# HVAC basics

## It's important that young (and seasoned) engineers remain familiar with the principles and equations behind HVAC design and modeling software.

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The design of HVAC systems is both an art and a science. While at one time it was dominated by hand calculations and “rules of thumb,” that has given way to computer models and simulations. Although these new methods are at once useful and powerful, they have caused practitioners to lose some “feel” for the basics.

This article presents some of the most fundamental and useful equations used for HVAC design along with example calculations for air and water systems.

Approximating the central plant

A seasoned engineer can make good progress on conceptual designs using estimates and rules of thumb that can subsequently be refined by computer simulations as a building’s design progresses. Waiting for an architect to develop the exact building footprint or draw wall sections to develop the computer model is not practical when mechanical rooms and shafts need to be roughed out and when the electrical engineer needs to plan service and work on one-line diagrams. Estimating the anticipated airflows and the sizes of chiller and boiler plants needs to be accomplished early in the conceptual design phase in order to make progress as an integrated design team.

Early in the design, architects will want to know how large mechanical equipment rooms will be, where they should be located, and how big louvers will be, which can be done with some level of precision simply by taking into account the type of facility planned and the climate in which it will be located. Office occupancies are the easiest to estimate because we can assume 1 cfm of airflow over the gross square footage and be very close to the final design value.

This rule of thumb is quite old, and even though lighting wattage has decreased over the years and building envelopes have made great improvements, the increase in plug loads due to computers, monitors, and printers have left this approximation intact. From 1 cfm per gross sq ft, apply the rule of thumb of 1 ton of refrigeration for every 400 cfm of system airflow. Considering both increased outdoor ventilation rates and climate, it is often good to hedge this by going with 350 cfm per ton for the office buildings.

Not only are these types of approximations useful for preliminary estimations, but they also can be used to review the output of computer simulations. If the output significantly varies from what is estimated, the reviewer should dig deeper into the input data to uncover and resolve potential errors.

Estimating preliminary HVAC central plant capacities is just as simple for other occupancy types. For example, laboratories and healthcare occupancies are typically driven by air change rates with patient rooms at 6 air changes per hour (ACH), operating rooms at 15 to 25 ACH, and laboratory spaces at 8 to 12 ACH.

Right-sizing and sustainability

Building owners have been critical of HVAC designers for years, and rightly so, for the over-design of building services where it appears that rules of thumb were used and never refined with definitive calculations. But it also should be noted that blindly accepting the output of a computer simulation without gauging it against some simple “back of the envelope” calculations before proceeding with design is not wise.

Right-sizing (using design software and models) and sustainability go hand-in-hand, but approximations and rules of thumb are important tools for engineers to use in the early phases of a project. These rules of thumb also are useful to building owners and operators to check the work of their consultants as well as to assess the capacities of HVAC systems in their existing building stock.

Air and water basics

The HVAC design profession uses a couple of important equations to calculate the thermal content of moving streams of air and water, which are used to heat and cool buildings. For example, the sensible heat carried by an airstream, expressed in Btu per hour, is equal to 1.085 multiplied by the cfm and the temperature difference between the airstream’s initial and final state (assuming standard conditions for the air). Many practitioners may not be aware of how the constant of 1.085 is derived and how to adjust its value for nonstandard conditions, such as when the system is operating at an altitude above sea level. The same goes for hydronic (water) systems, where the heat conveyed is equivalent to 500 multiplied by the gpm and temperature difference. The hydronic heat-transfer constant of 500 changes when a water/glycol solution is used as the heat transfer fluid. The sidebar equations show the derivations for these constants and how they are dependent upon fluid density.

The sidebar also includes useful guidelines for designing air and water distribution systems. While it is beneficial to gain the fundamentals of HVAC design from classroom learning, these types of guidelines are developed from designers’ collective experience over the years and can help in quickly assimilating the skills necessary to get up to speed as a productive designer. Design firms often take pride in their design manuals and have slightly different guidelines based upon experience, and these most likely will change over time. For example, at one time high-pressure duct distribution systems often were run at 3,000 to 4,000 fpm, which required large-horsepower fan systems and duct silencers. With modern energy codes, air distribution systems need to be operated at lower velocities, usually much less than 3,000 fpm, to comply with fan energy guidelines. It also should be noted that round duct systems usually can be operated at higher velocities than rectangular systems without noise being an issue.

Table 1. Water densities for different temperatures at sea level

Temperature (F) | Density (lbm/cu ft) |

32 | 62.42 |

40 | 62.43 |

50 | 62.41 |

60 | 62.37 |

68 | 62.31 |

70 | 62.30 |

80 | 62.22 |

90 | 62.11 |

100 | 62.00 |

140 | 61.38 |

160 | 61.00 |

180 | 60.58 |

200 | 60.12 |

212 | 59.53 |

While some of the more commonly used equations and guidelines are presented here, there are other sources of useful information such as the Mechanical Engineering Reference Manual and the ASHRAE Fundamentals Handbook. It is hoped that this information can be of use to new entrants to the HVAC design community as well the owners and operators of buildings that we design. If nothing else, it might get some discussion and debate going at the water cooler.

Table 2. Air densities for different temperatures at sea level

Temperature (F) | Density (lbm/cu ft) |

0 | 0.0862 |

20 | 0.0827 |

32 | 0.0810 |

40 | 0.0794 |

60 | 0.0763 |

68 | 0.0752 |

80 | 0.0735 |

100 | 0.0709 |

120 | 0.0684 |

250 | 0.0559 |

Table 3. Conversions for common engineering units

1 Watt | 3.413 Btu/hr |

1 EER | 0.293 COP |

1 HP | 0.7457 kW |

1 gallon | 0.1337 cu ft |

1 psi | 2.31 ft of head (H _{2} 0) |

1 cu ft/lb | 7.481 gal/lb |

1 lb | 7000 grains |

1 ton | 12,000 Btu/hr |

1 MBtu | 1,000 Btu |

Table 4. Selected constants

Specific heat of water | 1.000 Btu/(lbm x F) |

Specific heat of air | 0.240 Btu/(lbm x F) |

Density of water | 8.34 lb/gal at 60 F |

Table 5. Hydronic heat transfer constant values for water/ethylene glycol solutions. The constant is equal to 500 for 100% water at standard conditions.

Temperature (F) | 25 | 30 | 40 | 50 | 60 | 65 | 100 |

Ethylene glycol (% by volume) | |||||||

-40 | N/A | N/A | N/A | N/A | 381 | 398 | N/A |

0 | N/A | N/A | 449 | 429 | 402 | 392 | 313 |

40 | 479 | 471 | 452 | 433 | 412 | 400 | 322 |

80 | 479 | 473 | 456 | 439 | 419 | 407 | 334 |

120 | 481 | 475 | 460 | 443 | 425 | 414 | 341 |

Author Information |

Biada is a project manager with Advanced Engineering Consultants. He has four years of experience designing mechanical systems for commercial and institutional facilities. Schultz is a vice president of Advanced Engineering Consultants and has 20 years of mechanical design experience. |

Common rules of thumb

**HVAC hot water piping**

• Use either 20 F or 30 F

• Minimum return temperature 140 F to avoid condensation in boilers that are not designed for it.

• Use 4-ft/100-ft friction drop of 6 ft per sec (fps) to 8 fps.

• Use minimum of two boilers on most jobs. The boilers are usually sized at 2/3 capacity.

**HVAC chilled water**

Use a temperature difference of 10 F, 12 F, or 14 F

Use 4-ft /100-ft friction drop of 6 fps to 8 fps

**Ductwork**

Long radius elbows (1.5 times duct diameter)

Supply Air:

• Low pressure (2-in. or less): Pressure drop of 0.08-in./100-ft or a max velocity of 800 fpm to 1,200 fpm.

• High pressure (4-in. and up): Pressure drop of 0.2 in./100-ft or a max velocity of 2,200 fpm.

Return Air: Pressure drop of 0.08-in./100 ft or a velocity of 1,500 fpm.

Exhaust: Pressure drop of 0.08-in./100 ft or a velocity of 1,500 fpm.

Transfer ducts: Velocity of 250 to 500 fpm.

Door undercut: Max air transfer 150 cfm.

Louvers: Intake velocity 500 fpm.

Louvers: Relief/exhaust velocity 800 fpm.

Typical louver free area can vary from 15% to 65% depending on louver application; thus, it is important to divide air flow cfm by (area times free area %) to get louver velocity.

(i.e. (5,000 cfm / (louver area × free area percentage) = louver velocity)

General coil velocity: 500 fpm (this value may vary depending on the number of rows per coil).

Grilles, registers, diffusers: Max NC of 25 and velocity 800 fpm or less.

**Heating and cooling load info**

Heating: 35 Btu/sq ft.

Cooling: 400 cfm/ton.

Cooling: 250 to 450 sq ft/ton. (This depends on many factors, including space usage and building type.)

**Equipment info**

1 HP approx = 1 kW approx = 1 kVA (for exact values, use conversions).

Transformers: 2% to 3% of kVA is rejected into space as heat.

Exhaust fans: 3,000 cfm at 1 HP (at approximately 0.75 in. static pressure).

Chilled water: 2.4 gpm/ton at 10 F (i.e. 500 ton chiller = 1,200 gpm).

Condenser water: 3 gpm per ton at 10 F.

Cooling tower: 15 to 20 ton = 1 HP (axial fan).

Pumps: 50 gpm = 1 HP at 50-ft to 60-ft of head.

**Typical office equipment loads**

Computer: 75 W.

Monitor: 75 W.

Small printer: 120 W.

Copy machine: 800 W (idle 250 W)

People (light office work):

• 250 Btu/hr sensible heat.

• 200 Btu/hr latent heat.

Lighting: 0.9 to 1.4 W/sq ft: Can vary depending on space usage.

General equipment load: 1 to 2 W/sq ft

**Typical building construction U-values**

Wall U-value: 0.1 Btu/(hr × sq ft × F)

Window U-value: 0.5 Btu/(hr × sq ft × F)

Window solar heat gain coefficient (SHGC): 0.4

Roof U value: 0.06 Btu/(hr × sq ft × F)

Calculating hydronic flow rate

This equation is used to calculate the volumetric flow rate in gpm for a given amount of heat.

**Equation 1:**

V = Q / (C x %)

Where:

V= Volumetric flow rate (gpm)

Q = Heat flow rate (btu/hr)

C = Hydronic heat transfer constant: 500 ((Btu x min)/(gal x hr x F))

F = Difference in entering vs. leaving water temperature

While the hydronic heat transfer constant (C) is typically assumed to be 500 in most conditions, it can vary depending on fluid density and specific heat (Cp). Table 1 and Table 4 provide the density and specific heat of water at varying temperatures, which are then inserted in equation 2 to calculate the constant:

**Equation 2:**

C = x Cp x 60 (min/hr) x 0.134 (cu ft/gal)

Where:

C = Calculated hydronic heat transfer constant ((Btu x min)/(gal x hr x F))

F = Fluid density (lbm/cu ft)

Cp = Specific heat (Btu/(lbm x F))

However, because ethylene glycol is frequently mixed with water to lower the freezing point of a fluid when it is used outdoors, the hydronic heat transfer constant has to be determined based on the fraction of ethylene glycol in the solution. Table 5 provides the hydronic heat transfer constant value for various ethylene glycol/water solutions.

Assume a load of 750,000 Btu/hr and a %%TRANGL%%T = 12 F (Chilled water with no ethylene glycol):

Now, assume the same conditions, but this time, assume the chilled water will be a mixture of 50% ethylene glycol.

Note: In this particular case, the ethylene glycol caused about a 15% increase in required gpm, which can have an impact on pump and pipe sizing. If increasing the flow is not desired, the fluid %%TRANGL%%T or the amount of heat transfer can be varied to maintain the original gpm.

Sensible heat in an air stream

This equation is used to calculate the amount of heat that can be removed from a space with a certain amount of air at a certain

**Equation 3:**

Q = V / (C x T)

Where:

Q = Heat flow rate (Btu/hr)

V = Volumetric air flow (cfm)

C = Air heat transfer constant : 1.085 x ((Btu x min)/(cu ft x hr x F))

T = Difference in entering vs. leaving air temperature (F)

Sample calculation:

Assume air flow is 25,000 cfm with a T of 20 F (i.e., the desired room temperature is 75 F with a supply air temperature of 55 F).

Q = 25,000 cfm x 1.085 ((Btu x min)/(cu ft x hr x F)) x 20 F

Q = 542.5 MBtu / hr

The air heat transfer constant 1.085 is contingent on the density of air which can vary based on temperature and pressure. Table 2 provides the density of air at atmospheric pressure and varying temperatures, which can then be inserted into equation 4:

**Equation 4:**

C = %%PGREEK%% x Cp x 60 min/hr

Where:

C = Calculated air heat transfer constant ((Btu x min)/(cu ft x hr x F))

%%PGREEK%% = Fluid density (lbm/cu ft)

Cp = Specific heat (Btu/(lbm x F))

## Determining total heat

This equation is used to calculate the amount of heat a piece of equipment is capable of removing from a space (also known as the total cooling capacity of a piece of equipment such as a rooftop unit, air handling unit, fan coil, etc.).

**Equation 5:**

Q = C x V x %%TRANGL%%h

Where:

Q = Heat flow rate (Btu/hr)

C = Heat transfer constant ((lbm x min)/(cu ft x hr))

V = Volumetric air flow (cfm)

%%TRANGL%%h = Difference in enthalpy of mixed air and supply air (Btu/lbm)

Again, the heat transfer constant is dependent on the density of air; pick the appropriate value from Table 2 insert it into equation 5:

C = %%PGREEK%% x 60 min/hr

Where:

C = Heat transfer constant ((lbm x min)/(cu ft x hr))

%%PGREEK%% = Fluid density (lbm/cu ft)

Sample Calculation:

Assume the air flow is 42,000 cfm and a %%TRANGL%%h is 7.8 (i.e., the difference in enthalpy of mixed air and supply air):

Q = 4.5 ((lbm x min)/(cu ft x hr)) x 42,000 cfm x 7.8(btu/lbm)

Q = 1,474 MBtu/hr

## Mixed air temperature

Here’s how to calculate the dry-bulb mixed-air temperature based on outdoor and return air conditions.

**Equation 6:**

Where:

T _{MA} = Mixed-air temperature (F)

CFM _{OA} = Air flow rate of outside air (cfm)

T _{OA} = Outside air temperature (F)

CFM _{RA} = Air flow rate of return air (cfm)

T _{RA} = Return air temperature (F)

CFM _{T} = Aggregate supply air flow rate (cfm)

An alternative calculation can be performed using Equation 7:

**Equation 7:**

T _{MA} = (Outside air fraction (%) x T _{OA} ) + (Return air fraction (%) x T _{RA} )

Sample calculation:

Assume an outside air temperature of 95 F; a return-air temperature of 75 F, and an aggregate supply air flow rate of 25,000 cfm with an outside-air fraction of 15%.

T _{MA} = (25,000 cfm x 0.15 x 95 F + 25,000 x 0.85 x 75 F) / 25,000 cfm

T _{MA} = (3,750 cfm x 95 F + 21,250 cfm x 75 F) / 25,000 cfm

T _{MA} = 78 F (dry bulb)

Or, alternatively:

T _{MA} = (0.15 x 95 F + 0.85 x 75 F)

T _{MA} = 78 F (dry bulb)

## Air change rate

Use this equation to calculate the air flow rate needed in cfm to meet a desired air change rate in a space.

**Equation 8:**

Where:

L = Length of room (ft)

W = Width of room (ft)

H = Height of room (ft)

ACH = desired or given air change rate (air changes / hr)

Sample calculation:

Given a 14-ft by 20 ft space with a height of 12 ft and a desired ACH of 8:

CFM _{ACH} = 14-ft x 20-ft x 12-ft x 8 / 60

CFM _{ACH} = 450 cfm (approximately)

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