Voltage drop calculation

[Missing-spark]

Hello everyone just want a little help how you would calculate the volt drop if

230 single phase heater circuit
line & neutral conductors combined resistance of 0.68
Temperature at 20 degrees and load current of 16A?
show calculations


got this in my 2395 today wanna make sure I used the correct method
i ended up with a volt drop of 5.4 or 5.2 can't remember

please help thank you

i did 230 divided by 0.68 multiplied by 0.016

 

[Sintra]

Post up your workings and I'm sure someone will advise you.

 

[Farmelectrics]

Vd = r1 +rn x ib x 1.2

 

[Missing-spark]

Vd = r1 +rn x ib x 1.2

So 0.68 x 16 x 1.2 = 13.05v

is that right?

was not taught that method was only taught

mV/A/M x ib x length / 1000 = VD

thanks for reply

 

[Farmelectrics]

With your values that would give you a volt drop of 13 volts which is not acceptable are your r1 rn value correct seems a bit high if it's right you need to increase the case of your cable

- - - Updated - - -

Csa of your cable

 

[LeeH]

Vd = mv/A/m x Ib x L / 1000

Simples.


Edit. My bad I only read the title.....

 

[Farmelectrics]

Agree lee but there's 3 versions to working out volt drop he gave us the r1 rn value that's why I yoused that formula

 

[LeeH]

Yep, hence my edit mate. I never read the full queston.

 

[Farmelectrics]

Vd = ib x length x m volt per meter divided by 1000
vd = 2 x ib x length x m ohm per meter x 1.2 divide by 1000
vd = r1 rn x ib 1.2

 

[Missing-spark]

That's 3 marks lost then. Fingers crossed for my 2394 exam tonight. I'm good with the calculations just that one threw me cause I didn't have the length or mv/a/m

anyone else do the 2395 yday?

 

[LeeH]

Never seen the 2nd one before. What's with the 2 and 1.2?

Cheers.

 

[Farmelectrics]

There the factors that you use for that particular equation I've always used the first formula my self but in design questions sometimes you have to use one of the other formulas

 

[telectrix]

why complicate the issue. just use ohms law V=IR, i.e. 16 x 0.68

 

[Bick]

why complicate the issue. just use ohms law V=IR, i.e. 16 x 0.68

16 x .68 = 10.88V then adjust to 70 degrees (conductor operating temperature for 70 degree thermoplastic cable) by multiplying by 1.2 = 13.056V

To see if it falls within the allowable %
(13.056/230) x 100 = 5.68%