Question 10.6: Determine the cost-effectiveness of using a unit with an eff......
Determine the cost-effectiveness of using a unit with an efficiency of 98.50% rather than 98.95% (NEMA Premium) for a 225 kVA rated transformer. Assume that
• The cost of electricity is $0.12/kWh
• The installed costs of 98.5% and 98.95% efficient transformers are $5950 and $7100, respectively
• The average power factor of the load 0.90
For the analysis, consider that the transformer is used on average at 75% of its capacity for 8760 \ h/year
TABLE 10.10
Comparison of Energy Efficiency Values for Baseline, Department of Energy Regulations, and National Electrical Manufacturers Association Premium Classifications
| NEMA Premium Efficiency (%)^c | Efficiency DOE Regulations (%)^b | Baseline Efficiency (%)^a | Rating (kVA) |
| 97.90 | 97.89 | 97 | 15 |
| 98.25 | 98.23 | 97.50 | 30 |
| 98.23 | 98.40 | 97.70 | 45 |
| 98.60 | 98.60 | 98.80 | 75 |
| 98.74 | 98.74 | 98.20 | 112.5 |
| 98.81 | 98.83 | 98.30 | 150 |
| 98.95 | 98.94 | 98.50 | 225 |
| 99.02 | 99.02 | 98.60 | 300 |
| 99.09 | 99.14 | 98.70 | 500 |
| 99.16 | 99.23 | 98.80 | 750 |
| 99.23 | 99.28 | 98.90 | 1000 |
| ^a These values are based on NEMA TP-1 2002.
^b These values are based on DOE regulations of final rule of 10 CRR 431 effective January 2016. ^c These values are based on NEMA EL-3, CSL-3 |
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TABLE 10.11
Comparison of Typical Energy Efficiency Values and Losses for 1973 and 2013 Transformers
| 2013 Transformers (Full Load) | 1973 Transformers (Full Load) | Rating (kVA) | ||
| Total Losses (W) | Energy Efficiency (%) | Total Losses (W) | Energy Efficiency (%) | |
| 551 | 96.8 | 740 | 95.3 | 15 |
| 904 | 97.2 | 1250 | 96 | 30 |
| 1027 | 97.9 | 1620 | 96.6 | 45 |
| 1782 | 97.7 | 2570 | 96.8 | 75 |
| 2521 | 97.9 | 1400 | 97.6 | 112.5 |
| 2760 | 98.3 | 1830 | 97.6 | 150 |
| 4047 | 98.3 | 4775 | 98 | 225 |
| 5338 | 98.9 | 5400 | 98.3 | 300 |
| 5858 | 98.9 | 8300 | 98.4 | 500 |
To determine the cost-effectiveness of installing an energy-efficient transformer, a simplified economic analysis based on the payback period is used. The savings in energy losses in kWh for the high-efficiency transformer can be calculated using Equation 10.6:
kWh_{saved}=N_h\cdot kVA\cdot LF\cdot pf\cdot \left\lgroup\frac{1}{\eta _{std}}-\frac{1}{\eta _{eff}} \right\rgroup (10.6)
kWh_{saved}=8760*225*0.90*0.75*\left\lgroup\frac{1}{0.9850}-\frac{1}{0.9895} \right\rgroup =6143 \ kWh/yearTherefore, the payback period, SPP, for using the efficient transformer is
SPP=\frac{\$7100-\$5950}{6143 \ kWh*\$0.12/kWh}=1.56 \ yearsNote, for retrofit application, and if the existing one is operating normally (and thus does not need to be replaced), the payback would be estimated as follows:
SPP=\frac{\$7100}{6143 \ kWh*\$0.12/kWh}=9.63 \ yearsIt is clear that it can be cost-effective to consider investing in a more energy-efficient transformer for new constructions. However, it may not be cost-effective to replace an existing and functional transformer for retrofit applications.