CVEG 3213 Hydraulics
Lecture: Units and Quantities
Units and Quantities
mass
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kg, g, mg, mg(microgram)
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•
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1 kg = 1000 g = 103g ; 1 g = 0.001 kg = 10-3 kg
1 kg = 1000.1000 mg = 106 (a million) mg
1 mg = 10-6 g
1 kg = 2.2 lb
•
note: lb is actually a weight (force)
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F = ma
volume
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1 cubic foot = ft3 = 7.48 gal
1 gal = 3.8 L
1 m3 = 1000 L
1 mL = 1 cm3 = 1 cc
Units and Quantities
Density: r = mass per volume
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Water density =
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rw = 1 g/mL = 1 kg/L
= 62.4 lb/cubic foot*(see below)
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slightly temperature dependent
considered incompressible (very slightly compressible)
Unit weight of water: gw = weight per volume
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“specific weight”
weight is a force = mass x acceleration
•
= mass x g
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g = 9.8 m/s2 = 32.2 ft/s3
gw = rw.g
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•
= 1000 kg/m3 . 9.8 m/s2 = 9800 N/m3 = 9.8 kN/m3
= 1.94 slugs/ft3 . 32.2 ft/s2 = 62.4 lb/ft3
Units and Quantities
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Density of liquids
g (pcf)
SG
Ethyl Alcohol
49.3
0.79
Gasoline
42.5
0.68
Glycerin
78.6
1.26
Mercury
847
13.57
SAE 20 Oil
57
0.91
Seawater
64
1.03
Water
62.4
1.00
Units and Quantities
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Solid density
• density of sand = 2.6 g/cm3 = 2600 kg/m3
• concrete density = 2400 kg/m3 (150 lb/ft3),
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SG = 2.4
Dry bulk density of soil or solids
• Mass per volume (including voids)
• Approximately ~ 1.4 – 1.8 g/mL
Units and Quantities
 Pressure
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= force per area
N/m2 = Pa (Pascal)
• often use kPa
• atmospheric P: 1 atm = 101.325 kPa
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psi = lb/in2
• 1 atm = 14.7 psi
Units and Quantities
 Pressure
exerted by a 1-ft diameter
column of water 34 ft high
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Pressure = force (weight) per area
• wt = volume x unit wt (gw)
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volume = area x ht
• area = p/4 x dia2 = 3.14/4 x (1ft)2 = 0.785ft2
• Vol = 0.785ft2 x 34ft = 26.7ft3 = 200 gal
wt = 26.7ft3 x 62.4 lb/ft3 = 1666 lb
• Pressure = wt/area
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= 1666 lb / 0.785 ft2 = 2122 lb/ft2 = 14.7 psi = 1.00 atm
Units and Quantities
 Pressure
exerted by a 10-ft square column
of water 34 ft high
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Pressure = force (weight) per area
• wt = volume x unit wt (gw)
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volume = area x ht
• area = 10ft x 10 ft = 100ft2
• Vol = 100ft2 x 34ft = 3400ft3 = 25430 gal
wt = 3400ft3 x 62.4 lb/ft3 = 212,160 lb
• Pressure = wt/area
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= 212,160 lb / 100 ft2 = 2122 lb/ft2 = 14.7 psi = 1.00 atm
Units and Quantities
 Velocity
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and distance
If you drive 50 mph for two hours, how far do
you go?
• 2 hr x 50 mi/hr = 100 mi (duh)
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If you drive 120 miles in 3 hours, what’s your
average speed?
• 120 mi / 3 hr = 40 mph
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distance = v . t is important
• pipe flow
• stream flow
Units and Quantities
 Velocity
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and distance
If you drive 50 mph for two hours, how far do
you go?
• 2 hr x 50 mi/hr = 100 mi (duh)
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If you drive 120 miles in 3 hours, what’s your
average speed?
• 120 mi / 3 hr = 40 mph
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distance = v . t is important
• pipe flow
• stream flow
Units and Quantities
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Concentration: C
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Concentration is Mass per Volume
•
mass of chemical or solids per volume of solution
• mg/L (milligram per liter)
• 1 mg/L = 1 lb per _____ gal
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•
1 L/mg x 1000 mg/g x 1000 g/kg x 2.2 kg/lb x 1 gal/3.8 L
= 120,000 gal/lb or 1 lb per 120,000 gal
1 mg/L (microgram per liter) = 1/1000 x mg/L
Units and Quantities
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mg/L versus ppm (parts per million)
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ppm is mass/mass
• e.g., mg per million mg = mg/kg
• rwater = 1 kg/L,
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1 L of water weighs 1 kg
so a mg/L ~ ppm
Units and Quantities
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Flow: Q
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Flow is Volume per Time
• m3/s , ft3/s = cfs, gal/min = gpm
• MGD = million gallons per day
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note: 7.48 gal per ft3
• 1 cfs = _____ MGD
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1 ft3/s x 60 s/min x 60 min/hr x 24 hr/day
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x 7.48 gal/ft3 x 1 million/106
= 0.65 MGD
• Beaver Water Plant = ?
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40 – 60 MGD
Units and Quantities
Mass transported
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Mass/Time = Q.C
• vol/time x mass/vol = mass/time
If Fayetteville wastewater plant discharges 18 MGD of
water containing 2 mg/L phosphorus, how much
phosphorus (kg, lbs) is discharged per day and per year?
• Mass/Time = Q.C = 18 x 106 gal/day x 2 mg/L
x 3.8 L/gal x 1g/1000 mg x 1kg/1000 g
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•
= 137 kg/day
x 365 = ~ 50,000 kg/yr
18 MGD x 2 mg/L x (8.34 lb/MG)/mg/L
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= 300 lb/day
• x 1 kg/2.2 lb = 137 kg/day