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PUMPING 101 – TASK 2

PUMPING 101 – TASK 2. AFFINITY LAWS, FRICTION HEAD AND SYSTEM CURVES. PUMPING 101 – UNIT 2. PUMP AFFINITY LAWS. Efficiency. Pump Operation. Constant speed example. Pump Affinity Laws.

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PUMPING 101 – TASK 2

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  1. PUMPING 101 – TASK 2 AFFINITY LAWS, FRICTION HEAD AND SYSTEM CURVES

  2. PUMPING 101 – UNIT 2 PUMP AFFINITY LAWS

  3. Efficiency Pump Operation Constant speed example

  4. Pump Affinity Laws • FLOW changes DIRECTLY as a change in speed or diameter* • HEAD changes as the SQUARE of a change in speed or diameter* • HORSEPOWER changes as the CUBE of a change in speed or diameter * May not be true for higher specific speeds

  5. Pump Affinity Laws Important… Remember these:

  6. Pump Affinity Laws FLOW HEAD POWER %SPEED

  7. Efficiency Impeller Diameter Change Impact on Efficiency

  8. PUMPING 101 – TASK 2 SYSTEM FRICTION

  9. Friction of Water Velocity Chart and Friction of Water (new steel pipe) at 60° F 1 inch Source: Grundfos Technical Guide L-TG-PG-001

  10. Friction of Water Asphalt-dipped Cast Iron and New Steel Pipe (Based on Darcy’s Formula) 8 inch Pipe aging?

  11. Friction in Fittings Velocity Chart & Friction of Water Source: Grundfos Technical Guide L-TG-PG-001 The friction loss through one 1¼ inch standard 90° elbow is equal to the friction loss through how many feet of straight 1¼ inch pipe? These are NOT friction values!!! 3.6 ft

  12. S D Friction Head 17 ft. 10 gpm thru 250 ft – 1” Sched. 40 steel pipe 250 100 6.81 17 • What is the friction head in feet? • What is the pressure head in feet? • What is the elevation head in feet? • What is the total head? (4 = 1+2+3) 0 ft. 0 ft. 17 ft.

  13. Calculating Horsepower

  14. Where do we get “3960”? 1HP = 550Foot Pounds per Second x 60 Seconds per Minute 33,000Foot Pounds per Minute ÷ 8.333 Pounds per Gallon of Water 3960

  15. Practice Problems What is WHP(P3) for a pump moving 200 gpm of 60°F water against a TDH of 500’? (200 gpm X 500 feet) ÷ 3960 = 25.25 WHP

  16. Practice Problems What is BHP(P2) if the efficiency of the pump is 83%? 25.25 WHP ÷ 0.83 = 30.42 BHP

  17. Practice Problems What is EHP(P1) if the efficiency of the motor is 90%? 30.42 BHP ÷ 0.90 = 33.81 EHP(WWE)

  18. Practice Problems What is kW value if the WWE of the pump is 33.81? 33.81 WWE X 0.746 = 25.21 kW

  19. Practice Problems Calculate Energy Cost: 25.21 kW x 1000 hours per year 25,210kW/hrs per year x $0.10 per kWh $2,521.00 cost per year

  20. PUMPING 101 – TASK 2 SYSTEM CURVES

  21. System Curves Open System w/ Static Head Statichead Hf Totalhead System Head (Hspf) HEAD (FT.) OR PSI Hp Suctionlift Hs FLOW (GPM)

  22. Creating a System Curve

  23. Creating a System Curve Graph the head required through point for 50’ Equivalent Length of 2” Type L Copper Tubing for the following flows: .145 0.58 1.31 • 2.32 • • 3.65 • • •

  24. Creating a System Curve • Graph the head required through point for 200’ Equivalent Length of 2” Type L Copper Tubing for the following flows: .58 • 2.32 5.14 • 9.28 • 14.60 • •

  25. Creating a System Curve

  26. System and Pump Curves Operating Point • Operating point • Duty point • Options Duty Point

  27. Questions What are the Pump Affinity Laws? FLOWchanges DIRECTLY as a change in speed or diameter HEAD changes as the SQUARE of a change in speed or diameter HORSEPOWERchanges as the CUBEof a change in speed or diameter

  28. Questions What IS WWE? WWE is a calculation of the efficiency of pumping a liquid including corrections for Specific Gravity, Pump Efficiency, Motor Efficiency, and VSD Efficiency (if included) WWE = FlowxHead xSP GR / 3960 xEpxEmxEvsd

  29. www.grundfos.com

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