FIJACIÓN Y COBRO DE HONORARIOS

10.1 An isolated power system serving the community of Cantgettherefromhere uses a 100 kW diesel generator. The community plans to add 60 kW of wind power. The hourly average load and power from the wind turbine over a 24 hour period is detailed in Table B.11. The diesel generator has a no-load fuel usage of 3 liter per hour and an additional incremental fuel use of 1/4 liter per kilowatt-hour.

For the day in question, using the hybrid system design rules, determine:

a) The maximum renewable energy that can possibly be used in an ideal system.

b) The maximum renewables contribution without storage, and the maximum with storage.

c) The maximum fuel savings that can be achieved.

d) The minimum diesel fuel use that can be achieved with intelligent use of storage and controls.

Table B.11 Diesel system load data

Hour Load (kW) Wind (kW) Hour Load (kW) Wind (kW)

0 25 30 12 85 45

1 20 30 13 95 45

2 15 40 14 95 50

3 14 30 15 90 55

4 16 20 16 80 60

5 20 10 17 72 60

6 30 5 18 60 48

7 40 5 19 74 50

8 50 15 20 76 55

9 70 20 21 60 60

10 80 25 22 46 60

11 90 40 23 35 55

SOLUTION

The hybrid system design rules are:

Rule 1: The maximum renewable energy that can be used is limited by the load.

Rule 2: The use of renewable energy will be further limited by temporal mismatch between the load and the renewables.

Rule 3: The maximum possible benefit with improved controls or operating strategies is a system approaching the fuel use of the ideal diesel generator - fuel use proportional to the diesel-served load.

Rule 4: The maximum fuel savings arising from the use of renewables in an optimised system is never greater than the fuel savings of an ideal generator supplying the proportional reduction in load resulting from use of renewables.

a) Based on rule number 1, the maximum renewables contribution with any system cannot be larger than the load. Thus the maximum renewables contribution over the day would be 1338 kWh, the total daily energy need.

b) Based on the rules 1 and 2, the maximum renewables contribution in any hour without storage is the same as the load, if it is smaller than the available wind power, or the available wind power. An examination of the data shows that the maximum renewables contribution without storage would be 819 kWh. The maximum with storage would be the total daily load if it were less than the total available wind power, or the total available wind power. The total daily load is 1338 kWh; the total available wind power over the day is 913 kWh. Thus the maximum renewables contribution with storage is 913 kWh.

c) The total fuel use of the original system is 1338(1/4)+ 24(3) = 406.5 liters/day. The maximum fuel savings occurs with optimum use of storage, controls and renewables.

Under these conditions, the diesel load with would be 1338 – 913 = 425 kWh. The fuel use of an optimised system would be ¼ l/kWh or 106.25 l/day, a savings of 300.25 l/day.

d) Based on rule 3, the minimum diesel fuel use that can be achieved with intelligent use of storage and controls (and without renewables) is that part of the fuel use that is not proportional to the load: 24 hrs of 3 l/hr = 72 l/day.

10.2 Based on the analysis in Problem 10.1 and other input, the community of Cantgettherefromhere has upgraded its power system. The hybrid power system includes a 100 kW diesel generator, 100 kW of installed load, 60 kW of wind power, a 100 kW dump load, and energy storage with 100 kWh capacity. The hourly average load and power from the wind over a 24 hour period have remained the same and are listed in Table B.11.

Assume that the mean hourly data accurately describe the load and power from the wind and that fluctuations about the mean load are handled by the energy storage.

Determine the hourly energy flows in the system for the following system control and operating approaches.

a) Diesel-only system – In this baseline system, the diesel provides all of the power to the load. The diesel provides power down to 0 kW with no provision to limit the ensure a minimum diesel load to avoid diesel engine wear.

b) Minimum diesel – In this system minimizing diesel power takes priority, except that the diesel may not be shut off and must run at a minimum of 30 kW to ensure long diesel engine life. Thus, the system operates by the rules:

1. The minimum diesel power is 30 kW

2. Only the minimum diesel power that is needed above 30 kW is used (no power above 30 kW is used to fill up the storage)

3. Storage can only be used between 20 and 95 kWh of capacity, to maximize storage efficiency and battery life

4. Storage capacity starts at 50 kWh

5. The sum of the energy into the power sources in the system (diesel, wind, battery) must equal the sum of the energy into the power sinks in the system (the load, dump load, and battery).

6. If there is excess energy in the system it is first stored, if possible, and dumped only if necessary.

c) Diesel shut off – In this system a rotary inverter between the batteries and the grid provides reactive power and the diesel can be shut off. It shuts off whenever possible, but when running, needs to be at a minimum load of 30 kW. When running it is also used to fill up the storage to use the fuel most efficiently. The battery level is maintained between 20 and 90 kWh. This insures that there is adequate capacity to handle fluctuating loads when there is no diesel in the system. Thus, the system operates by the rules:

1. The diesel can be shut off

2. The minimum diesel power is 30 kW when the diesel is running

3. When the diesel is running, the storage is also filled up if possible to improve diesel fuel efficiency

4. Storage can only be used between 20 and 90 kWh of capacity, to maximize storage efficiency and battery life

5. Storage capacity starts at 50 kWh

6. The sum of the energy into the power sources in the system (diesel, wind, battery) must equal the sum of the energy into the power sinks in the system (the load, dump load, and battery)

7. If there is excess energy in the system it is first stored, if possible, and dumped only if necessary

Specifically, for each operating approach, determine how much energy is supplied by the diesel over the 24 hour period and the reduction in diesel power compared to the diesel only case.

SOLUTION