Electrical

Electrical Cost Estimation

Last updated: March 2026 · Beginner

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Electrical

Voltage drop, wire sizing, load balancing

Electrical cost estimation is essential for energy audits, customer proposals, and job pricing. Whether you are showing a customer how much they will save with LED lighting or estimating material costs for a bid, these calculations turn technical knowledge into dollars and cents.

The kWh Formula

The kilowatt-hour (kWh) is the standard unit for billing electrical energy:

$\text{kWh} = \frac{\text{Watts} \times \text{Hours}}{1{,}000}$

One kWh is the energy consumed by a 1,000W load running for 1 hour. Your electric bill is calculated by multiplying total kWh by the rate per kWh.

$\text{Cost} = \text{kWh} \times \text{Rate per kWh}$

Worked Examples

Example 1: Annual Cost of a 1,500W Heater

Scenario: A customer runs a 1,500W space heater 8 hours per day during winter (approximately 150 days per year). The electricity rate is 0.12 per kWh. What is the annual operating cost?

Step 1 — Daily kWh:

$\text{Daily kWh} = \frac{1{,}500 \times 8}{1{,}000} = 12 \text{ kWh}$

Step 2 — Annual kWh:

$\text{Annual kWh} = 12 \times 150 = 1{,}800 \text{ kWh}$

Step 3 — Annual cost:

$\text{Cost} = 1{,}800 \times 0.12 = 216.00$

Answer: The heater costs approximately 216.00 per year to operate.

Example 2: LED vs. Incandescent Cost Comparison

Scenario: A commercial office has 100 recessed light fixtures, each with a 60W incandescent bulb. The owner wants to switch to 9W LED bulbs. Lights run 10 hours per day, 260 business days per year. Rate is 0.11 per kWh.

Incandescent annual cost:

$\text{kWh} = \frac{60 \times 100 \times 10 \times 260}{1{,}000} = \frac{15{,}600{,}000}{1{,}000} = 15{,}600 \text{ kWh}$

$\text{Cost} = 15{,}600 \times 0.11 = 1{,}716.00$

LED annual cost:

$\text{kWh} = \frac{9 \times 100 \times 10 \times 260}{1{,}000} = \frac{2{,}340{,}000}{1{,}000} = 2{,}340 \text{ kWh}$

$\text{Cost} = 2{,}340 \times 0.11 = 257.40$

Annual savings:

$1{,}716.00 - 257.40 = 1{,}458.60 \text{ per year}$

Percentage reduction:

$\frac{1{,}458.60}{1{,}716.00} \times 100 = 85\%$

Answer: Switching to LED saves 1,458.60 per year — an 85% reduction in lighting energy costs. If LED bulbs cost 5.00 each (500.00 total), the payback period is about 4 months.

Example 3: Basic Material Takeoff

Scenario: You are bidding a job to wire a small addition with 6 new circuits. You need to estimate wire and conduit costs.

Material	Quantity	Unit Cost	Total
#12 AWG THHN (per 1,000 ft spool)	2 spools	85.00 each	170.00
3/4” EMT (10 ft sticks)	25 sticks	4.50 each	112.50
3/4” EMT connectors	50	0.85 each	42.50
3/4” EMT couplings	25	0.65 each	16.25
20A single-pole breakers	6	8.50 each	51.00
Single-gang device boxes	18	1.75 each	31.50
Duplex receptacles	12	2.25 each	27.00
Single-pole switches	6	2.50 each	15.00
Material subtotal			465.75

Adding markup and waste:

$\text{Material with 10% waste} = 465.75 \times 1.10 = 512.33$

$\text{Material with 20% markup} = 512.33 \times 1.20 = 614.79$

Answer: Material cost with waste and markup is approximately 614.79.

Monthly and Annual Cost Quick Reference

Appliance	Watts	Hours/Day	Daily kWh	Monthly kWh (30 days)	Monthly Cost at 0.12/kWh
LED bulb (60W equiv.)	9	6	0.054	1.62	0.19
Incandescent bulb	60	6	0.36	10.8	1.30
Refrigerator	150 (avg)	24	3.6	108	12.96
Window AC unit	1,200	8	9.6	288	34.56
Space heater	1,500	8	12.0	360	43.20
Electric water heater	4,500	3 (avg)	13.5	405	48.60
Electric dryer	5,000	1	5.0	150	18.00
EV charger (Level 2)	7,200	4	28.8	864	103.68

Demand Cost Estimation

Some commercial accounts are billed for both energy (kWh) and demand (kW). Demand is the peak power draw during a billing period, measured in 15-minute intervals.

$\text{Demand charge} = \text{Peak kW} \times \text{Demand rate}$

If a shop draws a peak of 45 kW and the demand rate is 8.50 per kW:

$\text{Demand charge} = 45 \times 8.50 = 382.50 \text{ per month}$

This is added on top of the energy charge.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: A 500W outdoor security light runs 12 hours per night, 365 days per year, at 0.10 per kWh. What is the annual cost?

$\text{kWh} = \frac{500 \times 12 \times 365}{1{,}000} = 2{,}190 \text{ kWh}$

$\text{Cost} = 2{,}190 \times 0.10 = 219.00$

Answer: The annual cost is 219.00. Switching to a 70W LED flood light would cost about 30.66 per year — a savings of 188.34.

Problem 2: A customer asks how much they will save by replacing twenty 100W incandescent bulbs with 15W LED bulbs. Lights run 8 hours/day, 365 days/year, at 0.13/kWh.

Incandescent: $\frac{100 \times 20 \times 8 \times 365}{1{,}000} = 5{,}840$ kWh, cost = $5{,}840 \times 0.13 = 759.20$

LED: $\frac{15 \times 20 \times 8 \times 365}{1{,}000} = 876$ kWh, cost = $876 \times 0.13 = 113.88$

Savings: $759.20 - 113.88 = 645.32$

Answer: Annual savings of 645.32.

Problem 3: You need 1,400 feet of #12 THHN for a job. Spools come in 500 ft. How many spools do you need, and what is the cost at 85.00 per spool with 10% waste allowance?

With 10% waste: $1{,}400 \times 1.10 = 1{,}540$ ft

Spools needed: $\frac{1{,}540}{500} = 3.08$ — round up to 4 spools

Cost: $4 \times 85.00 = 340.00$

Answer: Purchase 4 spools at a total cost of 340.00.

Problem 4: A 240V, 4,500W water heater runs an average of 3 hours per day. At 0.12/kWh, what does it cost per month (30 days)?

$\text{Monthly kWh} = \frac{4{,}500 \times 3 \times 30}{1{,}000} = 405 \text{ kWh}$

$\text{Cost} = 405 \times 0.12 = 48.60$

Answer: The water heater costs 48.60 per month.

Problem 5: A commercial building uses 12,000 kWh per month at 0.09/kWh energy rate, plus a peak demand of 60 kW at 9.00/kW demand rate. What is the total monthly electric bill?

Energy charge: $12{,}000 \times 0.09 = 1{,}080.00$

Demand charge: $60 \times 9.00 = 540.00$

Total: $1{,}080.00 + 540.00 = 1{,}620.00$

Answer: The total monthly bill is 1,620.00.

Common Mistakes to Avoid

Forgetting to divide by 1,000. The kWh formula uses watts, but kWh means kilowatt-hours. Always divide watts by 1,000.
Confusing power and energy. A 100W bulb uses 100 watts of power. If it runs for 10 hours, it consumes 1 kWh of energy. Power is the rate; energy is the total.
Using peak wattage for average consumption. Some appliances (like refrigerators) cycle on and off. Use the average wattage, not the compressor’s peak draw.
Not accounting for waste in material takeoffs. Always add 10-15% waste for wire and conduit. Cuts, mistakes, and scrap are unavoidable.
Forgetting demand charges on commercial bids. Energy cost alone does not tell the whole story for commercial customers.

Key Takeaways

The kWh formula ( $\frac{\text{W} \times \text{h}}{1{,}000}$ ) converts power usage to the billing unit
Multiply kWh by the rate to get cost — this works for any appliance, circuit, or building
LED retrofits typically save 75-85% on lighting energy costs with payback periods under 1 year
Material takeoffs should include 10-15% waste and your markup percentage
Commercial estimates must include both energy (kWh) and demand (kW) charges

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Last updated: March 28, 2026