# DAXPRO Academy | Reduce Construction Costs With Voltage Drop Checks

Release time:

2023-06-08 00:00

Source:

With the rapid development of the national economy, the demand for electricity is also increasing. Due to the tight control of the country's power project construction in the past few years, the power shortage situation is more severe, and power cuts are common, which restricts the development of the national economy. The progress has affected the daily lives of the people.

Electric power construction is a capital-intensive construction project with a long construction period and a high level of automation. With the construction of electric power projects, the market competition is becoming increasingly fierce. How to strengthen the cost management of electric power engineering construction projects and master the project cost expenditure is a problem that investors generally care about.

The cost management of electric power project construction is to ensure the project quality, project progress, construction safety, construction technology and other construction indicators on the premise of ensuring the equipment purchase cost, construction engineering cost, installation engineering cost, Reconnaissance and design fees, construction land fees, supervision fees, project construction management fees and other unamortized construction costs, through project budgets, construction organization, process control and scientific scheduling, etc., try to achieve the company's predetermined construction cost goals, and maximize It is a systematic and scientific management activity to minimize project construction expenditure, which mainly achieves the predetermined goal of project construction through technical management and economic management.

Today, let’s talk about how to reduce construction costs through voltage drop verification.

**Take a Daxpro Chongqing laboratory project as an example:**

The project is divided into Phase I and Phase II, among which the design and installation requirements for wastewater treatment equipment:

**30KW/380V** , power factor 0.8, cable specification 120*3+95*1, length 400 meters (the power supply of the distribution box is in the first phase, and the waste water treatment equipment is in the second phase);

**Obviously** : the first feeling is that **the 120mm² cable for 30KW/380V equipment is too large, and the 400-meter cable is a bit too long.** After on-site inspection, it was found that there is no problem with the design, but the problem is that it is too wasteful and the construction cost is too high;

**Calculation results** : section S = 120 square

bending radius R = 880 mm cable weight W = 4833 kg/km

calculated power Pjs = 30 kw calculated current Ijs = 56.98A

power factor Cos = 0.80 phase: three-phase correction factor = 0.70

**1. When selecting according to the voltage drop of the line** , the cross-section S = 35 square meets

the calculation voltage drop U% = 4.665% < the allowable voltage drop of the line = 5.00%

U%= (0.5013x0.80+0.0800x1)x56.98x400x1.732/380 *10

**2. When selecting according to the sensitivity** , the section S = 120 square meets

the allowable length of the line L = 220V/(line phase protection impedance x1.3x automatic switch setting current x instantaneous multiple> 400.0m

688.7m = 220V/(0.6174x1.3x63Ax10) > 400.0m

Note: This calculation does not take into account the upper system, the actual cable length may be shorter, and the cross-section should be increased by one level.

**3. When selecting according to the set current** , the cross-section S = 25 square meets

A, ampacity after correction = sample ampacity x correction factor x sample temperature/environment temperature

77.70 A=111.00Ax0.70x25.00/25

B, ampacity after correction /1.1 > Automatic switch setting current

77.70A/1.1=70.64A>63A

**Voltage drop principle and calculation formula**

Voltage drop formula: ∆U=2*I*R, where I is current ampere A, R is cable resistance: ohm Ω, allowable voltage drop ∆U≤5%U current I=P/

( K*U*COSϕ), where P is the power KW, U is the voltage V (380V: K takes √3/220V, K takes 1), COSϕ is the power factor 0.8, the calculated current in this case Ijs = 56.98A resistance R

= ρ*L, where ρ is the resistivity: copper is 0.018Ω.㎜²/m, aluminum is 0.0283Ω.㎜²/m, L is the cable length: m; the calculated resistance of this case is R = 7.2Ω

**. The voltage drop is: ∆U=2*I*R=41.0256>19V** , obviously the voltage drop exceeds the standard;

the cost reduction measures are: increase the voltage, increase the cable, and choose two smaller cables, but none of them are economical: According to the on-site inspection, the surplus power of the second phase is 95KW, and the power distribution room of the second phase is planned to be about 100 meters away from the wastewater treatment equipment.

After recalculation, we communicated with the owner about the new plan, and proposed that we only need to use 35㎜² cable * 100 meters instead of 120㎜² cable * 400 meters, which can greatly save the investment cost. The new plan was approved by the design school;

**Calculation results:** section S = 35 square

bending radius R = 572 mm cable weight W = 1591 kg/km

calculated power Pjs = 30 kw calculated current Ijs = 56.98A

power factor Cos = 0.80 phase: three-phase correction factor = 0.70

**1. When selecting according to the voltage drop of the line** , the cross-section S = 10 square meets

the calculation voltage drop U% = 3.791% < the allowable voltage drop of the line = 5.00%

U%= (1.7544x0.80+0.0940x1)x56.98x100x1.732/380 *10

**2. When selecting according to the sensitivity** , the section S = 35 square meets

the allowable length of the line L = 220V/(line phase protection impedance x1.3x automatic switch setting current x instantaneous multiple> 100.0m

688.7m = 220V/(2.4046x1.3x63Ax10) > 100.0m

Note: This calculation does not take into account the upper-level system, the actual cable length may be shorter, and the cross-section should be increased by one level.

**3. When selecting according to the set current** , the cross-section S = 25 square meets

A, ampacity after correction = sample ampacity x correction factor x sample temperature/environment temperature

77.70 A=111.00Ax0.70x25.00/25

B, ampacity after correction /1.1 > Automatic switch setting current

77.70A/1.1=70.64A>63A

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