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7154 VFD Presentation #4 May 2002 Paul Weingartner.

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Presentation on theme: "7154 VFD Presentation #4 May 2002 Paul Weingartner."— Presentation transcript:

1 7154 VFD Presentation #4 May 2002 Paul Weingartner

2 Load types Standard motors High instantaneous current Lumber cutting example Lumber cutting example Fans, pumps

3 Utility billing Energy cost Demand charge Power factor (can be)

4 Energy cost (approx) Residential $0.07/kWh Commercial $0.06/kWh Industrial $0.05/kWh

5 Real-time pricing vs. Demand You can be billed either on a standard Demand type arrangement or, Pay the real-time demand cost

6 Real-time cost Supply and demand set the cost of power. The typical cost of $0.06/kWh can skyrocket to $4-5/kWh for short periods $1/kWh for hours

7 Energy rebate programs For larger customers that are willing to shed an agreed upon number of kWh during peak times, if requested by Cinergy to do so, will get a credit for energy they DON’T USE.

8 Demand

9 Demand charges

10 Demand charges 2

11 Rachet – min bill

12 Summer period

13 Look at what happened to California last year

14 Demand kWh

15 Power factor

16 Yearly billing info

17 Pump application w/o VFD Without a VFD, a pump usually runs at 100% speed Flow is controlled with a control valve A flow controller reads the flow rate from a flow meter and compares it against the setpoint

18 Runs at 100% The pump motor is running full speed Lots of wasted electricity $$$$

19 Assumptions – Assume __ That the motor in a non-VFD pumping application is using 75% of the nameplate power That Demand charges are not applicable Power factor is not an issue All cost will be Energy alone Motor efficiency =92%

20 Motor efficiency

21 5 HP example – non VFD 5 HP x 746 W/HP = 3730 W 3730W / 0.92 (eff) = 4054 W 4054 W x 0.75 = 3040 W 3.040 kW * 8760 hrs = 26630 kWh 26630 kWh * $0.06/kWh = $1597

22 $1597 Almost $1600/year in energy cost for 7 x 24 operation

23 What is we used a low eff motor? Let the eff = 83%, then 3730 W / 0.83 = 4493 W 4493 W x 0.75 = 3370 W 3.37 kW x 8760 hrs = 29,525 kWh 29525 kWh x $0.06 = $1771/year Diff in cost is $174/year

24 5 HP eff motor Premium Efficiency Motor, 5 horsepower, 1755 RPM, 230/460 volts Premium Efficiency Motor, 5 horsepower, 1755 RPM, 230/460 volts $448.25

25 5 HP std motor General Electric Definite Purpose Motor, Rigid Base, Fan-Cooled, 3-Phase, Totally Enclosed, C-Face, 5 horsepower $381.25

26 Difference in cost between Std & Hi Eff $67

27 Pump Example with VFD Don’t need a control valve = -$1000 Need a VFD = +$500 Need a high eff, inverter rated motor = +$150

28 Less maintenance Control valves require more maintanence Hard to tell without added hardware how many strokes the valve has made Seals, etc

29 Less maintenance 2 VFDs provide diagnostics when problems occur

30 Energy cost Assume that if total usage is averaged that motor uses 33%

31 Fan basics Affinity law When the impeller diameter of a centrifugal pump is held constant the effect of changing the speed (RPM) of the pump is in accordance with the following: When the impeller diameter of a centrifugal pump is held constant the effect of changing the speed (RPM) of the pump is in accordance with the following: Capacity: Q1/Q2 = N1/N2 Capacity: Q1/Q2 = N1/N2 Head: H1/H2 = (N1/N2)2 Head: H1/H2 = (N1/N2)2 BHP: BHP1/BHP2 = (N1/N2)3 BHP: BHP1/BHP2 = (N1/N2)3 Q = Capacity, GPM Q = Capacity, GPM H = Head, Feet H = Head, Feet BHP = Brake Horsepower BHP = Brake Horsepower N = Pump Speed,RPM N = Pump Speed,RPM

32 Upshot Torque increases as a square of speed HP increases as the cube

33 Or look at it the other way If you reduce speed by 10% - new speed = 0.9 of old speed Torque = square (0.9) = ___ HP = cube (0.9) = ___

34 Centrifugal pumps http://www.pricepump.com/pumpschool/ps les.html http://www.pricepump.com/pumpschool/ps les.html


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