1 / 20

Objectives

Objectives. Finish with heat exchangers (ch.11) Start with Air-distribution systems. Overall Heat Transfer. Q = U 0 A 0 Δ t m. Need to find this. Resistance model. Q = U 0 A 0 Δ t m From eq. 1, 2, and 3: We can often neglect conduction through pipe walls

shiloh
Download Presentation

Objectives

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Objectives • Finish with heat exchangers (ch.11) • Start with Air-distribution systems

  2. Overall Heat Transfer Q = U0A0Δtm Need to find this

  3. Resistance model Q = U0A0Δtm From eq. 1, 2, and 3: • We can often neglect conduction through pipe walls • Sometime more important to add fouling coefficients R Internal R cond-Pipe R External

  4. Example The air to air heat exchanger in the heat recovery system from previous example has flow rate of fresh air of 200 cfm. With given: Calculate the needed area of heat exchanger A0=? Solution: Q = mcp,coldΔtcold = mcp,hotΔthot = U0A0Δtm From heat exchanger side: Q = U0A0Δtm→ A0 = Q/ U0Δtm U0 = 1/(RInternal+RCond+RFin+RExternal) = (1/10+0.002+0+1/10) = 4.95 Btu/hsfF Δtm = 16.5 F From air side: Q = mcp,coldΔtcold = = 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h Then: A0 = 3456 / (4.95·16.5) = 42 sf

  5. For Air-Liquid Heat Exchanger we need Fin Efficiency • Assume entire fin is at fin base temperature • Maximum possible heat transfer • Perfect fin • Efficiency is ratio of actual heat transfer to perfect case • Non-dimensional parameter tF,m

  6. Fin Theory k – conductivity of material hc,o – convection coefficient pL=L(hc,o /ky)0.5

  7. Fin Efficiency • Assume entire fin is at fin base temperature • Maximum possible heat transfer • Perfect fin • Efficiency is ratio of actual heat transfer to perfect case • Non-dimensional parameter

  8. Heat exchanger performance (11.3) • NTU – absolute sizing (# of transfer units) • ε – relative sizing (effectiveness)

  9. Air Distribution System Design • Describe room distribution basics • Select diffusers • Supply and return duct sizing

  10. Forced driven air flowDiffusers Grill (side wall) diffusers Linear diffusers Vertical Horizontal one side

  11. Diffusers types Valve diffuser swirl diffusers ceiling diffuser wall or ceiling floor

  12. Diffusers Perforated ceiling diffuser Jet nozzle diffuser Round conical ceiling diffuser Square conical ceiling diffuser Wall diffuser unit Swirl diffuser Floor diffuser Auditorium diffuser Linear slot diffuser DV diffuser External louvre Smoke damper http://www.titus-hvac.com/techzone/ http://www.halton.com/halton/cms.nsf/www/diffusers

  13. Low mixing Diffusers Displacement ventilation

  14. 18.7

  15. V = maximum volumetric flow rate (m3/s, ft3/min) Qtot = total design load (W, BTU/hr) Qsen = sensible design load (W,BTU/hr) ρ = air density (kg/m3, lbm/ft3) Δt = temperature difference between supply and return air (°C, °F) Δh = enthalpy difference between supply and return air (J/kg, BTU/lbm) Diffuser Selection Procedure • Select and locate diffusers, divide airflow amongst diffusers

More Related