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CVEG 3213 Hydraulics Lecture: Units and Quantities.

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Presentation on theme: "CVEG 3213 Hydraulics Lecture: Units and Quantities."— Presentation transcript:

1 CVEG 3213 Hydraulics Lecture: Units and Quantities

2 Units and Quantities  mass kg, g, mg,  g (microgram) kg, g, mg,  g (microgram) 1 kg = 1000 g = 10 3 g ; 1 g = kg = kg1 kg = 1000 g = 10 3 g ; 1 g = kg = kg 1 kg = mg = 10 6 (a million) mg1 kg = mg = 10 6 (a million) mg 1  g = g1  g = g 1 kg = 2.2 lb 1 kg = 2.2 lb note: lb is actually a weight (force)note: lb is actually a weight (force) F = ma F = ma  volume 1 cubic foot = ft 3 = 7.48 gal 1 cubic foot = ft 3 = 7.48 gal 1 gal = 3.8 L 1 gal = 3.8 L 1 m 3 = 1000 L 1 m 3 = 1000 L 1 mL = 1 cm 3 = 1 cc 1 mL = 1 cm 3 = 1 cc

3 Units and Quantities  Density:  = mass per volume Water density =  w = 1 g/mL = 1 kg/L Water density =  w = 1 g/mL = 1 kg/L = 62.4 lb/cubic foot *(see below) = 62.4 lb/cubic foot *(see below) slightly temperature dependent slightly temperature dependent considered incompressible (very slightly compressible) considered incompressible (very slightly compressible)  Unit weight of water:  w = weight per volume “specific weight” “specific weight” weight is a force = mass x acceleration weight is a force = mass x acceleration = mass x g= mass x g g = 9.8 m/s 2 = 32.2 ft/s 3 g = 9.8 m/s 2 = 32.2 ft/s 3  w =  w. g  w =  w. g = 1000 kg/m m/s 2 = 9800 N/m 3 = 9.8 kN/m 3= 1000 kg/m m/s 2 = 9800 N/m 3 = 9.8 kN/m 3 = 1.94 slugs/ft ft/s 2 = 62.4 lb/ft 3= 1.94 slugs/ft ft/s 2 = 62.4 lb/ft 3

4 Units and Quantities  Density of liquids  (pcf) SG Ethyl Alcohol Gasoline Glycerin Mercury SAE 20 Oil Seawater Water

5 Units and Quantities Solid density Solid density density of sand = 2.6 g/cm 3 = 2600 kg/m 3density of sand = 2.6 g/cm 3 = 2600 kg/m 3 concrete density = 2400 kg/m 3 (150 lb/ft 3 ),concrete density = 2400 kg/m 3 (150 lb/ft 3 ), SG = 2.4 SG = 2.4 Dry bulk density of soil or solids Dry bulk density of soil or solids Mass per volume (including voids)Mass per volume (including voids) Approximately ~ 1.4 – 1.8 g/mLApproximately ~ 1.4 – 1.8 g/mL

6 Units and Quantities  Pressure = force per area N/m 2 = Pa (Pascal) N/m 2 = Pa (Pascal) often use kPaoften use kPa atmospheric P: 1 atm = kPaatmospheric P: 1 atm = kPa psi = lb/in 2 psi = lb/in 2 1 atm = 14.7 psi1 atm = 14.7 psi

7 Units and Quantities  Pressure exerted by a 1-ft diameter column of water 34 ft high Pressure = force (weight) per area Pressure = force (weight) per area wt = volume x unit wt (  w )wt = volume x unit wt (  w ) volume = area x ht volume = area x ht area =  /4 x dia 2 = 3.14/4 x (1ft) 2 = 0.785ft 2area =  /4 x dia 2 = 3.14/4 x (1ft) 2 = 0.785ft 2 Vol = 0.785ft 2 x 34ft = 26.7ft 3 = 200 galVol = 0.785ft 2 x 34ft = 26.7ft 3 = 200 gal wt = 26.7ft 3 x 62.4 lb/ft 3 = 1666 lb wt = 26.7ft 3 x 62.4 lb/ft 3 = 1666 lb Pressure = wt/areaPressure = wt/area = 1666 lb / ft 2 = 2122 lb/ft 2 = 14.7 psi = 1.00 atm = 1666 lb / ft 2 = 2122 lb/ft 2 = 14.7 psi = 1.00 atm

8 Units and Quantities  Pressure exerted by a 10-ft square column of water 34 ft high Pressure = force (weight) per area Pressure = force (weight) per area wt = volume x unit wt (  w )wt = volume x unit wt (  w ) volume = area x ht volume = area x ht area = 10ft x 10 ft = 100ft 2area = 10ft x 10 ft = 100ft 2 Vol = 100ft 2 x 34ft = 3400ft 3 = galVol = 100ft 2 x 34ft = 3400ft 3 = gal wt = 3400ft 3 x 62.4 lb/ft 3 = 212,160 lb wt = 3400ft 3 x 62.4 lb/ft 3 = 212,160 lb Pressure = wt/areaPressure = wt/area = 212,160 lb / 100 ft 2 = 2122 lb/ft 2 = 14.7 psi = 1.00 atm = 212,160 lb / 100 ft 2 = 2122 lb/ft 2 = 14.7 psi = 1.00 atm

9 Units and Quantities  Velocity and distance If you drive 50 mph for two hours, how far do you go? If you drive 50 mph for two hours, how far do you go? 2 hr x 50 mi/hr = 100 mi (duh)2 hr x 50 mi/hr = 100 mi (duh) If you drive 120 miles in 3 hours, what’s your average speed? If you drive 120 miles in 3 hours, what’s your average speed? 120 mi / 3 hr = 40 mph120 mi / 3 hr = 40 mph distance = v. t is important distance = v. t is important pipe flowpipe flow stream flowstream flow

10 Units and Quantities  Velocity and distance If you drive 50 mph for two hours, how far do you go? If you drive 50 mph for two hours, how far do you go? 2 hr x 50 mi/hr = 100 mi (duh)2 hr x 50 mi/hr = 100 mi (duh) If you drive 120 miles in 3 hours, what’s your average speed? If you drive 120 miles in 3 hours, what’s your average speed? 120 mi / 3 hr = 40 mph120 mi / 3 hr = 40 mph distance = v. t is important distance = v. t is important pipe flowpipe flow stream flowstream flow

11 Units and Quantities  Concentration:  Concentration: C Concentration Concentration is is Mass per Volume massmass of chemical or solids per volume of solution mg/Lmg/L (milligram per liter) 1mg/L = 1 lb per _____ _____ gal 1L/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  g/L  g/L (microgram per liter) = 1/1000 x mg/L

12 Units and Quantities  mg/L versus ppm (parts per million) ppm is mass/mass ppm is mass/mass e.g., mg per million mg = mg/kge.g., mg per million mg = mg/kg  water = 1 kg/L,  water = 1 kg/L, 1 L of water weighs 1 kg 1 L of water weighs 1 kg so a mg/L ~ ppm so a mg/L ~ ppm

13 Units and Quantities  Flow: Q Flow is Volume per Time Flow is Volume per Time m 3 /s, ft 3 /s = cfs, gal/min = gpmm 3 /s, ft 3 /s = cfs, gal/min = gpm MGD = million gallons per dayMGD = million gallons per day note: 7.48 gal per ft 3 note: 7.48 gal per ft 3 1 cfs = _____ MGD1 cfs = _____ MGD 1 ft 3 /s x 60 s/min x 60 min/hr x 24 hr/day 1 ft 3 /s x 60 s/min x 60 min/hr x 24 hr/day x 7.48 gal/ft 3 x 1 million/10 6 x 7.48 gal/ft 3 x 1 million/10 6 = 0.65 MGD = 0.65 MGD Beaver Water Plant = ?Beaver Water Plant = ? 40 – 60 MGD 40 – 60 MGD

14 Units and Quantities  Mass transported Mass/Time = Q. C Mass/Time = Q. C vol/time x mass/vol = mass/timevol/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? 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 10 6 gal/day x 2 mg/L x 3.8 L/gal x 1g/1000 mg x 1kg/1000 gMass/Time = Q. C = 18 x 10 6 gal/day x 2 mg/L x 3.8 L/gal x 1g/1000 mg x 1kg/1000 g = 137 kg/day = 137 kg/day x 365 = ~ 50,000 kg/yr x 365 = ~ 50,000 kg/yr 18 MGD x 2 mg/L x (8.34 lb/MG)/mg/L18 MGD x 2 mg/L x (8.34 lb/MG)/mg/L = 300 lb/day = 300 lb/day x 1 kg/2.2 lb = 137 kg/dayx 1 kg/2.2 lb = 137 kg/day


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