10.7 System Curves, the Fan Laws, and Density Corrections

System Curves, the Fan Laws, and Density Corrections

Once the fan type is chosen, three questions decide whether the installation works: where the fan curve crosses the system resistance curve, what happens to flow and power when speed changes, and how non-standard air density distorts both.

1. System Characteristic Resistance Curves & Operating Point

The System Characteristic Parabolic Curve

In any duct system, total pressure loss is dominated by turbulent fluid friction and dynamic fitting losses, which scale with the square of volumetric flow rate (Q²):

$\mathbf{SP_{\text{system}} = k \times Q^2}$

Where k is the system resistance coefficient determined by duct diameter, length, elbow count, hood entry losses, and air cleaner resistance.

The Fan Operating Point

A fan manufacturer provides a Fan Characteristic Curve plotting developed static pressure (SP) versus airflow (Q) at a constant rotational speed (RPM). When the System Resistance Curve is plotted on the same axes, the intersection defines the Fan Operating Point:

   +-------------------------------------------------------------------------+
   |                  FAN CURVE & SYSTEM OPERATING POINT                     |
   +-------------------------------------------------------------------------+
   |  Static Pressure (in. w.g.)                                             |
   |    ^                                                                    |
   |    |         FAN PERFORMANCE CURVE (Constant RPM)                       |
   |    |       *'''--.                                                      |
   |    |              '--.  System Curve 2 (Plugged Duct / Dirty Filters)   |
   | SP2|------------------*--.      /                                       |
   |    |                 /|   '--. /   System Curve 1 (Clean / Design)      |
   | SP1|----------------/-+-------*--.    /                                 |
   |    |               /  |      /|   '--/                                  |
   |    |              /   |     / |                                         |
   |  0 +-------------+----+----+--+------------------------------------->   |
   |  0              Q2        Q1                         Airflow Q (cfm)    |
   +-------------------------------------------------------------------------+

System Dynamics and Deviations

1. Duct Plugging or Loaded Air Filters: Increases the system resistance coefficient (k2 > k1), steepening the parabolic system curve. The operating point shifts up and to the left, resulting in lower airflow (Q2 < Q1) and higher static pressure (SP2 > SP1).
2. Drive Belt Slippage or Speed Reduction: Shifts the entire fan characteristic curve downward, reducing both delivered airflow and developed static pressure.

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2. The Three Fundamental Fan Laws

The Fan Laws are mathematical similitude relationships governing changes in fan rotational speed (RPM), impeller diameter (D), and fluid density (ρ). For a fixed ventilation duct system and fixed fan size (D1 = D2), changing the fan rotational speed modulates flow, pressure, and power according to three exact laws:

   +-------------------------------------------------------------------------+
   |                   THE THREE FUNDAMENTAL FAN LAWS                        |
   +-------------------------------------------------------------------------+
   |                                                                         |
   |  FAN LAW 1 (Airflow is Linear):                                         |
   |     Q2 = Q1 * (RPM2 / RPM1)                                             |
   |     * Increasing RPM by 10% increases volumetric airflow by 10%.        |
   |                                                                         |
   |  FAN LAW 2 (Pressure is Quadratic):                                     |
   |     SP2 = SP1 * (RPM2 / RPM1)^2                                         |
   |     * Increasing RPM by 10% increases static pressure by 21% (1.10^2).  |
   |                                                                         |
   |  FAN LAW 3 (Brake Horsepower is Cubic):                                 |
   |     BHP2 = BHP1 * (RPM2 / RPM1)^3                                       |
   |     * Increasing RPM by 10% increases required power by 33.1% (1.10^3)! |
   |     * Increasing RPM by 50% increases power by 237.5% (1.50^3 = 3.375x)!|
   +-------------------------------------------------------------------------+

Exam Critical Warning: Fan Law 3 is the most common mathematical trap on the CIH exam. Because brake horsepower scales with the cube of the speed ratio ([RPM2 / RPM1]³), attempting to solve an LEV airflow deficiency by simply speeding up the fan sheave will dramatically overload and burn out the electric motor unless the motor horsepower is upsized accordingly!

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3. Brake Horsepower (BHP) and Mechanical Efficiency

Air Horsepower (AHP)

Air Horsepower is the theoretical fluid aerodynamic power required to move a given volume of air against a specified total or static pressure drop:

$\mathbf{AHP = \frac{Q \times FTP}{6356} \quad \text{or} \quad AHP_{\text{static}} = \frac{Q \times FSP}{6356}}$

(where Q is in cfm, FTP and FSP are in in. w.g., and 6356 is the imperial conversion factor derived from 33,000 ft·lbf/min / [5.1922 lbf/ft² / in. w.g.]).

Brake Horsepower (BHP)

Brake Horsepower (BHP) is the actual mechanical shaft power that must be delivered to the fan shaft by the motor, incorporating the fan's mechanical efficiency (η):

$\mathbf{BHP = \frac{Q \times FTP}{6356 \times \eta_{\text{total}}} = \frac{Q \times FSP}{6356 \times \eta_{\text{static}}}}$

Motor Electrical Input Horsepower (MHP)

To determine the actual electrical motor rating required, mechanical drive transmission losses (V-belts, ηdrive ≈ 0.90--0.95) and electric motor electrical efficiency (ηmotor ≈ 0.85--0.95) must be incorporated:

$MHP = \frac{BHP}{\eta_{\text{drive}} \times \eta_{\text{motor}}}$

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4. Non-Standard Air Density Corrections

Standard Air Properties

Fan catalog rating tables are published based on Standard Air Density:

• ρ0 = 0.075 lb/ft³ at standard sea level temperature of 70°F (530°R) and barometric pressure of 29.92 in. Hg (760 mmHg).

Air Density Correction Factor (d)

When an LEV system operates at elevated temperatures (e.g., furnace exhaust at 300°F) or high elevations (e.g., Denver, CO at 5,280 ft where PB ≈ 24.8 in. Hg), the actual air density (ρactual) decreases. The dimensionless Density Correction Factor (d) is:

$d = \frac{\rho_{\text{actual}}}{\rho_0} = \left(\frac{530}{460 + T(^\circ\text{F})}\right) \times \left(\frac{P_B(\text{in. Hg})}{29.92}\right)$

Aerodynamic Principles of Density Shifts

1. A fan is a constant volumetric displacement machine. At a given RPM, an industrial fan will move the exact same volumetric flow rate (Q, in cfm) regardless of air density!
2. Developed pressure and required shaft power are directly proportional to air density (d): $SP_{\text{actual}} = SP_{\text{standard}} \times d$ $BHP_{\text{actual}} = BHP_{\text{standard}} \times d$

Cold Startup Power Hazard

When designing an LEV system for a hot process (e.g., 400°F, where d ≈ 0.61), the fan requires much less power during steady-state hot operation (BHPhot = BHPstd × 0.61).

However, during morning cold startup when the system contains dense ambient air (70°F, d = 1.0), the fan will demand 1 / 0.61 = 1.64 times more power. If the electric motor was sized solely for the hot operating condition, the motor will trip thermal overloads or burn out during startup unless throttled by an inlet damper!

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5. Worked Step-by-Step Calculation Examples

Worked Example 9.3.1: Applying the Fan Laws to a Speed Increase

Problem: An existing industrial exhaust fan operates at a rotational speed of 1200 RPM, delivering a volumetric flow rate Q1 = 6000 cfm against a static pressure SP1 = 4.0 in. w.g. while consuming BHP1 = 6.0 hp of shaft power. To accommodate a new welding booth, the airflow must be increased by 25% to Q2 = 7500 cfm.

1. Calculate the required new fan rotational speed (RPM2).
2. Calculate the resulting new system static pressure (SP2).
3. Calculate the required new Brake Horsepower (BHP2).
4. Determine the percentage increase in motor horsepower required.

Solution Steps:

1. Calculate new speed (RPM2) using Fan Law 1 (Q2 = Q1 [RPM2 / RPM1]): $\frac{RPM_2}{RPM_1} = \frac{Q_2}{Q_1} = \frac{7500\text{ cfm}}{6000\text{ cfm}} = 1.25$ $RPM_2 = 1200\text{ RPM} \times 1.25 = 1500\text{ RPM}$

2. Calculate new static pressure (SP2) using Fan Law 2 (SP2 = SP1 [RPM2 / RPM1]²): $SP_2 = 4.0\text{ in. w.g.} \times (1.25)^2 = 4.0 \times 1.5625 = 6.25\text{ in. w.g.}$

3. Calculate new Brake Horsepower (BHP2) using Fan Law 3 (BHP2 = BHP1 [RPM2 / RPM1]³): $BHP_2 = 6.0\text{ hp} \times (1.25)^3 = 6.0 \times 1.953125 = 11.72\text{ hp}$

4. Calculate percentage power increase: $\Delta \% = \frac{11.72 - 6.0}{6.0} \times 100 = 95.33\%$

Result: Increasing airflow by 25% requires running the fan at 1500 RPM, increases static pressure to 6.25 in. w.g., and nearly doubles required power to 11.72 hp (+95.3%), requiring the existing motor to be replaced with a standard 15 hp motor.

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Worked Example 9.3.2: Calculating Brake Horsepower and Electrical Operating Power

Problem: A backward-inclined centrifugal exhaust fan delivers Q = 8500 cfm against a Fan Total Pressure FTP = 5.5 in. w.g. Standard air conditions apply. The fan has a total mechanical efficiency ηtotal = 78% (0.78), the V-belt drive efficiency is ηdrive = 92% (0.92), and the motor electrical efficiency is ηmotor = 90% (0.90).

1. Calculate the Air Horsepower (AHP).
2. Calculate the required fan shaft Brake Horsepower (BHP).
3. Calculate the total electrical input motor power (MHP) and select the appropriate standard NEMA motor size (10 hp, 15 hp, 20 hp).

Solution Steps:

1. Calculate Air Horsepower (AHP): $AHP = \frac{Q \times FTP}{6356} = \frac{8500\text{ cfm} \times 5.5\text{ in. w.g.}}{6356} = \frac{46750}{6356} = 7.355\text{ hp}$

2. Calculate Brake Horsepower (BHP): $BHP = \frac{AHP}{\eta_{\text{total}}} = \frac{7.355\text{ hp}}{0.78} = 9.429\text{ hp} \approx 9.43\text{ hp}$

3. Calculate total electrical motor horsepower (MHP): $MHP = \frac{BHP}{\eta_{\text{drive}} \times \eta_{\text{motor}}} = \frac{9.429}{0.92 \times 0.90} = \frac{9.429}{0.828} = 11.388\text{ hp}$ NEMA Motor Selection: A 10 hp motor is undersized (11.39 hp > 10 hp). The next standard NEMA commercial motor rating is 15 hp.

Result: The aerodynamic Air Horsepower is 7.36 hp, required shaft Brake Horsepower is 9.43 hp, and a 15 hp standard motor is required.

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Worked Example 9.3.3: High-Temperature and Elevation Air Density Corrections

Problem: A hot flue gas LEV system exhausts gas at T = 300°F from an industrial smelter located at an elevation of 5000 ft (barometric pressure PB = 24.9 in. Hg). The required actual operating volumetric flow is Q = 10,000 cfm against an actual operating static pressure of SPactual = 4.5 in. w.g.

1. Calculate the density correction factor (d).
2. Determine the equivalent standard static pressure (SPstd) required to select the fan from a manufacturer's standard air catalog.
3. If the catalog specifies BHPstd = 16.8 hp at standard conditions, calculate the actual operating horsepower (BHPhot) and the cold startup horsepower (BHPcold at 70°F and 5000 ft).

Solution Steps:

1. Calculate Density Correction Factor (d): $d = \left(\frac{530}{460 + 300}\right) \times \left(\frac{24.9\text{ in. Hg}}{29.92\text{ in. Hg}}\right) = \left(\frac{530}{760}\right) \times 0.83222 = 0.69737 \times 0.83222 = 0.5803$

2. Calculate Catalog Equivalent Static Pressure (SPstd): $SP_{\text{std}} = \frac{SP_{\text{actual}}}{d} = \frac{4.5\text{ in. w.g.}}{0.5803} = 7.755\text{ in. w.g.} \approx 7.76\text{ in. w.g.}$ (The fan is selected from the catalog to deliver 10,000 cfm at 7.76 in. w.g.)

3. Calculate Operating Power and Cold Startup Power:

   • Operating Hot Power (BHPhot): $BHP_{\text{hot}} = BHP_{\text{std}} \times d = 16.8\text{ hp} \times 0.5803 = 9.749\text{ hp} \approx 9.75\text{ hp}$
   • Cold Startup Power (BHPcold at 70°F, 5000 ft): $d_{\text{cold}} = \left(\frac{530}{460 + 70}\right) \times \left(\frac{24.9}{29.92}\right) = 1.00 \times 0.83222 = 0.8322$ $BHP_{\text{cold}} = BHP_{\text{std}} \times d_{\text{cold}} = 16.8\text{ hp} \times 0.8322 = 13.98\text{ hp}$

Result: The density factor is 0.580. While the fan only draws 9.75 hp during normal hot operations, it will draw 13.98 hp during a cold morning startup. A 15 hp motor must be installed to prevent tripping on cold startup.