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Heat Gain / Loss The goals of this exercise Examine contributing factors of heat gain / loss

Heat Gain / Loss The goals of this exercise Examine contributing factors of heat gain / loss Develop a sense of the magnitude of factors To see building energy performance as a critical design issue. The Dynamics of Energy Analysis. Forced ventilation.

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Heat Gain / Loss The goals of this exercise Examine contributing factors of heat gain / loss

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  1. Heat Gain / Loss • The goals of this exercise • Examine contributing factors of heat gain / loss • Develop a sense of the magnitude of factors • To see building energy performance as a critical design issue

  2. The Dynamics of Energy Analysis Forced ventilation Conduction walls & roof 30 deg. Solar radiation Lighting Conduction thru. glass Occupants & equipment

  3. Three means of heat transfer • Conduction • Convection • radiation convection radiation conduction

  4. Basic units of heat measurement • The BTU (British thermal unit) Roughly equal to a match • ConductionWe measure energy flow through the envelop in BTUs / Hr. - per square foot - per degree of temperature difference between interior and exterior • Resistance We measure the resistance to energy flow (R value) in Hrs. / BTU. - per square foot - per degree of temperature difference between interior and exterior

  5. Basic units of heat measurement Conduction vales Rates of energy flow through materials (1) inch thick are designated as “k” values Rates of energy flow through materials of a given thickness are designated as “C” values Material “k” “C” ½” glass .05 btu/hr Concrete .20 btu/hr Steel .50 btu/hr 8” CMU .03 btu/hr

  6. Conduction Units K value BTU / hr. for 1” thick material C value BTU / hr. for a given thickness 2’x2’x1” thick k=.6 btu/hr 2’x2’x3” thick C = .2 btu/hr 70 deg. 70 deg. 30 deg. 30 deg. Q= k x A x TD Q= .6 x 4 x 40 Q= c x A x TD Q= .2 x 4 x 40

  7. Calculating conduction through multiple layers of wall materials “U” values are rate of flow through layers of materials Each 1” panel has a K value of 1, 2, & 3 btu/hr respectively Conduction through the three panels can not be the sum of the k values U = k + k + k Wrong

  8. “R” Values – hrs/btu “R” values are the reciprocal of conduction values R= 1/k R=1/C “R” values accumulate 1 hr/btu + 3 hr/btu = 4hr/btu conduction rates R values (resistance rates) Material “k” “C” 1/k 1/C ½” glass .05 btu/hr 20 hrs/btu Concrete .20 btu/hr 2 hrs/btu Steel .50 btu/hr 8” CMU .03 btu/hr

  9. Calculating U: Conduction through multiple panels Conduction values (k or c) do not add. The resistance to energy flow does accumulate. Rtotal = R + R + R Convert conduction values to resistance R = 1/k + 1/k + 1/k R = 1/1 + 1/2 + 1/3 R = 1 + .5 + .33 = 1.83 hr/btu R(total) = 1.83 hr/btu K = 1 2 3 btu/hr U is the conduction value for an assembly of different materials. U = 1/R & R = 1/U U = 1/ R total = 1/ 1.83 U = .546 btu/hr Q = U x A x TD Q =.546 x A x TD

  10. Calculating U: Conduction through multiple panels Differences in material thickness K=4C=2 Resistance to energy flow can be added K=1 Convert to correct units R = 1/k + 1/C + 1/C R = 1/1 + 1/2 + 1/.33 R = 1 + .5 + 3.0 = 4.5 R(total) = 4.5 hr/btu C=.33 1” thick 2” thick 3” thick U is the conduction value for an assembly of different materials. U = 1/R & R = 1/U U = 1/ R total = 1/ 4.5 U = .222 btu/hr Q = U x A x TD Q =.222 x A x TD

  11. Energy flow through a wall Assume wall is 10’ x 20’ Summation of R values R = .44 R = 1.01 R = 1.5 R total 2.95 U = 1 / 2.95 = .338 btu/hr Q = U x A x TD Q = .388 x 120 sf. x 40 deg. Q = 1862 btus / hr. 30 70 deg.

  12. Energy flow through a wall Add 1” styrofoam Assume wall is 10’ x 20’ Summation of R values R = .44 R = 1.01 R = 5.0 (insulation) R = 1.5 R total 7.95 U = 1 / 7.95 = .125 btu/hr Q = U x A x TD Q = .125 x 120 sf. x 40 deg. Q = 600 btus / hr. 30 70 deg.

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