Voltage Drop Calculations

ME->EE · Mar 12, 2020

When calculating the effective impedance when we would we use Chapter 9 Table 9 Note 2: Ze = R x PF + XLsin[arccos(PF)] rather than just using the R and XL values from the table in the following formula: Z = (distance) * (R+jXL)

Most of my practice problems don't use the Note 2 equation at all so I'm not seeing when I should use that.

Thanks!

Orchid PE · Mar 13, 2020

Let me know if either of these help.

Orchid PE · Mar 13, 2020

@ME->EE Also this:

ME->EE · Mar 13, 2020

Thanks @Chattaneer PE, so if I'm understanding this correctly you never use the Note 2 equation for finding conductor impedance but you will use the Note 2 equation when you need to find the voltage drop and you are only given the circuit PF (not voltage or current PF separately)? Otherwise when finding the voltage drop you only use Z = (distance) * (R+jXL)?

Orchid PE · Mar 13, 2020

ME->EE said:
Thanks @Chattaneer PE, so if I'm understanding this correctly you never use the Note 2 equation for finding conductor impedance but you will use the Note 2 equation when you need to find the voltage drop and you are only given the circuit PF (not voltage or current PF separately)? Otherwise when finding the voltage drop you only use Z = (distance) * (R+jXL)?

Essentially.

If you know the current drawn by a load, and the load properties, you can calculate voltage drop.

For example, if a 3-ph 10kVA load is draws 20 amps @ 0.9 PF, what is the voltage drop across an uncoated copper 4 AWG wire that is 500ft long in PVC conduit?

R + X for 4 AWG per 1000 ft: 0.31 + 0.048

For 500ft: 0.155 + 0.024

Effective Z @ 0.9: 0.155*PF + 0.024*sin(acos(PF)) = 0.13175 + 0.012643 = 0.148\4

Voltage drop: 20 * 0.144 = 2.89

----

This is useful when calculating voltage drop for building electrical systems, because industrial equipment will generally give you the amps and PF in the specification sheets.

ME->EE · Mar 13, 2020

Wouldn't X = 0.048 since it's in a PVC conduit? I think the 0.06 comes from using a steel conduit.

R + X for 4 AWG per 1000 ft: 0.31 + 0.048

For 500ft: 0.155 + 0.024

theta = arccos(0.9) = 25.84 degrees

V = (20A)( @ 25.84 degrees)*(0.155 + 0.024)

V = 3.58 (@ 25.84 degrees)

Orchid PE · Mar 13, 2020

I could be wrong, but this is what I'm seeing:

But this is just a table I found online since I don't have the NEC in front of me. But regardless of the actual values, the principle remains the same.

Dude99 · Mar 13, 2020

Z = 0.9 x 0.155 x sin(acos0.9) x 0.024 = 0.150

0.150 x 20 = 3 V

4 awg ~42,000 cmil

BebeshKing PE · Mar 13, 2020

Chattaneer PE said:
Essentially.

If you know the current drawn by a load, and the load properties, you can calculate voltage drop.

For example, if a 3-ph 10kVA load is drawing 20 amps @ 0.9 PF, what is the voltage drop across an uncoated copper 4 AWG wire that is 500ft long in PVC conduit?

R + X for 4 AWG per 1000 ft: 0.31 + 0.06

For 500ft: 0.155 + 0.03

Effective Z @ 0.9: 0.155*PF + 0.03*sin(acos(PF)) = 0.13175 + 0.0158 = 0.148

Voltage drop: 20 * 0.148 = 2.96

----

This is useful when calculating voltage drop for building electrical systems, because industrial equipment will generally give you the amps and PF in the specification sheets.

@Chattaneer PE, what situation we can use the voltage drop equation in complex number? Vdrop =(R+jXL)(I<theta) ?

Orchid PE · Mar 13, 2020

BebeshKing said:
@Chattaneer PE, what situation we can use the voltage drop equation in complex number? Vdrop =(R+jXL)(I<theta) ?

Whenever you're given the values to do so. Actual voltage drop is defined exactly as you stated: Vdrop = (R + jX) * (I<theta). If you are given a current with an angle, go for it.

This effective Z method is used because most industrial equipment will generally give you the amps and PF in the specification. An angle is generally not given in equipment specifications, just the power factor. The effective Z method just gives an approximation of the voltage drop.

For example, this is from a datasheet:

Notice they give full load current and power factor. They do not give an angle for current. The effective Z method would be used to approximate Vdrop.

Dude99 · Mar 13, 2020

imho the NEC note is confusing

they give you R and X hence the line ph ang = atan x/r

assume 500' #4 R = 0.155 and X = 0.024

Z = R x cos(atan x/r) + j X sin(atan /r) = 0.153/8.8 deg

using the NEC method of 0.9 x R + X x sin(acos 0.9) = 0.150

moot on 100 A load: 15.3 vs 15.0 drop

BebeshKing PE · Mar 13, 2020

Chattaneer PE said:
Whenever you're given the values to do so. Actual voltage drop is defined exactly as you stated: Vdrop = (R + jX) * (I<theta). If you are given a current with an angle, go for it.

This effective Z method is used because most industrial equipment will generally give you the amps and PF in the specification. An angle is generally not given in equipment specifications, just the power factor. The effective Z method just gives an approximation of the voltage drop.

@Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:

An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.

Here's what I did:

Using complex number:

Vdrop= 2(R+jXL)(I<theta) = 2(0.15+j0.046)(90/1000)(40<-41.4)=1.13<-24.4 volts

Using approximate formula:

Vdrop=2(Rcos(theta)+XL(sin(theta))(I)=2(0.15(0.75)+0.046(0.66))(90/1000)(40)=1.03 volts

what's your thought about this?

Dude99 · Mar 13, 2020

The way I do it

X = 0.024, R = 0.155, Z Line phase = 8.8 deg, old pf = 0.85 31.79 deg, new 0.9 25.84 deg

Z at 0.85 = 0.145 from table at 0.85 pf

Z at 0.9 = Z x cos(Z line - Z load old/(cosZ line - Z load new)

plugging Z = 0.1506

any method will get you 'close enough)

if a motor load ph ang = acos(motor pf)

Dude99 · Mar 13, 2020

The real part of 1.13/-24.4 = 1.03

exactly 1.0291, same as the NEC 1.0291

BebeshKing PE · Mar 13, 2020

Dude99 said:
The real part of 1.13/-24.4 = 1.03

exactly 1.0291, same as the NEC 1.0291

@Dude99, So you were saying if we encounter this type of problem and we use the complex number formula, the answer would be the Real part of the Volt drop?

Dude99 · Mar 13, 2020

BebeshKing said:
@Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:

An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.

Here's what I did:

Using complex number:

Vdrop= 2(R+jXL)(I<theta) = 2(0.15+j0.046)(90/1000)(40<-41.4)=1.13<-24.4 volts

Using approximate formula:

Vdrop=2(Rcos(theta)+XL(sin(theta))(I)=2(0.15(0.75)+0.046(0.66))(90/1000)(40)=1.03 volts

what's your thought about this?

What were the answer choices?

BebeshKing PE · Mar 13, 2020

Dude99 said:
What were the answer choices?

This was not a multiple choice question, though. In any way, I just want to know the concept for this.

Also, In the NCEES #129 solution , they used the complex number formula but did not consider the real part of the voltage drop to get the voltage at the load side.

Little confuse.

Orchid PE · Mar 13, 2020

BebeshKing said:
@Chattaneer PE, I asked this because I encountered a problem that I used both formulas (complex number vs approximation formula) to compare each other but the answer is way far from each other. Here's the problem:

An 90 foot circuit provides power to a single phase load connected to a 120V panel. The conductor size is 1AWG uncoated copper ran in PVC conduit. Calculate the voltage drop in volts if the amps drawn by the load is 40 with lagging power factor at 0.75.

Here's what I did:

Using complex number:

Vdrop= 2(R+jXL)(I<theta) = 2(0.15+j0.046)(90/1000)(40<-41.4)=1.13<-24.4 volts

Using approximate formula:

Vdrop=2(Rcos(theta)+XL(sin(theta))(I)=2(0.15(0.75)+0.046(0.66))(90/1000)(40)=1.03 volts

what's your thought about this?

It's because the angle of the load cannot be directly translated to the angle of the "circuit." The phase angle of the circuit is different from the power factor of the load. The impedance of the conductors will change the power factor of the current supplying the load.

Orchid PE · Mar 13, 2020

Orchid PE · Mar 13, 2020

BebeshKing said:
So you were saying if we encounter this type of problem and we use the complex number formula, the answer would be the Real part of the Volt drop?

No, it only appears that way because with the smaller wires, the resistance has more of an impact. As the wire size gets larger, the effective Z becomes closer to XL.

Voltage Drop Calculations

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