Objectives

• Finish duct design

• Discuss HW4

• Learn about pumps

Frictional Losses

Non-circular Ducts

Dynamic (local) losses

• Losses associated with• Changes in velocity• Obstructions• Bends• Fittings and transitions

Loss CoefficientsΔPt = Co ΔPv,o

Duct Design Example

• Determine total pressure drop from in this duct system

l1=150ft

l2=50ft

l3=100ftF

AB

pdiffuserA = 10Pa

pdiffuserB = 15PaVA = 1000 CFMVA = 2000 CFM

VA = 1000 CFM

Solution

section L[ft] V [CFM] D[in] Dp/100ft [“wg] V [fpm] Dptot[“wg] Local Dp Coef. Sum Co Dp_loc[“wg] Tot_Dp[“wg]

L1 150 3000 18” 0.22 1700 0.33 Co=0.5 0.5 0.09 0.42L2 50 1000 12” 0.2 1300 0.10 Co=0.28 0.28 0.03 0.13

L3 100 2000 16” 0.18 1400 0.18 Co=13,Co=0.09 0.22 0.03 0.21

Equal friction method:

Two design methodsEqual friction method (Example 18.9):

Static regain method (Example 18.10):

Fan Performance

Controlling Air Flow

VAV Operation

How to use multiple pumps

Dampers

Summary of Air Distribution Systems

• Select diffuser based on ADPI

• Given a layout of duct system• Calculate total pressure drop• Deal with non-circular ducts and fitting losses

• Select the fan based on fan and duct curve

HW4 Design:

• Air handling unit (AHU) with components (cooling coil and fan) • Duct system• Diffusers• Chiller• Pump• Power requirement

Pumps and Plumbing

• Very similar to fans and ducts • Same principles based on fluid dynamics • With some specifics

Hydronic Terms

• Head loss

• Open-loop vs. closed loop

Head Loss

Fittings

g

Vkh

2

2

Pumps

• Driving force to move air water in buildings

• Raise pressure and produce flow

• One main type• Centrifugal

Pump curves

• NPSHR = Net Positive Suction Head Required

Use Multiple Pumps

Plumbing Arrangements

Example 19.7

• Similar to equal friction method

• Calculate head loss for common pieces of system

• Size pipes to have equal head loss between points 2 and 3