Loss reduction comprehensive application system based on cloud electricity intelligent cloud platform and analysis method
An application system and cloud platform technology, applied in the field of loss reduction comprehensive application system and analysis, can solve the problems of difficult and complicated data collection, hindering the accuracy of calculation, etc., to achieve the effect of improving security and helping management
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Embodiment 1
[0087] Embodiment 1: limited conditions, such as Figure 5 As shown, the high-voltage 110kV substation transfers power to the 35kV substation model, assuming that the reactive power Aq110=312kVar.h obtained from the Yundian Zhiyun platform in a certain 15 minutes, the active power is Ap110=900kW.h, and the active power of the 35kV substation As35 =852kW.h, the voltage U=110kV of this section of grid line, then according to the formula:
[0088] line current Calculated: The value from the model of the same line to the model database is r=0.0008, l=62.2km, according to the formula: Ri=r*l, then the theoretical line loss Ai=3*I2*Ri*t=0.0008×62.2×20 2 ×3×0.25=14.928kW.h;
[0089] Assuming that the load loss Pf of the S35 transformer is queried = 10kW, the theoretical line loss of the transformer At = iron loss Ar + copper loss Ac, where Ar = Pf*t = 10kW*0.25h = 2.5kW.h, the equivalent resistance of the transformer is according to the formula : Rt=(Pk×Ue 2 ) / Se 2 , can be ...
Embodiment 2
[0091] Embodiment 2: limited conditions, such as Image 6 As shown, in order to calculate the line loss of the power grid for a certain 15 minutes, first obtain the active power Ap = 40kW. The total power of the transformer is 37kW.h (A1=14.4kW.h, A2=22.6kW.h), the models of lines L1, L2, and L3 are the same, and the voltage U of this section of the grid line is U=10kV; then
[0092] Transformer distribution current is: I1=Itotal*A1 / (A1+A2)=3.89, I2=Itotal*A2 / (A1+A2)=6.11;
[0093] Look up the table to get unit resistance r=0.0001, l1=3km, l2=5km, l3=10km; according to the formula: Ri=r*l; line loss Ai3=3*I2*Ri*t=3×10 2 ×0.001×10×0.25=0.75kW.h, Ai1=3*I12*Ri*t=3×3.89 2 ×0.001×3×0.25=0.0088kW.h; Ai2=3*I22*Ri*t=3×6.11 2 ×0.001×5×0.25=0.064kW.h;
[0094] The load loss of transformer 1 and transformer 2 is Pf1=Pf2=0.32kw, and the total iron loss of the two transformers in this period is Ar total=(Pf1+Pf2)*t=(0.32+0.32)*0.25=0.16kW.h;
[0095] Calculate the equivalent resis...
Embodiment 3
[0097] Embodiment 3: limiting condition: such as Figure 7 As shown, in a certain 15-minute period of calculating the power grid, the output active power of S35 power station is Ap=40kW.h, the reactive power is Aq=14kW.h, the output active power of small hydropower is 10kW.h, and the reactive power is 3kW.h; the models of the two transformers are the same, the current power of transformer 1 is 24.2kW.h, and the current power of transformer 2 is 24.2kW.h. Assuming that the models of the lines are the same, the equivalent resistance of transformer 1 is Rt1=0.02, and the equivalent resistance of transformer 2 is Rt2=0.02,
[0098] The voltage U=10kV of the power grid line in this section, then first obtain the current on the main transmission line: Then distribute the current according to the principle of the nearest load: I1=(9.78+2.41)*24.2 / (24.2+24.2)=6.095(A), I2=6.095(A), then I4=I3-I1=3.685(A);
[0099] Assume that the models of the lines are the same, and the unit r...
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