A) Determine the wire size needed to limit voltage drop to 3% for a 100W, 24 V l

Question

A) Determine the wire size needed to limit voltage drop to 3% for a 100W, 24 V load at a distance of 75ft from the voltage source.

B) Using the wire size determined in part A, determine the actual voltage drop for the wiring.

Table 4.2 Properties of Copper Conductors with 90oC Insulation 12 10 16 18 Size dc kft 777 4.89 3.07 1.93 1.21 0.764 0.491 0.308 lma (A) 14 18 25 30 40 55 75 95 Size 000 0000 250 kcm dc okt 0.245 0.194 0.154 0.122 0.0967 0.0766 0.0608 0.0515 lma (A) 110 130 150 170 195 225 260 290 Note: Solid conductor wire sizes 18-8. Stranded larger sizes. Source: Based on datafrom NFPA 70 National Electrical Code, 2008 ed..National Fire Protection Quincy, MA, 2007

Explanation / Answer

A)

Given

(1).Drop To be Limited To: 3%

(2). Power at 24 V Voltage To be Transfered: 100W

(3) Value of Current from statement (2) is I = (P/V) => (100/24) => 4.167 Amps

(4) Let X be the Value of DC Ohm/ Kft for Selected wire size, Then Value of Resistance ( R in Ohms) offered by 75 Feets of Selected Wire is R = (Kft* DC ohm/kft) => ((75)/1000)*X => 0.075X

(5) From statement 2, Voltage Drop to be Limited to 3%, So Voltage Drop Across Selected Wire to be less than 3% of 24 V , i,e. less than 24*(3/100) = 0.72 V

i,e. I* R < 0.72

From Statment 3&4

4.167 * 0.075X < 0.72

0.3125 X < 0.72

X < 2.304 DC Ohm/ Kft

So the Sleceted Wire should have X closest to 2.304 . So Wire with 12 Size is selected having 1.93 DC Ohm/ Kft

B)

Selected Wire Size is A having DC ohm/Kft of 1.93.

So Voltage Drop from Statement 5,

= I*R

= 4.167* (0.075*X)

= 4.167 * (0.075 * 1.93)

=0.603 V i,e.2.51 % Drop

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