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NATURE OF ENGINEERING MECHANICS
Interaction
INTRODUCTION
Bodies nRest/MotioForces:
Tension
Compression
Shear
Solids:
Rigid
Deformable
Fluids:Liquids
Gases
- Velocity
- Acceleration
- Time
SIMILARITY IN THE BRANCHES OF MECHANICS
a. Basic Principles and Fundamental Concepts
b. Newtons Laws of Motion
c. Principle of Work and Energy
d. Principle of Impulse and Momentum
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UNITS OF MEASUREMENTS
System Mass ( M ) Length ( L ) Time ( T ) Force ( F )
SI Kilogram ( kg ) Meter ( m ) Second ( s ) Newton ( N ) *
English Slug * Foot ( ft ) Second ( s ) Pound ( lb )
Metric Kilogram Meter ( m ) Second ( s ) Kilogram (kgF)*
*Derived Units from Newtons Second Law of Motion
a. Standard Acceleration of Gravity ( g ):
SI: g = 9.81 m/sec2 English: g = 32.2 ft/sec2
b. Mass (M) and Weight (W): W = Mg
c. Conversion Factors:
1 kgF = 2.2 lb 1 kgF = 9.81 N 1 lb = 4.448 N
MaF 8.9/MakgF
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d. Common Prefixes of SI Units:
SI Prefix Abbreviation MultipleGiga G 109
Mega M 106
Kilo K 103
Centi c 10-2
Milli m 10-3
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CLASSIFICATION OF FLUIDS
a. Compressibility:
- incompressible
- compressible
b. Viscosity:
- Non-viscous
- Viscous
- Newtonian
- Non-Newtonian
c. Ideal Fluid:
- Incompressible and non-viscous
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FUNDAMENTAL CONCEPTS
Mechanics of Fluids and Hydraulics, these aresciences that deals with the laws of liquids and gases at restor in motion and of their practical applications. Fluidmechanics is based on the same fundamental principlesthat are employed in mechanics of solids. Hydraulics
utilized experimental techniques to developed empiricalformulas which supplied answers to practical problems.
DEFINITION OF A FLUID
Deforms continuously when subjected to shear stress
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Three Branches of Fluid Mechanics
1. Fluid Statics, is the study of the mechanics offluids at rest.
2. Kinematics, deals with velocities and streamlineswithout considering forces or energy involved.
3. Fluid Dynamics, is concerned with the relationsbetween velocities and accelerations and theforces exerted by or upon fluids in motion.
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1.1 DISTINCTIONS BETWEEN A GAS AND A LIQUID
GAS
1. Molecules are very far apart.
2. Very compressible.3. Expand indefinitely when external pressure is removed.
4. Can be held in equilibrium only when it is completely enclosed.
5. No free surface
LIQUID
1. Molecules are little closer.
2. Relatively incompressible.
3. Cohesion between molecules holds them together except of its
own vapor and does not expand indefinitely.
4. Can be held in equilibrium both in an open and close space.
5. May have a free surface (i.e a surface from which all pressure is
removed, except that of its own vapor) .
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A. FLUID STATICS
PART 1: PROPERTIES OF FLUIDS
1. Specific weight, y: the force exerted by gravity on a unit volume of fluid, or
Units:English System:
Metric System:
S.I. :
Note: This relationship is true for liquids where varies slightlyregardless of temperature and pressure. For gases, is
computed from Charles and Boyles laws which give
English:
R- gas constant
Tabsolute temperature, ( 460 + F)
V
W
3ft
lb
3cm
gr3
m
tons
3m
N
RT
p
or
where p - absolute pressure
R)lbf/lbmol(1545ft
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S.I.
T=K, K =273 + C
Alternative unit of R: J/kg-K; Btu/lb-R; ft-lbf/lb-R
2. Specific Volume,
The reciprocal of the density, or
English:
RT
p
2m
Np
KkgmNR
:
1
slugs
ft3
gr
cm3
kg
m3:Metric :.IS
K)m/kmol(8314N
R)lbmol(1.986Btu/
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3. Mass Density,
The mass per unit volume of the fluid, or
Units: English,
Metric,
S.I.,
:)(rho
VM
gVW
g
3
ft
slugs
3cm
gr
3
m
kg
The density of water at 4 C (39.2 F ) is 1.00 g/cc ( or 1.00 g/mL ), equivalent
to 1000 kg/cu. m.
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4. Specific Gravity; s: the dimensionless ratio of the fluid specific weight (or density) to
the standard reference fluid, water (for liquids),and air (for gases):
Note: At 39.2 F (4 C) the specific weight of water is approximately
or
5. Dynamic (Absolute) Viscosity, (mu): the property of a fluid which determines the
amount of its resistance to a shearing stress.
Note: The shearing stress, T (Tau), and the viscosity, (Mu) are related by,
Units of :
English System: S.I:
;
water
liquid
water
liquid
liquids
34.62ft
lbw ;9810 3
m
N
dx
dv
,,
22
mm
N
ft
lb
,dx
dv
2
sec
ft
lb 2m
sN
2)(
cm
sdynePpoise
:Metric
;
air
gas
air
gas
gass
3205.1 m
kgair
ml
g
cm
g
m
kgw 0.10.11000 33
s
mkg
is the velocity gradient
2
1.0m
sN At 20 C, = 1 cP
= 0.01P =mN-s/m2
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Newtons Law of Viscosity
Fluid between parallel plates:
a
F
y
b b
Velocity Distribution:
Assumptions/Observations:
a. Closely spaced plates, y is very small.
b. Lower plate is fixed.
c. Force F exerted on upper plate with area A imparting a velocity u to the plate.
d. No slip at the boundaries, velocity of fluid same as boundary.
e. Line ab moves to the line ab after time interval.
f. Velocity distribution is linear.
g. Based on experimental observations: The applied force is proportional to thearea and velocity gradient. is the constant of proportionality.
h. Differential form ( Newtons Law of Viscosity):
y
AuF
y
u or
dy
du
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Temp, C Specificweight
, N/m3
Density ,kg/ m
Viscosity ,
kg / (m-s)
10-3 =
KinematicViscosity, m2/s10-6 v =
SurfaceTension
, N / m
100 =
VaporPressure
head
p / , m
Bulk modulusof elasticity
K, N..m2
10-2 K =
0 9805 999.9 1.792 1.792 7.62 0.06 204
5 9806 1000.0 1.519 1.519 7.54 0.09 206
10 9803 999.7 1.308 1.308 7.48 0.12 211
15 9798 999.1 1.140 1.141 7.41 0.17 214
20 9789 998.2 1.005 1.007 7.36 0.25 220
25 9779 997.1 0.894 0.897 7.26 0.33 222
30 9767 995.7 0.801 0.804 7.18 0.44 223
35 9752 994.1 0.723 0.727 7.10 0.58 22440 9737 992.2 0.656 0.661 7.01 0.76 227
45 9720 990.2 0.599 0.605 6.92 0.98 229
,50 9697 988.1 0.549 0.556 6.82 1.26 230
'55 9679 985.7 0.506 0.513 6.74 1.61 231
60 9658 983.2 0.469 0.477 6.68 2.03 228
65 9635 980.6 0.436 0.444 6.58 2.56 226
70 9600 977.8 0.406 0.415 6.50 3.20 225
75 9589 974.9 0.380 0.390 6.40 3.96 223
80 9557 971.8 0.357 0.367 6.30 4.86 221
85 9529 968.6 0.336 0.347 6.20 5.93 217
90 9499 965.3 0.317 0.328 6.12 7.18 216
95 9469 961.9 0.299 0.311 6.02 8.62 211
100 9438 958.4 0.284 0.290 5.94 10.33 207
PHYSICAL PROPERTIES TABLE OF WATER IN SI UNITS
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Illustrative Examples:
a) The specific weight of water at ordinary pressure and temperature is
. The specific gravity of mercury is 13.55. Compute the
density of water and the specific weight and density of mercury.
Solution.
./81.93
mkN
gw
w
2
3
81.9
81.9
s
mm
kN
31000m
kg
Note : kgm
sN11
2
3313381.955.13
m
kN
m
kNs watermercurymercury
33550,13100055.13
m
kg
m
kgs watermercutymercury
2
3
81.9
100081.9
s
mm
Nx
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b) A gallon of water weighs 4.08 lb. Compute the following:
1) Mass in slugs
2) Mass in kg
3) Volume in cu. ft.
Solution.
1) Mass in slugs:
2) Mass in kg:
3) Volume in cu. ft. :
g
WM
22.32
08.4
s
ft
lbM )(1267.0
2
slugsft
slb
281.9
448.408.4
s
mlb
Nlb
M
)(85.1
2
kgm
sN
gal
ftgalxV
48.7
11
3
3134.0 ft
A i i h 16 N/ i d
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c) A certain gas weighs 16 N/cu. m. at a certain temperature and
pressure. What are the values of its density, specific volume, and
specific gravity relative to air weighing 12 N/cu. m. ?
Solution.
The density of the gas is,
The specific volume is,
The specific gravity is,
g
2
3
81.9
16
s
mm
N
3
631.1m
kg
1
3631.1
1
m
kg
kg
m3
613.0
a
s
3
3
12
16
m
Nm
N
333.1
)
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d) The specific gravity of glycerin is 1.26. Compute its density and
specific weight.
Solution.
The density is,
The specific weight is,
water
glycerins
33 26.100.126.1 cm
g
cm
gglycerin
or 33 1260100026.1 m
kg
m
kg
gglyceringlycerin
2381.91260
s
m
m
kg
3360,12
m
N or
3360.12
m
kN
( ) h f h d
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6. Kinematic Viscosity, v (nu): the ratio of the dynamic viscosity to its
mass density, or
Units of :
English System: S.I:
Metric System:7. Adhesion: the property of a liquid that enables it to stick to another
body.
8. Cohesion: the property of a liquid that resists tensile stress.
9. Surface Tension, (sigma): the effect of cohesion between particlesof the liquid at its surface.
Unit: English System, S.I,
sec
2ft)(
sec
2
ststokecm
sec
2cm
stoke
ft
lb
m
N
10 C ill i hi i d h i d dh i If dh i h10 C ill i hi i d h i d dh i If dh i h
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10. Capillarity: this is due to cohesion and adhesion. If adhesion has a
greater effect than cohesion, the liquid will rise at the point where it
comes in contact with another body. If cohesion is prevalent, the
liquid will depressed.
10. Capillarity: this is due to cohesion and adhesion. If adhesion has a
greater effect than cohesion, the liquid will rise at the point where it
comes in contact with another body. If cohesion is prevalent, the
liquid will depressed. The capillary rise in a tube can be expressed,
where
grrh
cos2cos2
= surface tension in units offorce per unit length = wetting angle = specific weight of liquid
r = radius of tubeh = capillary rise = density of liquidg = gravitational acceleration
hr
1111 V P h d b h
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11.11. Vapor Pressure, : the pressure exerted by the vapor as
evaporation of the liquid takes place within a closed space.
Units: English System,
S.I.,
Metric System,
vp
2ft
lb
or 2in
lb
2m
N
2cmgr
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Property Definition Dimension
(SI Units)
Common Values
For Water ( 4 C) *
, Mass Density mass per unit volume ML-3
(kg/m3)
1,000 kg/m3
, Specific Weight weight per unit
volume
FL-3
(N/m3)
9,800 N/m3
S, Specific gravity /water (for liquid) 0 1
p, Pressure Normal force per unit
area
FL-2
(N/m2)
, Absolute Viscosity Newtons law of
viscosity
FL-2T
(N s/m
2
)
1.52 x 10-3 N s/m2
, Kinematic Viscosity / L2T-1
(m2/s)
1.52 x 10-6 m2/s
* Note: The values of fluid properties vary with temperature
FLUID PROPERTIES MOST COMMONLY CONSIDERED
CCOMPRESSIBILITY OF LIQUID
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CCOMPRESSIBILITY OF LIQUID
The compressibility ( change in volume due to change in pressure ) of
a liquid is inversely proportional to its volume modulus of elasticity,
also known as the bulk modulus. This is defined as
where
Rearranging the definition of as an approximation, we may used
for the case of a fixed mass of liquid at constant temperature,
or
dpdv
v
dv
dpvE
v
v = specific volumep = pressure
,vE
vEp
vE
pp12
1
12
Coefficient of Compressibility
= coefficient of compressibility
vE
1
E l 1 R d 1 ti i t th E li h t If thi i th
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Example 1. Reduce 1 centipoise to the English system. If this is the
dynamic viscosity of water at 68 F (20C), what is the kinematic
viscosity in the English system?
Solution.
1 centipoise
1 lb = 444,800 dynes
a) The dynamic viscosity in English system
b) The Kinematic viscosity in the English system
scmg
cmsdynepoise
01.001.001.0
2
poise01.02
21254.2
800,444101.0
ftin
incm
dyneslb
cmsdynes
2
510088.2
ft
slbx
g where )(32.62 3 waterftlb at F68
sec
10079.1
32.62
2.3210088.22
5
5ft
xx
E l 2 Th ki ti i it it f 1 ft2/ i i l t t h
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Example 2. The kinematic viscosity unit of 1 ft2/sec is equivalent to how
many stokes?
Solution:
Example 3.A volume of 15.5 cu. ft of a certain fluid weighs 782 Ib.
Compute the mass density.Solution:
22
1254.2sec
1
ftin
incmft
sec929
2
cm
V
M
gV
W
32
5.15
sec
2.32
782
ftft
lb
3567.1
ft
slugs
or stokes929
Example 4 A barangay of 8 000 population has an average water
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Example 4. A barangay of 8,000 population has an average water
consumption per person per day of 110 gallons. Compute the average
total rate of consumption in cubic feet per sec. and in gallons per
minute. (1 cu. ft. = 7.48 gallons)
Solution:
a) Average Total Consumption: 8,000 persons x 110
b) Average Total Consumption:
dayperson
gallons
day
galx 41088
hrday
hr
ft
gal
day
galx
ATCsec3600
2448.7
1088
3
4
sec362.1
3ft
hrday
hr
day
galx
ATC
1
min6024
10884
mingal
min111.611
gal
Example 5 Compute the unit weight of dry air at 15 C and an absolute
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Example 5. Compute the unit weight of dry air at 15 C and an absolute
pressure of 14.7 psi (pounds per square inch). Also solve for the
kinematic viscosity.
Solution.
As determined by Bearden, the dynamic viscosity of air at 20 C is
0.0001819 poise and changes at that temperature at a rate of
0.0000005 poise /C
a)
where:
RT
p
FR
ft
ft
in
in
lb
594603.53
1447.14
2
2
2
30765.0
ft
lb
R
ftR
3.53 air
FC 5915
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b)
The Kinematic viscosity is,
at 0000005.050001819.015 C
)(0001794.0 Ppoise or2
sec
cm
dyne
22 54.212
800,444
sec0001794.0
in
cm
ft
in
lb
dynescm
dyne
27 sec1074.3
ftlbx
g
3
22
7
0765.0
sec2.32
sec1074.3
ft
lb
ft
ft
lbx
sec1058.1
24 ftx
Example 6 At a depth of 8 km in the ocean the pressure is 81 8 MPa
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Example 6. At a depth of 8 km in the ocean the pressure is 81.8 MPa.
Assume specific weight at the surface is 10.05 kN/cu.m and that the
average volume modulus is for the pressure
range.(a) What will be the change in specific volume between that at the
surface and at that depth?
(b) What will be the specific volume at that depth?
(c ) What will be the specific weight at that depth?
Solution.
(a) Change in specific volume,
2
91034.2
m
Nx
vEp
vv
vEpv
2
9
2
6
3
3
2
1034.2
108.81
1005.10
81.9
m
N
x
m
Nx
m
N
x
s
m
v
kg
mxv
35
10412.3
( b ) Specific volume at that depth of 8 km
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( b ) Specific volume at that depth of 8 km
kg
mxvvv
35
12 10412.3
kgmxv
3
5
1
210412.31
kg
m
x
g
v
35
12
10412.3
kg
mx
kg
Nx
s
m
v3
5
3
2
2 101412.3
1005.10
81.9
kg
mxv
34
2 10420.9
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(c) Specific weight at that depth 8 km
g22
gv
2
2
1
23
4
281.9
10420.9
1
sm
kg
mx
3414,10
m
N
Example 7 The radius of the tube as shown in the figure is 1 mm
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Example 7. The radius of the tube as shown in the figure is 1 mm.
The surface tension of water at 20 C is equal to 0.0728 N/m . For a
waterglass = 0
Solution.
( a) Capillary rise in the tube in mm
h
2r
grh
cos2
001.081.91000
0cos0728.02 .0148.0 m mm8.14
( b ) Total force due to surface tension
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( b ) Total force due to surface tension
( c ) Weight of water
))(cos( dF
002.0)0cos0728.0( F
NxF4
1057.4
VW mmmN 0148.0001.098102
3
Nx 41057.4
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Example 8. Calculate the density, specific weight and specific volume ofchlorine gas at 25 C and under a pressure of 600 kPa absolute. The gasconstant R for chlorine is 117 N m/ kg - K.
Solution:
1. The density of chlorine gas is
2. The specific weight of chlorine is
3. The specific volume of chlorine gas is
RT
p
27325117
106003
x
3
209.17m
kg
g 81.9209.173
820.168
m
N
1
209.17
1
kg
m3
058.0
E l 9 A f l il h i i i f 0 297 N / 2 fl h h
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Example 9. A fuel oil having a viscosity of 0.297 N s/m2 flows through acircular pipe 15 cm in diameter. At the center of the pipe the velocity is 1.20m/s and decreases to a minimum value at the pipe wall. The value of thevelocity at any point in the cross section a distance x from the center is
and r is the radius of the pipe. Compute the shear stress at a point midwaybetween the center and the wall ( x = 3.75 cm).
Solution:
22
4.63xr
V
1.2 m/s
15 cm V m/s
x (m)
V = f(x)
Substituting r = 7.5 cm or 0.075 mand = 0.296 Ns/m2 to V, weget
297.0
075.04.63
22x
V
2468.213201.1 xV
xdx
dV936.426
Note: The negative sign denotes that V is decreasing as x increases.
x (m)
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15 cm
x (m)
3.75 cm = 0.0375 m
The shearing stress at x = 0.0375 m is
dx
dV
0375.0936.426297.0
xdx
dV936.426Note :
2755.4
m
N
PROPERTY CHANGES IN IDEAL GAS
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PROPERTY CHANGES IN IDEAL GAS
For any ideal gas experiencing any process or changes, the equationof state is given by:
(Universal Gas Law) (1)
When temperature is held constant, equation 1 reduces to2
22
1
11
T
Vp
T
Vp
2211 VpVp (Boyles Law) (2)
When the volume of a confined gas is constant, the pressure isproportional to the absolute temperature, or
2
2
1
1
T
p
T
p (3)
If the pressure of a confined gas is unchanged, the volume is directlyproportional to the absolute temperature, or
2
2
1
1
T
V
T
V (4)
Equations (3) and (4) are called Charless laws.
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From equation (1)
k
T
pV R
W
m
T
pV nR
T
pV
nRTpV (General Gas Law)
where: n = number of molesm = mass of the gas in gramsW = atomic or molecular mass of the gas in grams/moleR = universal gas constant
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If a gas is compressed or expand without loss of heat through the wallsof its container, the change in volume is said to be adiabatic or isentropic. Thenthe special pressure volume relation is given as
tconsvpvpkk
tan2211
where: p = absolute pressurev = specific volume
k = adiabatic exponent, the ratio of specific heat at constantpressure to the specific heat at constant volume.
The value of k depends on the molecular structure of thegask = 1.4 for air, hydrogen, oxygen and nitrogen.
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SPECIFIC WEIGHT AND DENSITY OF AIR
TemperatureC
Density ( kg/m3)
Specific Weight (N/m3)
0 1.293 12.68
10 1.248 12.24
20 1.205 11.82
30 1.165 11.43
40 1.128 11.06
60 1.060 10.40
80 1.000 9.81
100 0.946 9.28
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DYNAMIC VISCOSITY OF AIR
Temperature
C
Viscosity
Pa s x 105
0 1.71
10 1.76
20 1.81
30 1.86
40 1.90
60 2.00
80 2.09
100 2.18
Holmans equation for finding of air is
23 00000034.000275.0110716.1 TTx where: is in Pa-s and T is the temperature in C
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Problems for exercise
1. If a certain gasoline weighs 7 kN/m3, what are the values of its density,specific volume, and specific gravity relative to water at 15 C?
2. A cubic meter of air at 101.3 kPa and 15 C weighs 12 N. What is itsspecific volume?
3. The density of alcohol is 790 kg/m3. Calculate its specific weight, specificweight, specific gravity and specific volume.
4. A certain gas weighs 16 N/m3 at a certain temperature and pressure.
What are the values of its density, specific volume, and specific gravityrelative to air weighing 12 N/m3?5. Compute the number of watts which are equivalent to one horsepower.
( 1 HP = 550 ft-lb/sec; W = 107 dynes-cm/sec; 1 lb = 444,800 dynes).6. A city of 6000 population has an average total consumption per person
per day of 100 gallons. Compute the daily total consumption of the city incubic meter per second ( 1 ft3 = 7.48 gallons ).
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