Saturday, April 22, 2017
Air conditioner Complete selection and installation guide
Why You Must Know Room Area Before Selecting an Air Conditioner
When it comes to buying an air conditioner, one of the most common questions in every customer’s mind is:
👉 “What size of AC is right for my room?”
Most people rely on sellers, friends, or other customers’ experiences. But here’s the truth: an incorrectly sized air conditioner will either waste electricity or fail to cool your room properly.
That’s why understanding the room area and cooling requirements is crucial before you make your purchase.
Why AC Sizing Matters
-
If AC capacity is higher than required:
-
Excessive power consumption
-
Cooling cycle stops early → humidity not removed → damp and clammy feeling
-
Lower efficiency and shorter equipment life
-
-
If AC capacity is lower than required:
-
Room will never reach comfortable cooling
-
AC runs continuously → higher electricity bills
-
Equipment wears out faster
-
Thus, the right-sized AC ensures effective cooling, proper dehumidification, and optimal electricity use.
Functions of an Air Conditioner
-
Cooling – lowers the room temperature to the desired level
-
Dehumidification – removes excess moisture from the air
If an oversized AC is installed, the room may cool quickly but without sufficient dehumidification, leaving it cool but humid, which feels uncomfortable.
How to Measure Room Area
-
Square or rectangular room: Length × Width
-
Triangular room: (Length × Width) ÷ 2
-
Irregular-shaped room: Divide into smaller sections, calculate area, then sum them up
For accurate measurement, take the help of an AC technician or sales expert.
Cooling Capacity Requirement
Air conditioners are rated in BTUs/hour (British Thermal Units).
-
1 Ton of cooling ≈ 12,000 BTU/hour
Here’s a simplified Room Size vs AC Size Guide:
| |||||||||||||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Additional Factors Affecting AC Capacity |
-
Room Orientation & Sunlight
-
Shaded rooms → reduce capacity by 10%
-
Sun-facing rooms → increase capacity by 10%
-
-
Number of Occupants
-
Standard: 2 people
-
Add 600 BTU/hour per extra person
-
-
Kitchen Usage
-
Cooking adds extra heat load
-
Increase capacity by 4,000–5,000 BTU/hour (≈30%)
-
Indoor & Outdoor Unit Installation
Indoor Unit
-
Install at 7–9 ft height for effective airflow
-
Place where air distributes evenly
-
Ideally above the bed (but not blowing directly on the body)
-
Adjust louvers for better airflow
Outdoor Unit
-
Install in an open space (terrace/external wall) with free airflow
-
Position above indoor unit for better refrigerant flow
-
Keep distance ≤15 meters between indoor & outdoor units
-
Ensure easy accessibility for maintenance
Single-Stage vs Two-Stage Compressor ACs
-
Single-Stage AC:
-
Works only at full capacity
-
Less efficient, higher energy bills
-
-
Two-Stage AC:
-
Operates at both high and low levels
-
Adjusts cooling as per outside temperature
-
Saves electricity and provides more comfort
-
Conclusion
Choosing the right size of air conditioner is not just about comfort, but also about energy savings, efficiency, and long-term durability. By measuring your room size, calculating BTU requirements, and considering additional factors (sunlight, occupants, kitchen use), you can select the most suitable AC for your home.
Disclaimer
The above information is for educational purposes only. Actual AC size requirements may vary based on room insulation, ceiling height, climate conditions, and manufacturer specifications. Always consult a qualified HVAC technician before making your final purchase.
Wednesday, April 19, 2017
Air conditioners ISEER rating; Power consumption calculations from ISEER rating
A complete guide for right rating selection and installation guide for air conditioners:-
http://electrialstandards.blogspot.com/2017/04/air-conditioner-complete-selection-and.html
Friday, April 14, 2017
Transmission conductors are Aluminum instead of copper
Why Aluminum Conductors Are Used in Transmission Instead of Copper Conductors
Key Requirements for Transmission Line Conductors
The materials selected for overhead transmission lines must satisfy the following essential properties:
-
Low cost of construction
-
Low electrical resistivity
-
High electrical conductivity
-
Low temperature coefficient of resistance
-
High current-carrying capacity
-
Adequate mechanical strength
-
Good weather resistance and corrosion resistance
-
Sufficient elasticity for mechanical stability
Among available metals, copper, aluminum, and steel are widely considered. In practice, these are often used individually or in composite forms (e.g., ACSR – Aluminum Conductor Steel Reinforced).
Copper vs. Aluminum – A Technical Comparison
-
Conductivity:
Copper has the highest conductivity among common conductor materials. Aluminum offers about 61% of copper’s conductivity. -
Weight:
For the same resistance, aluminum weighs nearly half that of copper. This reduced weight significantly impacts tower design and overall infrastructure cost. -
Resistance:
On a per-length basis, aluminum has about 1.6 times the resistance of copper of the same gauge. This implies 60% higher I²R losses for the same conductor cross-section. However, this disadvantage can be mitigated by increasing the cross-sectional area of aluminum conductors.
Why Aluminum is Preferred Over Copper
(i) Lower Cost
-
Aluminum is 5–6 times cheaper than copper on a per-kilogram basis.
-
For long-distance transmission (hundreds of kilometers), the raw material cost savings are enormous.
-
Copper’s higher weight demands stronger and costlier transmission towers, whereas aluminum’s lightweight nature reduces tower design and foundation costs.
(ii) Lower Density & Better Cross-Sectional Design
-
Aluminum’s density is about one-third that of copper.
-
By increasing the aluminum conductor’s cross-sectional area, resistance decreases, and I²R losses reduce significantly.
-
Example: Doubling the cross-sectional area of aluminum reduces I²R losses by about 20%. Even after increasing diameter, aluminum still weighs only about two-thirds of the equivalent copper conductor, keeping tower loads manageable.
-
Net result: Up to 90% savings in conductor costs when considering both material and infrastructure.
Advantages of Aluminum Conductors
-
Reduced Power Losses: Increasing conductor size decreases resistance, reducing I²R losses (from ~6% to ~5% in practical cases).
-
Lower Construction Cost: Lightweight conductors reduce tower strength requirements, cutting infrastructure cost by up to 33%.
-
Material Cost Savings: With aluminum being much cheaper, the total wire cost is around 10% of copper conductors for equivalent transmission capacity.
-
Ease of Handling: Aluminum is more elastic and easier to string and bend compared to copper.
Why Not Steel Alone?
While steel is inexpensive and has very high tensile strength, it is unsuitable as the sole conductor due to:
-
High magnetic permeability → reduces effective skin depth at 50/60 Hz.
-
Higher AC resistance → leads to excessive energy losses.
Thus, steel is mainly used as a reinforcement core (e.g., in ACSR conductors) rather than as the primary conducting medium.
Conclusion
Although copper offers superior conductivity and lower resistance, aluminum’s lightweight, lower cost, and adaptability in transmission design make it the preferred choice for long-distance power transmission. The trade-off of slightly higher losses is outweighed by the massive savings in material, construction, and maintenance costs.
Disclaimer
The above article is intended for educational and informational purposes only. While every effort has been made to ensure technical accuracy, actual conductor selection in transmission systems depends on multiple factors, including regional standards, cost variations, environmental conditions, and grid requirements. Engineers and project planners should always refer to national and international standards (such as IEC, IEEE, and IS codes) and perform detailed feasibility studies before finalizing material choices.
Wednesday, April 12, 2017
Energy Meter connection;Single Phase; Three Phase; CT operated Energy meters
Friday, April 7, 2017
Single Phase preventer working and installation Procedure
Tuesday, April 4, 2017
Energy Savings with IE1, IE2 & IE3 Motors
Energy savings with efficient motors — why it matters
Motors power virtually every industry and commercial facility. Because motors typically account for the majority of industrial electricity use, improving motor efficiency is one of the fastest ways to cut energy bills, reduce carbon emissions and improve profitability.
Key facts:
-
In many industries, motors account for ~70% of total electrical consumption.
-
Over a motor’s lifetime, ~88% of total lifecycle cost is energy cost (purchase, maintenance and downtime account for the rest).
→ This makes efficiency improvements extremely cost-effective.
How premium/high-efficiency motors reduce losses
Extra-premium / high-efficiency motors achieve lower losses by:
-
Reducing stator copper losses — more active copper/optimized winding designs.
-
Reducing core (iron) losses — higher-grade silicon steel and better lamination.
-
Reducing friction & windage losses — improved fan and bearing design.
-
Better thermal design and materials → lower thermal and electrical stress, longer life.
Advantages of extra-premium motors vs IE2 / IE3
-
Stable high efficiency from ~60%–100% load, yielding better real-world savings.
-
Lower electricity consumption (direct bill savings).
-
Better thermal and electrical stress resistance, enabling higher ambient operation.
-
Less frequent rewind-related efficiency degradation (and better OEM support).
Indian standards overview:
-
IS:12615 (1989) — initial energy-efficient motor standard (covers up to 37 kW, 4-pole).
-
Revision I (2004) — extended to 0.37 kW–160 kW (2–4 pole) plus 6 & 8 pole ranges; introduced Eff1/Eff2 and IEC 60034-2 test methodology.
-
Revision II (2011) — aligned with IEC 60034-30: introduced IE1/IE2/IE3 classes, extended motor range (0.37 kW–375 kW for 2/4/6 pole), and required test standard IS:15999 / IEC 60034-2-1; added parameters (breakaway torque, currents, etc.).
When should you replace motors? — Practical step-by-step
-
Build a motor database: nameplate details, rated efficiency, age, operating hours, location, duty cycle.
-
Record failures & rewinds: count of rewinds and repairs for each motor.
-
Compare OEM efficiency curves and document how rewinds affected efficiency (typical drop: 1%–5% per rewind).
-
Flag motors >8–10 years old for detailed review (higher chance of being inefficient).
-
If rewinding cost is high for small/medium motors (<50 kW), compare replacement vs repair.
-
Run a payback calculation: compare annual energy cost (using actual loading and hours) vs incremental purchase cost. Replace when payback is acceptable for your business.
Worked example — 20 kW motor (practical cost-benefit)
Assumptions:
-
Motor rated output = 20 kW
-
Operating load = 80% → output in-service = 20 × 0.8 = 16 kW
-
Annual run hours = 20 hr/day × 365 = 7,300 hr/year
-
Electricity cost = ₹8.00 / kWh
Energy consumed (input kWh/year) = (output kW × hours) / efficiency
Annual cost = input kWh/year × ₹8.00
Efficiencies used:
-
IE1 (Eff2) = 88.70% (0.887)
-
IE2 = 90.60% (0.906)
-
IE3 = 92.10% (0.921)
Calculated annual energy cost
| Motor class | Efficiency | Input energy (kWh/year) | Annual energy cost (₹) |
|---|---|---|---|
| IE1 | 88.70% | 131,679.82 kWh | ₹1,053,438.56 |
| IE2 | 90.60% | 128,918.32 kWh | ₹1,031,346.58 |
| IE3 | 92.10% | 126,818.68 kWh | ₹1,014,549.40 |
Annual savings (rounded)
-
IE1 → IE2 savings = ₹22,091.98 / year
-
IE1 → IE3 savings = ₹38,889.16 / year
-
IE2 → IE3 savings = ₹16,797.18 / year
Typical purchase prices (example)
-
IE1: ₹45,000
-
IE2: ₹55,000
-
IE3: ₹63,000
Payback (months)
-
IE1 → IE2: additional cost ₹10,000 → payback ≈ 5.43 months
-
IE1 → IE3: additional cost ₹18,000 → payback ≈ 5.55 months
-
IE2 → IE3: additional cost ₹8,000 → payback ≈ 5.72 months
Conclusion: With the assumptions above (heavy annual hours and ₹8/kWh), extra premium motors pay back within about 5–6 months — an excellent return on investment. (Adjust your inputs — run hours, energy price and actual motor efficiencies — to get plant-specific payback.)
Practical checklist before replacement
-
Confirm actual operating load profile (measure average % loading, not assumed).
-
Verify actual run hours and seasonal usage.
-
Check power tariff (energy charge, demand charge, time-of-day). Some savings affect demand charges too.
-
Include downtime, maintenance and scheduled life in lifecycle cost.
-
Validate OEM efficiency & warranty, and check service availability.
-
If motor is critically sized and variable-load, consider VFD + premium motor for maximum savings and process control.
Quick recommendations
-
For motors with high run-hours (>3,000–4,000 hr/yr) and older than 8 years, strongly consider replacement with IE3 or premium motors.
-
For frequent rewinds or where rewinding costs approach new-motor price, prefer replacement.
-
Use the site-specific payback model — plug actual tariffs, hours, and measured loading to get precise ROI.
-
Combine motor replacement with operational measures (VFDs, power factor correction, scheduled maintenance) for extra savings.








