Air Flow, Air Systems, Pressure, and Fan Performance

1、Designing Air Flow SystemsA theoretical and practical guide to the basics of designing air flow systems. HYPERLINK l AirFlow Air Flow HYPERLINK l Types_of_Flow Types of Flow HYPERLINK l Types_of_Pressure_Losses Types of Pressure Losses or Resistance to Flow HYPERLINK l TotalPress_VelocityPress_Stati

2、cPress Total Pressure, Velocity Pressure, and Static Pressure HYPERLINK l Air_Systems Air Systems HYPERLINK l Fan_Laws Fan Laws HYPERLINK l Air_Density Air Density HYPERLINK l System_Constant System Constant HYPERLINK l Pressure_Losses_of_System Pressure Losses of an Air System HYPERLINK l Sections_

3、in_Series Sections in Series HYPERLINK l Sections_in_Parallel Sections in Parallel HYPERLINK l System_Effect System Effect HYPERLINK l Fan_Performance_Spec Fan Performance Specification HYPERLINK l Fan_Total_Pressure Fan Total Pressure HYPERLINK l Fan_Static_Pressure Fan Static Pressure HYPERLINK l

4、Pressure_Calculations Pressure Calculations HYPERLINK l Methodology Methodology HYPERLINK l Assumptions_and_Corrections Assumptions and Corrections HYPERLINK l Problem1 Problem # 1 An Exhaust System HYPERLINK l Problem2 Problem # 2 A Change to the Systems Air Flow Rate HYPERLINK l Problem3 Problem #

5、 3 A Supply System HYPERLINK l Appendix1 Appendix 1 Equations HYPERLINK l Appendix2 Appendix 2 ASHRAE Fittings HYPERLINK l Appendix3 Appendix 3 Bullhead Tee Curves1. Air FlowFlow of air or any other fluid is caused by a pressure differential between two points. Flow will originate from an area of hi

6、gh energy, or pressure, and proceed to area(s) of lower energy or pressure.FLOWP0P1P1P0 Duct air moves according to three fundamental laws of physics: conservation of mass, conservation of energy, and conservation of momentum. Conservation of mass simply states that an air mass is neither created no

7、r destroyed. From this principle it follows that the amount of air mass coming into a junction in a ductwork system is equal to the amount of air mass leaving the junction, or the sum of air masses at each junction is equal to zero. In most cases the air in a duct is assumed to be incompressible, an

8、 assumption that overlooks the change of air density that occurs as a result of pressure loss and flow in the ductwork. In ductwork, the law of conservation of mass means a duct size can be recalculated for a new air velocity using the simple equation:V2 = (V1 * A1)/A2Where V is velocity and A is Ar

9、eaThe law of energy conservation states that energy cannot disappear; it is only converted from one form to another. This is the basis of one of the main expression of aerodynamics, the Bernoulli equation. Bernoullis equation in its simple form shows that, for an elemental flow stream, the differenc

10、e in total pressures between any two points in a duct is equal to the pressure loss between these points, or:(Pressure loss)1-2 = (Total pressure)1 - (Total pressure)2Conservation of momentum is based on Newtons law that a body will maintain its state of rest or uniform motion unless compelled by an

11、other force to change that state. This law is useful to explain flow behavior in a duct systems fitting.1.1. Types of FlowLaminar Flow Flow parallel to a boundary layer. In HVAC system the plenum is a duct.Turbulent Flow Flow which is perpendicular and near the center of the duct and parallel near t

12、he outer edges of the duct.Most HVAC applications fall in the transition range between laminar and turbulent flow.1.2. Types of Pressure Losses or Resistance to FlowPressure loss is the loss of total pressure in a duct or fitting. There are three important observations that describe the benefits of

13、using total pressure for duct calculation and testing rather than using only static pressure.Only total pressure in ductwork always drops in the direction of flow. Static or dynamic pressures alone do not follow this rule. The measurement of the energy level in an air stream is uniquely represented

14、by total pressure only. The pressure losses in a duct are represented by the combined potential and kinetic energy transformation, i.e., the loss of total pressure. The fan energy increases both static and dynamic pressure. Fan ratings based only on static pressure are partial, but commonly used. Pr

15、essure loss in ductwork has three components, frictional losses along duct walls and dynamic losses in fittings and component losses in duct-mounted equipment.Component PressureDue to physical items with known pressure drops, such as hoods, filters, louvers or dampers.Dynamic PressureDynamic losses

16、are the result of changes in direction and velocity of air flow. Dynamic losses occur whenever an air stream makes turns, diverges, converges, narrows, widens, enters, exits, or passes dampers, gates, orifices, coils, filters, or sound attenuators. Velocity profiles are reorganized at these places b

17、y the development of vortexes that cause the transformation of mechanical energy into heat. The disturbance of the velocity profile starts at some distance before the air reaches a fitting. The straightening of a flow stream ends some distance after the air passes the fitting. This distance is usual

18、ly assumed to be no shorter then six duct diameters for a straight duct. Dynamic losses are proportional to dynamic pressure and can be calculated using the equation:Dynamic loss = (Local loss coefficient) * (Dynamic pressure)where the Local loss coefficient, known as a C-coefficient, represents flo

19、w disturbances for particular fittings or for duct-mounted equipment as a function of their type and ratio of dimensions. Coefficients can be found in the ASHRAE Fittings diagrams.A local loss coefficient can be related to different velocities; it is important to know which part of the velocity prof

20、ile is relevant. The relevant part of the velocity profile is usually the highest velocity in a narrow part of a fitting cross section or a straight/branch section in a junction.Frictional PressureFrictional losses in duct sections are result from air viscosity and momentum exchange among particles

21、moving with different velocities. These losses also contribute negligible losses or gains in air systems unless there are extremely long duct runs or there are significant sections using flex duct. The easiest way of defining frictional loss per unit length is by using the Friction Chart ( HYPERLINK

22、 ://Design-Guide/DGHtm/references.distributionsystems.htm ASHRAE, 1997); however, this chart (shown below) should be used for elevations no higher of 500 m (1,600 ft), air temperature between 5C and 40C (40F and 100F), and ducts with smooth surfaces. The Darcy-Weisbach Equation should b

23、e used for “non-standard” duct type such as flex duct.Friction Chart (ASHRAE HANDBOOK, 1997)1.3. Total Pressure, Velocity Pressure, and Static PressureIt is convenient to calculate pressures in ducts using as a base an atmospheric pressure of zero. Mostly positive pressures occur in supply ducts and

24、 negative pressures occur in exhaust/return ducts; however, there are cases when negative pressures occur in a supply duct as a result of fitting effects.Airflow through a duct system creates three types of pressures: static, dynamic (velocity), and total. Each of these pressures can be measured. Ai

25、r conveyed by a duct system imposes both static and dynamic (velocity) pressures on the ducts structure. The static pressure is responsible for much of the force on the duct walls. However, dynamic (velocity) pressure introduces a rapidly pulsating load. Static pressure Static pressure is the measur

26、e of the potential energy of a unit of air in the particular cross section of a duct. Air pressure on the duct wall is considered static. Imagine a fan blowing into a completely closed duct; it will create only static pressure because there is no air flow through the duct. A balloon blown up with ai

27、r is a similar case in which there is only static pressure.Dynamic (velocity) pressure Dynamic pressure is the kinetic energy of a unit of air flow in an air stream. Dynamic pressure is a function of both air velocity and density: Dynamic pressure = (Density) * (Velocity)2 / 2The static and dynamic

28、pressures are mutually convertible; the magnitude of each is dependent on the local duct cross section, which determines the flow velocity. Total Pressure Consists of the pressure the air exerts in the direction of flow (Velocity Pressure) plus the pressure air exerts perpendicular to the plenum or

29、container through which the air moves. In other words:PT = PV + PSPT = Total PressurePV = Velocity PressurePS = Static PressureThis general rule is used to derive what is called the Fan Total Pressure. See the section entitled Fan Performance Specifications for a definition of Fan Total Pressure and

30、 Fan Static Pressure.2. Air SystemsFor kitchen ventilation applications an air system consists of hood(s), duct work, and fan(s). The relationship between the air flow rate (CFM) and the pressure of an air system is expressed as an increasing exponential function. The graph below shows an example of

31、 a system curve. This curve shows the relationship between the air flow rate and the pressure of an air system.Complex systems with branches and junctions, duct size changes, and other variations can be broken into sections or sub-systems. Each section or sub-system has its own system curve. See the

32、 diagram below for an illustration of this concept.2.1. Fan LawsUse the Fan Laws along a system curve. If you know one (CFM, S.P.) point of a system you could use Fan Law 2 to determine the static pressure for other flow rates. They apply to a fixed air system. Once any element of the system changes

33、, duct size, hood length, riser size, etc. the system curve changes. CFM x RPM xFan Law 1- = -CFM known RPM known SP x CFM2 x RPM2xFan Law 2 - = - = - SP known CFM2known RPM2known BHPx CFM3x RPM3xFan Law 3 - = - = - BHPknown CFM3known RPM3knownOther calculations can be utilized to maneuver around a

34、fan performance curve. For example, to calculate BHP from motor amp draw, use the following formula:1 phase motors3 phase motors BHP = V * I * E * PF BHP = 746746where:BHP = Brake HorsepowerV = Line VoltageI = Line CurrentE = Motor Efficiency (Usually about .85 to .9)PF = Motor Power Factor (Usually

35、 about .9)Once the BHP is known, the RPM of the fan can be measured. The motor BHP and fan RPM can then be matched on the fan performance curve to approximate airflow.2.2. Air DensityThe most common influences on air density are the effects of temperature other than 70 F and barometric pressures oth

36、er than 29.92” caused by elevations above sea level.Ratings found in fan performance tables and curves are based on standard air. Standard air is defined as clean, dry air with a density of 0.075 pounds per cubic foot, with the barometric pressure at sea level of 29.92 inches of mercury and a temper

37、ature of 70 F. Selecting a fan to operate at conditions other then standard air requires adjustment to both static pressure and brake horsepower. The volume of air will not be affected in a given system because a fan will move the same amount of air regardless of the air density. In other words, if

38、a fan will move 3,000 cfm at 70 F it will also move 3,000 CFM at 250 F. Since 250 F air weighs only 34% of 70F air, the fan will require less BHP but it will also create less pressure than specified.When a fan is specified for a given CFM and static pressure at conditions other than standard, the co

39、rrection factors (shown in table below) must be applied in order to select the proper size fan, fan speed and BHP to meet the new condition.The best way to understand how the correction factors are used is to work out several examples. Lets look at an example using a specification for a fan to opera

40、te at 600F at sea level. This example will clearly show that the fan must be selected to handle a much greater static pressure than specified.Example #1: A 20” centrifugal fan is required to deliver 5,000 cfm at 3.0 inches static pressure. Elevation is 0 (sea level). Temperature is 600F. At standard

41、 conditions, the fan will require 6.76 bhp1.Using the chart below, the correction factor is 2.00.2.Multiply the specified operating static pressure by the correction factor to determine the standard air density equivalent static pressure. (Corrected static pressure = 3.0 x 2.00 = 6”. The fan must be

42、 selected for 6 inches of static pressure.)3.Based upon the performance table for a 20 fan at 5,000 cfm at 6 inches wg, 2,018 rpm is needed to produce the required performance.4.What is the operating bhp at 600 F?Since the horsepower shown in the performance chart refers to standard air density, thi

43、s should be corrected to reflect actual bhp at the lighter operating air. Operating bhp = standard bhp 2.00 or 6.76 2.00 = bhp.2.3. System ConstantEvery air system or sub-system has a system constant. This constant can be calculated as long as you know one (CFM, Static Pressure) point. You use a var

44、iation of the fan laws to calculate the system constant. To calculate the system constant:K system = S.P./(CFM)2Once you have the system constant you can calculate the static pressure for any flow rate.S.P. = (CFM)2 * K system3. Pressure Losses of an Air SystemPressure losses are more easily determi

45、ned by breaking an air system into sections. Sections can be in series or in parallel.3.1. Sections in SeriesFor sections or components in series simply sum up all the sections. A single duct that has the same shape, cross section, and mass flow is called a duct section or just a section. Following

46、is the recommended procedure for calculating total pressure loss in a single duct section: Gather input data: air flow, duct shape, duct size, roughness, altitude, air temperature, and fittings; Calculate air velocity as a function of air flow and cross section; Calculate local C-coefficients for ea

47、ch fitting used; and Calculate pressure loss using the friction chartThe following is a simple example of how duct pressure accumulates and is totaled in a section.3.2. Sections in ParallelWhen designing sections that are parallel it is important to remember that the branches of a junction all have

48、the same total pressure. This is a fact. It is governed by a principle which states that areas of high energy move to areas of lower energy. We will see how this applies to air systems in parallel.To illustrate these concepts we will reference the diagram below. In this example we calculate the pres

49、sure losses for Section 1 to be -0.75” at the junction. We calculate the pressure losses for Section 2 to be -0.6” at the junction. (NOTE: For simplicitys sake we do not consider the pressure loss incurred by the junction.) These would be the actual pressure losses of the system were they operating

50、independently; however, they do not. They interact at the junction. This means that whenever air flow encounters a junction it will take the path of least resistance and the total pressure losses of each branch of the junction will be the same.For sections that run parallel, always use the section w

51、ith the higher pressure loss/gain to determine pressure losses/gains through a system. Adjust the branch with the lower pressure loss/gain by increasing the flow rate or decreasing the duct size to increase the pressure loss to that of the higher branch.If the flow rate or the duct size is not chang

52、ed the air flow through each branch will adjust itself so that each branch has the same total pressure loss/gain. In other words, more air flows through the branch with the lower pressure loss/gain or energy state.In the example below, the actual pressure loss would be somewhere between -0.75” and -

53、0.6”. Section 1 would pull less than 2000 CFM and Section 2 would pull more than 1800 CFM. 3.3. System EffectSystem Effect occurs in an air system when two or more elements such as fittings, a hood and a fitting, or a fan and a fitting occur within close proximity to one another. The effect is to in

54、crease the energy or pressure in a system as air flows through the elements. To calculate the pressure loss incurred by such a configuration, consider two elements at a time. For example, if two elbows occur 4 feet from one another this configuration will have a pressure loss associated with it. Cal

55、culate the pressure loss/gain associated with each fitting as if it occurs alone. Sum these and multiply them by a system effect coefficient (K). The system effect coefficient can be obtained from the ASHRAE Fitting Diagrams for only a limited number of configurations of elements. Configurations not

56、 listed must use estimates or best guesses. In many cases, you can use a listed configuration as a guide.One configuration not listed is an elbow within close proximity to the collar of a hood. As a rule of thumb, the chart below can offer some guidance for determining the system effect for this sit

57、uation. Remember the coefficients in the chart are only an estimate.System Effect TableDistance between Riser and ElbowSystem Effect Coefficient (K)2 feet3 feet4 feet5 feetThe diagrams below show system effect factors for straight through elements and turning elements. For rectangular ductwork, D =

58、(2HW)/(H+W). The following formula should be used to calculate the pressure caused by system effect:Pressure Loss = K * (Element A Resistance + Element B Resistance)Straight Through Flow Turning ElementsThe following diagrams show proper and improper methods of constructing ductwork:4. Fan Performan

59、ce SpecificationA fan performance spec is given as a Fan Total Pressure or a Fan Static Pressure which can handle a certain flow rate. Most manufacturers performance charts are based on Fan Static Pressure.4.1. Fan Total PressureFan total Pressure is the pressure differential between the inlet and t

60、he outlet of the fan. It can be expressed in these terms:P t fan = P t loss + P v system outlet + (P s system outlet + P s system entry + P v system entry)P t fan = Fan Total PressureP t loss = Dynamic, Component, and Frictional Pressure through the air system.P s system outlet = Static Pressure at