52nd North Carolina Industrial Ventilation Conference

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Owen Ferguson
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System Design II Objectives Module Understand use of the ACGIH Calc sheet for

balancing by design Balancing at a junction Air stream acceleration &

System Static Pressure System Static Pressure SP sys = SP out -SP in -VP in SP

static pressure losses throughout your system.

1 System Design II Objectives Module Understand use of the ACGIH Calc sheet for system design. Become familiar with new design topics: Use of blast gates vs. balancing by design Balancing at a junction Air stream acceleration & deceleration At a junction Duct expansions & contractions Fan selection 1 2 System Static Pressure System Static Pressure SP sys = SP out -SP in -VP in SP sys = SP out -SP in -VP in 3 4 is used to keep track of all air flow rates and static pressure losses throughout your system. Each Column is a Duct Segment Begins at hood, fitting, control device or Fan Ends at fitting, control device, Fan, or stack 5 6 1

2 Cells are identified by Coordinates - Column and Row System segment 1-B has a value of 1100 SCFM Each Row Contains data about that Segment 1-B Top of Sheet contains all of the Physical Data about the Segment Asterisk means data that is entered from tables or VS Plates or Process Asterisk (input information) Hood Information (Rows 15-27) No Asterisk (need to calculate) Equations Units 9 10 Balancing Methods Duct Information and Summary Lines With blast gates By design Also called H d

3 Blast Gate Method Balance By Design Advantages Adjustable Design flexibility Disadvantages Misadjustable Higher initial cost Advantages No misadjustment No blast gates to wear Disadvantages Higher airflows Balance at Junction Balance at Junction What happens if the static pressures of branch 1 and 2 don t balance at the junction point? Q 1 + Q 2 = Q 3 SP 1 = SP 2 = SP 3 The higher static pressure branch is considered the governing static pressure Balance at Junction Adjustment By Airflow If the ratio of the governing (higher) static pressure to the lower static pressure is > 1.2, then you will need to redesign the branch with the lower static pressure. This may include a change of duct size, selection of different fittings, and/or modifications to the hood design. If the ratio is less than 1.2, then balance can be obtained by increasing the airflow through the branch with the lower resistance. Q corr corr = Q design SP gov SP lower

Adjustment By Airflow Problem 1a Q corr = Q design SP gov SP lower How do we balance the two branches at point A? What is the governing static pressure at point A?

4 Adjustment By Airflow Problem 1a Q corr = Q design SP gov SP lower How do we balance the two branches at point A? What is the governing static pressure at point A? A Q design and Q corr represent the initial and corrected volumetric flow rate in the lower static pressure branch. 19 SP = -2.5 in. wg V = 3600 fpm D = SP = -2.3 in. wg V = 3200 fpm 20 Problem 1a - Solution Problem 1b 2.5 wg/2.3 wg = 1.09 Ratio is <1.20, so balance can be obtained by adjusting airflow. Q=VA Q 1 = (3600 fpm)(0.545 ft 2 ) = 1962 cfm Q 2 = (3200 fpm)(0.196 ft 2 ) = 628 cfm How do we balance the two branches at point A? What is the governing static pressure at point A? A Q corr = Q design (1.09) 1/2 Q corr = (628 cfm)(1.04) = 655 cfm Governing static pressure = 2.5 point A SP = -2.5 in. wg V = 3600 fpm D = SP = -1.9 in. wg V = 3200 fpm Problem 1b - Solution Problem 2a 2.5 wg/1.9 wg = 1.32 Ratio is >1.20, so we must redesign the lower static pressure branch (2 to A). Start by decreasing duct diameter by ½ or 1 inch. This will result in a higher velocity (and velocity pressure), hence increasing the hood, duct, elbow, and other losses in that branch. Use the ACGIH calc sheet

5 Problem 2b Problem 2c Use the ACGIH calc sheet. Use the ACGIH calc sheet Assume two ducts ( A & B ) joining together to form Duct C ; all ducts are identical size so all of the air from A & B is squeezed into C Lets assume that the flows (Q) from A and B are identical; what is the velocity in Duct C? If Velocity has increased, is there energy required?

How much? If we know some average Velocity Pressure going into the fitting and the VP coming out, we know we need at least that amount of energy to speed the air up through the fitting.

6 How much? If we know some average Velocity Pressure going into the fitting and the VP coming out, we know we need at least that amount of energy to speed the air up through the fitting Accounting For Acceleration Resultant or Weighted Average Velocity Pressure Accounting For Acceleration Resultant or Weighted Average Velocity Pressure Weighted average velocity pressure the velocity pressure we would expect at point C if there was no net acceleration or deceleration when the two air streams come together Accounting For Acceleration Resultant or Weighted Average Velocity Pressure Accounting For Acceleration Resultant or Weighted Average Velocity Pressure Fraction of total air flow coming from branch A Velocity pressure in Branch A

Find the correct static pressure (SPc) at point C Branch A-C V = 4075 fpm VP = 1.04 Q=800cfm SP ac = 2.0 w.g. Branch B-C D = 5 V = 3666 fpm VP = 0.84 Q = 500 cfm SP bc = 2.0 w.g. Problem 3 Point C D=7 V = 4864 fpm VP = 1.

7 Find the correct static pressure (SPc) at point C Branch A-C V = 4075 fpm VP = 1.04 Q=800cfm SP ac = 2.0 w.g. Branch B-C D = 5 V = 3666 fpm VP = 0.84 Q = 500 cfm SP bc = 2.0 w.g. Problem 3 Point C D=7 V = 4864 fpm VP = 1.48 Q = 1,300 cfm C =? Need to use the VP r equation to find the expected or weighted average velocity pressure for the two air streams coming together: VP r = (800 cfm/1300 cfm)(1.04 ) + (500 cfm/1300 cfm)(0.84 ) VP r = 0.96 inches w.g If there is no acceleration or deceleration of the two air streams coming together, the velocity pressure at point C should be 0.96 w.g. However, the measured velocity pressure at point C is 1.48 w.g., which means the air must accelerate at that point. Just as it s handled in the hood, the energy required to accelerate the air is expressed as an increase in static pressure w.g w.g. = 0.52 w.g. 39 Branch A-C V = 4075 fpm VP = 1.04 Q = 800 cfm Point C SP D=7 ac = 2.0 w.g. V = 4864 fpm VP = 1.48 Branch B-C D = 5 V = 3666 fpm VP = 0.84 Q = 500 cfm SP bc = 2.0 w.g. Q = 1,300 cfm C =? = 2.0 w.g w.g. = 2.52 w.g. 40 Branch A-C V = 4075 fpm VP = 1.04 Q = 800 cfm SP ac = 2.0 w.g. Static pressure at point C is 2.0 w.g. Branch A-C V = 4075 fpm VP = 1.04 Q = 800 cfm SP ac = 2.0 w.g. Static pressure at point C is 2.52 w.g. since the air has to be accelerated in the C D branch. Branch B-C D = 5 V = 3666 fpm VP = 0.84 Q = 500 cfm SP bc = 2.0 w.g. Branch B-C D = 5 V = 3666 fpm VP = 0.84 Q = 500 cfm SP bc = 2.0 w.g

System Components Transitions (IVM 9.49) VP r is checked in the segment after two branches come together. C-D Insert the VP r value in Row 39 of the ACGIH calc sheet 0.96 Line 40 is the added SP 1.

8 System Components Transitions (IVM 9.49) VP r is checked in the segment after two branches come together. C-D Insert the VP r value in Row 39 of the ACGIH calc sheet 0.96 Line 40 is the added SP = Issues: Transition length Angle of expansion or contraction Regain for expansions Conversion, SP into VP Contraction Conversion VP into SP Expansion To Fan 1 2 To Fan VP 1 = 0.9" SP 1 = -3.0" VP 2 = 1.5" SP 2 = VP 1 = 1.5" SP 1 = -3.0 " VP 2 = 0.9" SP 2 = Problem 4 Problem 5 Given: The following System, Standard Conditions Find: The static pressure (SP) at Pont F. Use the ACGIH calc sheet. Given: These two 15 diameter, 2 wide grinding wheels have good enclosures. Use the appropriate VS plate (VS 80-11) to design the LEV system. Find: Determine Q and the SP external to the fan/collector unit

Problem 6 Problem 7 Given:A system with the following

conditions: System Static Pressure (SSP) = 1.83 w.g.

Compact GI brochure pick a fan and determine the following.

the fan static pressure (FSP) selected for your .

9 Problem 6 Problem 7 Given:A system with the following design conditions: System Static Pressure (SSP) = 1.83 w.g. Volumetric flow rate (Q) = 560 cfm Inlet = 7-inch Using the Compact GI brochure pick a fan and determine the following. Find: The appropriate rpm and horsepower requirements for the fan static pressure (FSP) selected for your design Problem 8 Find: The total volumetric flow rate (Q) required and the appropriate slot width for good design and capture. 51 9

HVAC Clinic. Duct Design

HVAC Clinic. Duct Design HVAC Clinic Duct Design Table Of Contents Introduction... 3 Fundamentals Of Duct Design... 3 Pressure Changes In A System... 8 Example 1... 13 Duct Design Methods... 15 Example 2... 15 Introduction The