HVAC IMPORTANT FORMULAS | A Quick Reference for HVAC Engineers.🖊️📘
Understanding core HVAC formulas is essential for accurate system design, energy efficiency, and indoor comfort.
• Sensible Heat & Latent Heat calculations
• Total Heat (Enthalpy method)
• Air Changes per Hour (ACH)
• Fan Power calculation
•Coefficient of Performance (COP)
• Specific Volume of Air
• Velocity of Airflow
• Psychrometric relationships
These formulas are widely used in HVAC design, load calculation, airflow analysis, duct sizing, and energy analysis.
#HVAC #HVACDesign #AirConditioning #HVACTechnician #CoolingLoad #Airflow #REVIT #Navisworks #MEP #BIM #AEC #BIM_MODELER #ELECTRICAL ENGINEERING
Understanding core HVAC formulas is essential for accurate system design, energy efficiency, and indoor comfort.
• Sensible Heat & Latent Heat calculations
• Total Heat (Enthalpy method)
• Air Changes per Hour (ACH)
• Fan Power calculation
•Coefficient of Performance (COP)
• Specific Volume of Air
• Velocity of Airflow
• Psychrometric relationships
These formulas are widely used in HVAC design, load calculation, airflow analysis, duct sizing, and energy analysis.
#HVAC #HVACDesign #AirConditioning #HVACTechnician #CoolingLoad #Airflow #REVIT #Navisworks #MEP #BIM #AEC #BIM_MODELER #ELECTRICAL ENGINEERING
#CHILLER_SYSTEM_VS_VRV (VRF) SYSTEM
🔷 1. What is a Chiller System?
A chiller system cools water, not air directly.
How it works:
Water is cooled inside the chiller.
Chilled water is pumped to AHU or FCU units.
Air is cooled when it passes over the chilled water coil.
Cooling Medium
Chilled Water
🔷 Main Components of a Chiller System
Chiller unit (compressor, evaporator, condenser)
Chilled water pump
Condenser water pump
Cooling tower (for water cooled chillers)
AHU or FCU
Water piping
🔷 How a Chiller System Works (Simple)
The chiller cools water to about 5, 6, or 7°C.
The pump sends chilled water to AHU or FCU.
Air passes over the chilled water coil and becomes cool.
Water returns warmer, around 10, 11, or 12°C, to the chiller.
The cycle repeats continuously.
🔷 2. What is a VRV or VRF System?
A VRV or VRF system cools air directly using refrigerant.
There is no chilled water system.
Each indoor unit receives only the cooling it needs.
Cooling Medium
Refrigerant (DX system)
🔷 Main Components of a VRV System
VRV outdoor unit with inverter compressor
Indoor units (cassette, ducted, wall mounted)
Copper refrigerant piping
Electronic Expansion Valve (EEV)
Controller or BMS
🔷 How a VRV System Works (Simple)
The compressor pumps refrigerant.
Refrigerant flows to indoor units.
EEV controls the refrigerant flow.
Air is cooled directly over the DX coil.
Refrigerant returns to the outdoor unit.
🔷 3. Advantages and Disadvantages
Chiller System Advantages
Best for very large buildings
Long service life, around 20 to 25 years
Stable and consistent cooling
Easy integration with BMS
Can handle large AHU capacities
🔷 Chiller System Disadvantages
High initial cost
Large space required
Higher maintenance
Complex piping
Slow response time
🔷 VRV or VRF System Advantages
High energy efficiency
Individual room temperature control
No pumps or cooling tower
Requires less space
Fast cooling response
Clean installation
🔷 VRV or VRF System Disadvantages
High initial cost
Limit on refrigerant quantity
Requires skilled technicians
Not suitable for very large open areas
🔷 4. Where to Use Each System
Chiller Systems are suitable for
Airports
Shopping malls
IT parks
Large hospitals
Industrial buildings
🔷 VRV or VRF Systems are suitable for
Hotels
Offices
Residential towers
Hospital rooms
Schools
🔷 5. One Line Interview Answer
A chiller system uses chilled water for cooling, while a VRV or VRF system uses refrigerant directly to cool the air.
#HVAC #ChillerSystem #VRV #VRF #HVACTechnician
#AirConditioning #MechanicalEngineering #FacilitiesManagement
#BuildingServices #CoolingSystem #HVAC #bihar #usa #arwal #airdesigncoolingsystem #airconditioner #indian
🔷 1. What is a Chiller System?
A chiller system cools water, not air directly.
How it works:
Water is cooled inside the chiller.
Chilled water is pumped to AHU or FCU units.
Air is cooled when it passes over the chilled water coil.
Cooling Medium
Chilled Water
🔷 Main Components of a Chiller System
Chiller unit (compressor, evaporator, condenser)
Chilled water pump
Condenser water pump
Cooling tower (for water cooled chillers)
AHU or FCU
Water piping
🔷 How a Chiller System Works (Simple)
The chiller cools water to about 5, 6, or 7°C.
The pump sends chilled water to AHU or FCU.
Air passes over the chilled water coil and becomes cool.
Water returns warmer, around 10, 11, or 12°C, to the chiller.
The cycle repeats continuously.
🔷 2. What is a VRV or VRF System?
A VRV or VRF system cools air directly using refrigerant.
There is no chilled water system.
Each indoor unit receives only the cooling it needs.
Cooling Medium
Refrigerant (DX system)
🔷 Main Components of a VRV System
VRV outdoor unit with inverter compressor
Indoor units (cassette, ducted, wall mounted)
Copper refrigerant piping
Electronic Expansion Valve (EEV)
Controller or BMS
🔷 How a VRV System Works (Simple)
The compressor pumps refrigerant.
Refrigerant flows to indoor units.
EEV controls the refrigerant flow.
Air is cooled directly over the DX coil.
Refrigerant returns to the outdoor unit.
🔷 3. Advantages and Disadvantages
Chiller System Advantages
Best for very large buildings
Long service life, around 20 to 25 years
Stable and consistent cooling
Easy integration with BMS
Can handle large AHU capacities
🔷 Chiller System Disadvantages
High initial cost
Large space required
Higher maintenance
Complex piping
Slow response time
🔷 VRV or VRF System Advantages
High energy efficiency
Individual room temperature control
No pumps or cooling tower
Requires less space
Fast cooling response
Clean installation
🔷 VRV or VRF System Disadvantages
High initial cost
Limit on refrigerant quantity
Requires skilled technicians
Not suitable for very large open areas
🔷 4. Where to Use Each System
Chiller Systems are suitable for
Airports
Shopping malls
IT parks
Large hospitals
Industrial buildings
🔷 VRV or VRF Systems are suitable for
Hotels
Offices
Residential towers
Hospital rooms
Schools
🔷 5. One Line Interview Answer
A chiller system uses chilled water for cooling, while a VRV or VRF system uses refrigerant directly to cool the air.
#HVAC #ChillerSystem #VRV #VRF #HVACTechnician
#AirConditioning #MechanicalEngineering #FacilitiesManagement
#BuildingServices #CoolingSystem #HVAC #bihar #usa #arwal #airdesigncoolingsystem #airconditioner #indian
CARNOT ENGINE /THEOREM
A Carnot engine is an ideal heat engine that operates on a reversible thermodynamic cycle between two heat reservoirs at different absolute temperatures and converts heat into work with maximum possible efficiency.
➡️Carnot Principle / Carnot Theorem
Statement of Carnot Theorem:
No heat engine operating between two given temperatures can be more efficient than a reversible (Carnot) engine operating between the same temperatures.
All reversible heat engines operating between the same two temperatures have the same efficiency, irrespective of the working substance.
Implication:
The efficiency of a heat engine depends only on the temperatures of the heat reservoirs, not on the nature of the engine.
STRUCTURE:
Carnot engine consists of:
• A cylinder filled with an ideal gas.
• A frictionless piston.
• A hot reservoir at temperature Th.
• A cold reservoir at temperature Tc.
• All processes occur reversibly and without friction.
➡️Working of Carnot Engine (Carnot Cycle)
The Carnot cycle consists of four reversible processes:
Process 1: Isothermal Expansion at Temperature Th:
-The gas is kept in contact with the hot reservoir at temperature Th.
-Heat Qh is absorbed by the gas.
-Temperature remains constant.
-The gas expands and does work.
For an ideal gas:
Qh = W1
Qh = n R Th ln(V2 / V1)
where:
n = number of moles
R = gas constant
V1, V2 = initial and final volumes
Process 2:
(Adiabatic Expansion)
The cylinder is thermally insulated.
No heat exchange takes place.
Q = 0
The gas expands and its temperature drops from Th to Tc.
For adiabatic process:
T V^(γ − 1) = constant
or
Th V2^(γ − 1) = Tc V3^(γ − 1)
where:
γ = Cp / Cv
Process 3:
(Isothermal Compression at Temperature Tc):
-The gas is placed in contact with the cold reservoir.
-Heat Qc is rejected to the cold reservoir.
-Temperature remains constant.
-Gas is compressed.
For an ideal gas:
Qc = W3
Qc = n R Tc ln(V3 / V4)
(Note: Qc is negative since heat is rejected)
PROCESS 4:
(Adiabatic Compression);
-The gas is insulated again.
-No heat exchange occurs.
Q = 0
The gas is compressed and temperature rises from Tc back to Th.
Adiabatic relation:
Tc V4^(γ − 1) = Th V1^(γ − 1)
The system returns to its initial state.
➡️ Efficiency of Carnot Engine
Efficiency (η) is defined as:
η = Work output / Heat absorbed
η = W / Qh
Net work done:
W = Qh − Qc
So,
η = (Qh − Qc) / Qh
η = 1 − (Qc / Qh)
From thermodynamics of reversible engines:
Qc / Qh = Tc / Th
Therefore,
η = 1 − (Tc / Th)
where
Th = absolute temperature of hot reservoir (in Kelvin)
Tc = absolute temperature of cold reservoir (in Kelvin)
In general,
•Efficiency depends only on Th and Tc.
•Higher Th or lower Tc increases efficiency.
•Efficiency can never be 100% unless Tc = 0 K (impossible).
•No real engine can exceed Carnot efficiency.
APPLICATIONS OF CARNOT ENGINE:
Although a Carnot engine cannot be built in practice, it is used for:
-Determining maximum possible efficiency of real engines.
-Comparing performances of heat engines.
-Understanding the second law of thermodynamics.
-Designing power plants and thermal systems.
Basis of refrigerators and heat pumps (reversed Carnot cycle).
EXAMPLES
Q1.
Calculate the efficiency of an operating at:
Th = 600 K, Tc = 300 K.
Solution ⭐
Formula:
η = 1 − (Tc / Th)
Substitute values:
η = 1 − (300 / 600)
η = 1 − 0.5
η = 0.5
Efficiency = 50%
Q2.
A Carnot engine operating between two heat reservoirs at temperatures: Th = 500 K and
Tc = 300 K. If the heat absorbed, Qh is 2000J, calculate the work done.
Solution ⭐
Heat absorbed Qh = 2000 J
Step 1: Efficiency
η = 1 − (300 / 500)
η = 1 − 0.6
η = 0.4
Step 2: Work done
W = η × Qh
W = 0.4 × 2000
W = 800 J
Q3.
The heat absorbed, Qh by an engine whose efficiency is 25% is 4000J. Calculate the heat rejected.
Solution ⭐
Given:
Efficiency = 25% = 0.25
Heat absorbed Qh = 4000 J
Work done:
W = η × Qh
W = 0.25 × 4000
W = 1000 J
Heat rejected:
Qc = Qh − W
Qc = 4000 − 1000
Qc = 3000 J
A Carnot engine is an ideal heat engine that operates on a reversible thermodynamic cycle between two heat reservoirs at different absolute temperatures and converts heat into work with maximum possible efficiency.
➡️Carnot Principle / Carnot Theorem
Statement of Carnot Theorem:
No heat engine operating between two given temperatures can be more efficient than a reversible (Carnot) engine operating between the same temperatures.
All reversible heat engines operating between the same two temperatures have the same efficiency, irrespective of the working substance.
Implication:
The efficiency of a heat engine depends only on the temperatures of the heat reservoirs, not on the nature of the engine.
STRUCTURE:
Carnot engine consists of:
• A cylinder filled with an ideal gas.
• A frictionless piston.
• A hot reservoir at temperature Th.
• A cold reservoir at temperature Tc.
• All processes occur reversibly and without friction.
➡️Working of Carnot Engine (Carnot Cycle)
The Carnot cycle consists of four reversible processes:
Process 1: Isothermal Expansion at Temperature Th:
-The gas is kept in contact with the hot reservoir at temperature Th.
-Heat Qh is absorbed by the gas.
-Temperature remains constant.
-The gas expands and does work.
For an ideal gas:
Qh = W1
Qh = n R Th ln(V2 / V1)
where:
n = number of moles
R = gas constant
V1, V2 = initial and final volumes
Process 2:
(Adiabatic Expansion)
The cylinder is thermally insulated.
No heat exchange takes place.
Q = 0
The gas expands and its temperature drops from Th to Tc.
For adiabatic process:
T V^(γ − 1) = constant
or
Th V2^(γ − 1) = Tc V3^(γ − 1)
where:
γ = Cp / Cv
Process 3:
(Isothermal Compression at Temperature Tc):
-The gas is placed in contact with the cold reservoir.
-Heat Qc is rejected to the cold reservoir.
-Temperature remains constant.
-Gas is compressed.
For an ideal gas:
Qc = W3
Qc = n R Tc ln(V3 / V4)
(Note: Qc is negative since heat is rejected)
PROCESS 4:
(Adiabatic Compression);
-The gas is insulated again.
-No heat exchange occurs.
Q = 0
The gas is compressed and temperature rises from Tc back to Th.
Adiabatic relation:
Tc V4^(γ − 1) = Th V1^(γ − 1)
The system returns to its initial state.
➡️ Efficiency of Carnot Engine
Efficiency (η) is defined as:
η = Work output / Heat absorbed
η = W / Qh
Net work done:
W = Qh − Qc
So,
η = (Qh − Qc) / Qh
η = 1 − (Qc / Qh)
From thermodynamics of reversible engines:
Qc / Qh = Tc / Th
Therefore,
η = 1 − (Tc / Th)
where
Th = absolute temperature of hot reservoir (in Kelvin)
Tc = absolute temperature of cold reservoir (in Kelvin)
In general,
•Efficiency depends only on Th and Tc.
•Higher Th or lower Tc increases efficiency.
•Efficiency can never be 100% unless Tc = 0 K (impossible).
•No real engine can exceed Carnot efficiency.
APPLICATIONS OF CARNOT ENGINE:
Although a Carnot engine cannot be built in practice, it is used for:
-Determining maximum possible efficiency of real engines.
-Comparing performances of heat engines.
-Understanding the second law of thermodynamics.
-Designing power plants and thermal systems.
Basis of refrigerators and heat pumps (reversed Carnot cycle).
EXAMPLES
Q1.
Calculate the efficiency of an operating at:
Th = 600 K, Tc = 300 K.
Solution ⭐
Formula:
η = 1 − (Tc / Th)
Substitute values:
η = 1 − (300 / 600)
η = 1 − 0.5
η = 0.5
Efficiency = 50%
Q2.
A Carnot engine operating between two heat reservoirs at temperatures: Th = 500 K and
Tc = 300 K. If the heat absorbed, Qh is 2000J, calculate the work done.
Solution ⭐
Heat absorbed Qh = 2000 J
Step 1: Efficiency
η = 1 − (300 / 500)
η = 1 − 0.6
η = 0.4
Step 2: Work done
W = η × Qh
W = 0.4 × 2000
W = 800 J
Q3.
The heat absorbed, Qh by an engine whose efficiency is 25% is 4000J. Calculate the heat rejected.
Solution ⭐
Given:
Efficiency = 25% = 0.25
Heat absorbed Qh = 4000 J
Work done:
W = η × Qh
W = 0.25 × 4000
W = 1000 J
Heat rejected:
Qc = Qh − W
Qc = 4000 − 1000
Qc = 3000 J
Tape measure has two systems: metric (centimeters) and imperial (inches).
- Metric side: one centimeter equals ten millimeters.
- Imperial side: one inch is divided into fractions, not ten parts.
- Longest stripe represents one half inch.
- Next longest stripe represents one quarter inch.
- Smaller stripes represent one eighth inch and one sixteenth inch.
- Confusion between centimeters and inches causes mistakes in DIY projects.
- Correct reading prevents wood or material being cut short.
- Knowing fractions of an inch is a practical skill.
- Helps in fixing things, buying furniture, and doing crafts.
- Avoid saying “two inches and two stripes” by learning fractions.
- Understanding tape measure is a useful everyday tool skill.
#SiteEngineer #BuildingKnowledge #EngineeringTips #CivilWorks #TechnicalSkills #LearnCivil #SmartConstruction #EngineerLife #MeasurementTools #WorkSmart
- Metric side: one centimeter equals ten millimeters.
- Imperial side: one inch is divided into fractions, not ten parts.
- Longest stripe represents one half inch.
- Next longest stripe represents one quarter inch.
- Smaller stripes represent one eighth inch and one sixteenth inch.
- Confusion between centimeters and inches causes mistakes in DIY projects.
- Correct reading prevents wood or material being cut short.
- Knowing fractions of an inch is a practical skill.
- Helps in fixing things, buying furniture, and doing crafts.
- Avoid saying “two inches and two stripes” by learning fractions.
- Understanding tape measure is a useful everyday tool skill.
#SiteEngineer #BuildingKnowledge #EngineeringTips #CivilWorks #TechnicalSkills #LearnCivil #SmartConstruction #EngineerLife #MeasurementTools #WorkSmart
Field Duct Sizing Chart TABLES 👨🔧💯✅️
#ductwork #ducting #DUCTINGWORK #VRV #VRF #airconditioner #airconditioning #hvac #business #success #viral #1million #explore #followme #solutions
#ductwork #ducting #DUCTINGWORK #VRV #VRF #airconditioner #airconditioning #hvac #business #success #viral #1million #explore #followme #solutions
Lg Dual Inverter Split Ac Error Code List 💯🚀👨🔧
#LG #LGDualInverter #LGAirConditioner #HVAC #airconditioningrepair #airconditioner #Error #code #problemsolving #technician #business
#LG #LGDualInverter #LGAirConditioner #HVAC #airconditioningrepair #airconditioner #Error #code #problemsolving #technician #business
Diagrams of Conventional and Double Door Refrigerator Refrigeration Systems