Harder cable calc's question

[johninlondon1]

A 230v single-phase circuit is to be run in single core 900C thermosetting insul. cables with copper conductors, enclosed in trunking with 5 other similar circuits. Circuit protected by BS88-2-2gG fuse against overload and short circuit. The fuse giving protection against against indirect current with max disconnection time of five seconds. If Ib = 38amps, l = 90m, ta = 450, Ze= 0.8 ohms and the power factor of the load is 0.8 lagging, work out the minimum live and protective conductor CSA that can be used, the Voltage drop is not to exceed 2.5 % of the nominal voltage. Probably for more experienced electricians, show all calculations ie a) correction factors, B) vd taking account of PF and operating conductor temp. c) CCC of CPC, d) Zs, e) Ief f) thermal protection. Good luck

 

[steptoe]

is this for us to do John?

 

[kerching]

Well, the minimum that could be used would be 1mm..........wouldn't be right BUT would be the minimum....................sorry! BAd day, as you may gather from any of my other posts today!!! Sorry, ..................... :coat

 

[johninlondon1]

Protective Device

As In ≥ Ib BS88 fuse In = 40amps

Correction Factors

Table 4C1, Cg = 0.57. Table 4B1 Ca = 0.87, as providing short circuit and overload we use Protective device rating in correction factor calculation.

It = 40 x 1/0.57 x 1/0.87 = 80.66 amps

Table 4E1A col 4 min CSA = 16mm2 with Ita = 100 amps

Voltage Drop

Column 3, Table 4E1B m/Va/m = 2.9 mOhm.m

(2.9 x 38 x 90)/1000 = 9.918 Volt

% VD = +9.92/230 = 4.313%

therfore VD is too high

Try next highest conductors, being 25mm2 having VD of 1.85m/V/a/m for (m/V/a/m)r, (mvam)x = 0.31 mῼ and (m/Va/m)z = 1.9mῼ

Taking advantage of Power factor for VD use √1-Cos20.8 = 0.6

thus vd = (((0.8x1.9) + (0.6 x 0.31))/1000)x38*90 = 5.83452 v

=5.83452/230*100 = 2.5367%

This does not meet the VD requirement so we can use the actual conductor temperatures to make a more accurate calculation

Ct = (230 + 90 - ((0.872 x 0.572) - (382/402))(90-30))/(230+90) = 0.88

Recalculation is given by

((0.8x1.9x0.88) + (0.6x0.31))/1000 x38x90 = 5.2volt = 2.26%

The cable does meet requirements of VD taking account conductor temp.

CPC sizing

Now we calculate CPC

Table 41.4 the max Zs that can be tolerated is 1.35 ῼ, thus max value of (R1+R2) is (1.35-0.8) = 0.55ῼ

Thus max value of (R1 +R2)/m is given by (0.55*1000)/90 = 6.1111mῼ/m m

Table 54.3 applies as the circuit is grouped with other circuuits Table 3.1 Col 6 for the R1/m for 25mm2 is 0.931 can be used with 3.94mῼ/m.

(R1 + R2) = ((0.931 + 3.94) x 90)/1000 = 0.43839 ῼ

Therefore the max R2/m that can be used is 5.17mῼ/m and therefore 6mm2

Zs = (0.8 + 0.44) = 1.24 ῼ

Thermal Protection

Earth Fault current Ief = 230/(0.8+.43839) = 185.73 amps

Appendix 3 the disconnection time is apprx. 4s and K = 143

a. K2S2 = 1432 x 62 = 736,164A2s

b. Ief2t = 185.732 x 4 s = 137982.5316 a2s

as a > b cpc is thermally protected

This circuit can use 25mm2 conductors with minimum CPC of 6mm2

Any errors/queries/comments, feel free to message/critisise etc, lol

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Excellent - The students are setting questions now :slap @John - I hope you have the correct answer ROTFWL as there seems to be something in the question missing to enable it to be answered (Thanks Sidey - Think i have had too much Red !)

What's missing, everything is there to calculate, lol