1. Flowmeters, Basic Hydraulics of Pipe Flow, Carrying Capacity and Continuity Equation Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 7; (pgs. 319-341) Math for Water Technology MTH 082 Lecture 5 Hydraulics Chapter 7; (pgs. 319-341)

2. Objectives Review Flow meters Pipe flow Continuity Equation Finish Basic Hydraulics Review Flow meters Pipe flow Continuity Equation Finish Basic Hydraulics

3. A wall or plate placed in an open channel and used to measure flow: 1.Baffle 2.Weir 3.Parshall Flume 4.Flow board 1.Baffle 2.Weir 3.Parshall Flume 4.Flow board

4. Weirs are most often used to measure flows in 1.Treatment plant intakes 2.Open channels 3.Pipelines 4.Underground pipes 1.Treatment plant intakes 2.Open channels 3.Pipelines 4.Underground pipes

5. Which of the following is not an example of a flow measuring device? 1.Magnetic meter 2.Parshall flume 3.Weirs 4.Manometer 5.Venturi 1.Magnetic meter 2.Parshall flume 3.Weirs 4.Manometer 5.Venturi A manometer measures pressure near atmospheric. The term manometer is often used to refer specifically to liquid column hydrostatic instruments.

6. Which of the following flow measuring devices is the most accurate? 1.Magnetic meter 2.Parshall flume 3.Weirs 4.Manometer 5.Venturi 1.Magnetic meter 2.Parshall flume 3.Weirs 4.Manometer 5.Venturi “The in line type magnetic flow meters offer a higher accuracy. They can be as accurate as 0.5% of the flow rate. The insertion styles offer a 0.5 to 1% accuracy.”

7. Magnetic flow meters work on which of the following principles of operation? 1.The volume of water required to separate two magnets. 2.The reduction in magnetic pull as the volume of water separates a magnet and plug. 3.Magnetic induction where voltage is generated in a magnetic field and converted to a velocity. 4.The volume of water that can be moved by an electromagnet. 1.The volume of water required to separate two magnets. 2.The reduction in magnetic pull as the volume of water separates a magnet and plug. 3.Magnetic induction where voltage is generated in a magnetic field and converted to a velocity. 4.The volume of water that can be moved by an electromagnet. “The operation of a magnetic flowmeter or mag meter is based upon Faraday's Law, which states that the voltage induced across any conductor as it moves at right angles through a magnetic field is proportional to the velocity of that conductor.”

8. A thin plate with a hole in the middle used to measure flow is called _________. 1.An orifice plate 2.A parshall flume 3.A pinhole weir 4.A venturi restriction 1.An orifice plate 2.A parshall flume 3.A pinhole weir 4.A venturi restriction “Orifices are the most popular liquid flowmeters in use today. An orifice is simply a flat piece of metal with a specific-sized hole bored in it. Most orifices are of the concentric type, but eccentric, conical (quadrant), and segmental designs are also available.”

9. The effluent weir of a sedimentation basin should be level in order to prevent: 1.Clogging of the “V notch” 2.Corrosion of the weir material 3.Uneven flows and short circuiting 4.They need not be kept level 1.Clogging of the “V notch” 2.Corrosion of the weir material 3.Uneven flows and short circuiting 4.They need not be kept level

10. What calibrated device developed for measuring flow in an open channel consists of a contracting length, a throat with a sill, and an expanding length? 1.An orifice plate 2.A Parshall flume 3.A v-notched weir 4.A venturi restriction 1.An orifice plate 2.A Parshall flume 3.A v-notched weir 4.A venturi restriction

11. The difference in pressure between high- and low-pressure taps is proportional to the square of the flow rate through the Venturi. Therefore, a differential-pressure sensor with a square root output signal can be used to indicate flow. 1.True 2.False 1.True 2.False

12. A centrifugal untreated raw water pump starts pumping at 25 gal/min and has a maximum pumping capacity of 100 gal/min. A Venturi flowmeter can be used to measure flow from this pump. 1.True 2.False 1.True 2.False

13. Venturi flowmeters can measure flow when partially full of liquid. 1.True 2.False

14. Carrying Capacity Carrying Capacity = (D2) 2 (D1) 2 Carrying Capacity = (D2) 2 (D1) 2 Capacity ratio = (new pipe diameter) 2 (old pipe diameter) 2 Capacity ratio = (new pipe diameter) 2 (old pipe diameter) 2 Capacity ratio = (Big pipe diameter) 2 (Little pipe diameter) 2 Capacity ratio = (Big pipe diameter) 2 (Little pipe diameter) 2

15. Carrying Capacity Capacity ratio = (D2) 2 (D1) 2 Capacity ratio = (D2) 2 (D1) 2 Capacity ratio = (12 in) 2 (6 in) 2 Capacity ratio = (12 in) 2 (6 in) 2 Capacity ratio = 144 in 2 36 in 2 Capacity ratio = 144 in 2 36 in 2 Capacity ratio = 4 times more A = 0.785 (Diameter) 2 ; Q= VA or V=Q/A Assuming the same flow rate and velocity. A 12 inch pipe carries how much more water then a six inch pipe?

16. When the flow rate increases (Q) the flow velocity increases (V) and so does the friction or resistance to flow caused by the liquid viscosity and the head loss 1.True 2.False Q = V A

17. Carrying Capacity When the inside diameter is **made larger** the flow area increases and the liquid velocity and head loss for a given capacity is reduced When the inside diameter is made smaller the flow area decreases and the liquid velocity and head loss for a given capacity is increased

18. Determine the relationship between carrying capacity and flow rate. What is the velocity (ft/min) in a pipe that is 12 inches in diameter and currently has a flow rate of 50 gal/min (gpm)? 1.8.5 FT/MIN 2.5.2 FT/MIN 3.39.2 Ft/Min 4.64 Ft/MIN 1.8.5 FT/MIN 2.5.2 FT/MIN 3.39.2 Ft/Min 4.64 Ft/MIN DRAW: Given: D1= 1ft ; Q= 50 gpm conversions: (1ft 3 /7.48 gal) Formula: A = 0.785 (Diameter) 2 ; Q/A= V Solve:Q= 50 gal/min (1ft 3 /7.48 gal)=6.68 ft 3 /min A = 0.785 (Diameter) 2 A = 0.785 (1ft) 2 A= 0.785 (1ft 2 ) A= 0.785 ft 2 Q/A= V V= (6.68FT 3 /MIN)/(0.785 FT 2 )= 8.5 FT/MIN DRAW: Given: D1= 1ft ; Q= 50 gpm conversions: (1ft 3 /7.48 gal) Formula: A = 0.785 (Diameter) 2 ; Q/A= V Solve:Q= 50 gal/min (1ft 3 /7.48 gal)=6.68 ft 3 /min A = 0.785 (Diameter) 2 A = 0.785 (1ft) 2 A= 0.785 (1ft 2 ) A= 0.785 ft 2 Q/A= V V= (6.68FT 3 /MIN)/(0.785 FT 2 )= 8.5 FT/MIN

19. Determine the relationship between carrying capacity and flow rate. What is the velocity (ft/min) in a pipe that is 4 inches in diameter and currently has a flow rate of 50 gal/min (gpm)? DRAW: Given: D1= 4”=0.33ft;Q= 50 gpm Conversions: (1ft 3 /7.48 gal) Formula: A = 0.785 (Diameter) 2 ; Q/A= V Solve:Q= 50 gal/min (1ft 3 /7.48 gal)=6.68 ft 3 /min A = 0.785 (Diameter) 2 A = 0.785 (.33ft) 2 A= 0.785 (.11ft 2 ) A= 0.085 ft 2 Q/A= V V= (6.68FT 3 /MIN)/(0.085 FT 2 )= 78.6 FT/MIN DRAW: Given: D1= 4”=0.33ft;Q= 50 gpm Conversions: (1ft 3 /7.48 gal) Formula: A = 0.785 (Diameter) 2 ; Q/A= V Solve:Q= 50 gal/min (1ft 3 /7.48 gal)=6.68 ft 3 /min A = 0.785 (Diameter) 2 A = 0.785 (.33ft) 2 A= 0.785 (.11ft 2 ) A= 0.085 ft 2 Q/A= V V= (6.68FT 3 /MIN)/(0.085 FT 2 )= 78.6 FT/MIN 1.4.25FT/MIN 2.0.58 FT/MIN 3.588 FT/Min 4.79 FT/MIN 1.4.25FT/MIN 2.0.58 FT/MIN 3.588 FT/Min 4.79 FT/MIN

20. Assuming both are flowing full at the same FLOW RATE (Q). The velocity in a 4 inch pipe relative to a 12 inch pipe is????? A 12 in pipe with a Q of 50 (gpm) has a velocity of 8.5 ft/min. A smaller 4 inch pipe with the same Q (50 gpm) has a velocity of 79 ft/min. Thus water is moving (79/8.5= 9 times faster). 1.~9 times faster 2.~3 times faster 3.~632 times faster 4.The same rate

21. The flow velocity in a 6-in. diameter pipe is twice that in a 12-in diameter pipe if both are carrying 50 gal/min of water. 1.True 2.False V= Q/A = 50 gpm/.785 = 64 V=Q/A = 50 gpm/0.19 = 255 Decreasing the pipe diameters increases the flow velocity if all else is held equal. Going from a 12 inch to a 6 inch pipe speeds up the water 4 times. V= Q/A = 50 gpm/.785 = 64 V=Q/A = 50 gpm/0.19 = 255 Decreasing the pipe diameters increases the flow velocity if all else is held equal. Going from a 12 inch to a 6 inch pipe speeds up the water 4 times.

22. “The bigger the pipe the more water it can carry. Increase the pipe size increase the carrying capacity. For a double in pipe size you increase its carrying capacity 4 fold.” “If two pipes have the same flow rate (Q) the smaller diameter pipe has a faster flow velocity (V). You are moving the same flow volume of (Q) water through a smaller hole so it goes faster.”

23. Increasing thisTo thisIncreases the capacity pipe diameterdiameterby a factor of (inches) 462.25 484.00 681.78 6102.78 6124.00 8101.56 8122.25 8153.52 10121.44 10152.25 12151.56

24. Job Interview Clean Water Service ?: “A 12 inch pipeline is flowing full of water and is necked down to a four inch pipeline, does the flow velocity of the water in the 4 inch line increase or decrease? 1.Increases 2.Decreases 3.Flow is not impacted 1.Increases 2.Decreases 3.Flow is not impacted

25. Job Interview Clean Water Service ?: “A 12 inch pipeline is flowing full of water and is necked down to a four inch pipeline, does the velocity of the water in the 4 inch line increase or decrease and by a factor of ________________ 1.Increases, 9 fold 2.Decreases it 9 fold 3.Flow is not impacted 1.Increases, 9 fold 2.Decreases it 9 fold 3.Flow is not impacted

26. Job Interview Clean Water Service ?: “You need to replace a 4 inch sewer pipe with a 6 inch sewer pipe. If velocity is the same in both pipes the new pipe will be able to carry 2.25 times as much material.” 1.True 2.False 3.Cannot determine with the info given. 1.True 2.False 3.Cannot determine with the info given.

27. DRAW: Given: D1= 1ft ; (CC or CR)=2; D2=? Formula: Solve: DRAW: Given: D1= 1ft ; (CC or CR)=2; D2=? Formula: Solve: A 12 in water main must be replaced with a new main that has double the carrying capacity. What is the diameter of the new main, rounded to the nearest inch? D1=12 in =1 ft D1=12 in =1 ft Capacity ratio = D 2 2 /D 1 2 D 1 2 (CR)=D 2 2 D 1 2 (2)=D 2 2 (12in) 2 (2)=D 2 2 144in 2 (2)=D 2 2 288 in 2 =D 2 2 √288 in 2 =D 16.97 inches =D Capacity ratio = D 2 2 /D 1 2 D 1 2 (CR)=D 2 2 D 1 2 (2)=D 2 2 (12in) 2 (2)=D 2 2 144in 2 (2)=D 2 2 288 in 2 =D 2 2 √288 in 2 =D 16.97 inches =D Old New D2= ?? CR=2 1.12 inches 2.15 inches 3.17 inches 4.24 inches 1.12 inches 2.15 inches 3.17 inches 4.24 inches

28. Definitions Continuity rule states that flow (Q) entering a system must equal flow that leaves a system. Q 1 =Q 2 Or A 1 V 1= A 2 V 2 Flow of water in a system is dependant on the amount of force causing the water to move. Pressure is the amount of force acting (pushing) on a unit area. Continuity rule states that flow (Q) entering a system must equal flow that leaves a system. Q 1 =Q 2 Or A 1 V 1= A 2 V 2 Flow of water in a system is dependant on the amount of force causing the water to move. Pressure is the amount of force acting (pushing) on a unit area.

29. Example 9. Different diameter pipe & velocities (ft/time) If the velocity in the 10 in diameter section of pipe is 3.5 ft/sec, what is the ft/sec velocity in the 8 in diameter section? D=diameter (10 inches) Convert! (10in)(1ft/12in) D=0.83 ft D=diameter (10 inches) Convert! (10in)(1ft/12in) D=0.83 ft V 1 = 3.5 ft/sec A 1 V 1 =A 2 V 2 V 2 = A 1 V 1 /A 2 = (0.54ft 2 )(3.5 ft/sec)/(0.35ft 2 ) =5.37 ft/sec A 1 V 1 =A 2 V 2 V 2 = A 1 V 1 /A 2 = (0.54ft 2 )(3.5 ft/sec)/(0.35ft 2 ) =5.37 ft/sec V 1 = 3.5 ft/sec d1=10 in Q 1 = Q 2 and A 1 V 1 =A 2 V 2 Pipe Area = 0.785 (diameter) 2 Area 1 (pipe)= 0.785 (0.833ft) 2 = 0.54 ft 2 Area 2 (pipe)= 0.785 (0.67ft) 2 = 0.35 ft 2 Q 1 = Q 2 and A 1 V 1 =A 2 V 2 Pipe Area = 0.785 (diameter) 2 Area 1 (pipe)= 0.785 (0.833ft) 2 = 0.54 ft 2 Area 2 (pipe)= 0.785 (0.67ft) 2 = 0.35 ft 2 V 2 = ?ft/sec d2=8 in D=diameter (8 inches) Convert! (8in)(1ft/12in) D=0.67 ft D=diameter (8 inches) Convert! (8in)(1ft/12in) D=0.67 ft V 2 = ? ft/sec

30. Example 10. Different flows & Continuity Rule (ft 3 /time) A flow entering the leg of a tee connection is 0.25 m 3 /sec. If the flow is 0.14 m 3 /sec in one branch what is the flow through the other branch? Q 1 = Q 2 + Q 3 Q 3 = Q 1 – Q 2 Q 3 =0.25 m 3 /sec- 0.14 m 3 /sec Q 3 =0.11 m 3 /sec Q 1 = Q 2 + Q 3 Q 3 = Q 1 – Q 2 Q 3 =0.25 m 3 /sec- 0.14 m 3 /sec Q 3 =0.11 m 3 /sec Q 1 = 0.25 m 3 /sec Q 2 = 0.14 m 3 /sec Q 3 = ? m 3 /sec CR-states that flow (Q) entering a system must equal flow that leaves a system.

31. Example 11. Different velocities & Continuity Rule (ft/time) Determine the velocities at the different points (A,B, and C) in ft/sec. Q a = Q b + Q c Q c = Q a – Q b Q c =910 gpm- 620 gpm Q c =290 gpm Q a = Q b + Q c Q c = Q a – Q b Q c =910 gpm- 620 gpm Q c =290 gpm CR-states that flow (Q) entering a system must equal flow that leaves a system. Q 2 = 0.14 m 3 /sec A A B B C C dB=4 in DB=diameter (4 inches) Convert! (4in)(1ft/12in) DB=0.33 ft DB=diameter (4 inches) Convert! (4in)(1ft/12in) DB=0.33 ft V=620 gpm V=??? gpm V=910 gpm dA=6 in DA=diameter (6 inches) Convert! (6in)(1ft/12in) DA=0.5 ft DA=diameter (6 inches) Convert! (6in)(1ft/12in) DA=0.5 ft DC=diameter (3 inches) Convert! (3in)(1ft/12in) DC=0.25 ft DC=diameter (3 inches) Convert! (3in)(1ft/12in) DC=0.25 ft dC=3 in

32. Example 11. Different velocities & Continuity Rule (ft/time) Determine the velocities at the different points (A,B, and C) in ft/sec. Convert gpm to ft 3 /sec Q a =910 gpm (1min/60 sec)(1 gal/7.48 ft 3 )= 2.03 ft 3 /sec Q b = 620 gpm(1min/60 sec)(1 gal/7.48 ft 3 )= 1.38 ft 3 /sec Q c =290 gpm(1min/60 sec)(1 gal/7.48 ft 3 )= 0.65 ft 3 /sec Convert gpm to ft 3 /sec Q a =910 gpm (1min/60 sec)(1 gal/7.48 ft 3 )= 2.03 ft 3 /sec Q b = 620 gpm(1min/60 sec)(1 gal/7.48 ft 3 )= 1.38 ft 3 /sec Q c =290 gpm(1min/60 sec)(1 gal/7.48 ft 3 )= 0.65 ft 3 /sec CR-states that flow (Q) entering a system must equal flow that leaves a system.

33. Example 11. Different velocities & Continuity Rule (ft/time) Determine the velocities at the different points (A,B, and C) in ft/sec. CR-states that flow (Q) entering a system must equal flow that leaves a system. Q 2 = 0.14 m 3 /sec A A B B C C dB=4 in DB=diameter (4 inches) Convert! (4in)(1ft/12in) DB=0.33 ft DB=diameter (4 inches) Convert! (4in)(1ft/12in) DB=0.33 ft V=620 gpm V=??? gpm V=910 gpm dA=6 in DA=diameter (6 inches) Convert! (6in)(1ft/12in) DA=0.5 ft DA=diameter (6 inches) Convert! (6in)(1ft/12in) DA=0.5 ft DC=diameter (3 inches) Convert! (3in)(1ft/12in) DC=0.25 ft DC=diameter (3 inches) Convert! (3in)(1ft/12in) DC=0.25 ft dC=3 in Pipe Area = 0.785 (diameter) 2 Area a (pipe)= 0.785 (0.5ft) 2 = 0.19 ft 2 Area b (pipe)= 0.785 (0.33ft) 2 = 0.09 ft 2 Areac (pipe)= 0.785 (0.25ft) 2 = 0.05 ft 2 Pipe Area = 0.785 (diameter) 2 Area a (pipe)= 0.785 (0.5ft) 2 = 0.19 ft 2 Area b (pipe)= 0.785 (0.33ft) 2 = 0.09 ft 2 Areac (pipe)= 0.785 (0.25ft) 2 = 0.05 ft 2

34. Example 11. Different velocities & Continuity Rule (ft/time) Determine the velocities at the different points (A,B, and C) in ft/sec. Solve Q=VA at Each Point V a =Q a/ A a =2.03 ft 3 /sec/ (0.19 ft 2 )=10.34 ft/sec V b = Q b /A b = 1.38 ft 3 /sec/ (0.09 ft 2 )= 16.14 ft/sec V c = Q c /A c = 0.65 ft 3 /sec/ (0.05 ft 2 )= 13.25 ft/sec Solve Q=VA at Each Point V a =Q a/ A a =2.03 ft 3 /sec/ (0.19 ft 2 )=10.34 ft/sec V b = Q b /A b = 1.38 ft 3 /sec/ (0.09 ft 2 )= 16.14 ft/sec V c = Q c /A c = 0.65 ft 3 /sec/ (0.05 ft 2 )= 13.25 ft/sec CR-states that flow (Q) entering a system must equal flow that leaves a system.

35. What is the Continuity Equation? Flow in = flow out Q 1 = Q 2 and A 1 V 1 =A 2 V 2 Q 1 = Q 2 + Q 3

36. Syllabus Objective: Flowmeters, Flow rates and the continuity equation were discussed this evening? 1.Strongly Agree 2.Agree 3.Neutral 4.Disagree 5.Strongly Disagree 1.Strongly Agree 2.Agree 3.Neutral 4.Disagree 5.Strongly Disagree