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Now that we have examined the thermal side and cooling system of cooling tower economy. Energy consumption in mechanical draft cooling towers is confined to the fan and pumps. Minimizing annual energy consumption in the fan drive equipment is achieved not only by proper specifications, but also by prudent operation.

  • ●  Cooling water bypass (manual)
  • ●  On-off fan operation (manual)
  • ●  Louvers or shutters (automatic)
  • ●  High efficiency motors
  • ●  Two-speed fan speed (manual or automatic)
  • ●  Variable speed fans (automatic)
  • ●  Variable pitch fans (automatic)

The several major different methods to drive the fans shall be discussed hereto. The oldest and simplest is probably the cooling water bypass method. This is merely bypassing the thruput if the cooling water is overcooling. On-off fan control is simple and is often used if there are a large number of fans in an identical service. Automatic louvers are the first step to modulated air flow. The only problem is that the fan horse-power is wasted as the flow is throttled by the louver. At complete shut-off the fan is stalled and horsepower actually increases. The details for the other methods are described in below.

1) Efficiency Motor

Using a high efficient motor is a simplest way to be able to reduce energy consumption. The motors called "High Efficiency", "Energy Saving Motor", or "Premium Energy Efficient Motors" were introduced in the late 1970's. The technology and manufacturing changes during that time were allowed to increase and ongoing the efficiencies.

Figure graphically depicts a 89% efficient 175 HP motor coupled to a fan load requiring 175 HP (130 KW) of mechanical input power. The total electric input power is 146.7 KW of which 146.7 KW is converted to mechanical power required by the fan load. This motor actually consumes 16.7 KW of power in the process converting the electrical energy to the mechanical energy to do useful work.

The emphasis on motor efficiency and the energy conservation potential has spawned new lines of "Energy Efficient" motors. These premium efficiency motors offer substantial reduction in motor losses and, when applied, can reduce the industrial power consumption significantly. In most cases, the motor's initial price premium can be totally justified in energy savings and, in many cases using total life cycle costing considerations, the entire motor cost, not just the initial price premium, can be justified.

Useful cost justification methods associated with the application of the energy efficient motors shall be discussed below for users of electric motors.

(1) Efficiency: Motor efficiency is a measure of the effectiveness with which a motor converts electrical energy to mechanical energy. It is defined as the ratio of power output to power input or, in terms of electrical power, watts output to watts input and can be restated as the ratio of output to output + losses. The difference-watts loss - is due to electrical losses plus friction and windage. Even though higher horsepower motors are typically more efficient, their losses are significant and should not be ignored. In fact, higher horsepower motors offer the greatest savings potential for the least analysis effort, since just one motor can save more energy than several smaller motors.

(2) Watts Loss: Every AC motor has five components of watts losses which are the reasons for its inefficiency. Watts losses are converted into heat which is dissipated by the motor frame aided by internal or external fans. Stator and rotor I2r losses are caused by current flowing through the motor winding and are proportional to the current squared times the winding resistance (I2r). Iron losses are mainly confined to the laminated core of the stator and rotor and can be reduced by utilizing steels with low core loss characteristics found in high grade silicon steel. Friction and windage loss is due to all sources of friction and air movement in the motor and may be appreciable in large high-speed or totally enclosed fan-cooled motors. The stray load loss is due mainly to high frequency flux pulsations caused by design and manufacturing variations.

(3) Improvements of Watts Losses: Improvements in motor efficiency can be achieved without compromising motor performance - at higher cost - within the limits of existing design and manufacturing technology.

Watts Loss Area

Efficiency Improvement

1. Iron

Use of thinner gauge, lower loss core steel reduces eddy current losses. Longer core adds more steel to the design, which reduces losses due to lower operating flux densities.

2. Stator I2R

Use of more copper and larger conductors increases cross sectional area of stator windings. This lowers resistance (R) of the windings and reduces losses due to current flow (I)

3. Rotor I2R

Use of larger rotor conductors bars increases size of cross section, lowering conductor resistance (R) and losses due to current flow (I)

4. Friction & Windage

Use of low loss fan design reduces losses due to air movement.

5. Stray Load Loss

Use of optimized design and strict quality control procedures minimizes stray load losses.

(4) Evaluating Motor Efficiency: Today's motor user faces the evaluation of higher efficiency motors at premium prices with standard designs. In many applications, an initial purchase price premium can be justified based on energy cost savings. The basis for this justification depends on the individual user's situation. Factors such as running hours, cost of electricity, payback period, cost of capital and service life affect the premium price justifications, and will vary with the individual user.

There are two methods for user determination as to how much additional capital investment or price premium is justified per kilowatt of power load reduction. If, for example, an additional capital investment of $1,000/KW is determined, the motor user would be justified in paying up to a $1,000 price premium for every KW saved because of the watts loss reduction by using a more efficient motor. An example of the power or watts loss reduction provided by a higher efficiency 175 HP motor follows:

  • ●  Watts Losses = Input - Output = (HP x 0.746 / Motor Efficiency) - (HP x 0.746)
  • ●  Watts Losses with 89% efficiency motor = (175 x 0.746 / 0.89) - (175 x 0.746) = 16.135 KW
  • ●  Watts Losses with 95% premium efficiency motor = (175 x 0.746 / 0.95) - (175 x 0.746) = 6.871 KW

Operating the 95% efficiency motor vs. the 89% efficiency motor thus saves 9.264 KW. In other words, the power loss reduction is 9.264 Kilowatts. Kilowatt saved can be calculated using a simple formula below.

Kilowatts Saved = HP x 0.746 [ (1 / Standard Motor Efficiency) - (1 / Premium Motor Efficiency) ]

The method of calculating the additional capital investment to justify the price premium per KW of loss reduction are shown next:

  • ●  Simple Payback: The first method is a simple payback analysis where the user specifies how long he is willing to wait for his after tax power cost savings to equal the premium paid. The user must specify: P = Justified price premium/KW saved = HRS/YR x $/KWH x YRS x (1-T)

    Where,
    HRS/YR: hours of motor operation per year
    $/KWH: Cost of electricity (dollars or Korean Won per kilowatt hour)
    YRS: Maximum acceptable years to payback or break even
    T: Tax rate

    A motor user that averages three - 8 hour shifts per day; 6 days a week; 50 weeks per year has a power cost of $0.46/KWH and considers two years breakdown or payback period and has a tax rate of 22.5%.

    Then,
    P = HRS/YR x $/KWH x YRS x (1-T)
    = (8 x 3 x 6 x 50) x $0.46 x 2 x (1 - 0.225) = $5,133.60

    This user could justify up to a $5,133.60 price premium for a motor that saved or reduced the KW load by 1 KW. Using the two alternative efficiencies in the previous 175 HP motor example, this user could justify paying an initial price premium for the high efficiency motor of up to $47,557.7. (P x loss reduction = $5,133.60 x 9.264 KW = $47,557.67)

    There are three basic components of industrial power cost: cost of Real Power used; power factor penalties and demand charges. To understand these three charges and how they are determined, a review of the power vector diagram identifies each component of electrical energy and its corresponding energy charge.
  • ●  Real Power: The Real Power-KW is the energy consumed by the load. Real Power-KW is measured by a watthour meter and is billed at a given rate ($/KWH). It is the Real Power component that performs the useful work and which is affected by motor efficiency.
  • ●  Power Factor: Power factor is the ratio of Real Power-KW to Total KVA. Total KVA is the vector sum of the Real Power and reactive KVAR. Reactive power is required to support the magnetic field of the induction motor. Although Reactive KVAR performs no actual work, an electric utility must maintain an electrical distribution systems, (i.e. power transformers, transmission lines, etc.) to accommodate this additional electrical energy. To recoup this cost burden, utilities may pass this cost on to industrial customers in the form of a power factor penalty, for power factor below a certain value.

    Power factors in industrial plants are usually low due to the inductive or reactive nature of induction motors, transformers, lighting and certain other industrial process equipment. Low power factor is costly, and requires an electric utility to transmit more total KVA than would be required with an improved power factor. Low power factor also reduces the amount of Real Powerthat a plant's electrical distribution system can handle, and increased line currents will increase losses in a plant's distribution system.

    A method to improve power factor, that is typically expensive, is to use a unity or leading power factor synchronous motor or generator in the power system. A less expensive method is to connect properly sized capacitors with induction motors provides lower first cost and reduced maintenance expense. Below figure graphically shows how to the total KVA vector approaches the size of Real Power vector as Reactive KAVR is reduced by corrective capacitors. Because of Power Factor correction, less power need be generated and distributed the same amount of useful energy to the motor.

    Just as the efficiency of an induction motor may be reduced as its load decreases, the same is true for the power factor only at a faster rate of decline. A typical 10 HP, 1800 RPM, 3-phase, NEMA Design B motor with a full load power factor of about 80 percent decreases to about 65 percent as half-load. Therefore, it is important not to oversize motors. Select the right size motor for the right job.
  • ●  Demand Charges: The third energy component affecting cost is demand charge and is based on the peak or maximum power consumed or demanded by an industrial customer during a specified time interval. Because peak power demands may require an electric utility to increase generating equipment capacity, a penalty is assessed when demand exceeds a certain level. This energy demand is measured by a demand meter and a multiplier is applied to the Real Power KW consumed. Cooling towers with varying load requirements may be able to affect demand charges by: 1) load cycling - stagger the starting and use of all electrical equipment, and discontinue use during peak power intervals, and 2) use of either electrical or mechanical "soft start" hardware which limits power in rush and permits a gradual increase in power demand.

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