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I—_ <br />FRANK LI_N <br />ENGINEERING <br />MEO 1AN1( Al ((l:'nl il. i <br />025 Feorth Avenue <br />Suite 202 <br />Kirkland WA 98033 <br />c� <br />PH 425 827-3324 <br />FAQ( 425 827-6252 <br />hvaC0frarkhrr %340m <br />U-value Calculation for "Vented" T-bar Ceiling with Light Fixtures <br />R-19 Batts <br />U-value I rea UA <br />Vent Gap 1131 0.2915 0.330 <br />Insulation R-19 0.05 93.0084 4.650 <br />Total UA 4.980 <br />Average U-value = U, = UAIA U, = 0.053 <br />R-19 balls <br />Fi-30 Batts <br />U-value Area UA <br />Vent Gap 1.13 0.2916 0.330 <br />Insulation R-30 0.034 93.0084 3.162 <br />Total UA 3.492 <br />Average U-value = U, = UA/A U, = 0.037 <br />R•30 batts <br />ons <br />R-25 Batts <br />U-value <br />Area UA <br />Vent Gap 1.13 <br />Insulation R-25 0.04 <br />0.2916 0,336 <br />93.0084 3.720 <br />Total UA 4.050 <br />Average U-value = U. = UA/A <br />U, = 0.043 <br />i•25 baits <br />I Batts <br />Gap <br />U-value = U, = UA'A U. = 0.029 <br />R-38 belts <br />Vent Gap <br />Size: 7/8" wido by 48" long U-value of vent gap: <br />Area = 7/8 x 48 = 42 in` Assume R=0.88, i/R=U=1.13 <br />42 in2/144 in' = 0,2916 It' <br />Vent Gap Area = 0,2916 It U•value of Vent <br />s <br />Assume: <br />112 watts/fixture = 93.3 It <br />1.2 wattsHl <br />1 LI ht Fixture er 93.3 It <br />Area of insulation <br />= area of Light Fixtures - area of Vent Gap <br />Area of insulation = 93.3 - 0.2916 <br />Insulation Area = 93.0084 It <br />