Method for solving difficulties of peak load regulation of power grid, installation and electricity and coal deficiency
A power grid and peak-shaving technology, which is applied in the field of power systems, can solve problems such as excessive water waste during peak-shaving, difficulty in filling valleys in power grid troughs, and high on-grid electricity prices
- Summary
- Abstract
- Description
- Claims
- Application Information
AI Technical Summary
Problems solved by technology
Method used
Image
Examples
example 1
[0032] Example one: 3,290,000 kw of hydropower installed capacity in a certain provincial power grid, 3,000,000 kw of local hydropower installed capacity, totally 4632 places, the present invention installs additional peak-valley watt-hour meter at the local hydropower plant (network) gateway. 365 days a year, the local peak on that day is in line with the requirements of the present invention. The present invention purchases local hydropower at a high price to adjust the peak according to the installed hydropower capacity of the power grid of 0.08 billion kw.h / 10,000 kw. Without purchasing, local hydropower utilizes 10,000 reservoirs to store water in low valleys. Because the localities generate more high-priced electricity in peaks and reduce loads in low valleys to fill valleys, the invention promotes 800 to 2,000 more utilization hours of power grid generation per year. 3.29 million kw×0.08 billion kw.h / 10,000 kw=2.632 billion kw.h, 3.29 million kw×2000 utilization hours=6....
example 3
[0054] Example 3: The present invention purchases local water and electricity to adjust the peak benefit calculation as follows:
[0055] National hydropower installed capacity of 77 million KW (see China Electric Power News 2001.02.03 "Hydropower installed capacity...") National local small hydropower 23 million KW (see Nanguo Zaobao 2000.06.18)
[0056] 7700-2300=54 million KW (National Power Grid), the present invention is purchased at the rate of 12 million KW / 10,000 KW
[0057]
[0058]
[0059]
[0060]
[0061] 1 to 20 years to create benefits:
[0062] 486 billion yuan ÷ 5 = 97.2 billion yuan 972×20 = 1.944 billion yuan
[0063] 4860+11589.23+27635.85+65900.87=109985.95 (100 million yuan)
[0064] According to 5,000 yuan / KW, the hydropower installed capacity can be calculated:
[0065] 10998.595 billion yuan ÷ 5000 yuan / KW=2199719000KW≈2.2 billion KW
[0066] According to 2500 yuan / KW expansion hydropower installed capacity:
[0067] 10998.595 billion ...
PUM
Login to View More Abstract
Description
Claims
Application Information
Login to View More 