Moisture and Oxygen Barrier Properties of ... - ACS Publications

Nov 26, 2017 - Forest Products Laboratory, U.S. Forest Service, 1 Gifford Pinchot Drive, Madison, Wisconsin 53726, United States. ‡. Advanced Struct...
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REPLACE THIS LINE WITH YOUR PAPER IDENTIFICATION NUMBER (DOUBLE-CLICK HERE TO EDIT) < When I1 is small, iron core of current transformer is

im.2 (t )

unsaturation. Z m is big and Im is small. When I1 increases, current transformer iron core start to saturate. Z decreases and I increases. According to m

in.2 (t ) in.1 (t )

im.1 (t )

(a)

m

Δim.2 (t )

equation (3), I 2′ decreases.

Δin.2 (t )

Δim.1 (t )

When Zm decreases to zero, I1 = Im and I 2′ =0. When faults occur in transmission lines, fault current increases quickly and current transformer will saturate. DC component of fault current exacerbates current transformer saturation. Because the flux of iron core can’t change suddenly, the relation between current transformer primary current and secondary current is linear at fault start time. The length of linear zone is related to current transformer saturation speed. The wave distortion of secondary current is related to CT saturation degree.

Δin.1 (t )

ZN

ZM

U F (t )

(b) Fig. 3. (a) The internal fault network and (b) its fault part. Δin.1 (t )

Δim.1 (t )

III. CURRENT SAMPLE DATA CHARACTERISTICS OF CURRENT TRANSFORMER SATURATION A. Normal Condition The normal condition of transmission line is shown in Fig. 2, im.2(t) and in.2(t) are secondary currents sample data of current transformer. im.1(t) and in.1(t) are primary currents sample data of current transformer. im.2 (t )

2

(a) Δin.2 (t )

in.2 (t ) im.1 (t )

Δim.2 (t )

in.1 (t )

Fig. 2. Normal power system network

In normal condition, CT is unsaturated. The relations between primary currents and secondary currents are linear. B. Internal Faults For internal faults of transmission lines (Fig. 3(a)), the fault part is shown in Fig. 3(b). In the fault part, ᇞim.2(t) and ᇞin.2(t) are secondary fault currents sample data of current transformer 1 and 2. (6) ᇞim.1(t)= im.1(t)- im.1(t-N) (7) ᇞin.1(t)= in.1(t)- in.1(t-N) (8) ᇞim.2(t)= im.2(t)- im.2(t-N) (9) ᇞin.2(t)= in.2(t)- in.2(t-N) where N is sample data number in a power frequency cycle. For internal faults of lines, ᇞim.1(t)×ᇞin.1(t)>0 and ᇞim.2(t)×ᇞin.2(t)>0. ᇞim.1(t) and ᇞin.1(t) can be placed in the plane shown in Fig. 4(a). ᇞim.2(t) and ᇞin.2(t) can be placed in the plane shown in Fig. 4(b). For an internal fault in F1, ᇞim(t) and ᇞin(t) are in zone I or III of plane ᇞim.1(t)-ᇞin.1(t) and plane ᇞim.2(t)-ᇞin.2(t). C. External Faults For internal faults of transmission lines (Fig. 5(a)), the fault

‹,(((

(b) Fig. 4. (a) ᇞim.1(t)-ᇞin.1(t) plane; (b) ᇞim.2(t)-ᇞin.2(t) plane.

im.2 (t )

in.2 (t )

in.1 (t )

im.1 (t )

(a) Δim.2 (t ) Δim.1 (t ) ZM

Δin.2 (t ) Δin.1 (t )

U F (t )

ZN

(b) Fig. 5. (a) The external fault network and (b) its fault part.

part is shown in Fig. 5(b). For external faults of lines, ᇞim.1(t)=ᇞin.1(t) and ᇞim.2(t)=ᇞin.2(t). When external faults cause current transformer saturation, ᇞim.1(t)×ᇞin.1(t)