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🔥 内容介绍
Price is an important element of market behavior and is closely related to energy management. From consumer’s perspective, the demand increase/decrease as the marginal benefit higher/lower than the market price. From the producer’s perspective, the generation increase/decrease as the production cost lower/higher
than the market price. This paper considers the energy imbalance management in a microgrid system. Suppose in a microgrid, the demand and supply change according to the price at the time. Then by properly designing the pricing scheme, the energy imbalance in the microgrid can be controlled. Unlike the traditional power grid, microgrid can obtain energy from a renewable energy sources (RES)
such as solar panel of wind turbine. Microgrid contain components include, dynamical load, RES and dynamic generation. The energy in the microgrid should be balanced in anytime, that is the sum of demand equals the sum of generation. However, the intermittency of RES brings uncertainty to the system and make the system imbalanced. To balance the system, a pricing scheme termed area control error (ACE) pricing has been studied.The ACE pricing uses the rate of energy imbalance to control the rate of change of the price. However, in this paper, simulation result shows that the performance of ACR pricing will degrade when additional intermittent power input is involved. In this paper, a novel robust pricing scheme is proposed to deal with the uncertainty caused by RES. Then ovel pricing scheme is based on fuzzy interpolation techniques. The uncertainty caused by intermittent RES is addressed by 𝐻∞ performance criterion design; that’s to say, over all the possible disturbances, the uncertainty and the fluctuating effects is less than a fixed attenuation level. The pricing parameters can be solved by a linear matrix inequality, which is convex and can be solved efficiently.
📣 部分代码
function fm2 = FM2(pg, Fm2)
Fuzzy logic setting for power demoand, see Figure 1b
Fm2 = Fm2 - 1;
k = [5 11.67 18.33 25];
if Fm2 == 0
if pg < k(1)
fm2 = 1;
else
fm2 = max([1/(k(2)-k(1))*(k(2)-pg) 0]);
end
elseif Fm2 == 1
fm2 = max([min([1/(k(2)-k(1))*(pg-k(1)), 1/(k(3)-k(2))*(k(3)-pg)]) 0]);
elseif Fm2 == 2
fm2 = max([min([1/(k(3)-k(2))*(pg-k(2)), 1/(k(4)-k(3))*(k(4)-pg)]) 0]);
elseif Fm2 == 3
if pg > k(4)
fm2 = 1;
else
fm2 = max([1/(k(4)-k(3))*(pg-k(3)) 0]);
end
end
⛳️ 运行结果
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