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Showing posts with label Voltage Drop Calculation. Show all posts
Showing posts with label Voltage Drop Calculation. Show all posts

Sunday, November 29, 2020

Voltage Drop in Consumer Installations According to BS 7671

 



Voltage drop in a consumer’s installation can be a contentious issue. However, it is an important aspect of installation design since if it is too high some certain equipment will not function correctly or will not function at all.


Rules Applied: 

  • 525.1 In the absence of any other consideration, under normal service conditions the voltage at the terminals of any fixed current-using equipment shall be greater than the lower limit corresponding in the product standard relevant to the equipment.
  • 525.100 Where fixed current-using equipment is not the subject of a product standard the voltage at the terminals shall be such as not to impair the safe functioning of that equipment.
  • 525.101 The above requirements are deemed to be satisfied if the voltage drop between the origin of the installation (usually the supply terminals) and a socket-outlet or the terminals of fixed current-using equipment, does not exceed that stated in Appendix 4 Section 6.4

Calculating Voltage Drop 


When calculating voltage drop due consideration should be given to the following: 
  • motor starting currents; in-rush currents
  • control voltages (particularly those associated with computerized systems).

Notes: 
  1. Motor control contactors and relays can ‘drop out’ if the coil voltages fall towards 80% of the operating voltage.
  2. The effects of harmonic currents may also need to be considered and included in the calculation.
  3. Voltage transients and voltage variations due to an abnormal operation can be ignored.


The maximum voltage drop values can be taken from the table shown below:

For 230 Volts Network

Allowable Voltage Drop Based on BS 7671

*The voltage drop within each final circuit on Private Networks, should not exceed the values given in (i) above for Public Networks

For 400 Volts Network

Allowable Voltage Drop Based on BS 7671

*The voltage drop within each final circuit on Private Networks, should not exceed the values given in (i) above for Public Networks


When calculating the voltage drop in a circuit, the design current can be taken as being either the equipment rated current or, where there are a number of loads, the total connected load multiplied by a diversity factor. 

Note: If the total circuit length exceeds 100 meters, the limits given in Table 4Ab may be increased by 0.005% per meter up to a maximum of 0.5%.

The voltage drop can be apportioned throughout the system circuits as the designer wishes, but the final circuit voltage drop is limited to the values given for Public Networks, regardless of whether it is a Public Network or a Private Network.

In case that the supply voltage at the origin is lower than the nominal 230/400V, the designer needs to consider the effect of the minimum permissible supply voltage. This is a maximum of 6% below the nominal supply voltage, which equates to 216.2V (for 230 V network ) and 376V (for 400 V network) respectively.

Reference: 

Electrical Contractors Association Fact Sheet | Download

Thursday, November 19, 2020

Voltage Drop Calculation for Single Phase and Three Phase Systems

 

Electric Panel Board

When it comes to voltage drop calculations, there are several valuable individual approach that can be used when working with electrical systems. Thus, it is important for an engineer to determine to match his calculations against the actual scenario in the type system or type of installation. The following scenario covers comprehensive calculation involving electric circuits in different scenario. 




Example 1. Calculate voltage drop in a single-phase dc circuit.


A solid #14 copper two-wire cable that is 750 feet long supplies a 125 ohm load resistor for heating purposes at 75 deg. C. Find the voltage drop in the cable; and find the resulting voltage across the load resistor. 


Diagram: DC Circuit 

Solution: 


Calculate the cable resistance, for the sake of illustration, the overall length need to be multiplied by two to consider the cable resistance (out) and cable resistance (back). Assume that the cable has a uniform resistance of 75 Deg. C. 


Solve for cable resistance, 


Cable resistance = (RESISTANCE FOR 1000 FT.) x (750 / 1000) x 2

  • R = (3.07 ohms per M ft.) X (0.750) x 2
  • R = 2.3 ohms x 2
  • R = 4.6 ohms.

Total resistance  = 125ohms + 2.3 ohms  = 129.6 ohms

Current = 125/ 129.6 = 0.9645 Amperes 

Voltage drop in cable = current x cable resistance

  • Vd = (0.9645) X (4.6)

Therefore, 

  • Voltage Drop in the Cable = 4.436 V
  • Voltage at the load terminal = 0.9645 x 125 ohms =  120.56 Volts

Example 2. Calculate the approximate voltage drop in a single-phase ac circuit at unity power factor in a plastic conduit.

From a 115VAC circuit breaker, a solid #12 copper two-wire cable in plastic conduit that is 750 feet long supplies a motor load that requires 1.3 kW. Find the approximate voltage drop in the cable?; and find the resulting voltage supplied to the motor load? (Note: ignore the constant kVA motor action). 


Diagram: AC circuit


Solution: 

Cable resistance = (RESISTANCE FOR 1000 FT.) x (750 / 1000) x 2

  • R = (1.70 ohms per M ft.) X (0.750) x 2
  • R = 1.275 ohms x 2
  • R = 2.55 ohms.

Calculate the total current flowing in the circuit.

  • I = 1,300/ 115 (ignoring power factor)
  • I = 11.3 Ampere

Voltage drop in cable = current x cable resistance

  • Vd = (11.3) X (2.55 ohms)
  • Vd = 28.82 Volts

Therefore, 

  • Voltage Drop in the Cable = 28.82 Volts
  • Voltage at the load terminal = 115 - 28.82 = 86.18 Volts

Example 3. Calculate voltage drop in a three-phase ac circuit at less than unity power factor. 

From a 480/3P circuit breaker, a stranded #0000 copper three-wire Type TC (non-armored) cable that is 280 ft. long and is laid into in aluminum cable tray. The cable supplies an AC load operating at a power factor of 85% that draws 200 amperes. Find the voltage drop in the cable; and find the resulting voltage supplied to the load. 

Diagram: 3- phase AC circuit

Solution: 

Cable resistance = (RESISTANCE FOR 1000 FT.) x (280 / 1000) - per line

  • R = (0.078 ohms per M ft.) X (0.0.28) 
  • R = 0.0218 ohms - per line

The current is given at 200 Amperes. 

Voltage drop in cable = current x cable resistance

  • Vd = (200) X (0.0218 ohms) x 1.73 (multiple a factor 1.73 for 3 phase systems)
  • Vd = 7.55 Volts

Therefore, 

  • Voltage Drop in the Cable = 7.55 Volts
  • Voltage at the motor terminals = 480 - 7.55 = 472.45 Volts

Friday, March 06, 2015

How to Calculate Transformer Voltage Drop


Introduction

Transformer voltage drop is important to know since it is one of the factors that affect the performance of an electrical system where it is installed. Obviously high voltage drop in transformer could lead into low voltage at the load side of the system.

Formula 

Single Phase Transformer: Vd = I (R cos theta + X sin theta)
Three Phase Transformer: Vd =  sqrt(3) x I (R cos theta + X theta)

where: 

Vd = voltage drop
R = Resistance 
X = Reactance
theta = power factor angle


Read: What is the Importance of X/R Ratio?


Example 1 (Single Phase Transformer)

Find the voltage drop of the single phase transformer supplying a 50 HP motor with a power factor of 0.70. The transformer has manufacturer rating given below.
  • voltage rating = 12.7KV /  230V
  •  KVA rating = 100 KVA
  • % R =  2.24 %
  • % X = 3.34 %
Solution:

In this case the resistance and reactance of the transformer is given in its percentage form, therefore we need to determine the actual value of these quantities. In determinin23g the actual values we need to use the following formula,

R actual = 10 (%R) (KV secondary) ^2 
                        KVA transformer

R actual = 10 (2.24%) (0.23 KV) ^ 2     >> use 230 V secondary as base voltage
                          100 KVA 

R actual = 0.01185 ohms >> value of the actual resistance

X actual  = 10 (%X) (KV secondary) ^ 2
                           100 KVA

X actual = 10 (3.44%) (0.23 KV) ^2 >> use 230 V secondary as base voltage
                              100 KVA

X actual = 0.0182 ohms >> value of the actual reactance

Determine the value of the current

P = 50 HP x  746 W = 37,300 watts
                         HP 

I = P / VL * pf

I = 37, 300 Watts / 230 V * 0.7
I = 231 Amperes

cos theta = 0.7 
sin theta = 0.7

Vd = I (R cos theta + X sin theta)
Vd = 231 A x [ (0.01185)( 0.7) + (0.0182) (0.7) ]

Vd = 4.85 Volts or

%Vd = (4.85 V) x 100 = 2.11 %
             230 Volts Base


Example 2 (Three Phase Transformer) 

Find the voltage drop of the three phase transformer supplying a 100 KVA load with a power factor of 0.80. The transformer has manufacturer rating given below.
  • voltage rating = 12.7KV /  230V
  •  KVA rating = 150 KVA
  • % R =  1.08 %
  • % X = 1.63 %
Solution: 
The process is still the same above but we differ only in the final calculation of the current since the value that we can get is to be multiplied with sqrt(3) or 1.73.

R actual = 10 (%R) (KV secondary) ^2 
                        KVA transformer

R actual = 10 (1.08%) (0.23 KV) ^ 2     >> use 230 V secondary as base voltage
                          100 KVA 

R actual = 0.0031 ohms >> value of the actual resistance

X actual  = 10 (%X) (KV secondary) ^ 2
                           100 KVA

X actual = 10 (1.63%) (0.23 KV) ^2 >> use 230 V secondary as base voltage
                              100 KVA

X actual = 0.0047 ohms >> value of the actual reactance

Determine the value of the current

I = S / (1.73 x VL)
I = 100,000 VA / (1.73 x 230)
I = 251 Amperes

cos theta = 0.8
sin theta = 0. 6

Vd = 1.73 x I (R cos theta + X sin theta)
Vd = 1.73 x 251 A x [ (0.0031)( 0.8) + (0.0047) (0.6) ]

Vd = 2.30 Volts or

%Vd = (2.30 V) x 100 = 1.0 %
             230 Volts Base

How to Calculate Voltage Drop of Distributed Loads

Voltage drop calculation is necessary in designing electrical system in order to keep our equipment operate normally. Failure to calculate voltage drop properly would result into under-voltage that can damage our equipment. 

In other article we discuss about voltage drop calculation based on national electrical code in an ideal way. It is ideal since we are assuming a system of only one load neglecting the possibility that in reality we could encounter a system where load is distributed and the voltage drop is a critical factor.

For example if we are asked to design a lighting of a recreational park where each light is 50 ft apart. Normally we cannot generalized the computation of the voltage drop since it would result into higher cost of conductors.

Example:

Consider figure 1, seven lighting loads each are separated by a distance of 50 ft and carries a load current of 3 amperes. The distance of the source from the first load (Load A) is 20 feet.


a. Determine the size of the conductor that can be used in the given circuit.

b. Perform voltage drop calculation and check if  each load has proper voltage level.



Figure 1. Load Arrangement


Solution:

a. By inspection we can say that the total current is the sum of the individual current of each load, thus 3 amperes x 7 loads = 21 amperes. From NEC table 310-16 the right size of conductor to serve 21 amperes is #12 AWG.

b. Perform voltage drop calculation
  • by looking at NEC chapter 9 table 8, Z of #12 is 1.7 per 1000 ft.
  • From the other article we know that for single phase installations Vd = (2 x Z x I x L) / 1000

Point A: From the source to load A

Vd = (2 x 1.7 x 21 x  20 ft) / 1000
Vd = 1.4 Volts

Point B: From load A to load B

Vd = (2 x 1.7 x 18 x 50ft ) / 1000
Vd =  3 volts

Point C: From load B to load C

Vd = (2 x 1.7 x 15 x 50ft ) / 1000
Vd =  2.6 Volts

Point D: From load C to load D

Vd = (2 x 1.7 x 12 x 50ft ) / 1000
Vd =  2 Volts

Point E: load E

Vd = (2 x 1.7 x 3 x 50ft ) / 1000
Vd =  0.51 Volts

Point F: Load D to load E

Vd = (2 x 1.7 x 6 x 50ft ) / 1000
Vd =  1 Volt

Point G: load G

Vd = (2 x 1.7 x 3 x 50ft ) / 1000
Vd =  0.51 Volts

Based on the attained value evaluate each point of the circuit and determine if the voltage level at that point is still suitable to supply its load. Choose 3% allowable voltage drop recommended by the NEC to determine if we need to choose the next higher conductor size or not.

With 3% tolerance the lowest voltage value from 220 V is 213.4 Volts.

Voltage at Point A: 220 V - 1.4V = 218.6 Volts
%Vd @ point A = 1.4V / 220V =0.63% ---> allowed

Voltage at Point B: 220V - 1.4 - 3V =  215.6 Volts
%Vd @ point B = 4.4V / 220V = 2% ---> allowed

Voltage at Point C: 220V - 1.4 - 3V - 2.6V =  213 Volts
%Vd @ point B = 7V / 220V = 3.18% ---> more than 3% is not allowed
therefore lets choose the next higher size of the conductor that will connect load load B to C. In this case lets choose #10 that has Z = 1.1 ohm per 1000 ft.

Recalculating values: Vd  at point C = (2 x 1.1 x 15 x 50) / 1000 = 1.65 V

Then it follows that at,
point D = 1.32 V
point E = .33 V
point F = .66 V
point G = .33 V

Thus,

Vd at point C = 220V - 1.4V - 3V - 1.65V = 213.95V
%Vd at point C = 6.05V / 220V = 2.76% ---> allowed

Vd at point D = 213.95 V - 1.32 V = 212.63 V --> not allowed since it is lower than 213.4 Volts at 3% voltage drop tolerance. Discard all succeeding values from point D. Choose next highersize #8 with Z = 0.69

Vd at point D = (2 x 0.69 x 12 x 50) / 1000 = 0.828 Volts
Voltage at point D = 213.95 V - 0.828 V = 213.122 V
% Vd = 6.878/220 = 3.12% ---> more than 3% select next higher, #6 with Z = 0.45

Vd at point D = (2 x 0.45 x 12 x 50) / 1000 = 0.54 Volts
Voltage at point D = 213.95 V - 0.54 V = 213.41 V
% Vd = 6.6V /220V = 3% ---> allowed

At this point we will expect that the next load is above 3% then it is needed to select the next higher conductor size which is #4 with Z = 0.29

Vd at point E = (2 x 0.29 x 3 x 50) / 1000 = 0.087 V
Voltage at point G = 213.41 V - 0.087 V = 213.32 Volts
%Vd = 6.7V / 220V = 3.04% ---> allowed

Vd at point F = (2 x 0.29 x 6 x 50) / 1000 = .174 V
Voltage at point G = 213.32 V - 0.174V = 213.146 V
%Vd = 6.85V / 220V = 3.11% ---> not allowed select next higher size, #3 with Z= 0.23

Recalculation,

Vd at point F = (2 x 0.23 x 6 x 50) / 1000 = 0.138 V
Voltage at point G = 213.41 V - 0.138 V = 213.27 Volts
%Vd = 6.7V / 220V = 3.04% ---> allowed

Vd at point G = (2 x 0.23 x 3 x 50) / 1000 = 0.069 V
Voltage at point G = 213.27 V - 0. 069 V = 213.2 Volts
%Vd = 6.8 V / 220V = 3.11% ---> not allowed select next higher size, #1 with Z = 0.16

Recalculation,

Vd at point G = (2 x 0.16 x 3 x 50) / 1000 = 0.039 V
Voltage at point G = 213.27 V - 0. 069 V = 213.2 Volts
%Vd = 6.7 V / 220V = 3.08% ---> approximately equal to 3% then allow.

Note:


The calculation above was made for the sake of discussion and demonstration on how the manual voltage drop calculation works. According to the National Electrical Code 3% is just a recommendation and it is not enforceable. Therefore to allow or not to allow the conductor sizes based on the deviation of the voltage drop is still on the discretion of the engineer. 
Conclusion:

Proper voltage drop calculation has economic rewards. Instead of using uniform large conductors to compensate the voltage losses we can be guided on which portion of the system we can use large sizes and small sizes.

However this is a tedious task especially when we are dealing with systems larger than this example. There are computer software that can help us to make our job easy especially when it comes to repetitive tasks. But before we will seek the assistance of computer for further applications an engineer need to understand first the manual calculation and know how the flow of the computation works.

Click here to access the online voltage drop calculator for street lights.

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