin
g Po
int(s
)
Sour
ce: A
CRP
Pum
ps B
ook
Pum
p Ef
ficie
ncy
Curv
e
Sour
ce: A
CRP
Pum
ps B
ook
122
Effic
ienc
y Cu
rves
Effic
ienc
y Cu
rves
Whe
neve
r pos
sibl
e, se
lect
a p
ump
for m
axim
um
Whe
neve
r pos
sibl
e, se
lect
a p
ump
for m
axim
W effic
ienc
y at
the
syst
em o
pera
ting
poi
nt(s
)So
urce
: AC
RP P
umps
Boo
k
Brak
e Ho
rsep
ower
Cur
ve
Sour
ce: A
CRP
Pum
ps B
ook
125
Brak
e Ho
rsep
ower
Cur
veCu
rve
Fam
ilies
A w
hole
lot g
oing
on!
A w
hole
lot g
oing
on!
A Find
the
desi
red
or e
xist
ing
impe
ller d
iam
eter
Find
the
desi
red
or e
xist
ing
F Loca
te th
e sy
stem
poi
nts
Sour
ce: A
CRP
Pum
ps B
ook
NPSH
Req
uire
d Gr
aph
As fl
ow a
nd h
ead
incr
ease
, so
does
NPS
Hre
q.So
urce
: AC
RP P
umps
Boo
k12
8
Head
Cap
acity
Cur
ve
Curv
e fo
r a M
ulti-
Spee
d Pu
mp
Spee
d ch
ange
s p
gSp
eed
chan
gS in
stea
d of
in
stea
d of
im
pelle
r siz
e
Find
the
syst
em
Find
the
syst
emF po
int o
verl
ays
Find
dBH
pH
p, F
low
, Fi
ndF
d HBHH
pF, F
and
Hea
d
Sour
ce: A
CRP
Pum
ps B
ook
Grou
p Di
scus
sion
Pum
p to
pics
to b
e di
scus
sed
by th
e gr
oup
Impa
ct o
f rep
laci
ng v
alve
in th
e di
scha
rge
line
wit
h a
valv
e of
diff
eren
t siz
e or
type
Impa
ct o
f inc
reas
ing/
decr
easi
ng s
ucti
on p
ipe
diam
eter
Impa
ct o
f inc
reas
ing/
decr
easi
ng d
isch
arge
pip
e di
amet
erIm
pact
of i
ncre
asin
g/de
crea
sing
mot
or s
peed
Wha
t if t
here
isn’
t eno
ugh
NPS
H?
Quiz
Click
the Q
uizbu
tton t
o ed
it th
is ob
ject
Pum
ps a
nd P
umpi
ngE.
E. S
keet
Ara
smit
hA
CR
Publ
icat
ions
, Inc
.w
ww.
acrp
.com
